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CHAPTER 15 Introduction to Simulation Modeling

REAL APPLICATIONS OF SIMULATION WITH @RISK This chapter introduces Palisade’s @RISK add-in for simulation modeling in Excel. @ RISK is not only for academic use. Palisade has trained numerous well-known compa- nies in the use of @RISK, and its website chronicles how many of these companies have used @RISK in their businesses. Here are a few of these applications.

• Merck, the multinational pharmaceutical company, recognizes the importance of value-at-risk in its risk management programs. (Value-at-risk, defined in the next chap- ter, is nearly the worst that can happen.) Merck is also aware that exchange rate volatil- ity is one of the largest components of its value-at-risk. @RISK provides the flexibility to fit and evaluate alternative distributions of currency rates. The company must man- age currency exposures in both the balance sheet and in future revenues. Evidently, simulating currency risks on the balance sheet is relatively straightforward. However, simulating hedged cash flow currency risk presents challenges because of accounting standards for derivative investments. It requires a model that can project economic and accounting hedge performance through time. This involves many uncertain variables, including option time decay and the volatility of option price components. @RISK has the power to handle this complexity, and for this reason, @RISK is Merck’s analytic tool of choice.

• Benjamin Waisbren uses @RISK in all his negotiations. Waisbren is president of LSC Film Corporation, a company that provides the funding for major film productions, such as V for Vendetta, Blood Diamond, and 300. He recently used @RISK to complete a $200 million deal with Sony, giving his team, LStar Capital, a stake in nearly all mov- ies produced by Sony. He estimates that because of @RISK modeling, closing costs on this deal were about $20 million lower than on similar deals. Since then, he has used @RISK to perform statistical analysis on risks in the motion picture business. This has led to surprisingly accurate predictions of how much a film will make on its opening day or weekend, as well as the amount film production will cost in any given number of months. Trained as a lawyer, Waisbren urges other law firms to train their employees in @RISK, arguing that this can save them huge amounts in complex negotiations.

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Copyright 2020 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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• Amway, the global direct sales giant with annual sales over $10 billion and manufac- turer of more than 450 different products in 18 plants in the U.S. and Southeast Asia, must make many detailed capacity planning decisions on a regular basis. In the past, a team from the IE (industrial engineering) team would gather data and use traditional modeling tools to evaluate various scenarios. This was time-consuming, and the results were often less than accurate by the time they were available. In 2014, faced with a planned expansion of five new manufacturing sites, the IE team sought a better solution process and turned to @RISK. They developed a simulation tool called Long Range Capacity Planning (LRCP) that could quickly evaluate thousands of what-if scenar- ios, accounting for the vast number of uncertainties faced by Amway’s manufacturing teams. This tool allows plant managers to change variables such as demand, output rates, new products, and run sizes for a selected plant. Then it can simulate up to 20 scenarios and provide real-time results on which configurations work best.

• Deloitte, the global consulting firm, has used @RISK to help its cell captive insurance clients. This growing form of insurance occurs when a host insurer, the “cell captive,” allows other companies, the “cell owners,” to piggyback on the host’s insurer’s license, so that the cell owners don’t have to deal with the costs and regulations of buying their own licenses. The cell owners can then perform some of the functions on behalf of the host insurer to make their own profits. However, the host insurer takes on significant risks in such an arrangement, and it is exposed to huge financial risks if disastrous events occur. For this reason, the cell owners are required to capitalize the cell at the outset, providing funds for the host insurer in case of a disaster. The big question is how much capital is required, and this is where @RISK enters the picture. It can be used to simulate many possible scenarios over a future time period such as a year, and its results can predict how bad things could be. In this case, Deloitte uses the 99.5th percentile as the relevant value, the amount of exposure faced by the host insurer in a “one-in-200-event.”

15-1 Introduction A simulation model is a computer model that imitates a real-life situation. It is like other mathematical models, but it explicitly incorporates uncertainty in one or more input vari- ables. When you run a simulation, you allow these random input variables to take on var- ious values, and you keep track of any resulting output variables of interest. In this way, you are able to see how the outputs vary as a function of the varying inputs.

The fundamental advantage of a simulation model is that it provides an entire distri- bution of results, not simply a single bottom-line result. As an example, suppose an auto- mobile manufacturer is planning to develop and market a new model car. The company is ultimately interested in the net present value (NPV) of the cash flows from this car over the next 10 years. However, there are many uncertainties surrounding this car, including the yearly customer demands for it, the cost of developing it, and others. The company could develop a spreadsheet model for the 10-year NPV, using its best guesses for these uncertain quantities. It could then report the NPV based on these best guesses. However, this analysis would be incomplete and probably misleading because there is no guarantee that the NPV based on best-guess inputs is representative of the NPV that will actually occur. It is much better to treat the uncertainty explicitly with a simulation model. This involves entering probability distributions for the uncertain quantities and seeing how the NPV varies as the uncertain quantities vary.

Each different set of values for the uncertain quantities is a scenario. Simulation allows the company to generate many scenarios, each leading to a particular NPV. In the end, it sees a whole distribution of NPVs, not a single best guess. The company can see what the NPV will be on average, and it can also see worst-case and best-case results.

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These approaches are summarized in Figures 15.1 and 15.2. Figure 15.1 indicates that the deterministic (nonsimulation) approach, using best guesses for the uncertain inputs, is generally not the appropriate method. It leads to the “flaw of averages,” as we will discuss later in the chapter. The problem is that the outputs from the deterministic model are often not representative of the true outputs. The appropriate method is shown in Figure 15.2. Here the uncertainty is modeled explicitly with random inputs, and the end result is a probability distribution for each of the important outputs.

Best guesses for uncertain inputs

Deterministic (nonsimulation) model

Best guesses for important outputs

Usually not correct: the “flaw of averages”

Figure 15.1 Inappropriate Deterministic Model

Figure 15.2 Appropriate Simulation Model Probability distributions for

uncertain inputs Simulation model Probability distributions for important outputs

Simulation models are also useful for determining how sensitive a system is to changes in operating conditions. For example, the operations of a supermarket could be simulated. Once the simulation model has been developed, it could then be run (with suit- able modifications) to ask a number of what-if questions. For example, if the supermarket experiences a 20% increase in business, what will happen to the average time customers must wait for service?

A huge benefit of computer simulation is that it enables managers to answer these types of what-if questions without actually changing (or building) a physical system. For example, the supermarket might want to experiment with the number of open registers to see the effect on customer waiting times. The only way it can physically experiment with more registers than it currently owns is to purchase more equipment. Then if it determines that this equipment is not a good investment—customer waiting times do not decrease appreciably—the company is stuck with expensive equipment it doesn’t need. Computer simulation is a much less expensive alternative. It provides the company with an electronic replica of what would happen if the new equipment were purchased. Then, if the simula- tion indicates that the new equipment is worth the cost, the company can be confident that purchasing it is the right decision. Otherwise, it can abandon the idea of the new equip- ment before the equipment has been purchased.

Spreadsheet simulation modeling is similar to the other modeling applications in this book. You begin with input variables and then relate these with appropriate Excel® for- mulas to produce output variables of interest. The main difference is that simulation uses random numbers to drive the process. These random numbers are generated with special functions that we will discuss in detail. Each time the spreadsheet recalculates, all the ran- dom numbers change. This provides the ability to model the logical process once and then use Excel’s recalculation to generate many different scenarios. By collecting the data from these scenarios, you can see the most likely values of the outputs and the best-case and worst-case values of the outputs.

In this chapter we begin by illustrating spreadsheet models that can be developed with built-in Excel functionality. However, because simulation is such an important tool for analyzing real problems, add-ins to Excel have been developed to streamline the process of developing and analyzing simulation models. Therefore, we then introduce @RISK, one of the most popular simulation add-ins. This add-in not only augments the simula- tion capabilities of Excel, but it also enables you to analyze models much more quickly and easily.

Like the other Palisade add-ins, @RISK works only with Excel for Windows, not with Excel for Mac.

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The purpose of this chapter is to introduce basic simulation concepts, show how sim- ulation models can be developed in Excel, and demonstrate the capabilities of @RISK. Then in the next chapter, armed with the necessary simulation tools, we will explore a variety of simulation models.

Before proceeding, you might ask whether simulation is really used in the business world. The answer is a resounding “yes.” The chapter opener described an airline example, and many other examples can be found online. For example, if you visit www.palisade .com, you will see descriptions of interesting @RISK applications from companies that regularly use this add-in. Simulation has always been a powerful tool, but until the intro- duction of Excel add-ins such as @RISK, it had limited use for several reasons. It typically required specialized software that was either expensive and difficult to learn, or it required tedious computer programming. Fortunately, in the past two decades, spreadsheet simula- tion, together with Excel add-ins such as @RISK, has put this powerful methodology in the hands of the masses—people like you and the companies you are likely to work for. Many businesses now understand that there is no longer any reason to ignore uncertainty; they can model it directly with spreadsheet simulation.

15-2 Probability Distributions for Input Variables In this section we discuss the building blocks of spreadsheet simulation models: prob- ability distributions for input variables that capture uncertainty. All spreadsheet sim- ulation models are similar to the spreadsheet models from previous chapters. They have a number of cells that contain values of input variables. The other cells then contain formulas that embed the logic of the model and eventually lead to the output variable(s) of interest. The primary difference between the spreadsheet models you have developed so far and simulation models is that at least one of the input variable cells in a simulation model contains random numbers. Each time the spreadsheet recalculates, the random numbers change, and the new random values of the inputs produce new values of the outputs. This is the essence of simulation—it enables you to see how outputs vary as random inputs change.

In spreadsheet simulation models, input cells can contain random numbers. Any output cells then vary as these random inputs change.

Recalculation Key

The easiest way to make a spreadsheet recalculate is to press the F9 key. This is often called the “recalc” key.

Excel Tip

Technically speaking, input cells do not contain random numbers; they contain prob- ability distributions. In general, a probability distribution indicates the possible values of a variable and the probabilities of these values. As a very simple example, you might indi- cate by an appropriate formula (to be described later) that you want a probability distri- bution with possible values 50 and 100, and corresponding probabilities 0.7 and 0.3. If you force the sheet to recalculate repeatedly and watch this input cell, you will see the value 50 about 70% of the time and the value 100 about 30% of the time. No other values besides 50 and 100 will appear.

When you enter a given probability distribution in a random input cell, you are describ- ing the possible values and the probabilities of these values that you believe mirror reality. There are many probability distributions to choose from, and you should always attempt to

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choose an appropriate distribution for each specific problem. This is not necessarily easy. Therefore, we address it in this section by answering several key questions:

• What types of probability distributions are available, and why do you choose one probability distribution rather than another in any particular simulation model?

• Which probability distributions can you use in simulation models, and how do you invoke them with Excel formulas?

In later sections we address one additional question: Does the choice of input probabil- ity distribution really matter—that is, are the outputs from the simulation sensitive to this choice?

Basic elements of Spreadsheet Simulation

A spreadsheet simulation model requires three elements: (1) a method for entering random quantities from specified probability distributions in input cells, (2) the usual types of Excel formulas for relating outputs to inputs, and (3) the ability to make the spreadsheet recalculate many times and capture the resulting outputs for statistical analysis. Excel has some capabilities for performing these steps, but Excel add-ins such as @RISK provide excellent tools for automating the process.

Fundamental Insight

15-2a Types of Probability Distributions Imagine a toolbox that contains the probability distributions you know and understand. As you obtain more experience in simulation modeling, you will naturally add probability distributions to your toolbox that you can then use in future simulation models. We begin by adding a few useful probability distributions to this toolbox. However, before adding any specific distributions, it is useful to provide a brief review of some important general characteristics of probability distributions.1 These include the following distinctions:

• Discrete versus continuous • Symmetric versus skewed • Bounded versus unbounded • Nonnegative versus unrestricted

1 Much of this material was covered in Chapters 2 and 5, but it is reviewed here.

Choosing probability Distributions for Uncertain Inputs

In simulation models, it is important to choose appropriate probability distributions for uncertain inputs. These choices can strongly affect the results. However, there are no “right answers.” You need to choose the probability distributions that best describe the uncertainty, and this is usually not easy. However, the properties discussed in this section provide useful guidelines for making reasonable choices.

Fundamental Insight

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Discrete Versus Continuous A probability distribution is discrete if it has a finite number of possible values.2 For example, if you throw two dice and look at the sum of the faces showing, there are only 11 discrete possibilities: the integers 2 through 12. In contrast, a probability distribution is continuous if its possible values are essentially a continuum. An example is the amount of rain that falls during a month in Indiana. It could be any decimal value from 0 to, say, 15 inches.

The graph of a discrete distribution is a series of spikes, as shown in Figure 15.3.3 The height of each spike is the probability of the corresponding value.

2 It is possible for a discrete variable to have a countably infinite number of possible values, such as all the nonnegative integers. However, this is not an important distinction for practical applications. 3 This figure and several later figures are from Palisade’s @RISK add-in.

Figure 15.3 Typical Discrete Probability Distribution

In contrast, a continuous distribution is characterized by a density function, a smooth curve as shown in Figure 15.4. Recall from Chapter 5 that the height of the density func- tion above any value indicates the relative likelihood of that value, and probabilities can be calculated as areas under the curve.

Sometimes it is convenient to treat a discrete probability distribution as continuous, and vice versa. For example, consider a student’s random score on an exam that has 1000 possible points. If the grader scores each exam to the nearest integer, then even though the score is discrete with many possible integer values, it is probably more con- venient to model its distribution as a continuum. Continuous probability distributions are typically more intuitive and easier to work with than discrete distributions when there are many possible values. In contrast, continuous distributions are sometimes dis- cretized for simplicity. In this case, the continuum of possible values is replaced by a few typical values.

Symmetric Versus Skewed A probability distribution can be symmetric or skewed to the left or right. Figures 15.4, 15.5, and 15.6 provide examples of these. You typically choose between a symmetric and skewed distribution on the basis of realism. For example, if you want to model a student’s score on a 100-point exam, you will probably choose a left-skewed distribution. This is because a few poorly prepared students typically “pull down the curve.” On the other hand, if you want to model the time it takes to serve a customer at a bank, you will proba- bly choose a right-skewed distribution. This is because most customers take only a minute or two, but a few customers take a long time. Finally, if you want to model the monthly return on a stock, you might choose a distribution symmetric around zero, reasoning that

The heights above a density function are not probabilities, but they still indicate relative likelihoods of the possible values.

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Figure 15.4 Typical Continuous Probability Distribution

Figure 15.5 Positively Skewed Probability Distribution

Figure 15.6 Negatively Skewed Probability Distribution

the stock return is just as likely to be positive as negative and there is no obvious reason for skewness in either direction.

Bounded Versus Unbounded A probability distribution is bounded if there are values A and B such that no possible value can be less than A or greater than B. The value A is then the minimum possible value,

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and the value B is the maximum possible value. The distribution is unbounded if there are no such bounds. Of course, it is possible for a distribution to be bounded in one direction but not the other. As an example, the distribution of scores on a 100-point exam is bounded between 0 and 100. In contrast, the distribution of the amount of damages Mr. Jones submits to his insurance company in a year is bounded on the left by 0, but there is no natural upper bound. Therefore, you might model this amount with a distribution that is bounded by 0 on the left but is unbounded on the right. Alternatively, if you believe that no damage amount larger than $20,000 can occur, you could model this amount with a distribution that is bounded in both directions.

Nonnegative Versus Unrestricted One important special case of bounded distributions is when the only possible values are nonnegative. For example, if you want to model the random cost of manufacturing a new product, you know that this cost must be nonnegative. There are many other such examples. In these cases, you should model the randomness with a probability distribution that is bounded below by 0. This prevents negative values that make no practical sense.

15-2b Common Probability Distributions Now that you know the types of probability distributions available, you can add some common probability distributions to your toolbox. The file Probability Distributions.xlsx was developed to help you learn and explore the distributions discussed in this section, plus others. Each sheet in this file illustrates a particular probability distribution. It describes the general characteristics of the distribu- tion, indicates how you can generate random numbers from the distribution with Excel’s built-in functions, with @RISK functions, or with Albright’s RandGen add-in (freely available at https://kelley.iu.edu/albrightbooks/free_downloads. htm) and it includes histograms of these distributions from simulated data to illustrate their shapes.4

Each of the following distributions is really a family of distributions. Each member of the family is specified by one or more parameters. For example, there is a normal distribution for each possible mean and standard deviation you spec- ify. Therefore, when you try to find an appropriate input probability distribution for a simulation model, you first have to choose an appropriate family, and then you have to select the appropriate parameters for that family.

Uniform Distribution The uniform distribution is the “flat” distribution illustrated in Figure 15.7. It is bounded by a minimum and a maximum, and all values between these two extremes are equally likely. You can think of this as the “I have no idea” distribu- tion. For example, a manager might realize that a building cost is uncertain. If she can state only that, “I know the cost will be between $20,000 and $30,000, but other than this, I have no idea what the cost will be,” then a uniform distribution from $20,000 to $30,000 is a natural choice. However, even though some peo- ple do sometimes use the uniform distribution in such situations, this choice is usually not very realistic. If the manager really thinks about it, she can probably provide more information about the uncertain cost, such as, “The cost is more

Think of the Probability Distributions.xlsx file as a “dictionary” of the most commonly used distributions. Keep it handy for reference.

A family of distributions has a common name, such as “normal.” Each member of the family is specified by one or more numerical parameters.

4 Later sections of this chapter and all the next chapter discuss much of @RISK’s functionality. For this section, the only functionality used is @RISK’s collection of functions, such as RISKNORMAL and RISKTRIANG, for gener- ating random numbers from various probability distributions. You can skim the details of these functions for now and refer back to them as necessary in later sections.

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Figure 15.7 Uniform Distribution

likely to be close to $25,000 than to either of the extremes.” Then some distribution other than the uniform is more appropriate.

Nevertheless, the uniform distribution is important for another reason. All simulation software packages, including Excel, are capable of generating random numbers uniformly distributed between 0 and 1. These are the building blocks of most simulated random numbers, in that random numbers from other probability distributions are generated from them.

In Excel, you can generate a uniformly distributed random number between 0 and 1 by entering the formula

=RAND()

in any cell. (The parentheses to the right of RAND indicate that this is an Excel function with no arguments. These parentheses must be included.)

The RAND function is Excel’s “building block” function for generating random numbers.

RAND

To generate a random number equally likely to be anywhere between 0 and 1, enter the formula =RAND() into any cell. Press the F9 key, or recalculate in any other way, to make it change randomly.

Excel Function RAND and RANDBETWEEN Functions

In addition to being between 0 and 1, the numbers created by this function have two important properties.

• Uniform property. Each time you enter the RAND function in a cell, all numbers between 0 and 1 have the same chance of occurring. This means that approximately 10% of the numbers generated by the RAND function will be between 0.0 and 0.1; 10% of the numbers will be between 0.65 and 0.75; 60% of the numbers will be

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between 0.20 and 0.80; and so on. This property explains why the random numbers are said to be uniformly distributed between 0 and 1.

• Independence property. Different random numbers generated by RAND functions are probabilistically independent. This implies that when you generate a random num- ber in cell A5, for example, it has no effect on the values of any other random numbers generated in the spreadsheet. If one call to the RAND function yields a large random number such as 0.98, there is no reason to suspect that the next call to RAND will yield an abnormally small (or large) random number; it is unaffected by the value of the first random number.

RANDBETWEEN

There is one other Excel function that generates random numbers, the RANDBETWEEN function. It takes two integer arguments, as in =RANDBETWEEN(1,6), and returns a random integer between these values (including the two values themselves) so that all integers in this interval are equally likely.

Excel Function

To illustrate the RAND function, open a new workbook, enter the formula =RAND() in cell A4, and copy it to the range A4:A503. This generates 500 random numbers. Figure 15.8 displays a few of them. However, when you try this on your PC, you will undoubtedly obtain different random numbers. This is an inherent characteristic of simulation—no two answers are ever exactly alike. Now press the F9 recalc key. All the random numbers will change. In fact, each time you press the F9 key or do anything to make your spreadsheet recalculate, all cells containing the RAND function will change.

Figure 15.8 Uniformly Distributed Random Numbers Generated by the RAND Function

1 2 3 4 5 6 7 8 9

10 11

A B C 500 random numbers from RAND func�on

Random number 0.4023 0.0978 0.1494 0.7144 0.1751 0.2523 0.3183 0.8947

A histogram of the 500 random numbers appears in Figure 15.9. (Again, if you try this on your PC, the shape of your histogram will not be identical to the one shown in Figure 15.9, because it will be based on different random numbers.) From property 1, you would expect equal numbers of observations in the 10 categories. Obviously, the heights of the bars are not exactly equal, but the differences are due to chance—not to a faulty random number generator.

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Figure 15.9 Histogram of the 500 Random Numbers Generated by the RAND Function

Histogram

0.05 0.15 0.25 0.35 0.45 0.55 0.65 0.75 0.85 0.95

Pseudo-random Numbers

The “random” numbers generated by the RAND function (or by the random number generator in any simulation software package) are not really random. They are sometimes called pseudo-random numbers. Each successive random number follows the previous random number by a complex arithmetic opera- tion. If you happen to know the details of this arithmetic operation, you can predict ahead of time exactly which random numbers will be generated by the RAND function. This is quite different from using a “true” random mechanism, such as spinning a wheel, to get the next random number—a mechanism that would be impractical to implement on a computer. Mathematicians and com- puter scientists have studied many ways to produce random numbers that have the two properties we just discussed, and they have developed many competing random number generators such as the RAND function in Excel. The technical details are not important here. The important point is that these random number generators produce numbers that appear to be random and are useful for simu- lation modeling.

Technical Note

It is simple to generate a uniformly distributed random number with a minimum and maximum other than 0 and 1. For example, the formula

=200+100*RAND()

generates a number uniformly distributed between 200 and 300. (Make sure you see why.) Alternatively, you can use the @RISK formula5

=RiskUniform(200,300)

You can take a look at this and other properties of the uniform distribution on the Uniform sheet in the Probability Distributions.xlsx file.

5 As with built-in Excel functions, case is irrelevant with @RISK functions. This function could be spelled as RISKUNIFORM, RiskUniform, riskuniform, or any other variation.

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RiskUniform

To generate a random number from any uniform distribution, enter the formula =RiskUniform(MinVal,MaxVal) in any cell. Here, MinVal and MaxVal are the minimum and maximum possible values. If MinVal is 0 and MaxVal is 1, this function is equivalent to Excel’s RAND function.

@RISK Function

Freezing Random Numbers The automatic recalculation of random numbers is usually useful, but sometimes it can be annoying. There are situations when you want the random numbers to stay fixed—that is, you want to freeze random numbers at their current values. The following three-step method does this.

1. Select the range that you want to freeze, such as A4:A503 in Figure 15.8. 2. Press Ctrl1c to copy this range. 3. With the same range still selected, select the Paste Values option from the Paste drop-

down menu on the Home ribbon. This procedure pastes a copy of the range onto itself, except that the entries are now numbers, not formulas. Therefore, whenever the spread- sheet recalculates, these numbers do not change.

15-2c Using @RISK to Explore Probability Distributions The Probability Distributions.xlsx file illustrates a few frequently used probability dis- tributions, and it shows the formulas required to generate random numbers from these distributions. Another option is to use Palisade’s @RISK add-in, which allows you to experiment with probability distributions with its distribution functions. Essentially, it allows you to see the shapes of various distributions and calculate probabilities for them, all in a user-friendly graphical interface.

To run @RISK, click the Windows Start button, go to the Programs tab, locate the Palisades DecisionTools® Suite, and select @RISK. After a few seconds, you will see the welcome screen, which you can close. At this point, you should have an @RISK tab and corresponding ribbon. (The Project button in the Tools group will be present only if you have Microsoft Project installed on your computer.) Select a blank cell in your worksheet and click the Define Distributions button on the @RISK ribbon (see Figure 15.10). You will see one of several galleries of distributions, depending on the tab you select. For example, Figure 15.11 shows the gallery of common distributions. Highlight one of the distributions and click Select Distribution. For example, choose the uniform distribution and enter 75 and 150 as the Min and Max parameters. You will see the shape of the distribution and a list of summary measures to the right, as shown in Figure 15.12. For example, it indicates that the mean and standard deviation of this uniform distribu- tion are 112.5 and 21.65.

Everything in this window is interactive. Suppose you want to find the probability that a value from this distribution is less than 95. You can drag the left-hand “slider” in the diagram (the vertical line with the triangle at the top) to the position 95, as shown in

Random numbers that have been frozen do not change when you press the F9 key.

Exploring Distributions with @RISK

Figure 15.10 @RISK Ribbon

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Figure 15.12. You see immediately that the left-hand probability is 0.267. Similarly, if you want the probability that a value from this distribution is greater than 125, you can drag the right-hand slider to the position 125 to see that the required probability is 0.3333. (Rather than sliding, you can enter the numbers, such as 95 and 125, directly in the areas above the sliders.)

You can also enter probabilities instead of values. For example, if you want the value such that there is probability 0.10 to the left of it—the 10th percentile—you can enter 10% in the left space above the chart. You will see that the corresponding value is 82.5. Similarly, if you want the value such that there is probability 0.10 to the right of it, you can enter 10% in the right space above the chart, and you will see that the corresponding value is 142.5.

Figure 15.11 @RISK Gallery of Common Distributions

The interactive capabilities of @RISK’s Define Distribu- tion window, with its sliders, make it perfect for finding probabilities or percentiles for any given distribution.

Figure 15.12 Uniform Distribution (from @RISK)

The Define Distribution window in @RISK is quick and easy. We urge you to use it and experiment with some of its options. By the way, you can click the second button from the left at the bottom of the window to copy the chart into an Excel worksheet. However, you then lose the interactive capabilities, such as moving the sliders.

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Discrete Distribution A discrete distribution is useful for many situations, either when the uncertain quantity is not really continuous (the number of televisions demanded, for example) or when you want a discrete approximation to a continuous variable. You need to specify the possible values and their probabilities, making sure that the probabilities sum to 1. Because of this flexibility in specifying values and probabilities, discrete distributions can have practically any shape.

For example, suppose a manager estimates that the demand for a particular brand of television during the coming month will be 10, 15, 20, or 25, with respective probabilities 0.1, 0.3, 0.4, and 0.2. This typical discrete distribution is illustrated in Figure 15.13.

The Discrete sheet of the Probability Distributions.xlsx file indicates how to work with a discrete distribution. As you will see, a lookup table is required if you aren’t using an add-in. We discuss this in detail in Section 15-4. For now, we simply mention that this is one case (of many) where it is much easier to generate random numbers with @RISK functions than with built-in Excel functions. Assuming that @RISK is loaded, you enter the function RiskDiscrete with two arguments: a list of possible values and a list of their probabilities, as in

=RiskDiscrete(B11:B14,C11:C14)

The Excel way, which requires cumulative probabilities and a lookup table, takes more work and is harder to remember.

@RISK’s way of generating a discrete random number is much simpler and more intuitive than Excel’s method, which requires cumulative probabilities and a lookup function.

Generating Random Numbers with Excel Functions

Figure 15.13 Discrete Distribution (from @RISK)

RiskDiscrete

To generate a random number from any discrete probability distribution, enter the formula =RiskDiscrete(vals,probs) into any cell. Here, vals is a list of pos- sible values and probs is a list of their probabilities. You can list the values and probabilities inside curly brackets, as in the top of Figure 15.13, or you can reference ranges with these values and probabilities.

@RISK Function

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15-2 probability Distributions for Input Variables 7 3 1

You might ask why a manager would choose this particular discrete distribution. First, it is clearly an approximation. After all, if it is possible to have demands of 20 and 25, it should also be possible to have demands between these values. Here, the manager approximates a discrete distribution with many possible values—all integers from 0 to 50, say—with a discrete distribution with a few typical values. This is fairly common in simulation modeling. Second, where do the probabilities come from? They are probably a blend of historical data (perhaps demand was near 15 in 30% of previous months) and the manager’s subjective feelings about demand next month.

Normal Distribution The normal distribution is the familiar bell-shaped curve that was discussed in detail in Chapter 5. (See Figure 15.14.) It is useful in simulation modeling as a continuous input distribution. However, it is not always the most appropriate distribution. It is symmetric, which can be a drawback when a skewed distribution is more realistic. Also, it allows neg- ative values, which are not appropriate in many situations.

The selected input distri- butions for any simulation model reflect historical data and an analyst’s best judg- ment as to what will happen in the future.

Figure 15.14 Normal Distribution (from @RISK)

A tip-off that a normal distribution might be an appropriate candidate for an input variable is a statement such as, “We believe the most likely value of demand is 100, and the chances are about 95% that demand will be no more than 40 units on either of side of this most likely value.” Because a normally distributed value is within two standard deviations of its mean with probability 0.95, this statement translates easily to a mean of 100 and a standard deviation of 20. This does not imply that a normal distribution is the only candidate for the distribution of demand, but the statement suggests that the normal distribution is a good candidate.

The Normal sheet in the Probability Distributions.xlsx file indicates how you can generate normally distributed random numbers in Excel. This is one case where an add-in is not really necessary. The formula

=NORM.INV(RAND(),Mean,Stdev)

always works. Still, this is not as easy to remember as @RISK’s formula

=RiskNormal(Mean,Stdev)

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Triangular Distribution The triangular distribution is somewhat similar to the normal distribution in that its den- sity function rises to some point and then falls, but it is more flexible and intuitive than the normal distribution. Therefore, it is an excellent candidate for many continuous input variables. The shape of a triangular density function is literally a triangle, as shown in Figure 15.15. It is specified by three easy-to-understand parameters: the minimum possi- ble value, the most likely value, and the maximum possible value. The high point of the triangle is above the most likely value. Therefore, if a manager states, “We believe the most likely development cost is $1.5 million, and we don’t believe the development cost could possibly be less than $1.2 million or greater than $2.1 million,” the triangular distribution with these three parameters is a natural choice. As in this numerical example, the triangular distribution can be skewed if the mostly likely value is closer to one extreme than another. Of course, it can also be symmetric if the most likely value is right in the middle.

RiskNormal

To generate a normally distributed random number, enter the formula =RiskNormal(Mean,Stdev) in any cell. Here, Mean and Stdev are the mean and standard deviation of the normal distribution.

@RISK Function

A triangular distribution is a good choice in many simu- lation models because it can have a variety of shapes and its parameters are easy to understand.

Figure 15.15 Triangular Distribution (from @RISK)

The Triangular sheet of the Probability Distributions.xlsx file indicates how to gen- erate random values from this distribution. As it indicates, there is no easy way to do it with Excel functions only. However, it is easy with @RISK, using the RiskTriang function, as in

=RiskTriang(B10,B11,B12)

This function takes three arguments: the minimum value, the most likely value, and the maximum value—in this order and separated by commas. You will see this function in

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15-2 probability Distributions for Input Variables 7 3 3

many of our examples. Just remember that it has an abbreviated spelling: RiskTriang, not RiskTriangular.

RiskTriang

To generate a random number from a triangular distribution, enter the formula =RiskTriang(MinVal,MLVal,MaxVal) in any cell. Here, MinVal is the minimum possible value, MLVal is the most likely value, and MaxVal is the maximum value.

@RISK Function

Binomial Distribution The binomial distribution is a discrete distribution that was discussed extensively in Chapter 5. Recall that the binomial distribution applies to a very specific situation: when a number of independent and identical trials occur, and each trial results in a success or failure. Then the binomial random number is the number of successes in these trials. The two parameters of this distribution, n and p, are the number of trials and the probability of success on each trial.

As an example, suppose an airline company sells 170 tickets for a flight and estimates that 80% of the people with tickets will actually show up for the flight. How many people will actually show up? It is tempting to state that exactly 80% of 170, or 136 people, will show up, but this neglects the inherent randomness. A more realistic way to model this situation is to say that each of the 170 people, independently of one another, will show up with probability 0.8. Then the number of people who actually show up is then bino- mially distributed with n 5 170 and p 5 0.8. (This assumes independent behavior across passengers, which might not be the case, for example, if whole families either show up or don’t.) This distribution is illustrated in Figure 15.16.

A random number from a binomial distribution indicates the number of successes in a certain number of identical trials.

Figure 15.16 Binomial Distribution (from @RISK)

The Binomial sheet of the Probability Distributions.xlsx file indicates how to gen- erate random numbers from this distribution. Although it is possible to do this with Excel

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using the built-in BINOM.INV function and the RAND function, it is not very intuitive or easy to remember. The @RISK way is easier. In the airline example, you would generate the number who show up with the formula

=RiskBinomial(170,0.8)

Note that the histogram in this figure is approximately bell-shaped. This is no accident. When the number of trials n is reasonably large and p isn’t too close to 0 or 1, the binomial distribution can be approximated well by the normal distribution.

RiskBinomial

To generate a random number from a binomial distribution, enter the formula =RiskBinomial(n, p) in any cell. Here, n is the number of trials, and p is the probability of a success on each trial.

@RISK Function

It is natural to ask which distribution to use for a given uncertain quantity such as the price of oil, the demand for laptops, and so on. Admittedly, the choices we make in later examples are sometimes fairly arbitrary. However, in real business situations the choice is not always clear-cut, and it can make a difference in the results. Stanford professor Sam Savage and two of his colleages discuss this choice in a series of two articles on “Proba- bility Management.” They argue that with the increasing importance of simulation models in today’s business world, input distributions should not only be chosen carefully, but they should be kept and maintained as important corporate assets. They shouldn’t just be cho- sen in some ad hoc fashion every time they are needed. For example, if the price of oil is an important input in many of a company’s decisions, then experts within the company should assess an appropriate distribution for the price of oil and modify it as necessary when new information arises. The authors even suggest a new company position, Chief Probability Officer, to control access to the company’s probability distributions.

As you are reading these final two chapters, keep Savage’s ideas in mind. The choice of probability distributions for inputs is not easy, but it is also not arbitrary. The choice can make a difference in the results. This is the reason why you want as many families of prob- ability distributions in your toolbox as possible. You then have more flexibility in choosing a distribution that is appropriate for your situation.

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 1. Use the RAND function and the Copy command to gen-

erate a set of 100 random numbers. a. What fraction of the random numbers are smaller than 0.5? b. What fraction of the time is a random number less than

0.5 followed by a random number greater than 0.5? c. What fraction of the random numbers are larger than 0.8? d. Freeze these random numbers. However, instead of

pasting them over the original random numbers, paste them onto a new range. Then press the F9 recalculate key. The original random numbers should change, but the pasted copy should remain the same.

2. Use Excel’s functions (not @RISK) to generate 1000 random numbers from a normal distribution with mean 100 and standard deviation 10. a. Calculate the mean and standard deviation of these random

numbers. Are they approximately what you would expect? b. What fraction of these random numbers are within k

standard deviations of the mean? Answer for k 5 1; for k 5 2; for k 5 3. Are the answers close to what they should be (according to the empirical rules you learned in Chapters 2 and 5)?

c. Create a histogram of the random numbers. Does this his- togram have approximately the shape you would expect?

3. Use @RISK’s Define Distributions tool to show a uni- form distribution from 400 to 750. Then answer the fol- lowing questions. a. What are the mean and standard deviation of this

distribution? b. What are the 5th and 95th percentiles of this distribution?

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15-2 probability Distributions for Input Variables 7 3 5

c. What is the probability that a random number from this distribution is less than 450?

d. What is the probability that a random number from this distribution is greater than 650?

e. What is the probability that a random number from this distribution is between 500 and 700?

4. Use @RISK’s Define Distributions tool to draw a nor- mal distribution with mean 500 and standard deviation 100. Then answer the following questions. a. What is the probability that a random number from

this distribution is less than 450? b. What is the probability that a random number from

this distribution is greater than 650? c. What is the probability that a random number from

this distribution is between 500 and 700? 5. Use @RISK’s Define Distributions tool to show a tri-

angular distribution with parameters 300, 500, and 900. Then answer the following questions. a. What are the mean and standard deviation of this

distribution? b. What are the 5th and 95th percentiles of this distribution? c. What is the probability that a random number from

this distribution is less than 450? d. What is the probability that a random number from

this distribution is greater than 650? e. What is the probability that a random number from

this distribution is between 500 and 700? 6. Use @RISK’s Define Distributions tool to show a bino-

mial distribution that results from 50 trials with proba- bility of success 0.3 on each trial, and use it to answer the following questions. a. What are the mean and standard deviation of this

distribution? b. You have to be more careful in interpreting @RISK

probabilities with a discrete distribution such as this binomial. For example, if you move the left slider to 11, you find a probability of 0.139 to the left of it. But is this the probability of “less than 11” or “less than or equal to 11”? One way to check is to use Excel’s BINOM.DIST function. Use this function to interpret the 0.139 value from @RISK.

c. Using part b to guide you, use @RISK to find the prob- ability that a random number from this distribution will be greater than 17. Check your answer by using the BINOM.DIST function appropriately in Excel.

7. Use @RISK’s Define Distributions tool to draw a tri- angular distribution with parameters 200, 300, and 600. Then superimpose a normal distribution on this draw- ing, choosing the mean and standard deviation to match those from the triangular distribution. (Click the Add Overlay button at the bottom of the window and then choose the distribution to superimpose.) a. What are the 5th and 95th percentiles for these two

distributions? b. What is the probability that a random number from

the triangular distribution is less than 400? What is this probability for the normal distribution?

c. Experiment with the sliders to answer questions similar to those in part b. Would you conclude that these two distributions differ most in the extremes (right or left) or in the middle? Explain.

8. We all hate to keep track of small change. By using ran- dom numbers, it is possible to eliminate the need for change and give the store and the customer a fair deal. This problem indicates how it could be done. a. Suppose that you buy something for $0.20. How could

you use random numbers (built into the cash register sys- tem) to decide whether you should pay $1.00 or nothing?

b. If you bought something for $9.60, how would you use random numbers to eliminate the need for change?

c. In the long run, why is this method fair to both the store and the customers? Would you personally (as a customer) be willing to abide by such a system?

Level B 9. A company is about to develop and then market a new

product. It wants to build a simulation model for the entire process, and one key uncertain input is the devel- opment cost. For each of the following scenarios, choose an appropriate distribution together with its parameters, justify your choice in words, and use @RISK’s Define Distributions tool to show your chosen distribution. a. Company experts have no idea what the distribution

of the development cost is. All they can state is “we are 95% sure it will be at least $450,000, and we are 95% sure it will be no more than $650,000.”

b. Company experts can still make the same statement as in part a, but now they can also state: “We believe the distribution is symmetric, reasonably bell-shaped, and its most likely value is about $550,000.”

c. Company experts can still make the same statement as in part a, but now they can also state: “We believe the distribution is skewed to the right, and its most likely value is about $500,000.”

10. Continuing the preceding problem, suppose that another key uncertain input is the development time, which is measured in an integer number of months. For each of the following scenarios, choose an appropriate distribu- tion together with its parameters, justify your choice in words, and use @RISK’s Define Distributions tool to show your chosen distribution. a. Company experts believe the development time will

be from 6 to 10 months, but they have absolutely no idea which of these will result.

b. Company experts believe the development time will be from 6 to 10 months. They believe the probabilities of these five possible values will increase linearly to a most likely value at 8 months and will then decrease linearly.

c. Company experts believe the development time will be from 6 to 10 months. They believe that 8 months is twice as likely as either 7 months or 9 months and that either of these latter possibilities is three times as likely as either 6 months or 10 months.

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EXAMPLE

15.1 ORDERING CALENDARS AT WALTON BOOKSTORE In August, Walton Bookstore must decide how many of next year’s nature calendars to order. Each calendar costs the bookstore $7.50 and sells for $10. After January 1, all unsold calendars will be returned to the publisher for a refund of $2.50 per calendar. Walton believes that the number of calendars it can sell by January 1 follows some probability distribution with mean

200. Walton believes that ordering to the average demand, that is, ordering 200 calendars, is a good decision. Is it?

Objective To illustrate the difference between a deterministic model with a best guess for uncertain inputs and a simulation model that incorporates uncertainty explicitly.

Where Do the Numbers Come From? The monetary values are straightforward. The mean demand is probably an estimate based on historical demands for similar calendars.

Solution The variables for this model are shown in Figure 15.17. (See the file Ordering Calendars Big Picture.xlsx.) Note that in addition to the “Big Picture” conventions illustrated in the two previous chapters, we use a green rectangle with a rounded top for uncertain quantities. An order quantity is chosen and demand is then observed. The ordering cost is based on the order quantity, the revenue is based on the smaller of the order quantity and demand, and there is a refund if demand is less than the order quantity.

15-3 Simulation and the Flaw of Averages To help motivate simulation modeling in general, we present a simple example in this section. It will clearly show the distinction between Figure 15.1 (a deterministic model with best-guess inputs) and Figure 15.2 (an appropriate simulation model). In doing so, it illustrates a pitfall called the “flaw of averages.”6

6 As far as we know, the term “flaw of averages” was coined by Sam Savage, the same Stanford professor quoted earlier.

The Flaw of Averages

Profit

Order quantityDemand distribution Demand

Unit refund

Unit cost

Order cost

Revenue from sales Refund from leftovers

Unit price

Figure 15.17 Big Picture for Ordering Model

A deterministic model appears in Figure 15.18. (See the file Ordering Calendars - Flaw of Averages Finished.xlsx.) Assuming the best guess for demand, Walton orders to this average value, and it appears that the company’s best guess for profit is $500.

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15-3 Simulation and the Flaw of averages 7 3 7

(The formulas in cells B16:F16 are straightforward and are listed in row 18. Before reading further, do you believe the average profit will be $500 when uncertainty in demand is introduced explicitly (and the company still orders 200 calendars)? Think what happens to profit when demand is less than 200 and when it is greater than 200. Are these two cases symmetric?

Figure 15.18 Deterministic Model 1

2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18

A B C D E F Walton’s bookstore - determinis�c model

Cost data Unit cost $7.50 Unit price $10.00 Unit refund $2.50

Uncertain quan�ty Demand (average shown) 200

Decision variable Order quan�ty

Formulas

200

Profit model Demand Revenue Cost Refund Profit

200

=B11 =B6*B14 =C18-D18+E18=B7*MIN(B11,B14) =B8*MAX(B14-B11,0)

$2,000.00 $1,500.00 $0.00

This determinis�c model gives no hint of what will happen when demand is treated explicitly as random. From the simula�on model on the next sheet, we see that the average profit is nowhere near the $500 value in this sheet.

$500.00

We now contrast this with a simulation model where the demand in cell B9 is replaced by a random number. For this example, we assume that demand is normally distributed with mean 200 and standard deviation 40, although these spe- cific assumptions are not crucial for the qualitative aspects of the example. Specifically, cell B9 should contain the formula =ROUND(RiskNormal(200,40),0) where the ROUND function has been used to round to the nearest integer. (We assume that @RISK has been loaded.) Now the model appears as in Figure 15.19.

The random demand in cell B9 is now live, as are its dependents in row 16, so each time you press the F9 key, you get a new demand and associated profit. (This assumes that the @RISK “dice” button is toggled to colored, its “random” setting. More will be said about this setting later in the chapter.) Do you get about $500 in profit on average? Absolutely not! The sit- uation isn’t symmetric. The largest profit you can get is $500, which occurs about half the time, whenever demand is greater than 200. A typical example appears in the figure, where the excess demand of 54 is simply lost. However, when demand is less than 200, the profit is less than $500, and it keeps decreasing as demand decreases.

Figure 15.19 Simulation Model 1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16

A B C D E F Walton’s bookstore - simula�on model

Cost data Unit cost $7.50 Unit price $10.00 Unit refund $2.50

Uncertain quan�ty (assumed normal with mean 200, stdev 40) Demand (random) 254

Decision variable Order quan�ty 200

Profit model Demand Revenue Cost Refund Profit

254 $2,000.00 $1,500.00 $0.00 $500.00

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We ran @RISK with 1000 iterations (which will be explained in detail in Section 15-5) and found the resulting histogram of 1000 simulated profits shown in Figure 15.20. The large spike on the right is due to the cases where demand is 200 or more and profit is $500. All the little spikes to the left are where demand is less than 200 and profit is less than $500, sometimes considerably less. You can see on the right that the mean profit, the average of the 1000 simulated profits, is only about $380, well less than the $500 suggested by the deterministic model.

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Figure 15.20 Histogram of Simulated Profits (from @RISK)

The point of this simple example is that a deterministic model can be very misleading. In particular, the output from a deterministic model that uses best guesses for uncertain inputs is not necessarily equal to, or even close to, the average of the outputs from a simulation. This is exactly what “the flaw of averages” means.

the Flaw of averages

If a model contains uncertain inputs, it can be misleading to build a deterministic model by using the means of the inputs to predict an output. The resulting output value can be considerably different—lower or higher—than the mean of the output values obtained from running a simulation with uncertainty incorporated explicitly.

Fundamental Insight

15-4 Simulation with Built-in Excel Tools In this section, we show how spreadsheet simulation models can be developed and analyzed with Excel’s built-in tools without using add-ins. As you will see, this is certainly possible, but it presents two problems. First, the @RISK functions illustrated in the Probability Distributions.xlsx file are not available. You are able to use only Excel’s RAND function and transformations of it to generate random numbers from various probability distributions. (You can also use the RANDBETWEEN function, but except for special cases, this doesn’t help much.) Second, there is a bookkeeping problem. Once you build an Excel model with output cells linked to appropriate random input cells,

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15-4 Simulation with Built-in excel tools 7 3 9

you can press the F9 key as often as you like to see how the outputs vary. However, there is no quick way to keep track of these output values and summarize them. This bookkeeping feature is the real strength of a simulation add-in such as @RISK. It can be done with Excel, usually with data tables, but summarizing the resulting data is completely up to the user—you. Therefore, we strongly recommend that you use the “Excel-only” method described in this section only if you don’t have an add-in such as @RISK.

To illustrate the Excel-only procedure, we continue to analyze the calendar ordering problem from Example 15.1. This general problem occurs when a company (such as a news vendor) must make a one-time purchase of a product (such as a newspaper) to meet customer demands for a certain period of time. If the company orders too few newspapers, it will lose potential profit by not having enough on hand to satisfy its customers. If it orders too many, it will have newspapers left over at the end of the day that, at best, can be sold at a loss. More generally, the problem is to match supply to an uncertain demand, a very common problem in business. In much of the rest of this chapter, we will discuss variations of this problem, generally referred to as the newsvendor problem.

EXAMPLE

15.2 SIMULATING WITH EXCEL ONLY AT WALTON BOOKSTORE

Recall that Walton Bookstore must decide how many of next year’s nature calendars to order. Each calendar costs the book- store $7.50 and sells for $10. After January 1, all unsold calendars will be returned to the publisher for a refund of $2.50 per calendar. In this version, we assume that demand for calendars (at the full price) is given by the probability distribution shown in Table 15.1. Walton wants to develop a simulation model to help it decide how many calendars to order.

Demand probability

100 0.30

150 0.20

200 0.30

250 0.15

300 0.05

Table 15.1 Probability Distribution of Demand for Walton Example

Objective To use built-in Excel tools—including the RAND function and data tables, but no add-ins—to simulate profit for several order quantities and ultimately choose the “best” order quantity.

Where Do the Numbers Come From? The numbers in Table 15.1 are the key to the simulation model. They are discussed in more detail next.

Solution We first discuss the probability distribution in Table 15.1. It is a discrete distribution with only five possible values: 100, 150, 200, 250, and 300. In reality, it is clear that other values of demand are possible. For example, there could be demand for exactly 187 calendars. In spite of its apparent lack of realism, we use this discrete distribution for two reasons. First, its simplicity is a nice feature to get you started with simulation modeling. Second, discrete distributions are often used in real business simu- lation models. Even though the discrete distribution is only an approximation to reality, it can still provide important insights.

As for the probabilities listed in Table 15.1, they are typically drawn from historical data or (if historical data are lacking) educated guesses. In this case, the manager of Walton Bookstore has presumably looked at demands for calendars in previous years, and he has used any information he has about the market for next year’s calendars to estimate, for example, that the probability of a demand for 200 calendars is 0.30. The five probabilities in this table must sum to 1. Beyond this requirement, they should be as reasonable and consistent with reality as possible.

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It is important to realize that this is really a decision problem under uncertainty. Walton must choose an order quantity before knowing the demand for calendars. Unfortunately, Solver cannot be used because of the uncertainty.7 Therefore, we develop a simulation model for any fixed order quantity. Then we run this simulation model with various order quantities to see which one appears to be best.

Developing the Simulation Model Now we discuss the ordering model. For any fixed order quantity, we show how Excel can be used to simulate 1000 replications (or any other number of replications). Each replication is an independent replay of the events that occur. To illustrate, suppose you want to simulate profit if Walton orders 200 calendars. Figure 15.21 illustrates the results obtained by simulating 1000 independent replications for this order quantity. (See the file Ordering Calendars - Excel Only 1 Finished.xlsx.) Note that there are many hidden rows in Figure 15.21. To develop this model, use the following steps.

7 @RISK contains a tool called RISKOptimizer that can be used for optimization in a simulation model, but we will not discuss it here.

Developing the Walton Model with Excel Tools Only

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1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23

1016 1017 1018 1019

A B C Simula�on of Walton’s bookstore Range names used:

Cost data Unit cost Unit price Unit refund

Demand distribu�on

Decision variable Order quan�ty

Simula�on Distribu�on of profit

Summary measures for simula�on below Average profit Stdev of profit Minimum profit Maximum profit

$204.13 $328.04

�$250.00 $500.00

Demand 150 100 100 200 100 100 150 150

Revenue $1,500 $1,000 $1,000 $2,000 $1,000 $1,000 $1,500 $1,500

Cost $1,500 $1,500 $1,500 $1,500 $1,500 $1,500 $1,500 $1,500

Refund $125 $250 $250

$0 $250 $250 $125 $125

Value �250

125 500

Frequency 299 191 510

Rel. Freq. 0.299 0.191

0.51

200

D E

LookupTable Order_quan�ty Profit Unit_cost Unit_price Unit_refund

=Model!$D$5:$F$9 =Model!$B$9 =Model!$G$19:$G$1018 =Model!$B$4 =Model!$B$5 =Model!$B$6

F G H I J K

Replica�on 1 2 3 4 5

998 999

1000

$7.50 $10.00

$2.50

Cum Prob 0.00 0.30 0.50 0.80 0.95

Demand 100 150 200 250 300

Probability 0.30 0.20 0.30 0.15 0.05

Random # 0.4695 0.0022 0.2614 0.6220 0.1417 0.1005 0.3798 0.4530

Profit $125

�$250 �$250

$500 �$250 �$250

$125 $125

Figure 15.21 Walton Bookstore Simulation Model

1. Inputs. Enter the cost data in the range B4:B6, the probability distribution of demand in the range E5:F9, and the proposed order quantity, 200, in cell B9. Pay particular attention to the way the probability distribution is entered (and compare to the Discrete sheet in the Probability Distributions.xlsx file). Columns E and F contain the possible demand values and the probabilities from Table 15.1. It is also necessary (see step 2 for the reasoning) to have the cumulative probabilities in column D. To calculate these, first enter the value 0 in cell D5. Then enter the formula

=F5+D5

in cell D6 and copy it to the range D7:D9. 2. Generate random demands. The key to the simulation is the generation of a customer demand in column C from a

random number generated by the RAND function in column B and the probability distribution of demand. Here is how it works. The interval from 0 to 1 is split into five segments: 0.0 to 0.3 (length 0.3), 0.3 to 0.5 (length 0.2), 0.5 to 0.8 (length 0.3), 0.8 to 0.95 (length 0.15), and 0.95 to 1.0 (length 0.05). These lengths are the probabilities of the various demands. Then a demand is associated with each random number, depending on which interval the random number falls in. For example, if a random number is 0.5279, this falls in the third interval, so it is associated with the third possible demand value, 200.

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To implement this procedure, you use a VLOOKUP function based on the range D5:F9 (range-named LookupTable). This table has the cumulative probabilities in column D and the possible demand values in column E. In fact, the whole purpose of the cumulative probabilities in column D is to allow the use of the VLOOKUP function. To generate the simulated demands, enter the formula

=VLOOKUP(RAND(),LookupTable,2)

in cell C19. This formula compares any RAND value to the values in D5:D9 and returns the appropriate demand from E5:E9. (In the file, you will note that random cells are colored green. This coloring convention is not required, but we use it consistently to identify the random cells.)

This step is the key to the simulation, so make sure you understand exactly what it entails. The rest is bookkeeping, as indicated in the following steps.

3. Revenue. Once the demand is known, the number of calendars sold is the smaller of the demand and the order quantity. For example, if 150 calendars are demanded, 150 will be sold. But if 250 are demanded, only 200 can be sold (because Walton orders only 200). Therefore, to calculate the revenue in cell D19, enter the formula

=Unit_price*MIN(C19,Order_quantity)

4. Ordering cost. The cost of ordering the calendars does not depend on the demand; it is the unit cost multiplied by the number ordered. Calculate this cost in cell E19 with the formula

=Unit_cost*Order_quantity

5. Refund. If the order quantity is greater than the demand, there is a refund of $2.50 for each calendar left over; otherwise, there is no refund. Therefore, calculate the refund in cell F19 with the formula

=Unit_refund*MAX(Order_quantity-C19,0)

For example, if demand is 150, then 50 calendars are left over, and this MAX is 50, the larger of 50 and 0. However, if demand is 250, then no calendars are left over, and this MAX is 0, the larger of 250 and 0. (This calculation could also be accomplished with an IF function instead of a MAX function.)

6. Profit. Calculate the profit in cell G19 with the formula

=D19+F19-E19

7. Copy to other rows. This is a “one-line” simulation, where all of the logic is captured in a single row, row 19. For one- line simulations, which are fairly rare, you can replicate the logic with new random numbers very easily by copying down. Copy row 19 down to row 1018 to generate 1000 replications.

8. Summary measures. Each profit value in column G corresponds to one randomly generated demand. You usually want to see how these vary from one replication to another. First, calculate the average and standard deviation of the 1000 profits in cells B12 and B13 with the formulas

=AVERAGE(G19:G1018)

and

=STDEV.S(G19:G1018)

Similarly, calculate the smallest and largest of the 1000 profits in cells B14 and B15 with the MIN and MAX functions. 9. Distribution of simulated profits. There are only three possible profits, 2$250, $125, or $500 (depending on whether

demand is 100, 150, or at least 200—see the following discussion). You can use the COUNTIF function to count the num- ber of times each of these possible profits is obtained. To do so, enter the formula

=COUNTIF($G$19:$G$1018,I19)

in cell J19 and copy it down to cell J21.

Checking Logic with Deterministic Inputs It can be difficult to check whether the logic in your model is correct, because of the random numbers. The reason is that you usually get different output values, depending on the particular random numbers generated. Therefore, it is sometimes useful to enter well-chosen fixed values for the random inputs, just to see whether your logic is correct. We call these deterministic checks. In the present example, you might try several fixed demands, at least one of which is less than the order quantity and at least one of which is greater than the order quantity. For example, if you enter a fixed demand of 150, the revenue, cost,

This rather cumbersome procedure for generating a discrete random number is not necessary when you use @RISK.

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refund, and profit should be $1500, $1500, $125, and $125, respectively. Or if you enter a fixed demand of 250, these outputs are $2000, $1500, $0, and $500. There is no randomness in these values; every correct model should get these same values. If your model doesn’t get these values, there must be a logic error in your model that has nothing to do with random numbers or simulation. Of course, you should fix any such logical errors before reentering the random demand and running the simulation.

You can make a similar check by keeping the random demand, repeatedly pressing the F9 key, and watching the outputs for the different random demands. For example, if the refund is not $0 every time demand exceeds the order quantity, you know you have a logical error in at least one formula. The advantage of deterministic checks is that you can compare your results with those of other users, using agreed-upon test values of the random quantities. You should all get exactly the same outputs.

Discussion of the Simulation Results At this point, it is a good idea to stand back and see what you have accomplished. First, in the body of the simulation, rows 19 through 1018, you randomly generated 1000 possible demands and the corresponding profits. Because there are only five possible demand values (100, 150, 200, 250, and 300), there are only five possible profit values: 2$250, $125, $500, $500, and $500. Also, for the order quantity 200, the profit is $500 regardless of whether demand is 200, 250, or 300. (Make sure you understand why.) A tally of the profit values in these rows, including the hidden rows, indicates that there are 299 rows with profit equal to 2$250 (demand 100), 191 rows with profit equal to $125 (demand 150), and 510 rows with profit equal to $500 (demand 200, 250, or 300). The average of these 1000 profits is $204.13, and their standard deviation is $328.04. (Again, however, remember that your answers will probably differ from these because of different random numbers.)

Typically, a simulation model should capture one or more output variables, such as profit. These output variables depend on random inputs, such as demand. The goal is to estimate the probability distributions of the outputs. In the Walton simulation the estimated probability dis- tribution of profit is

P1Profit 5 2$2502 5 299>1000 5 0.299

P1Profit 5 $1252 5 191>1000 5 0.191

P1Profit 5 $5002 5 510>1000 5 0.510

The estimated mean of this distribution is $204.13 and the estimated standard deviation is $328.04. It is important to realize that if the entire simulation is run again with different random numbers, such as the ones you might have generated on your PC, the answers will probably be slightly different. For illustration, we pressed the F9 key five times and got the following average profits: $213.88, $206.00, $212.75, $219.50, and $189.50. This is truly a case of “answers will vary.”

Notes about Confidence Intervals It is common in computer simulations to estimate the mean of some distribution by the average of the simulated observations. The usual practice is then to accompany this estimate with a con- fidence interval, which indicates the accuracy of the estimate. You should recall from Chapter 8 that to obtain a confidence interval for the mean, you start with the estimated mean and then add and subtract a multiple of the standard error of the estimated mean. If the estimated mean (that is, the average) is X, the confidence interval is given in the following formula.

For this model, the output distribution is also discrete: There are only three possible profits for an order quantity of 200.

The confidence interval provides a measure of accuracy of the mean profit, as estimated from the simulation.

Confidence Interval for the Mean

X { Multiple 3 Standard Error of X

Standard Error of X

s>!n

The standard error of X is the standard deviation of the observations divided by the square root of n, the number of observations:

We repeat these basic facts about confidence intervals from Chapter 8 here for your convenience.

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Here, s is the standard deviation of the observations. You can obtain it with the STDEV.S function in Excel. The multiple in the confidence interval formula depends on the confidence level and the number of observations. If the

confidence level is 95%, for example, the multiple is very close to 2, so a good guideline is to go out two standard errors on either side of the average to obtain an approximate 95% confidence interval for the mean.

Approximate 95% Confidence Interval for the Mean

X { 2s>!n

Analysts often plan a simulation so that the confidence interval for the mean of some important output will be sufficiently narrow. The reasoning is that narrow confidence intervals imply more precision about the estimated mean of the output variable. If the confidence level is fixed at some value such as 95%, the only way to narrow the confidence interval is to simu- late more replications. Assuming that the confidence level is 95%, the following value of n is required to ensure that the resulting confidence interval will have a half-length approximately equal to some specified value B:

The idea is to choose the number of iterations large enough so that the resulting confidence interval will be sufficiently narrow.

Sample Size Determination

n 5 4 3 (Estimated standard deviation)2

This formula requires an estimate of the standard deviation of the output variable. For example, in the Walton simula- tion you can check (from the values in Figure 15.21) that the 95% confidence interval for mean profit with n 5 1000 has half-length ($224.46 2 $183.79) >2 5 $20.33. Suppose that you want to reduce this half-length to $12.50—that is, you want B 5 $12.50. You do not know the exact standard deviation of the profit distribution, but you can estimate it from the simula- tion as $328.04. Therefore, to obtain the required confidence interval half-length B, you need to simulate n replications, where

n 5 4(328.04)2

12.502 ^ 2755

(When this formula produces a noninteger, it is common to round upward.) The claim, then, is that if you rerun the simu- lation with 2755 replications rather than 1000 replications, the half-length of the 95% confidence interval for the mean profit will be close to $12.50.

Finding the Best Order Quantity We are not yet finished with the Walton example. So far, the simulation has been run for only a single order quantity, 200. Walton’s ultimate goal is to find the best order quantity. Even this statement must be clarified. What does “best” mean? As in Chapter 6, one possibility is to use the expected profit—that is, EMV—as the optimality criterion, but other characteristics of the profit distribution could influence the decision. You can obtain the required outputs with a data table. Specifically, you can use a data table to rerun the simulation for other order quantities. This data table and a corresponding chart are shown in Figure 15.22. (This is still part of the finished version of the Ordering Calendars - Excel Only 1 Finished.xlsx file.)

To create this table, enter the trial order quantities shown in the range M20:M28, enter the link 5B12 to the average profit in cell N19, and select the data table range M19:N28. Then select Data Table from the What-If Analysis dropdown list on the Data ribbon, specifying that the column input cell is B9 (see Figure 15.21). Finally, create a column chart of the average prof- its in the data table. An order quantity of 150 appears to maximize the average profit. Its average profit of $258.00 is slightly larger than the average profits from nearby order quantities and much larger than the profit gained from an order of 200 or more calendars. However, again keep in mind that this is a simulation, so that all these average profits depend on the particular random numbers generated. If you rerun the simulation with different random numbers, it is conceivable that some other order quantity could be best.

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Using a Data Table to Repeat Simulations The Walton simulation is a particularly simple one-line simulation model. All the logic—generating a demand and calcu- lating the corresponding profit—can be captured in a single row. Then to replicate the simulation, you can simply copy this row down as far as you like. Many simulation models are significantly more complex and require more than one row to capture the logic. Nevertheless, they still result in one or more output quantities (such as profit) that you want to rep- licate. We now illustrate another method of replicating with Excel tools only that is more general (still using the Walton example). It uses a data table to generate the replications. Refer to Figure 15.23 and the file Ordering Calendars - Excel Only 2 Finished.xlsx.

Through row 19, the only difference between this model and the previous model is that the RAND function is embed- ded in the VLOOKUP function for demand in cell B19. This makes the model slightly more compact. As before, it uses the given data at the top of the spreadsheet to construct a typical “prototype” of the simulation in row 19. This time, however, you do not copy row 19 down. Instead, you create a data table in the range A23:B1023 to replicate the simulation 1000 times.

Figure 15.22 Data Table for Walton Bookstore Simulation 17

18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

M N O P Q R S T Data table for average profit versus order quan�ty

Order quan�ty

100 125 150 175 200 225 250 275 300

Average Profit $204.13 $250.00 $256.81 $258.00 $237.44 $209.75 $118.50

$12.63 ($95.88)

($196.13)

Average Profit

($250.00) ($200.00) ($150.00) ($100.00)

($50.00) $0.00

$300.00 $250.00 $200.00 $150.00 $100.00

$50.00

100 125 150 175 200 225 250

Order Quan�ty

275 300

To optimize in simulation models, try various values of the decision variable(s) and run the simulation for each of them.

Calculation Settings with Data Tables

Sometimes you will create a data table and the values will be constant the whole way down. This could mean you did something wrong, but more likely it is due to a calculation setting. To check, go to the Formulas ribbon and click the Calculation Options dropdown arrow. If it isn’t Automatic (the default setting), you need to click the Calculate Now (or Calculate Sheet) button or press the F9 key to make the data table calculate correctly. (The Calculate Now and F9 key recalculate everything in your workbook. The Calculate Sheet option recalculates only the active sheet.) Note that the Automatic Except for Data Tables setting is there for a reason. Data tables, especially those based on complex simulations, can take a lot of time to recalculate, and with the default setting, this recalculation occurs every time anything changes in your workbook. So the Automatic Except for Data Tables setting is handy to prevent data tables from recalculating until you force them to by pressing the F9 key or clicking one of the Calculate buttons.

Excel Tip

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In column A, you list the replication numbers, 1 to 1000. Next, you enter the formula 5F19 in cell B23. This forms a link to the profit from the prototype row for use in the data table. Then you create a data table and enter any blank cell (such as C23) as the column input cell. (No row input cell is necessary, so its box should be left empty.) This tricks Excel into repeating the row 19 calculations 1000 times, each time with a new random number, and reporting the profits in column B of the data table. (If you wanted to see other simulated quantities, such as revenue, for each replication, you could add extra output columns to the data table.)

Figure 15.23 Using a Data Table to Simulate Replications 17

18 19 20 21 22 23 24 25 26 27

1021 1022 1023

A B C D E F

Demand Revenue Cost Refund Profit 200 $2,000 $1,500 $0 $500

Data table for replica�ons, each shows profit from that replica�on

Simula�on

Replica�on Profit $500

1 $125 2 –$250 3 $500 4 –$250 5 $125

998 $125 999 $500

1000 $500

28 The key to simulating many replications in Excel (without an add-in) is to use a data table with any blank cell as the column input cell.

How Data Tables Work

To understand this procedure, you must understand exactly how data tables work. When you create a data table, Excel takes each value in the left column of the data table (here, column A), substitutes it into the cell desig- nated as the column input cell, recalculates the spreadsheet, and returns the output value (or values) you have requested in the top row of the data table (such as profit). It might seem silly to substitute each replication num- ber from column A into a blank cell such as cell C23, but this part is really irrelevant. The important part is the recalculation. Each recalculation leads to a new random demand and corresponding profit, and these profits are the quantities you want to keep track of. Of course, this means that you should not freeze the quantity in cell B19 before forming the data table. The whole point of the data table is to use a different random number for each replication, and this will occur only if the random demand in row 19 is “live.”

Excel Tip

Using a Two-Way Data Table You can carry this method one step further to see how the profit depends on the order quantity. Here you use a two-way data table with the replication number along the side and possible order quantities along the top. See Figure 15.24 and the file Ordering Calendars - Excel Only 3 Finished.xlsx. Now the data table range is A23:J1023, and the driving formula in cell A23 is again the link 5F19. The column input cell should again be any blank cell, and the row input cell should be B9 (the order quantity). Each cell in the body of the data table shows a simulated profit for a particular replication and a particular order quantity, and each is based on a different random demand.

By averaging the numbers in each column of the data table (see row 14 in the file), you can see which is the best order quantity. It is also helpful to construct a column chart of these averages, as in Figure 15.25. Now, however, assuming you have not frozen anything, the data table and the corresponding chart will change each time you press the F9 key. To see whether 150 is always the best order quantity, you can press the F9 key and see whether the bar above 150 continues to be the highest. (It usually is, but not always.)

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Figure 15.24 Using a Two-Way Data Table for the Simulation Model

17 18 19 20 21 22 23 24 25 26 27 28

1021 1022 1023

A B C D E F G H I J Simulation

Demand Revenue Cost Refund Profit 100 $1,000 $1,500 $250

Data table showing profit for replications with various order quantities Order quantityReplication

100 $250 $250 $250 $250 $250 $250 $250 $250

125 $313 $125 $313 $125 $125 $125 $313 $125

150 $375 $375

$0 $375

$0 $375 $375 $375

175 –$125

$250 $438 $438

–$125 $438

200 $500

–$250 $500 $500

–$250 –$250

$125 $500

225 –375 –375

375 0

250 625

–500 625 250 250

–500 –125

625

275 125 125

–250 –625 –625 –625

125 –625

300 –375 –375 –750 –750 –750 –375 –750 –750

($250.00) 1 2 3 4 5

998 999

1000

–$250

–$300.00

–$200.00

–$100.00

$0.00

$400.00

$300.00

$200.00

$100.00

100 125 150 175 200 225

Order Quan�ty

Average Profit

275 300250

Figure 15.25 Column Chart of Average Profits for Different Order Quantities

By now you should appreciate the usefulness of data tables in spreadsheet simulations. They allow you to take a simula- tion model and replicate its key results as often as you like. This method makes summary statistics (over the entire group of replications) and corresponding charts fairly easy to obtain. Nevertheless, it takes some work to create the data tables, sum- mary measures, and charts. In the next section you will see how the @RISK add-in does much of this work for you.

Decisions Based on Simulation results

Given the emphasis in Chapter 6 on the EMV criterion for decision making under uncertainty, you might believe that the mean of an output from a simulation is the only summary measure of the output relevant for decision making. However, this is not necessarily true. When you run a simulation, you approximate the entire distribution of an output, including its mean, its standard deviation, its percentiles, and more. As a decision maker, you could base your decision on any of these summary measures, not just the mean. For example, you could focus on making the standard deviation small, making the 5th percentile large, or others. The point is that the results from a simulation provide a lot more information about an output than simply its mean, and this information can be used for decision making.

Fundamental Insight

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Problems

Level A 11. Suppose you own an expensive car and purchase auto

insurance. This insurance has a $1000 deductible, so that if you have an accident and the damage is less than $1000, you pay for it out of your pocket. However, if the damage is greater than $1000, you pay the first $1000 and the insurance pays the rest. In the current year there is probability 0.025 that you will have an accident. If you have an accident, the damage amount is normally distrib- uted with mean $3000 and standard deviation $750. a. Use Excel to simulate the amount you have to pay for

damages to your car. This should be a one-line sim- ulation, so run 5000 iterations by copying it down. Then find the average amount you pay, the standard deviation of the amounts you pay, and a 95% confi- dence interval for the average amount you pay. (Note that many of the amounts you pay will be 0 because you have no accidents.)

b. Continue the simulation in part a by creating a two- way data table, where the row input is the deduct- ible amount, varied from $500 to $2000 in multiples of $500. Now find the average amount you pay, the standard deviation of the amounts you pay, and a 95% confidence interval for the average amount you pay for each deductible amount.

c. Do you think it is reasonable to assume that damage amounts are normally distributed? What would you criticize about this assumption? What might you sug- gest instead?

12. In August of the current year, a car dealer is trying to determine how many cars of the next model year to order. Each car ordered in August costs $20,000. The demand for the dealer’s next year models has the prob- ability distribution shown in the file P15_12.xlsx. Each car sells for $25,000. If demand for next year’s cars exceeds the number of cars ordered in August, the dealer must reorder at a cost of $22,000 per car. Excess cars

can be disposed of at $17,000 per car. Use simulation to determine how many cars to order in August. For your optimal order quantity, find a 95% confidence interval for the expected profit.

13. In the Walton Bookstore model, suppose that Walton receives no money for the first 50 excess calendars returned but receives $2.50 for every calendar after the first 50 returned. Does this change the optimal order quantity?

14. A sweatshirt supplier is trying to decide how many sweatshirts to print for the upcoming NCAA basketball championships. The final four teams have emerged from the quarterfinal round, and there is now a week left until the semifinals, which are then followed in a couple of days by the finals. Each sweatshirt costs $10 to produce and sells for $25. However, in three weeks, any leftover sweatshirts will be put on sale for half price, $12.50. The supplier assumes that the demand for his sweat- shirts during the next three weeks (when interest in the tournament is at its highest) has the distribution shown in the file P15_14.xlsx. The residual demand, after the sweatshirts have been put on sale, has the distribution also shown in this file. The supplier, being a profit max- imizer, realizes that every sweatshirt sold, even at the sale price, yields a profit. However, he also realizes that any sweatshirts produced but not sold (even at the sale price) must be thrown away, resulting in a $10 loss per sweatshirt. Analyze the supplier’s problem with a simu- lation model.

Level B 15. In the Walton Bookstore model with a discrete demand

distribution, explain why an order quantity other than one of the possible demands cannot maximize the expected profit. (Hint: Consider an order of 190 cal- endars, for example. If this maximizes expected profit, then it must yield a higher expected profit than an order of 150 or 100. But then an order of 200 calendars must also yield a larger expected profit than 190 calendars. Why?)

15-5 Simulation with @RISK Spreadsheet simulation modeling has become extremely popular in the past few decades, both in the academic and corporate communities. Much of the reason for this popularity is due to simulation add-ins such as @RISK. There are two primary advantages to using such an add-in. First, an add-in gives you easy access to many probability distributions you might want to use in your simulation models. You already saw in Section 15-2 how the RiskDiscrete, RiskNormal, and RiskTriang functions, among others, are easy to use and remember. Second, an add-in allows you to perform simulations much more easily than is possible with Excel alone. To replicate a simulation in Excel, you typically need to build a data table. Then you have to calculate summary statistics, such as averages, standard deviations, and percentiles, with built-in Excel functions. If you want graphs to enhance

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the analysis, you have to create them. In short, you have to perform a number of time- consuming steps for each simulation. Simulation add-ins such as @RISK perform much of this work automatically.

Although we focus only on @RISK in this book, it is not the only simulation add-in available for Excel. Two worthy competitors are Crystal Ball, available from Oracle, and Risk Solver Platform, available from Frontline Systems, the developer of Solver. Both Crystal Ball and Risk Solver Platform have much of the same functionality as @RISK. However, the authors have a natural bias for @RISK—we have been permitted by its developer, Palisade Corporation, to provide the academic version with this book. (Instruc- tions for accessing the academic version of this software can be found in the preface and on the cengage.com website.) If it were not included, you would have to purchase it from Palisade at a fairly steep price. Indeed, Microsoft Office does not include @RISK, Crystal Ball, Risk Solver Platform, or any other simulation add-in—you must purchase them separately.

15-5a @RISK Features Here is an overview of some of @RISK’s features. We will discuss these in more detail in this section.

• @RISK contains a number of functions such as RiskNormal and RiskDiscrete that make it easy to generate observations from a wide variety of probability distributions. You saw some of these in Section 15-2.

• You can designate any cell or range of cells in your simulation model as output cells. When you run the simulation, @RISK automatically keeps summary measures (averages, standard deviations, percentiles, and others) from the values generated in these output cells across the replications. It also creates graphs such as histograms of these values. In other words, @RISK takes care of tedious bookkeeping operations for you.

• @RISK has a special function, RiskSimtable, that allows you to run the same simu- lation several times, using a different value of some key input variable each time. This input variable is often a decision variable. For example, suppose that you want to sim- ulate an inventory ordering policy as in the Walton Bookstore example. Your ultimate purpose is to compare simulation outputs across a number of possible order quantities such as 100, 150, 200, 250, and 300.If you use an appropriate formula involving the RiskSimtable function, the entire simulation is performed for each of these order quantities separately—with one click of a button. You can then compare the outputs to choose the best order quantity.

15-5b Loading @RISK To build simulation models with @RISK, you must have Excel open with @RISK loaded. The first step, if you have not already done so, is to install the Palisade Deci- sionTools suite with its Setup program. Then you can load @RISK by clicking the Windows Start button, selecting the Programs group, selecting the Palisade Decision- Tools group, and selecting @RISK. If Excel is already open, this loads @RISK inside Excel. If Excel is not yet open, this launches Excel and @RISK simultaneously. After @RISK is loaded, you see an @RISK tab and the corresponding @RISK ribbon in Figure 15.26.

@RISK provides a number of functions for simulating from various distributions, and it takes care of all the bookkeeping in spreadsheet simulations. Excel simula- tions without @RISK require much more work for the user.

Figure 15.26 @RISK Ribbon

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Learning the Software

When you launch @RISK, you will see a Welcome screen. (This Welcome screen is also available at any time from the @RISK Help menu.) Among other things, this Welcome screen has a Quick Start link, which guides beginners through the basic features of @RISK. It also has a Guided Tours link to a series of videos on @RISK’s basic to advanced features.

@RISK Tip

The majority of the work (and thinking) goes into developing the model. Setting up @RISK and then running it are relatively easy.

15-5c @RISK Models with a Single Random Input In the remainder of this section we will illustrate some of @RISK’s functionality by revis- iting the Walton Bookstore example. The next chapter demonstrates the use of @RISK in a number of interesting simulation models. Throughout this discussion, you should keep one very important idea in mind. The development of a simulation model is basically a two-step procedure. The first step is to build the model itself. This step requires you to enter the logic that transforms inputs (including @RISK functions such as RiskDiscrete) into outputs such as profit. This is where most of the work and thinking go, exactly as in models from previous chapters, and @RISK cannot do this for you. It is your job to enter the formulas that link inputs to outputs appropriately. However, once this logic has been incorporated, @RISK takes over in the second step. It automatically replicates your model, with different random numbers on each replication, and it reports any summary measures that you request in tabular or graphical form. Therefore, @RISK greatly decreases the amount of busy work you need to do, but it is not a magic bullet.

We begin by analyzing an example with a single random input variable.

EXAMPLE

15.3 USING @RISK AT WALTON BOOKSTORE Recall that Walton Bookstore buys calendars for $7.50, sells them at the regular price of $10, and gets a refund of $2.50 for all calendars that cannot be sold. In contrast to Example 15.2, we assume now that Walton estimates a triangular probability distribution for demand, where the minimum, most likely, and maximum values of demand are 100, 175,and 300, respectively. The company wants to use this probability distribution, together with @RISK, to simulate the profit for any particular order quantity, with the ultimate goal of finding the best order quantity.

Objective To learn @RISK’s basic functionality by revisiting the Walton Bookstore problem.

Where Do the Numbers Come From? The monetary values are the same as before. The parameters of the triangular distribution of demand are probably Walton’s best subjective estimates, possibly guided by its experience with previous calendars. As in many simulation examples, the triangular distribution is chosen for simplicity. In this case, the manager would need to estimate only three quantities: the min- imum possible demand, the maximum possible demand, and the most likely demand.

Solution We use this example to illustrate important features of @RISK. We first show how it helps you to implement an appropriate input probability distribution for demand. Then we show how it can be used to build a simulation model for a specific order quantity and generate outputs from this model. Finally, we show how the RiskSimtable function can be used to simultaneously generate outputs from several order quantities so that you can choose the optimal order quantity.

This is the same Walton Bookstore model as before, except that a triangular distribution for demand is used.

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Developing the Simulation Model The spreadsheet model for profit is essentially the same model developed previously without @ RISK, as shown in Figure 15.27. (See the file Ordering Calendars - Basic Model Finished. xlsx.) There are only a few new things to be aware of. Developing the Walton

Model with @RISK

Settings When Opening a Workbook

When you open a workbook with an @RISK model, such as those that accompany this chapter, you might be asked whether you want to change the current @RISK settings to match those stored in the workbook. You should generally click Yes. This changes settings, such as the number of iterations, to those you stored previ- ously with the workbook instead of using @RISK default settings.

@RISK Tip

Figure 15.27 Simulation Model with a Fixed Order Quantity

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23

A B C D E F G H I J Simula�on of Walton’s Bookstore using @RISK Range names used:

Order_quan�ty =Model!$B$9 Cost data Demand distribu�on - triangular

Unit_cost =Model!$B$4Unit cost =Model!$B$5Unit_price

Minimum Unit price Most likely

=Model!$B$6Unit_refundUnit refund Maximum

Decision variable Order quan�ty 200

Simula�on Demand Revenue Cost Refund Profit

255 $2,000 $1,500 $0 $500

Summary measures of profit from @RISK - based on 1000 itera�ons Minimum –$242.50 Maximum $500.00 Average $337.51 Standard devia�on $189.06 5th percen�le –$47.50 95th percen�le $500.00 P(profit <=300) 0.360 P(profit > 400) 0.515

Profit =Model!$F$13 100$7.50

$10.00 175 $2.50 300

1. Input distribution. To generate a random demand, enter the formula

=ROUND(RiskTriang(E4,E5,E6),0)

in cell B13 for the random demand. This uses the RiskTriang function to generate a demand from the triangular distribu- tion. (As before, our convention is to color random input cells green.) Excel’s ROUND function is used to round demand to the nearest integer. Recall from the discussion in Section 15-3 that Excel has no built-in functions to generate random numbers from a triangular distribution, but this is easy with @RISK.

2. Output cell. When the simulation runs, you want @RISK to keep track of profit. In @RISK’s terminology, you need to designate the Profit cell, F13, as an output cell. To do this, select cell F13 and then click the Add Output button on the @RISK ribbon. (See Figure 15.26.) This adds RiskOutput(“label”)1 to the cell’s formula. (Here, “label” is a label that @RISK uses for its reports. In this case it makes sense to use “Profit” as the label.) The formula in cell F13 changes from

=C13+E13-D13

=RiskOutput(‘‘Profit’’)+C13+E13-D13

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15-5 Simulation with @rISK 7 5 1

The plus sign following RiskOutput is simply @RISK’s way of indicating that you want to keep track of the value in this cell (for reporting reasons) as the simulation progresses. Any number of cells can be designated in this way as output cells. They are typically the “bottom line” values of primary interest. Our convention is to color such cells gray.

3. Summary functions (optional). There are several places where you can store @RISK results. One of these is to use @RISK statistical functions to place results in your model worksheet. @RISK provides several functions for summarizing output values. Some of these are illustrated in the range B16:B23 of Figure 15.27. They contain the formulas

=RiskMin(F13)

=RiskMax(F13)

=RiskMean(F13)

=RiskStddev(F13)

=RiskPercentile(F13,0.05)

=RiskPercentile(F13,0.95)

=RiskTarget(F13,300)

and

=1-RiskTarget(F13,400)

The values in these cells are not meaningful until you run the simulation (so do not be alarmed if they contain errors when you open the file). However, once the simulation runs, these formulas capture summary statistics of profit. For example, RiskMean calculates the average of the 1000 simulated profits, RiskPercentile finds the value such that the specified per- centage of simulated profits are less than or equal to this value, and RiskTarget finds the percentage of simulated profits less than or equal to the specified value. Although these same summary statistics also appear in other @RISK reports, you might like to have them in the same worksheet as the model. (You can find a list of all @RISK statistical functions from the Simulation Result group in the Insert Function dropdown list on the @RISK ribbon.)

The RiskOutput function indicates that a cell is an output cell, so that @RISK will keep track of its values throughout the simulation.

These @RISK summary functions allow you to show simulation results on the same sheet as the model. However, they are totally optional.

Color Coding

@RISK has an optional color coding feature. This option is in the Utilities group of the @RISK ribbon. It is a toggle. If it is toggled off, you see our blue/red/gray/green coloring. If it is toggled on, you see @RISK’s color coding: blue for random input cells, red for output cells, green for statistical functions, and yellow for decision cells (for RISKOptimizer models). @RISK even allows you to change the coloring scheme if you prefer.

@RISK Feature

Running the Simulation After you develop the model, the rest is straightforward. The procedure is always the same: (1) specify simulation settings, (2) run the simulation, and (3) examine the results.

1. Simulation settings. You must first choose some simulation settings. To do so, the buttons on the left in the Simulation group (see Figure 15.28) are useful. We typically do the following:

• Set Iterations to a number such as 1000. (@RISK calls replications “iterations.”) Any number can be used, but because the academic version of @RISK allows only 1000 uninterrupted iterations, we typically choose 1000.

• Set Simulations to 1. In a later section, we will explain when you would request multiple simulations.

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• The “dice” button is a toggle for what appears in your worksheet. If it is colored ( Random), all random cells appear random (they change when you press the F9 key). If it is white (Static), only the means appear in random input cells and the F9 key has no effect. We tend to prefer the Random setting, but this setting is irrelevant when you run the simulation.

• Many more settings are available by clicking the Simulation Settings button to the left of the “dice” button, but the ones we mentioned should suffice. In addition, more permanent settings can be chosen from Application Settings in the Utilities dropdown list on the @RISK ribbon. You can experiment with these, but the only one we like to change is the Place Reports In setting in the Reports group. The default is to place reports in a new work- book. If you like the reports to be in the same workbook as your model, you can change this setting to Active Workbook.

Simulation and Application Settings in @RISK

Figure 15.28 Simulation Group on @RISK Ribbon

Leave Latin Hyper-cube sampling on. It produces more accurate results.

Latin Hypercube Sampling and Mersenne Twister Generator

Two settings you shouldn’t change are the Sampling Type and Generator settings (available from the Simulation Settings button and then the Sampling tab). They should remain at the default Latin Hypercube and Mersenne Twister settings. The Mersenne Twister is one of many algorithms for generating random numbers, and it has been shown to have very good statistical properties. (Not all random number generators do.) Latin Hypercube sampling is a more efficient way of sampling than the other option (Monte Carlo) because it produces a more accurate estimate of the output distribution. In fact, we were surprised how accurate it is. In repeated runs of this model, always using different random numbers, we virtually always got a mean profit within a few pennies of $337.50. It turns out that this is the true mean profit for this input distribution of demand. Amazingly, simulation estimates it correctly—almost exactly—on virtually every run. However, this means that a confidence interval for the mean, based on @RISK’s outputs and the usual confidence inter- val formula (which assumes Monte Carlo sampling), is much wider (more pessimistic) than it should be. Therefore, we do not even calculate such confidence intervals from here on. However, it is not impossible. The accompanying video explains a method called Batch Means for calculating confidence intervals when Latin Hypercube sampling is used.

@RISK Technical Issues

2. Run the simulation. To run the simulation, click the Start Simulation button on the @RISK ribbon. When you do so, @RISK repeatedly generates a random number for each random input cell, recalculates the worksheet, and keeps track of all output cell values. You can watch the progress at the bottom left of the screen. Also, if the Automatically Show Output Graph button (to the right of the dice button) is toggled to colored, you will see a histogram of the cur- rently selected input or output cell being built as the simulation runs. If you find this annoying, you can toggle this button to white.

3. Examine the results. You must decide which results you want and where you want them. @RISK provides a lot of possi- bilities, and we mention the most frequently used.

• You can ask for summary measures in your model worksheet by using the @RISK statistical functions, such as RiskMean, discussed earlier.

• The quickest way to get results is to select an input or output cell (we chose the profit cell, F13) and then click the Browse Results button in the Results group of the @RISK ribbon. (See Figure 15.29.) This provides an interactive graph of the selected value, as shown in Figure 15.30. You can move the “sliders” (the two vertical bars) on this graph to see probabilities of various outcomes. The window you see from Browse Results is temporary—it goes away when you click Close. You can make a per- manent copy of the chart by clicking the second button from the left (see the bottom of Figure 15.30) and choosing one of the copy options.

Latin Hypercube vs Monte Carlo Sampling

For a quick graph of the distribution of an output or input, select the output or input cell and click @RISK’s Browse Results button.

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15-5 Simulation with @rISK 7 5 3

Figure 15.29 Results Group on @RISK Ribbon

Figure 15.30 Interactive Graph of Profit Distribution

Graph Type

Once you get an @RISK graph such as the one in Figure 15.30, you can display it in several ways by clicking the fourth button’s dropdown arrow at the bottom of the graph window. In particular, if your graph doesn’t look quite like the ones shown here, try changing from Discrete Probability to Probability Density.

@RISK Tip

Percentiles Displayed on Graphs

The graph in Figure 15.35 has the right slider on 500 and shows 5% to the right of it. By default, @RISK puts the sliders at the 5th and 95th percentiles, so that 5% is on either side of them. For this example, 500 is indeed the 95th percentile (why?), but the picture is a bit misleading because there is no chance of a profit greater than 500. If you manually move the right slider away from 500 and back again, it will correctly indicate that there is no probability to the right of 500.

@RISK Tip

Saving Graphs and Tables

When you run a simulation with @RISK and then save your file, it asks whether you want to save your graphs and tables. We suggest that you save them. This makes your file slightly larger, but when you reopen it, the temporary graphs and tables, such as the histogram in Figure 15.30, are still available. Otherwise, you will have to rerun the simulation.

@RISK Tip

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• You can click the Summary button (see Figure 15.29) to see the window in Figure 15.31 with the summary measures for Profit. In general, this report shows the summary for all des- ignated inputs and outputs. By default, this Results Summary window shows a mini graph for each output and a number of numerical summary measures. It is also easy to customize. If you right-click anywhere on this table and choose Columns for Table, you can check or uncheck various options. For most of the later screenshots in this book, we elected not to show the Error column, but instead to show the standard deviation column.

For a quick (and customizable) report of the results, click @RISK’s Summary button.

Figure 15.31 Summary Table of Profit Output

• You can click the Excel Reports button (see Figure 15.29) to choose from a number of reports that are placed on new worksheets. This is a good option if you want permanent (but non-interactive) copies of reports in your workbook. As an example, Figure 15.32 shows part of the Quick Reports option you can request. It has the same information as the sum- mary report in Figure 15.31, plus more.

If you want permanent copies of the simulation results, click on @RISK’s Excel Reports buttons and check the reports you want. They will be placed in new worksheets.

Figure 15.32 @RISK Quick Report Profit

Simulation Summary Information

Summary Statistics for Profit

Workbook Name Ordering Calendars - Basi 1 1000 1 1 Latin Hypercube 7/24/2018 10:58 00:00:04 Mersenne Twister 123

Number of Simulations

Simulation Start Time Simulation Duration Random # Generator Random Seed

Number of Iterations Number of Inputs Number of Outputs Sampling TypeProfit

Minimum −$242.50 $500.00 $337.51 $189.06

1000

Maximum Mean Std Dev Values

−$48

−$ 30

$3 00

$4 00

$5 00

$6 00

−$ 20

$2 00

−$ 10

$1 00$0

$500

5.0% 45%

40%

35%

30%

25%

20%

15%

10%

5% 0%

Profit

Statistics Minimum Maximum Mean Std Dev Variance Skewness Kurtosis Median Mode Left X

Right X

Diff X Diff P #Errors Filter Min Filter Max #Filtered

Right P

Left P

Percentile −$243 5% −$48

$43 $103 $163 $208 $253 $290 $328 $373

10% 15% 20% 25% 30% 35% 40% 45%

$410 $455 $500 $500 $500 $500 $500 $500 $500 $500

50% 55% 60% 65% 70% 75% 80% 85% 90% 95%

$500 $338 $189 35744.95743 −0.948717482

−$48 5%

95% $548 90% 0

Off Off

$500

2.797215761 $410 $500

Profit

Minimum −$242.50 $500.00 $337.51 $189.06

1000

Maximum Mean Std Dev Values

−$48

−$ 30

$3 00

$4 00

$5 00

$6 00

−$ 20

$2 00

−$ 10

$1 00$0

$500

5.0% 1.0

0.0

0.2

0.4

0.6

0.8

90.0%

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15-5 Simulation with @rISK 7 5 5

Discussion of the Simulation Results The strength of @RISK is that it keeps track of any outputs you designate and then allows you to show the corresponding results as graphs or tables, in temporary windows or in permanent worksheets. As you have seen, @RISK provides several options for displaying results, and we encourage you to explore the possibilities. However, don’t lose sight of the overall goal: to see how outputs vary as random inputs vary, and to generate reports that tell the story most effectively. For this particular example, the results in Figures 15.27, 15.30, 15.31, and 15.32 allow you to conclude the following:

• The smallest simulated profit (out of 1000) was 2$242.50, the largest was $500, the average was $337.51, and the standard deviation of the 1000 profits was $189.06. Of all simulated profits, 5% were 2$47.50 or below, 95% were $500 or above, 36% were less than or equal to $300, and 51.5% were larger than $400. (See Figure 15.27. These results are also available from the summary table in Figure 15.31 or the quick report in Figure 15.32.)

• The profit distribution for this order quantity is extremely skewed to the left, with a large bar at $500. (See Figure 15.30.) Do you see why? It is because profit is exactly $500 if demand is greater than or equal to the order quantity, 200. In other words, the probability that profit is $500 equals the probability that demand is at least 200. (This probability is 0.4.) Lower demands result in decreasing profits, which explains the gradual decline in the histogram from right to left.

Using RiskSimtable Walton’s ultimate goal is to choose an order quantity that provides a large average profit. You could rerun the simulation model several times, each time with a different order quantity in the order quantity cell, and compare the results. However, this has two drawbacks. First, it takes a lot of time and work. The second drawback is more subtle. Each time you run the simulation, you get a different set of random demands. Therefore, one of the order quantities could win the contest just by luck. For a fairer comparison, it is better to test each order quantity on the same set of random demands.

The RiskSimtable function in @RISK enables you to obtain a fair comparison quickly and easily. This function is illustrated in Figure 15.33. (See the file Ordering Calendars - RiskSimtable Finished.xlsx.) There are two modifications to the previous model. First, the order quantities to test are listed in row 9. (We chose these as representative order quantities. You could change, or add to, this list.) Second, instead of entering a number in cell B9, you enter the formula

=RiskSimtable(D9:H9)

Note that the list does not need to be entered in the spreadsheet (although it is a good idea to do so). You could instead enter the formula

=RiskSimtable({150,175,200,225,250})

where the list of numbers must be enclosed in curly brackets. In either case, the worksheet displays the first member of the list, 150, and the corresponding calculations for this first order quantity. However, the model is now set up to run the simulation for all order quantities in the list.

The RiskSimtable function allows you to run several simulations at once—one for each value of some variable (often a decision variable).

Figure 15.33 Model with a RiskSimtable Function

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23

A B C D E F G H I J K Simula�on of Walton’s Bookstore using @RISK Range names used:

Order_quan�ty =Model!$B$9 Cost data Demand distribu�on - triangular

Unit_cost =Model!$B$4 Profit =Model!$F$13

Unit cost =Model!$B$5Unit_price

100Minimum$7.50 Unit price $10.00 Most likely

=Model!$B$6Unit_refund 175

Unit refund $2.50 Maximum 300

Decision variable Order quan��es to try Order quan�ty 150 150 175 200 225 250

Simulated quan��es Demand Revenue Cost Refund Profit

187 $1,500 $1,125 $0 $375

Summary measures of profit from @RISK - based on 1000 itera�ons for each simula�on 54321Simula�on

Order quan�ty 150 175 200 225 250 Minimum $7.50 –$117.50 –$242.50 –$367.50 –$492.50 Maximum $375.00 $437.50 $500.00 $562.50 $625.00 Average $354.14 $367.17 $337.48 $270.29 $174.96 Standard devia�on $59.04 $121.93 $189.12 $247.10 $287.01 5th percen�le $202.50 $77.50 –$47.50 –$172.50 –$297.50 95th percen�le $375.00 $437.50 $500.00 $562.50 $625.00

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To implement this, only one setting needs to be changed. As before, enter 1000 for the number of iterations, but also enter 5 for the number of simulations. @RISK then runs five simulations of 1000 iterations each, one simulation for each order quantity in the list, and it uses the same 1000 random demands for each simulation. This provides a fair comparison.

RiskSimtable

To run several simulations all at once, enter the formula =RiskSimtable(Inputs) in any cell. Here, Inputs refers to a list of the values to be simulated, such as various order quantities. This can be a range reference or a list of values in curly brackets. Before running the simulation, make sure the number of simulations is set to the num- ber of values in the Inputs list.

@RISK Function

You can again get results from the simulation in various ways. Here are some possibilities.

• You can enter the same @RISK statistical functions in cells in the model work sheet, as shown in rows 18–23 of Figure 15.33. The trick is to realize that each such function has an optional last argument that specifies the simulation number. For example, the formulas in cells C20 and C22 are

=RiskMean($F$13,C16)

and

=RiskPercentile($F$13,0.05,C16)

Remember that the results in these cells are meaningless (or show up as errors) until you run the simulation.

• You can select the profit cell and click the Browse Results button to see a histogram of profits, as shown in Figure 15.34. By default, the histogram shown is for the first simulation, where the order quantity is 150. However, if you click the red histo- gram button with the pound sign, you can select any of the simulations. As an example, Figure 15.35 shows the histogram of profits for simulation 5, where the order quantity is 250. (Do you see why these two histograms are so different? When the order quantity is 150, there is a high probability of selling out, so the spike on the right is large. But the probability of selling out with an order quantity of 250 is much lower, so its spike at the right is much less dominant.)

• You can click the Summary button to get the results from all simulations shown in Figure 15.36. (These results match those in Figure 15.33.)

Figure 15.34 Histogram of Profit with Order Quantity 150

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15-5 Simulation with @rISK 7 5 7

For this example, the results in Figures 15.33–15.36 are illuminating. You can see that an order quantity of 175 provides the largest mean profit. However, is this necessarily the “best” order quantity? This depends on the company’s attitude toward risk. Larger order quantities incur more risk (their histograms are more spread out, their 5th and 95th percentiles are more extreme), but they also have more upside potential. On the other hand, a smaller order quantity, while having a somewhat smaller mean, might be preferable because of less variability. It is not an easy choice, but at least the simulation results provide plenty of information for making the decision.

Figure 15.35 Histogram of Profit with Order Quantity 250

Figure 15.36 Summary Report for All Five Simulations

• You can click the Excel Reports button to get any of several reports on permanent worksheets. Again, Quick Reports is a good choice. This produces several graphs and summary measures for each simulation, each on a different worksheet. This provides a lot of information with almost no work.

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To avoid potential problems, close all other workbooks when running an @RISK model.

15-5d Some Limitations of @RISK The academic version of @RISK included has some limitations you should be aware of. (The commercial version of @RISK doesn’t have these limitations. Also, the exact limita- tions could change as newer academic versions become available.)

• The simulation model must be contained in a single workbook with at most four worksheets, and each worksheet is limited to 300 rows and 100 columns.

• The number of @RISK input probability distribution functions, such as RiskNormal, is limited to 100.

• The number of unattended iterations is limited to 1000. You can request more than 1000, but you have to click a button after each 1000 iterations.

• All @RISK graphs contain a watermark. • The Distribution Fitting tool can handle only 250 observations.

The first limitation shouldn’t cause problems, at least not for the fairly small models discussed in this book. However, we recommend that you close all other workbooks when you are running an @RISK simulation model, especially if they also contain @ RISK functions. @RISK does a lot of recalculation, both in your active worksheet and in all other worksheets or workbooks that are open. So if you are experiencing slow simulations, this is probably the reason.

The second limitation can be a problem, especially in multiperiod problems. For example, if you are simulating 52 weeks of a year, and each week requires two ran- dom inputs, you are already over the 100-function limit. One way to get around this is to use built-in Excel functions for random inputs rather than @RISK func- tions whenever possible. For example, if you want to simulate the flip of a fair coin, the formula =IF(RAND()<0.5,”Heads”,”Tails”) works just as well as the formula =IF(RiskUniform(0, 1)<0.5,”Heads”,”Tails”), but the former doesn’t count against the 100-function limit.

15-5e @RISK Models with Several Random Inputs We conclude this section with another modification of the Walton Bookstore example. To this point, there has been a single random variable, demand. Often there are several ran- dom variables, each reflecting some uncertainty, and you want to include each of these in the simulation model. Example 15.4 illustrates how this can be done, and it also illustrates a very useful feature of @RISK, its sensitivity analysis.

EXAMPLE

15.4 ADDITIONAL UNCERTAINTY AT WALTON BOOKSTORE As in the previous Walton Bookstore example, Walton needs to place an order for next year’s calendar. We continue to assume that the calendars sell for $10 and customer demand for the calendars at this price is triangularly distributed with minimum value, most likely value, and maximum value equal to 100, 175, and 300. However, there are now two other sources of uncer- tainty. First, the maximum number of calendars Walton’s supplier can supply is uncertain and is modeled with a triangular distribution. Its parameters are 125 (minimum), 200 (most likely), and 250 (maximum). Once Walton places an order, the sup- plier will charge $7.50 per calendar if it can supply the entire Walton order. Otherwise, it will charge only $7.25 per calendar. Second, unsold calendars can no longer be returned to the supplier for a refund. Instead, Walton will put them on sale for $5 apiece after January 1. At that price, Walton believes the demand for leftover calendars is triangularly distributed with param- eters 0, 50, and 75. Any calendars still left over, say, after March 1, will be thrown away. Walton again wants to use simulation to analyze the resulting profit for various order quantities.

Objective To develop and analyze a simulation model with multiple sources of uncertainty using @RISK, and to introduce @RISK’s sensitivity analysis features.

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Where Do the Numbers Come From? As in Example 15.3, the monetary values are straightforward, and the parameters of the triangular distributions are probably educated guesses, possibly based on experience with previous calendars.

Solution The variables for this model, including the three sources of uncertainty, are shown in Figure 15.37. (See the file Ordering Calendars - More Uncertainty Big Picture.xlsx.) Using this as a guide, the first step, as always, is to develop the model. Then you can run the simulation with @RISK and examine the results.

15-5 Simulation with @rISK 7 5 9

Figure 15.37 Big Picture for Ordering Model with More Uncertainty

Profit

Order quantity

Unit cost if less than full supply

Unit cost if full supply

Order cost

Actual supply

Supplier capacity

Regular price demand

Sale price demand

Number leftover

Revenue from sale price sales

Revenue from regular price sales

Regular price Sale price

Developing the Simulation Model The completed model is shown in Figure 15.38. (See the file Ordering Calendars - More Uncertainty Finished.xlsx.) The model itself requires a bit more logic than the previous Walton model. It can be developed with the following steps.

1. Random inputs. There are three random inputs in this model: the maximum supply the supplier can provide Walton, the customer demand when the selling price is $10, and the customer demand for sale-price calendars. Generate these in cells B14, E14, and H14 (using the ROUND function to obtain integers) with the formulas

=ROUND(RiskTriang(I5,I6,I7),0)

=ROUND(RiskTriang (E5,E6,E7),0)

and

=ROUND(RiskTriang (F5,F6,F7),0)

The formula in cell H14 generates the random potential demand for calendars at the sale price, even though there might not be any calendars left to put on sale.

2. Actual supply. The number of calendars supplied to Walton is the smaller of the number ordered and the maximum the supplier is able to supply. Calculate this value in cell C14 with the formula

=MIN(B14,Order_quantity)

3. Order cost. Walton gets the reduced price, $7.25, if the supplier cannot supply the entire order. Otherwise, Walton must pay $7.50 per calendar. Therefore, calculate the total order cost in cell D14 with the formula (using the obvious range names)

=IF(B14>=Order_quantity,Unit_cost_1,Unit_cost_2)*C14

The Walton Model with Multiple Uncertain Inputs

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4. Other quantities. The rest of the model is straightforward. Calculate the revenue from regular-price sales in cell F14 with the formula

=Regular_price*MIN(C14,E14)

Calculate the number left over after regular-price sales in cell G14 with the formula

=MAX(C14-E14,0)

Calculate the revenue from sale-price sales in cell I14 with the formula

=Sale_price*MIN(G14,H14)

Finally, calculate profit in cell J14 with the formula

=F14+I14-D14

Then designate this cell as an @RISK output cell. If you like, you can also designate other cells (the revenue cells, for example) as output cells.

5. Order quantities. As before, enter the following RiskSimtable formula in cell B10 so that Walton can try different order quantities:

=RiskSimtable(D10:H10)

Running the Simulation As usual, the next steps are to specify the simulation settings (we chose 1000 iterations and 5 simulations), and run the simulation. It is important to realize what @RISK does when it runs a simulation when there are several random input cells. In each iteration, @RISK generates a random value for each input variable independently. In this example, it generates a maximum supply in cell B14 from one triangular distribution, it generates a regular-price demand in cell E14 from another triangular distribution, and it generates a sale-price demand in cell H14 from a third triangular distribution. With these input values, it then calculates profit. For each order quantity, it then iterates this procedure 1000 times and keeps track of the corresponding profits.8

8 It is also possible to correlate the inputs, as we demonstrate in the next section.

Figure 15.38 @RISK Simulation Model with Three Random Inputs

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

A B C Simula�on of Walton’s Bookstore with more uncertainty

Cost data Unit cost 1 Unit cost 2 Regular price Sale price

Demand distribu�on: triangular

Decision variable Order quan�ty

Simulated quan��es

Range names used:Summary measures of profit from @RISK - based on 1000 itera�ons for each simula�on Simula�on Order quan�ty Minimum Maximum Average Standard devia�on 5th percen�le 95th percen�le

1 150

$55.00 $409.75 $361.40

$44.00 $265.00 $375.00

2 175

�$102.50 $478.50 $389.50

$95.87 $167.50 $459.25

3 200

�$290.00 $547.25 $393.48 $149.25

$49.00 $525.25

4 225

�$347.50 $616.00 $395.10 $175.02

$1.75 $577.50

5 250

�$371.00 $676.50 $397.71 $178.30

$11.75 $593.50

Order_quan�ty Regular_price Sale_price Unit_cost_1 Unit_cost_2

=Model!$B$10 =Model!$B$6 =Model!$B$7 =Model!$B$4 =Model!$B$5

150150 175 200 225 250

Maximum supply 203

Actual supply 150

Cost $1,125

Demand 208

Revenue $1,500

Le� over 0

Demand 56

Revenue $0

Profit $375

D E F G H I J

$7.50 $7.25

$10.00 $5.00

Minimum Most likely Maximum

Order quan��es to try

Minimum Most likely Maximum

Regular price 100 175 300

Supply distribu�on: triangular 125 200 250

Sale price 0

50 75

At regular price At sale price

On each iteration, @RISK generates a new set of random inputs and calcu- lates the corresponding output(s).

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15-5 Simulation with @rISK 7 6 1

Discussion of the Simulation Results Selected results are listed in Figure 15.38 (at the bottom), and the profit histogram for an order quantity of 200 is shown in Figure 15.39. (The histograms for the other order quantities are similar to what you have seen before, with more skewness to the left and a larger spike to the right as the order quantity decreases.) For this order quantity, the results indicate an average profit of $393.48, a 5th percentile of $49.00, a 95th percentile of $525, and a distribution of profits that is again skewed to the left.

Figure 15.39 Histogram of Simulated Profits for Order Quantity 200

Sensitivity Analysis We now demonstrate a useful feature of @RISK when there are several random input cells. This feature lets you see which of these inputs has the most effect on an output cell. To perform this analysis, select the profit cell, J14, and click the Browse Results button. You will see a histogram of profit, as we have already discussed, with a number of buttons at the bottom of the window. Click the red button with the pound sign to select a simulation. We chose #3, where the order quantity is 200. Then click the “tornado” button (the fifth button from the left) and choose Change in Output Mean from its dropdown list. This pro- duces the graph in Figure 15.40. (The Regression and Correlation options produce similar results.)

Figure 15.40 Tornado Graph for Sensitivity Analysis

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This figure shows graphically and numerically how each of the random inputs affects profit: the longer the bar, the stronger the relationship between that input and profit. Specifi- cally, each bar shows how the mean profit varies as each input varies over its range (and the other inputs are held constant). In this sense, you can see that the regular-price demand has by far the largest effect on profit. The other two inputs, maximum supply and sale-price demand, have much smaller effects. Identifying important input variables is important for real applications. If a random input has a large effect on an important output, then it is probably worth the time and money to learn more about this input and possibly reduce the amount of uncertainty involving it.

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A tornado graph indicates which of the random inputs have large effects on an output.

Problems

Level A 16. If you add several normally distributed random numbers,

the result is normally distributed, where the mean of the sum is the sum of the individual means, and the vari- ance of the sum is the sum of the individual variances. (Remember that variance is the square of standard devi- ation.) This is a difficult result to prove mathematically, but it is easy to demonstrate with simulation. To do so, run a simulation where you add three normally distrib- uted random numbers, each with mean 100 and standard deviation 10. Your single output variable should be the sum of these three numbers. Verify with @RISK that the distribution of this output is approximately normal with mean 300 and variance 300 (hence, standard deviation !300 5 17.32).

17. In Problem 11 from the previous section, we stated that the damage amount is normally distributed. Suppose instead that the damage amount is triangularly distrib- uted with parameters 500, 1500,and 7000.That is, the damage in an accident can be as low as $500 or as high as $7000, the most likely value is $1500,and there is definite skewness to the right. (It turns out, as you can verify in @RISK, that the mean of this distribution is $3000, the same as in Problem 11.) Use @RISK to sim- ulate the amount you pay for damage. Then answer the following questions. In each case, explain how the indi- cated event would occur. a. What is the probability that you pay a positive amount

but less than $750? b. What is the probability that you pay more than $600? c. What is the probability that you pay exactly $1000

(the deductible)? 18. Continuing the previous problem, assume, as in Problem

11, that the damage amount is normally distributed with mean $3000 and standard deviation $750. Use @RISK to simulate the amount you pay for damage. Compare your results with those in the previous problem. Does it appear to matter whether you assume a triangular dis- tribution or a normal distribution for damage amounts? Why isn’t this a totally fair comparison? (Hint: Use

@RISK’s Define Distributions tool to find the standard deviation for the triangular distribution.)

19. In Problem 12 of the previous section, suppose that the demand for cars is normally distributed with mean 100 and standard deviation 15. Use @RISK to determine the “best” order quantity—in this case, the one with the largest mean profit. Using the statistics and/or graphs from @RISK, discuss whether this order quantity would be considered best by the car dealer. (The point is that a decision maker can use more than just mean profit in making a decision.)

20. Use @RISK to analyze the sweatshirt situation in Prob- lem 14 of the previous section. Do this for the discrete distributions given in the problem. Then do it for nor- mal distributions. For the normal case, assume that the regular demand is normally distributed with mean 9800 and standard deviation 1300 and that the demand at the reduced price is normally distributed with mean 3800 and standard deviation 1400.

Level B 21. Although the normal distribution is a reasonable input

distribution in many situations, it does have two potential drawbacks: (1) it allows negative values, even though they may be extremely improbable, and (2) it is a symmetric distribution. Many situations are modeled better with a distribution that allows only positive values and is skewed to the right. Two of these that have been used in many real applications are the gamma and lognormal distributions. @RISK enables you to generate observations from each of these distributions. The @RISK function for the gamma distribution is RiskGamma, and it takes two arguments, as in 5RiskGamma 13,10 2 . The first argument, which must be positive, determines the shape. The smaller it is, the more skewed the distribution is to the right; the larger it is, the more symmetric the distribution is. The second argument determines the scale, in the sense that the product of it and the first argument equals the mean of the distribution. (The mean in this example is 30.) Also, the product of the second argument and the square root of the first argument is the standard deviation of the distribution. (In this example, it is !3(10) 5 17.32.) The @RISK function for the lognormal distribution is RiskLognorm.

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15-6 the effects of Input Distributions on results 7 6 3

It has two arguments, as in =RiskLognorm(40,10). These arguments are the mean and standard deviation of the distribution. Rework Example 15.2 for the following demand distributions. Do the simulated outputs have any different qualitative properties with these skewed distributions than with the triangular distribution used in the example?

a. Gamma distribution with parameters 2 and 85 b. Gamma distribution with parameters 5 and 35 c. Lognormal distribution with mean 170 and standard

deviation 60

15-6 The Effects of Input Distributions on Results In Section 15-2, we discussed input distributions. The randomness in input variables causes the variability in the output variables. We now briefly explore whether the choice of input distribution(s) makes much difference in the distribution of an output variable such as profit. This is an important question. If the choice of input distributions doesn’t matter much, then you do not need to agonize over this choice. However, if it does make a difference, then you have to be more careful about choosing an appropriate input distri- bution. Unfortunately, it is impossible to answer the question definitively. The best we can say in general is, “It depends.” Some models are more sensitive to changes in the shape or parameters of input distributions than others. Still, the issue is worth exploring.

We discuss two types of sensitivity analysis in this section. First, we check whether the shape of the input distribution matters. In the Walton Bookstore example, we assumed a triangularly distributed demand with some skewness. Are the results basically the same if a symmetric distribution such as the normal distribution is used instead? Second, we check whether the independence of input variables that has been assumed implicitly to this point is crucial to the output results. Many random quantities in real situations are not independent; they are positively or negatively correlated. Fortunately, @RISK enables you to build correlation into a model. We analyze the effect of this correlation.

15-6a Effect of the Shape of the Input Distribution(s) We first explore the effect of the shape of the input distribution(s). As Example 15.5 indi- cates, if parameters that allow for a fair comparison are used, the shape can have a rela- tively minor effect.

EXAMPLE

15.5 EFFECT OF DEMAND DISTRIBUTION AT WALTON BOOKSTORE

We continue to explore the demand for calendars at Walton Bookstore. We keep the same unit cost, unit price, and unit refund for leftovers as in Example 15.3. However, in that example we assumed a triangular distribution for demand with parameters 100, 175, and 300. Assuming that Walton orders 200 calendars, is the distribution of profit affected if a normal distribution of demand is used instead?

Objective To see whether a triangular distribution with some skewness gives the same profit distribution as a normal distribution for demand.

Where Do the Numbers Come From? The numbers here are the same as in Example 15.3. However, as discussed next, the parameters of the normal distribution are chosen to provide a fair comparison with the triangular distribution used earlier.

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Solution It is important in this type of analysis to make a fair comparison. When you select a normal distribution for demand, you must choose a mean and standard deviation for this distribution. Which values should you choose? It seems only fair to choose the same mean and standard deviation that the triangular distribution has. To find the mean and standard deviation for a triangular distribution with given minimum, most likely, and maximum values, you can take advantage of @RISK’s Define Distributions tool. Select any blank cell, click the Define Dis- tributions button, select the triangular distribution, and enter the parameters 100, 175, and 300. You will see that the mean and standard deviation are 191.67 and 41.248, respectively. Therefore, for a fair comparison you should use a normal distribution with mean 191.67 and standard deviation 41.248. @RISK allows you to see a comparison of these two distributions, as in Figure 15.41. To get this chart, click the Add Overlay button, select the normal distribution from the gallery, and enter 191.67 and 41.248 as its mean and standard deviation.

For a fair comparison of alternative input distri- butions, the distributions should have (at least approx- imately) equal means and standard deviations.

Figure 15.41 Triangular and Normal Distributions for Demand

Developing the Simulation Model The logic in this model is almost exactly the same as before. (See Figure 15.42 and the file Ordering Calendars - Different Demand Distributions Finished.xlsx.) However, a clever use of the RiskSimtable function allows you to run two simulations at once, one for the triangular dis- tribution and one for the corresponding normal distribution. The following two steps are required.

The Walton Model with Alter- native Input Distributions

Figure 15.42 @RISK Model for Comparing Two Input Distributions

Simula�on of Walton’s Bookstore using @RISK - two possible demand distribu�ons

Range names used: Order_quan�ty =Model!$B$9 Profit =Model!$F$15

Cost data Demand distribu�on 1 - triangular Demand distribu�on 2 - normal

Unit_cost =Model!$B$4

Unit cost Mean100$7.50

=Model!$B$5Unit_price

191.67 Unit price $10.00

=Model!$B$6Unit_refund

41.248Stdev175 Unit refund 300$2.50

Decision variable Order quan�ty 200

Demand distribu�on to use 1 Formula is = RiskSimtable({1,2})

Simulated quan��es Demand Revenue Cost Refund Profit

149 $1,490 $1,500 $128 $118

Summary measures of profit from @RISK - based on 1000 itera�ons for each simula�on 21Simula�on

Minimum –$242.50 –$730.00 $500.00$500.00Maximum $342.64$337.47

Distribu�on Triangular Normal

$189.13 $202.41 –$47.50 –$77.50

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

A B C D E F G H I

Average Standard devia�on 5th percen�le

$500.00 $500.0025 95th percen�le

Minimum Most likely Maximum

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15-6 the effects of Input Distributions on results 7 6 5

1. RiskSimtable function. It is useful to index the two distributions as 1 and 2. To indicate that you want to run the simulation with both of them, enter the formula

=RiskSimtable({1,2})

in cell B11. Remember that when you enter actual numbers in this function, rather than cell references, you must put curly brackets around the list.

2. Demand. When the value in cell B11 is 1, the demand distribution is triangular. When it is 2, the distribution is normal. Therefore, enter the formula

=ROUND(IF(B11=1,RiskTriang(E4,E5,E6),RiskNormal(H4,H5)),0)

in cell B15. The effect is that the first simulation will use the triangular distribution, and the second will use the normal distribution.

Running the Simulation The only @RISK setting to change is the number of simulations. It should now be set to 2, the number of values in the RiskSimtable formula. Other than this, you run the simulation exactly as before.

Discussion of the Simulation Results The comparison is shown numerically in Figure 15.43 and graphically in Figure 15.44. As you can see, there is more chance of really low profits when the demand distribution is normal, but each simulation results in the same maximum profit. Both of these statements make sense. The normal distribution, being unbounded on the left, allows for very low demands, and these occasional low demands result in very low profits. On the other side, Walton’s maximum profit is $500 regardless of the input distribution (provided that it allows demands greater than the order quantity). This occurs when Walton’s sells all it orders, in which case excess demand has no effect on profit. Note that the mean profits for the two distributions differ by only about $5.

Look for ways to use the RiskSimtable function. It can really improve efficiency because it runs several simulations at once.

Figure 15.43 Summary Results for Comparison Model

Figure 15.44 Graphical Results for Comparison Model

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It is probably safe to conclude that the profit distribution in this model is not greatly affected by the choice of demand distribution, at least not when (1) the candidate input distributions have the same mean and standard deviation, and (2) their shapes are not too dissimilar. We would venture to guess that this general conclusion about insensitivity of output distributions to shapes of input distributions can be made in many simulation models. However, it is always worth checking, as we have done here, especially if there is a lot of money at stake.

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Shape of the Output Distribution

Predicting the shape of the output distribution from the shape(s) of the input distribution(s) is difficult. For example, normally distributed inputs don’t nec- essarily produce normally distributed outputs. It is also difficult to predict how sensitive the shape of the output distribution is to the shape(s) of the input dis- tribution(s). For example, normally and triangularly distributed inputs (with the same means and standard deviations) are likely to lead to similar output distribu- tions, but there could be differences in the tails of the output distributions. In any case, you should examine the entire output distribution carefully, not just a few of its summary measures.

Fundamental Insight

Input variables in real- world problems are often correlated, which makes the material in this section important.

15-6b Effect of Correlated Inputs Until now, all random numbers generated with @RISK functions have been probabilis- tically independent. This means, for example, that if a random value in one cell is much larger than its mean, the random values in other cells are completely unaffected. They are no more likely to be abnormally large or small than if the first value had been average or below average. Sometimes, however, independence is unrealistic. In such cases, correlated inputs are more appropriate. If they are positively correlated, then large numbers will tend to go with large numbers, and small with small. If they are negatively correlated, then large will tend to go with small and small with large. As an example, you might expect daily stock price changes for two companies in the same industry to be positively correlated. If the price of one oil company increases, you might expect the price of another oil company to increase as well. You can create correlated inputs in @RISK with the RiskCorrmat function, as we illustrate in the following continuation of the Walton example.

EXAMPLE

15.6 CORRELATED DEMANDS AT WALTON BOOKSTORE Suppose Walton Bookstore must order two different calendars. To simplify the example, we assume the calendars each have the same unit cost, unit selling price, and unit refund value as in previous examples. Also, we assume that each has a triangularly distributed demand with parameters 100, 175, and 300. However, we now assume they are “substitute” prod- ucts, so that their demands are negatively correlated. This means that if a customer buys one, she is not likely to buy the other. Specifically, we assume a correlation of 20.9 between the two demands. How do these correlated inputs affect the distribution of profit, as compared to the situation where the demands are uncorrelated (correlation 0) or highly positively correlated (correlation 0.9)?

Objective To see how @RISK enables us to simulate correlated demands, and to see the effect of correlated demands on profit.

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15-6 the effects of Input Distributions on results 7 6 7

Where Do the Numbers Come From? The only new input here is the correlation. It is probably negative because the calendars are substitute products, but it is a dif- ficult number to estimate accurately. This is a good candidate for a sensitivity analysis.

Solution The key to building in correlation is @RISK’s RiskCorrmat (correlation matrix) function. To use this function, you must include a correlation matrix in the model, as shown in the range J5:K6 of Figure 15.45. (See the file Ordering Calendars - Correlated Demands Finished.xlsx.) A correlation matrix must always have 1’s along its diagonal (because a variable is always perfectly correlated with itself) and the correlations between variables elsewhere. Also, the matrix must be symmetric, so that the correlations above the diagonal are a mirror image of those below it. (You can enforce this by entering the formula 5J6 in cell K5. Alternatively, @RISK allows you to enter the correlations only below the diagonal, or only above the diago- nal, and it then infers the mirror images.)

Figure 15.45 Simulation Model with Correlated Demands

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

KJIHGFEDCBA Simulation of Walton’s Bookstore using @RISK - correlated demands

Cost data - same for each product Demand distribution for each product - triangular Correlation matrix between demands Unit cost Product 1100Minimum$7.50 Product 2 Unit price Most likely$10.00 Product 1175 1 �0.9 Unit refund Product 2300Maximum$2.50 �0.9 1

Decision variables Possible correlations to try Order quantity 1 200 �0.9 0 0.9 Order quantity 2 200

Range names used: Simulated quantities Order_quantity_1

Order_quantity_2 =Model!$B$9 =Model!$B$10

Profit =Model!$F$16 ProfitRefundCostRevenueDemand

Product 1 =Model!$B$4Unit_cost

$335$55$1,500$1,780178 Product 2

=Model!$B$5Unit_price $110$130$1,500$1,480148

=Model!$B$6Unit_refund $445$185$3,000$3,260326Totals

Summary measures of profit from @RISK - based on 1000 iterations Simulation 1 2 3 Correlation �0.9 0 0.9

$265.00Minimum �$267.50 �$440.00 $1,000.00$1,000.00$1,000.00Maximum

$675.06$675.06$675.06Average Standard deviation $158.73 $267.83 $365.64 5th percentile $392.50 $182.50 �$72.50 95th percentile $932.50 $1,000.00 $1,000.00

Note RiskSimtable function in cell J6.

To enter random values in any cells that are correlated, you start with a typical @RISK formula, such as

=RiskTriang(E4,E5,E6)

Then you add an extra argument, the RiskCorrmat function, as follows:

=RiskTriang(E4,E5,E6,RiskCorrmat(J5:K6,1))

The first argument of the RiskCorrmat function is the correlation matrix range. The second is an index of the variable. In this example, the first calendar demand has index 1 and the second has index 2.

The RiskCorrmat function is “tacked on” as an extra argument to a typical random @RISK function.

RiskCorrmat

This function enables you to correlate two or more input variables. The function has the form RiskCorrmat(CorrMat,Index), where CorrMat is a matrix of correlations and Index is an index of the variable being correlated to others. For example, if there are three correlated variables, Index is 1 for the first variable, 2 is for the second, and 3 is for the third. The RiskCorrmat function is not entered by itself. Rather, it is entered as the last argument of a random @RISK function, such as =RiskTriang(10,15,30,RiskCorrmat(CorrMat,2)).

@RISK Function

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Developing the Simulation Model Armed with this knowledge, the simulation model in Figure 15.45 is straightforward and can be developed as follows.

1. Inputs. Enter the inputs in the blue ranges in columns B and E. 2. Correlation matrix. For the correlation matrix in the range J5:H6, enter 1’s on the diagonal, and enter the formula

=J6

in cell K5 (or leave cell K5 blank). Then enter the formula

=RiskSimtable(I9:K9)

in cell J6. This allows you to simultaneously simulate negatively correlated demands, uncorrelated demands, and posi- tively correlated demands.

3. Order quantities. Assume for now that the company orders the same number of each calendar, 200, so enter this value in cells B9 and B10. However, the simulation is set up so that you can experiment with any order quantities in these cells, including unequal values.

4. Correlated demands. Generate correlated demands by entering the formula

=ROUND(RiskTriang(E4,E5,E6,RiskCorrmat(J5:K6,1)),0)

in cell B14 for demand 1 and the formula

=ROUND(RiskTriang(E4,E5,E6,RiskCorrmat(J5:K6,2)),0)

in cell B15 for demand 2. The only difference between these is the index of the variable being generated. The first has index 1; the second has index 2.

5. Other formulas. The other formulas in rows 14 and 15 are identical to ones developed in previous examples, so they aren’t presented again here. The quantities in row 16 are sums of rows 14 and 15. Also, the only @RISK output we desig- nated is the total profit in cell F16, but you can designate others as output cells if you like.

Running the Simulation You should set up and run @RISK exactly as before. For this example, set the number of iterations to 1000 and the number of simulations to 3 (because three different correlations are being tested).

Discussion of the Simulation Results Selected numerical and graphical results are shown in Figures 15.46 and 15.47. You will probably be surprised to see that the mean total profit is the same, regardless of the correlation. This is no coincidence. In each of the three simulations, @RISK uses the same random numbers but “shuffles” them in different orders to get the correct correlations. This means that averages are unaffected. (The idea is that the average of the numbers 30, 26, and 48 is the same as the average of the numbers 48, 30, and 26.)

The Walton Model with Correlated Demands

Correlations in @RISK Models

Figure 15.46 Summary Results for Correlated Model

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15-6 the effects of Input Distributions on results 7 6 9

However, the correlation has a definite effect on the distribution of total profit. You can see this in Figure 15.46, for exam- ple, where the standard deviation of total profit increases as the correlation goes from negative to zero to positive. This same increase in variability is apparent in the histograms in Figure 15.47. Do you see intuitively why this increase in variability occurs? It is basically the “Don’t put all of your eggs in one basket” effect. When the correlation is negative, high demands for one product tend to cancel low demands for the other product, so extremes in profit are rare. However, when the correlation is positive, high demands for the two products tend to go together, as do low demands. These make extreme profits on either end much more likely.

This same phenomenon would occur if you simulated an investment portfolio containing two stocks. When the stocks are positively correlated, the portfolio is much riskier (more variability) than when they are negatively correlated. Of course, this is the reason for diversifying a portfolio.

Modeling Issues We illustrated the RiskCorrmat function for triangularly distributed values. However, it can be used with any of @RISK’s distributions by tacking on RiskCorrmat as a last argument. You can even mix them. For example, assuming CMat is the range name for a 2 3 2 correlation matrix, you could enter the formulas

5RiskNormal 110,2,RiskCorrmat 1CMat,1 2 2 and

5RiskUniform 1100,200,RiskCorrmat 1CMat,2 2 2 into any two cells. When you run the simulation, @RISK generates a sequence of normally distributed random numbers based on the first formula and another sequence of uniformly distributed random numbers based on the second formula. Then it shuffles them in some complex way until their correlation is approximately equal to the specified correlation in the correlation matrix.

Figure 15.47 Graphical Results for Correlated Model

With the RiskCorrmat function, you can correlate random numbers from any distributions.

Correlated Inputs

When you enter random inputs in an @RISK simulation model and then run the simulation, each iteration generates independent values for the random inputs. If you know or suspect that some of the inputs are positively or negatively cor- related, you should build this correlation structure into the model explicitly with the RiskCorrmat function. This function might not change the mean of an output, but it can definitely affect the variability and shape of the output distribution.

Fundamental Insight

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Problems

Level A 22. Fizzy Company produces six-packs of soda cans.

Each can is supposed to contain at least 12 ounces of soda. If the total weight in a six-pack is less than 72 ounces, Fizzy is fined $100 and receives no sales rev- enue for the six-pack. Each six-pack sells for $3.00. It costs Fizzy $0.02 per ounce of soda put in the cans. Fizzy can control the mean fill rate of its soda-filling machines. The amount put in each can by a machine is normally distributed with standard deviation 0.10 ounce. a. Assume that the weight of each can in a six-pack has a

0.8 correlation with the weight of the other cans in the six-pack. What mean fill quantity maximizes expected profit per six-pack? Try mean fill rates from 12.00 to 12.35 in increments of 0.05.

b. If the weights of the cans in the six-pack are prob- abilistically independent, what mean fill quantity maximizes expected profit per six-pack? Try the same mean fill rates as in part a.

c. How can you explain the difference in the answers to parts a and b?

23. When you use @RISK’s correlation feature to generate correlated random numbers, how can you verify that they are correlated? Try the following. Use the RiskCorrmat function to generate two normally distrib- uted random numbers, each with mean 100 and stan- dard deviation 10, and with correlation 0.7. To run a simulation, you need an output variable, so sum these two numbers and designate the sum as an output vari- able. Now run @RISK. Click @RISK’s Excel Reports button and check the Simulation Data option to see the actual simulated data. a. Use Excel’s CORREL function to calculate the cor-

relation between the two input variables. It should be close to 0.7. Then create a scatterplot of these two input variables. The plot should indicate a definite positive relationship.

b. Are the two input variables correlated with the out- put? Use Excel’s CORREL function to find out. Inter- pret your results intuitively.

24. Work the previous problem, but make the correlation between the two inputs equal to 20.7. Explain how the results change.

25. Work Problem 23, but now make the second input vari- able triangularly distributed with parameters 50, 100, and 500. This time, verify not only that the correlation between the two inputs is approximately 0.7, but also that the shapes of the two input distributions are approx- imately what they should be: normal for the first and triangular for the second. Do this by creating histo- grams in Excel. The point is that you can use @RISK’s

RiskCorrmat function to correlate random numbers from different distributions.

26. Suppose you are going to invest equal amounts in three stocks. The annual return from each stock is normally distributed with mean 0.01 11%2 and standard deviation 0.06. The annual return on your portfolio, the output variable of interest, is the average of the three stock returns. Run @RISK, using 1000 iterations, on each of the following scenarios. a. The three stock returns are highly correlated. The

correlation between each pair is 0.9. b. The three stock returns are practically independent.

The correlation between each pair is 0.1. c. The first two stocks are moderately correlated. The

correlation between their returns is 0.4. The third stock’s return is negatively correlated with the other two. The correlation between its return and each of the first two is 20.8.

d. Compare the portfolio distributions from @RISK for these three scenarios. What do you conclude?

e. You might think of a fourth scenario, where the correlation between each pair of returns is a large negative number such as 20.8. But explain intuitively why this makes no sense. Try to run the simulation with these negative correlations and see what happens.

27. The effect of the shapes of input distributions on the distribution of an output can depend on the output function. For this problem, assume there are 10 input variables. The goal is to compare the case where these 10 inputs each have a normal distribution with mean 1000 and standard deviation 250 to the case where they each have a triangular distribution with parameters 600, and 1700. (You can check with @RISK’s Define Distributions window that even though this triangular distribution is very skewed, it has the same mean and approximately the same standard deviation as the nor- mal distribution.) For each of the following outputs, run two @RISK simulations, one with the normally dis- tributed inputs and one with the triangularly distributed inputs, and comment on the differences between the resulting output distributions. For each simulation run 1000 iterations. a. Let the output be the average of the inputs. b. Let the output be the maximum of the inputs. c. Calculate the average of the inputs. Then the output is

the minimum of the inputs if this average is less than 1000; otherwise, the output is the maximum of the inputs.

Level B 28. The Business School at State University currently has

three parking lots, each containing 155 spaces. Two hundred faculty members have been assigned to each lot. On a peak day, an average of 70% of all lot 1 park- ing sticker holders show up, an average of 72% of all

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15-7 Conclusion 7 7 1

15-7 Conclusion For years, simulation did not receive the attention it deserved in management science courses (or in business). The primary reason for this was the lack of easy-to-use simulation software. Now, with Excel’s built-in simulation capabilities, plus powerful and affordable add-ins such as @RISK, simulation is receiving its rightful emphasis. The world is full of uncertainty, which is what makes simulation so valuable. Simulation models provide important insights that are missing in models that do not incorporate uncertainty explicitly. In addition, simulation models are relatively easy to understand and develop. In this chapter we have illustrated the basic ideas of simulation, how to perform simulation with Excel built-in tools, and how @RISK greatly enhances Excel’s basic capabilities. In the next chapter we will build on this knowledge to develop and analyze simulation models in a variety of business areas.

Summary of Key Terms TERM EXPLANATION EXCEL PAGE

Simulation model Model with random inputs that affect one or more outputs, where the randomness is modeled explicitly

760

F9 key The “recalc” key, used to make a spreadsheet recalculate

762

Probability distributions for input variables

Specification of the possible values and their prob- abilities for random input variables; these distribu- tions must be specified in any simulation model

762

Uniform distribution The flat distribution, where all values in a bounded continuum are equally likely

766

RAND function Excel’s built-in random number generator; gen- erates uniformly distributed random numbers between 0 and 1

=RAND() 767

RANDBETWEEN function Excel’s built-in function for generating equally likely random integers over an indicated range

=RANDBETWEEN (min,max)

768

Freeze random numbers Change “volatile” random numbers into “fixed” numbers

Copy range, paste it onto itself with the Paste Values option

770

@RISK random functions A set of functions, including RiskNormal and RiskTriang, for generating random numbers from various distributions

771

Discrete distribution A general distribution where a discrete number of possible values and their probabilities are specified

772

Triangular distribution Literally a triangle-shaped distribution, specified by a minimum value, a most likely value, and a maxi- mum value

775

@RISK A simulation add-in developed by Palisade @RISK ribbon 794

RiskSimtable function Used to run an @RISK simulation model for several values of some variable, often a decision variable

795

lot 2 parking sticker holders show up, and an average of 74% of all lot 3 parking sticker holders show up. a. Given the current situation, estimate the probability

that on a peak day, at least one faculty member with a sticker will be unable to find a spot. Assume that the number who show up at each lot is independent of the number who show up at the other two lots. Compare

two situations: (1) each person can park only in the lot assigned to him or her, and (2) each person can park in any of the lots (pooling). (Hint: Use the RiskBino- mial function.)

b. Now suppose the numbers of people who show up at the three lots are highly correlated (correlation 0.9). How are the results different from those in part a?

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TERM EXPLANATION EXCEL PAGE

RiskOutput function Used to indicate that a cell contains an output that will be tracked by @RISK

798

Latin Hypercube sampling An efficient way of simulating random numbers for a simulation model, where the results are more accurate than with other sampling methods

799

RiskCorrmat function Used to correlate two or more random input variables

815

Correlated inputs Random quantities, such as returns from stocks in the same industry, that tend to go together (or possibly go in opposite directions from one another)

815

Key Terms (continued)

Problems

Conceptual Questions C.1. You are making several runs of a simulation model, each

with a different value of some decision variable (such as the order quantity in the Walton calendar model), to see which decision value achieves the largest mean profit. Is it possible that one value beats another simply by ran- dom luck? What can you do to minimize the chance of a “better” value losing out to a “poorer” value?

C.2. If you want to replicate the results of a simulation model with Excel functions only, not @RISK, you can build a data table and let the column input cell be any blank cell. Explain why this works.

C.3. Suppose you simulate a gambling situation where you place many bets. On each bet, the distribution of your net winnings (loss if negative) is highly skewed to the left because there are some possibilities of really large losses but not much upside potential. Your only simulation out- put is the average of the results of all the bets. If you run @RISK with many iterations and look at the resulting histogram of this output, what will it look like? Why?

C.4. You plan to simulate a portfolio of investments over a multiyear period, so for each investment (which could be a particular stock or bond, for example), you need to simulate the change in its value for each of the years. How would you simulate these changes in a realistic way? Would you base it on historical data? What about correlations? Do you think the changes for different investments in a particular year would be cor- related? Do you think changes for a particular invest- ment in different years would be correlated? Do you think correlations would play a significant role in your simulation in terms of realism?

C.5. Big Hit Video must determine how many copies of a new video to purchase. Assume that the company’s goal is to purchase a number of copies that maximizes

its expected profit from the video during the next year. Describe how you would use simulation to shed light on this problem. Assume that each time a video is rented, it is rented for one day.

C.6. Many people who are involved in a small auto accident do not file a claim because they are afraid their insur- ance premiums will be raised. Suppose your insurance company has three rates. If you file a claim, you are moved to the next higher rate. How might you use sim- ulation to determine whether you should file a claim?

C.7. A building contains 1000 lightbulbs. Each bulb lasts at most five months. The company maintaining the building is trying to decide whether it is worthwhile to practice a “group replacement” policy. Under a group replacement policy, all bulbs are replaced every T months (where T is to be determined). Also, bulbs are replaced when they burn out. Assume that it costs $0.05 to replace each bulb during a group replacement and $0.20 to replace each burned-out bulb if it is replaced individually. How would you use simulation to determine whether a group replacement policy is worthwhile?

C.8. Why is the RiskCorrmat function necessary? How does @RISK generate random inputs by default, that is, when RiskCorrmat is not used?

C.9. Consider the claim that normally distributed inputs in a simulation model are bound to lead to normally distributed outputs. Do you agree or disagree with this claim? Defend your answer.

C.10. It is very possible that when you use a correlation matrix as input to the RiskCorrmat function in an @RISK model, the program will inform you that this is an invalid correlation matrix. Provide an example of an obviously invalid correlation matrix involving at least three variables, and explain why it is invalid.

C.11. When you use a RiskSimtable function for a deci- sion variable, such as the order quantity in the Walton model, explain how this provides a “fair” comparison across the different values tested.

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C.12. Consider a situation where there is a cost that is either incurred or not. It is incurred only if the value of some random input is less than a specified cutoff value. Why might a simulation of this situation give a very dif- ferent average value of the cost incurred than a deter- ministic model that treats the random input as fixed at its mean? What does this have to do with the “flaw of averages”?

Level A 29. Six months before its annual convention, the American

Medical Association must determine how many rooms to reserve. At this time, the AMA can reserve rooms at a cost of $150 per room. The AMA believes the number of doctors attending the convention will be normally dis- tributed with a mean of 5000 and a standard deviation of 1000. If the number of people attending the conven- tion exceeds the number of rooms reserved, extra rooms must be reserved at a cost of $250 per room. a. Use simulation with @RISK to determine the num-

ber of rooms that should be reserved to minimize the expected cost to the AMA. Try possible values from 4100 to 4900 in increments of 100.

b. Redo part a for the case where the number attend- ing has a triangular distribution with minimum value 2000, maximum value 7000, and most likely value 5000. Does this change the substantive results from part a?

30. You have made it to the final round of the show Let’s Make a Deal. You know that there is a $1 million prize behind either door 1, door 2, or door 3. It is equally likely that the prize is behind any of the three doors. The two doors without a prize have nothing behind them. You randomly choose door 2. Before you see whether the prize is behind door 2, host Monty Hall opens a door that has no prize behind it. Specifically, suppose that before door 2 is opened, Monty reveals that there is no prize behind door 3. You now have the opportunity to switch and choose door 1. Should you switch? Simulate this situation 1000 times. For each replication use an @RISK function to generate the door that leads to the prize. Then use another @RISK function to generate the door that Monty will open. Assume that Monty plays as follows: Monty knows where the prize is and will open an empty door, but he cannot open door 2. If the prize is really behind door 2, Monty is equally likely to open door 1 or door 3. If the prize is really behind door 1, Monty must open door 3. If the prize is really behind door 3, Monty must open door 1.

31. A new edition of a very popular textbook will be pub- lished a year from now. The publisher currently has 2000 copies on hand and is deciding whether to do another printing before the new edition comes out. The publisher estimates that demand for the book during the next year is governed by the probability distribution in

the file P15_31.xlsx. A production run incurs a fixed cost of $10,000 plus a variable cost of $15 per book printed. Books are sold for $130 per book. Any demand that cannot be met incurs a penalty cost of $20 per book, due to loss of goodwill. Up to 500 of any leftover books can be sold to Barnes & Noble for $35 per book. The publisher is interested in maximizing expected profit. The following print-run sizes are under consideration: 0 (no production run) to 16,000 in increments of 2000. What decision would you recommend? Use simulation with 1000 replications. For your optimal decision, the publisher can be 90% certain that the actual profit asso- ciated with remaining sales of the current edition will be between what two values?

32. A hardware company sells a lot of low-cost, high- volume products. For one such product, it is equally likely that annual unit sales will be low or high. If sales are low 160,0002, the company can sell the product for $10 per unit. If sales are high 1100,0002 , a competitor will enter and the company will be able to sell the prod- uct for only $8 per unit. The variable cost per unit has a 25% chance of being $6, a 50% chance of being $7.50, and a 25% chance of being $9. Annual fixed costs are $30,000. a. Use simulation to estimate the company’s expected

annual profit. b. Find a 95% interval for the company’s annual profit,

that is, an interval such that about 95% of the actual profits are inside it.

c. Now suppose that annual unit sales, variable cost, and unit price are equal to their respective expected val- ues—that is, there is no uncertainty. Determine the company’s annual profit for this scenario.

d. Can you conclude from the results in parts a and c that the expected profit from a simulation is equal to the profit from the scenario where each input assumes its expected value? Explain.

33. A direct marketer of women’s clothing, must deter- mine how many telephone operators to schedule during each part of the day. The company estimates that the number of phone calls received each hour of a typi- cal eight-hour shift can be described by the probability distribution in the file P15_33.xlsx. Each operator can handle 15 calls per hour and costs the company $20 per hour. Each phone call that is not handled is assumed to cost the company $6 in lost profit. Considering the options of employing 6, 8, 10, 12, 14, or 16 operators, use simulation to determine the number of operators that minimizes the expected hourly cost (labor costs plus lost profits).

34. Assume that all of a company’s job applicants must take a test, and that the scores on this test are normally distributed. The selection ratio is the cutoff point used by the company in its hiring process. For example, a selection ratio of 20% means that the company will accept applicants for jobs who rank in the top 20% of

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all applicants. If the company chooses a selection ratio of 20%, the average test score of those selected will be 1.40 standard deviations above average. Use simulation to verify this fact, proceeding as follows. a. Show that if the company wants to accept only the

top 20% of all applicants, it should accept applicants whose test scores are at least 0.842 standard deviation above average. (No simulation is required here. Just use the appropriate Excel normal function.)

b. Now generate 1000 test scores from a normal dis- tribution with mean 0 and standard deviation 1.The average test score of those selected is the average of the scores that are at least 0.842. Calculate this with an AVERAGEIF function. Then press the F9 key sev- eral times to see how this average changes with new random numbers. Is it much larger than the average of all applicants?

35. Lemington’s is trying to determine how many Jean Hudson dresses to order for the spring season. Demand for the dresses is assumed to follow a nor- mal distribution with mean 400 and standard devi- ation 100. The contract between Jean Hudson and Lemington’s works as follows. At the beginning of the season, Lemington’s reserves x units of capac- ity. Lemington’s must take delivery for at least 0.8x dresses and can, if desired, take delivery on up to x dresses. Each dress sells for $160 and Hudson charges $50 per dress. If Lemington’s does not take delivery on all x dresses, it owes Hudson a $5 penalty for each unit of reserved capacity that is unused. For example, if Lemington’s orders 450 dresses and demand is for 400 dresses, Lemington’s will receive 400 dresses and owe Jean 4001$502 1 501$52 . How many units of capacity should Lemington’s reserve to maximize its expected profit?

36. A department store is trying to determine how many Hanson T-shirts to order. Currently the shirts are sold for $21, but at later dates the shirts will be offered at a 10% discount, then a 20% discount, then a 40% discount, then a 50% discount, and finally a 60% dis- count. Demand at the full price of $21 is believed to be normally distributed with mean 1800 and stan- dard deviation 360. Demand at various discounts is assumed to be a multiple of full-price demand. These multiples, for discounts of 10%, 20%, 40%, 50%, and 60% are, respectively, 0.4, 0.7, 1.1, 2, and 50. For example, if full-price demand is 2500, then at a 10% discount customers would be willing to buy 1000 T-shirts. The unit cost of purchasing T-shirts depends on the number of T-shirts ordered, as shown in the file P15_36.xlsx. Use simulation to determine how many T-shirts the store should order. Model the problem so that the store first orders some quantity of T-shirts, then discounts deeper and deeper, as necessary, to sell all of the shirts.

Level B 37. The annual return on each of four stocks for each of

the next five years is assumed to follow a normal dis- tribution, with the mean and standard deviation for each stock, as well as the correlations between stocks, listed in the file P15_37.xlsx. You believe that the stock returns for these stocks in a given year are correlated, accord- ing to the correlation matrix given, but you believe the returns in different years are uncorrelated. For example, the returns for stocks 1 and 2 in year 1 have correlation 0.55, but the correlation between the return of stock 1 in year 1 and the return of stock 1 in year 2 is 0, and the correlation between the return of stock 1 in year 1 and the return of stock 2 in year 2 is also 0. The file has the formulas you might expect for this situation in the range C20:G23. You can check how the RiskCorrmat function has been used in these formulas. Just so that there is an @RISK output cell, calculate the average of all returns in cell B25 and designate it as an @RISK output. (This cell is not really important for the problem, but it is included because @RISK requires at least one output cell.) a. Using the model exactly as it stands, run @RISK

with 1000 iterations. The question is whether the cor- relations in the simulated data are close to what they should be. To check this, go to @RISK’s Report Set- tings and check the Input Data option before you run the simulation. This gives you all of the simulated returns on a new sheet. Then calculate correlations for all pairs of columns in the resulting Inputs Data Report sheet. (StatTools can be used to create a matrix of all correlations for the simulated data.) Comment on whether the correlations are different from what they should be.

b. Recognizing that this is a common situation (cor- relation within years, no correlation across years), @RISK allows you to model it by adding a third argument to the RiskCorrmat function: the year index in row 19 of the P15_37.xlsx file. For exam- ple, the RiskCorrmat part of the formula in cell C20 b e c o m e s 5RiskNormal 1$B5,$C5, RiskCorrmat 1$B$12:$E$15,$B20,C$19 2 2 . Make this change to the formulas in the range C20:G23, rerun the simula- tion, and redo the correlation analysis in part a. Verify that the correlations between inputs are now more in line with what they should be.

38. It is surprising (but true) that if 23 people are in the same room, there is about a 50% chance that at least two people will have the same birthday. Suppose you want to estimate the probability that if 30 people are in the same room, at least two of them will have the same birthday. You can proceed as follows. a. Generate random birthdays for 30 different people.

Ignoring the possibility of a leap year, each person has

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a 1/365 chance of having a given birthday (label the days of the year 1 to 365). You can use the RANDBE- TWEEN function to generate birthdays.

b. Once you have generated 30 people’s birthdays, how can you tell whether at least two people have the same birthday? One way is to use Excel’s RANK function. (You can learn how to use this function in Excel’s online help.) This function returns the rank of a num- ber relative to a given group of numbers. In the case of a tie, two numbers are given the same rank. For example, if the set of numbers is 4, 3, 2, 5, the RANK function returns 2, 3, 4, 1. (By default, RANK gives 1 to the largest number.) If the set of numbers is 4, 3, 2, 4, the RANK function returns 1, 3, 4, 1.

c. After using the RANK function, you should be able to determine whether at least two of the 30 people have the same birthday. What is the (estimated) probability that this occurs?

39. United Electric (UE) sells refrigerators for $400 with a one-year warranty. The warranty works as follows. If any part of the refrigerator fails during the first year after purchase, UE replaces the refrigerator for an average cost of $100. As soon as a replacement is made, another one-year warranty period begins for the customer. If a refrigerator fails outside the warranty period, we assume that the customer immediately purchases another UE refrigerator. Suppose that the amount of time a refrigera- tor lasts follows a normal distribution with a mean of 1.8 years and a standard deviation of 0.3 year. a. Estimate the average profit per year UE earns from a

customer. b. How could the approach of this problem be used to

determine the optimal warranty period? 40. A Flexible Savings Account (FSA) plan allows you

to put money into an account at the beginning of the calendar year that can be used for medical expenses. This amount is not subject to federal tax. As you pay medical expenses during the year, you are reimbursed by the administrator of the FSA until the money is exhausted. From that point on, you must pay your medi- cal expenses out of your own pocket. On the other hand, if you put more money into your FSA than the medical expenses you incur, this extra money is lost to you. Your annual salary is $80,000 and your federal income tax rate is 30%. a. Assume that your medical expenses in a year are

normally distributed with mean $2000 and standard deviation $500. Build an @RISK model in which the output is the amount of money left to you after paying taxes, putting money in an FSA, and paying any extra medical expenses. Experiment with the amount of money put in the FSA, using a RiskSimtable function.

b. Rework part a, but this time assume a gamma distri- bution for your annual medical expenses. Use 16 and 125 as the two parameters of this distribution. These

imply the same mean and standard deviation as in part a, but the distribution of medical expenses is now skewed to the right, which is probably more realistic. Using simulation, see whether you should now put more or less money in an FSA than in the symmetric case in part a.

41. At the beginning of each week, a machine is in one of four conditions: 1 5 excellent; 2 5 good; 3 5 average; 4 5 bad. The weekly revenue earned by a machine in state 1, 2, 3, or 4 is $100, $90, $50, or $10, respectively. After observing the condition of the machine at the beginning of the week, the company has the option, for a cost of $200, of instantaneously replacing the machine with an excel- lent machine. The quality of the machine deteriorates over time, as shown in the file P15_41.xlsx. Four main- tenance policies are under consideration: n Policy 1: Never replace a machine. n Policy 2: Immediately replace a bad machine. n Policy 3: Immediately replace a bad or average

machine. n Policy 4: Immediately replace a bad, average, or good

machine. Simulate each of these policies for 50 weeks (using at

least 250 iterations each) to determine the policy that maximizes expected weekly profit. Assume that the machine at the beginning of week 1 is excellent.

42. Simulation can be used to illustrate a number of results from statistics that are difficult to understand with nonsimulation arguments. One is the famous central limit theorem, which says that if you sample enough values from any population distribution and then average these values, the resulting average will be approximately normally distributed. Confirm this by using @RISK with the following population dis- tributions (run a separate simulation for each): (a) discrete with possible values 1 and 2 and probabili- ties 0.2 and 0.8; (b) exponential with mean 1 (use the RiskExpon function with the single argument 1); (c) triangular with minimum, most likely, and maximum values equal to 1, 9, and 10. (Note that each of these distributions is very skewed.) Run each simulation with 10 values in each average, and run 1000 itera- tions to simulate 1000 averages. Create a histogram of the averages to see whether it is indeed bell-shaped. Then repeat, using 30 values in each average. Are the histograms based on 10 values qualitatively different from those based on 30?

43. In statistics we often use observed data to test a hypothesis about a population or populations. The basic method uses the observed data to calculate a test statistic (a single number), as discussed in Chapter 9. If the mag- nitude of this test statistic is sufficiently large, the null hypothesis is rejected in favor of the research hypothe- sis. As an example, consider a researcher who believes teenage girls sleep longer than teenage boys on average.

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She collects observations on n 5 40 randomly selected girls and n 5 40 randomly selected boys. (Each observation is the average sleep time over several nights for a given person.) The averages are X1 5 7.9 hours for the girls and X2 5 7.6 hours for the boys. The standard deviation of the 40 observations for girls is s1 5 0.5 hour; for the boys it is s2 5 0.7 hour. The researcher, consulting Chapter 9, then calculates the test statistic

X1 2 X2

"s2 1>40 1 s2

2>40 5

7.9 2 7.6

!0.25>40 1 0.49>40 5 2.206

Based on the fact that 2.206 is “large,” she claims that her research hypothesis is confirmed—girls do sleep longer than boys.

You are skeptical of this claim, so you check it out by running a simulation. In your simulation you assume that girls and boys have the same mean and standard deviation of sleep times in the entire population, say, 7.7 and 0.6. You also assume that the distribution of sleep times is normal. Then you repeatedly simulate observa- tions of 40 girls and 40 boys from this distribution and calculate the test statistic. The question is whether the observed test statistic, 2.206, is “extreme.” If it is larger than most or all of the test statistics you simulate, then the researcher is justified in her claim; otherwise, this large a statistic could have happened easily by chance, even if the girls and boys have identical population means. Use @RISK to see which of these possibilities occurs.

44. A technical note in the discussion of @RISK indi- cated that Latin Hypercube sampling is more effi- cient than Monte Carlo sampling. This problem allows you to see what this means. The file P15_44. xlsx gets you started. There is a single output cell, B5. You can enter any random value in this cell, such as RISKNORMAL 1500,100 2 . There are already @RISK statistical formulas in rows 9–12 to calculate summary measures of the output for each of 10 simulations. On the @RISK ribbon, click on the button to the left of the “dice” button to bring up the Simulation Settings dialog box, click on the Sampling tab, and make sure the Sampling Type is Latin Hypercube. Run 10 sim- ulations with at least 1000 iterations each, and then paste the results in rows 9–12 as values in rows 17–20. Next, get back in Simulations Settings and change the Sampling Type to Monte Carlo, run the 10 simulations again, and paste the results in rows 9–12 as values into rows 23–26. For each row, 17–20 and 23–26, sum- marize the 10 numbers in that row with AVERAGE and STDEV. What do you find? Why do we say that Latin Hypercube sampling is more efficient? (Thanks

to Harvey Wagner at University of North Carolina for suggesting this problem.)

45. We are continually hearing reports on the nightly news about natural disasters—droughts in Texas, hurricanes in Florida, floods in California, and so on. We often hear that one of these was the “worst in over 30 years,” or some such statement. Are natural disasters getting worse these days, or does it just appear so? How might you use simulation to answer this question? Here is one possible approach. Imagine that there are N areas of the country (or the world) that tend to have, to some extent, various types of weather phenomena each year. For example, hurricanes are always a potential problem for Florida, and fires are always a potential problem in southern California. You might model the severity of the prob- lem for any area in any year by a normally distributed random number with mean 0 and standard deviation 1, where negative values are interpreted as good years and positive values are interpreted as bad years. (We suggest the normal distribution, but there is no reason other distributions couldn’t be used instead.) Then you could simulate such values for all areas over a period of several years and keep track, say, of whether any of the areas have worse conditions in the current year than they have had in the past several years, where “several” could be 10, 20, 30, or any other number of years you want to test. What might you keep track of? How might you interpret your results?

46. The Weibull distribution is often used in engineering studies to model the random lifetime of a device. It is a right-skewed distribution and can be generated with the RiskWeibull function. This function takes two positive parameters, alpha and beta, that determine the shape and location of the distribution. Consider a piece of equipment that contains a battery with a Weibull-distributed lifetime (in days) with parame- ters alpha 5 2 and beta 5 80. (It can be shown that the mean and standard deviation of this distribution are about 70 and 37.) NASA is about to send this piece of equipment into space, where its mission is to last at least 600 days. The equipment will be accom- panied by one battery and n-1 standby spare batteries, that is, n batteries total. When a battery fails, it will automatically be replaced by a spare, at least as long as there are any spares left. a. Use simulation to estimate the probability that the

equipment will accomplish its mission if n 5 10 bat- teries are supplied.

b. Although the probability from part a is reasonably large, NASA wants it to be even higher. Use simula- tion to find the value of n so that the equipment will accomplish its mission with probability at least 0.99.

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15-7 Conclusion 7 7 7

Egress, Inc., is a small company that designs, produces, and sells ski jackets and other coats. The creative design team has labored for weeks over its new design for the coming winter season. It is now time to decide how many ski jackets to produce in this production run. Because of the lead times involved, no other production runs will be possible during the season. Predicting ski jacket sales months in advance of the selling season can be quite tricky. Egress has been in operation for only three years, and its ski jacket designs were quite successful in two of those years. Based on real- ized sales from the last three years, current economic condi- tions, and professional judgment, 12 Egress employees have independently estimated demand for their new design for the upcoming season. Their estimates are listed in Table 15.2.

Table 15.2 Estimated Demands

14,000 16,000

13,000 8000

14,000 5000

14,000 11,000

15,500 8000

10,500 15,000

To assist in the decision on the number of units for the production run, management has gathered the data in Table 15.3. Note that S is the price Egress charges retailers. Any ski jackets that do not sell during the season can be sold by Egress to discounters for V per jacket. The fixed cost of

plant and equipment is F. This cost is incurred regardless of the size of the production run.

Table 15.3 Monetary Values

Variable production cost per unit 1C2: $80

Selling price per unit 1S2: $100

Salvage value per unit 1V2: $30

Fixed production cost 1F2: $100,000

Questions 1. Egress management believes that a normal distribution

is a reasonable model for the unknown demand in the coming year. What mean and standard deviation should Egress use for the demand distribution?

2. Use a spreadsheet model to simulate 1000 possible out- comes for demand in the coming year. Based on these scenarios, what is the expected profit if Egress produc- es Q 5 7800 ski jackets? What is the expected profit if Egress produces Q 5 12,000 ski jackets? What is the standard deviation of profit in these two cases?

3. Based on the same 1000 scenarios, how many ski jackets should Egress produce to maximize expected profit? Call this quantity Q.

4. Should Q equal mean demand or not? Explain. 5. Create a histogram of profit at the production level Q.

Create a histogram of profit when the production level Q equals mean demand. What is the probability of a loss greater than $100,000 in each case?

CASE 15.1 Ski Jacket Production

CASE 15.2 Ebony Bath Soap Management of Ebony, a leading manufacturer of bath soap, is trying to control its inventory costs. The weekly cost of hold- ing one unit of soap in inventory is $30 (one unit is 1000 cases of soap). The marketing department estimates that weekly demand averages 120 units, with a standard deviation of 15 units, and is reasonably well modeled by a normal distribu- tion. If demand exceeds the amount of soap on hand, those sales are lost—that is, there is no backlogging of demand. The production department can produce at one of three levels: 110, 120, or 130 units per week. The cost of changing the produc- tion level from one week to the next is $3000.

Management would like to evaluate the following pro- duction policy. If the current inventory is less than L 5 30 units, they will produce 130 units in the next week. If the current inventory is greater than U 5 80 units, they will pro- duce 110 units in the next week. Otherwise, Ebony will con- tinue at the previous week’s production level.

Ebony currently has 60 units of inventory on hand. Last week’s production level was 120.

Questions 1. Develop a simulation model for 52 weeks of operation

at Ebony. Graph the inventory of soap over time. What is the total cost (inventory cost plus production change cost) for the 52 weeks?

2. Run the simulation for 500 iterations to estimate the av- erage 52-week cost with values of U ranging from 30 to 80 in increments of 10. Keep L 5 30 throughout.

3. Report the sample mean and standard deviation of the 52- week cost under each policy. Using the simulated results, is it possible to construct valid 90% confidence intervals for the average 52-week cost for each value of U? In any case, graph the average 52-week cost versus U. What is the best value of U for L 5 30?

4. What other production policies might be useful to investigate?

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7 7 8 C h a p t e r 1 5 I n t r o d u c t i o n t o S i m u l a t i o n M o d e l i n g

APPENDIX Simulation with DADM_Tools This chapter has illustrated two ways to run simulations in Excel: by using Excel-only tools and by using @RISK. The former takes a lot of time for creating data tables and summarizing, and the latter uses an add-in that is almost too powerful with its large number of options. Albright has written a simulation program in his DADM_Tools add-in that you might want to try as an alternative. (It is freely available at https://kelley.iu.edu/albrightbooks/free_downloads.htm.) This program does much of what @RISK does, but it is much more straightforward, allowing many fewer options than @RISK. You build a simulation model in the same way as illustrated in this chapter, you fill out one dialog box to indicate the number of replications, the outputs, and optionally a decision variable (the counterpart to RiskSimtable), and you get the results in a new nicely formatted worksheet. For illustration, we have included versions of many of the examples and problem solutions that use this simulation program.

If you decide to use the DADM_Tools add-in for simulation, you should also install Albright’s RandGen add-in (freely available at the same website). This add-in provides functions such as Normal_ and Triangular_ for generating random num- bers. These functions are recognized by the simulation program.

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EFFECTS OF MERIT PAY ON PAYROLL GROWTH AT DOD The National Security Personnel System (NSPS), a Depart- ment of Defense (DoD) civilian personal-management system, was designed to provide more merit-based pay raises than the existing civil service system. However, a U.S. Army organization was concerned that increased merit raises would eventually result in an unaffordable total payroll. As reported by Huntsinger et al. (2012), he and a team of analysts were asked by the Army to develop a quantitative model that could determine the effect of merit pay increases on total payroll growth.

The team quickly realized that forecasting eventual payroll growth was a complex problem. Payroll growth

depends on many uncertainties, including the number and salary of employees in each pay pool, annual employee ratings, employees leaving and being hired, cost-of-living raises, funds designated for merit raises, and the split between bonuses and pay raises. Nevertheless, they believed a simulation model could be developed to determine the most important factors affecting payroll growth. If a simulation using current policies showed an unacceptable level of payroll growth, the model could then be used to see whether pol- icy changes could be made to bring payroll increases to more acceptable levels.

The current NSPS policy grouped employees into pay pools of 35 to over 300 members, usually within one organization. Employees in a given pay pool can belong to different pay bands. The base salaries move within a pay band in three ways: annual across-the-board raises, in-band pay raises for increased responsibilities, and merit raises based on performance ratings. The merit raises can be raises in base pay or one-time bonuses. Each organization can determine the split between the two, and this split can depend on the performance rating. For example, lower performers might receive a 50-50 split, whereas higher performers might receive a 75-25 split. The latter split benefits the employee because the level of his or her base pay has increased for future years, but it has an adverse effect on the long-term total payroll.

Another important effect that had to be captured in a simulation model is attrition. If employees who receive lower performance ratings become discouraged and leave the organization, while those who receive higher ratings stay, the organization could become top-heavy with higher paid employees. In addition, who should be hired to replace those who leave? It was traditional to replace employees at given salary levels with others at approximately the same salary levels, but there was no rule that it had to be done this way.

The team was eventually able to develop a simulation model that: (1) captured the important drivers of payroll increases; (2) was sufficiently flexible to allow policy changes; (3) was straightforward enough to be understood by those who requested the study; and (4) had a graphical user interface for exploring policy changes. The simulation model starts by generating random initial base salaries for the employees, based on current data. Then each year, it uses probabilities, again based on historical data, to generate performance ratings and hence merit raises and bonuses, depending on the merit raise-bonus split. It also randomly chooses which employees will leave and hires new people at randomly generated salaries to replace them.

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CHAPTER 16 Simulation Models

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7 8 0 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

Once the simulation model was completed, the analysts made 1000 simulation runs and judged policies based on the average percent that base salaries increased above the assumed 3% cost-of-living increases over 20 years. Using a base case from using current policies, the simulation showed an average increase in base salaries of approx- imately 30% over the cost of living percentage. This was deemed unacceptable. Then running various sensitivity analyses, they discovered two factors that could lower this increase. First, in the base case, the fraction of merit pay given as a raise (as opposed to a bonus) ranged from 45% for the lowest performers to 70% for the highest perform- ers. By decreasing these percentages to 25% and 50%, the 30% overrun was reduced to 21%. Second, the base case hired replacements at approximately the same base salaries as those they replaced. By instead replacing them with employees earning about 20% less, the 30% overrun was reduced to 12%. Finally, by implementing both of these policy changes simultaneously, the 30% overrun was reduced to zero. Of course, the results of a simulation vary from run to run because of randomness. However, a statistical analysis of the results showed that with the implementation of these two policy changes, the actual 20-year pay increases were within 3% of the cost-of-living increases in about 750 of the 1000 replications.

In summary, the simulation model provided the Army with the information it needed. It showed that the current policy would indeed result in unacceptable totally payroll increases in the long run, and it discovered several policy changes the Army could make to bring these payroll increases down to acceptable levels.

16-1 Introduction The previous chapter introduced most of the important concepts for developing and analyzing spreadsheet simulation models. It also discussed many of the features avail- able in the powerful simulation add-in, @RISK, that accompanies this book. Now we apply the tools to a wide variety of problems that can be analyzed with simulation. We group the applications into four general areas: (1) operations models, (2) financial models, (3) marketing models, and (4) games of chance. The overriding theme in this chapter is that simulation models can yield important insights in all these areas. You do not need to cover all the models in this chapter or cover them in any particular order. You can cover the ones of most interest to you in practically any order.

16-2 Operations Models Whether we are discussing the operations of a manufacturing or a service company, there is likely to be uncertainty that can be modeled with simulation. In this section, we look at examples of bidding for a government contract (uncertainty in the bids by competitors), warranty costs (uncertainty in the time until failure of an appliance), and drug production (uncertainty in the yield and timing).

16-2a Bidding for Contracts In situations where a company must bid against competitors, simulation can often be used to determine the company’s optimal bid. Usually the company does not know what its competitors will bid, but it might have an idea about the range of the bids its competitors will choose. In this section, we show how to use simulation to determine a bid that maxi- mizes the company’s expected profit.

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16-2 Operations Models 7 8 1

EXAMPLE

16.1 BIDDING FOR A GOVERNMENT CONTRACT AT MILLER CONSTRUCTION

Miller Construction Company must decide whether to make a bid on a construction project. Miller believes it will cost the company $10,000 to complete the project (if it wins the contract), and it will cost $350 to prepare a bid. However, there is uncertainty about each of these. Upon further reflection, Miller assesses that the cost to complete the project has a triangular distribution with minimum, most likely, and maximum values $9000, $10,000, and $15,000. Similarly, Miller assesses that the cost to prepare a bid has a triangular distribution with parameters $300, $350, and $500. (Note the skewness in these distri- butions. Miller recognizes that cost overruns are much more likely than cost underruns.) Four potential competitors are going to bid against Miller. The lowest bid wins the contract, and the winner is then given the winning bid amount to complete the project. Based on past history, Miller believes that each potential competitor will bid, independently of the others, with prob- ability 0.5. Miller also believes that each competitor’s bid will be a multiple of its (Miller’s) most likely cost to complete the project, where this multiple has a triangular distribution with minimum, most likely, and maximum values 0.9, 1.3, and 1.8, respectively. If Miller decides to prepare a bid, its bid amount will be a multiple of $500 in the range $10,500 to $15,000. The company wants to use simulation to determine which strategy to use to maximize its expected profit.

Objective To simulate the profit to Miller from any particular bid, and to see which bid amount is best.

Where Do the Numbers Come From? The data required here are the parameters of the distributions of Miller’s costs, those of the competitors’ bids, and the prob- ability that a given competitor will place a bid. Triangular distributions are chosen for simplicity, although Miller could try other types of distributions. The parameters of these distributions are probably educated guesses, possibly based on previous contracts and bidding experience against these same competitors. The probability that a given competitor will place a bid can be estimated from these same competitors’ bidding history.

Solution The logic is shown in Figure 16.1. (See the file Contract Bidding Big Picture.xlsx.) You first simulate the number of compet- itors who will bid and then simulate their bids. Then for any bid Miller makes, you see whether Miller wins the contract, and if so, what its profit is.

Miller’s bid

Number of potential competitors

Number of competing bids

Miller’s cost to prepare a bid

Miller’s cost to complete project

Competing bids (if any)

Minimum competing bid (if any)

Miller wins bid?

Miller’s profit

Probability a given competitor bids

Figure 16.1 Big Picture for Bidding Simulation Model

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7 8 2 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

Developing the Simulation Model The simulation model appears in Figure 16.2. (See the file Contract Bidding Finished.xlsx.) It can be developed with the following steps. (Note that this model does not check the possibility of Miller not bidding at all. But this case is easy. If Miller opts not to bid, the profit is a certain $0.)

1. Inputs. Enter the inputs in the blue cells. 2. Miller’s bid. You can test all of Miller’s possible bids simultaneously with the RiskSimtable func-

tion. To set up for this, enter the formula

5RiskSimtable(D16:M16)

in cell B16. As with all uses of this function, the spreadsheet shows the simulated values for the first bid, $10,500. However, when you run the simulation, you see outputs for all of the bids.

3. Miller’s costs. Generate Miller’s cost to prepare a bid in cell B19 with the formula

5RiskTriang(B5,C5,D5)

Then copy this to cell B20 to generate Miller’s cost to complete the project. 4. Competitors and their bids. First, generate the random number of competitors who bid. This has a binomial distribution

with four trials and probability of “success” equal to 0.5 for each trial, so enter the formula

5RiskBinomial(B8,B9)

in cell B21. Then generate random bids for the competitors who bid in row 23 by entering the formula

5IF(B22*5$B$21,RiskTriang($B$12,$B$13,$B$14)*$C$6,””)

in cell B23 and copying across. This generates a random bid for all competitors who bid, and it enters a blank for those who don’t. (Remember that the random value is the multiple of Miller’s most likely cost to complete the project.) Calcu- late the smallest of these (if there are any) in cell B24 with the formula

5IF(B21+51,MIN(B23:E23),””)

Of course, Miller will not see these other bids until it has submitted its own bid. 5. Win contract? See whether Miller wins the bid by entering the formula

5IF(OR(B16*B24,B2150),1,0)

in cell B26. Here, 1 means that Miller wins the bid, and 0 means a competitor wins the bid. If there are no competing bids, Miller wins for sure. Then designate this cell as an @RISK output cell. Recall that to designate a cell as an @RISK output

Contract Bidding Model

Recall that the RiskSimtable function allows you to run a separate simulation for each value in its list.

Figure 16.2 Bidding Simulation Model

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

MLKJIHGFEDCBA Bidding for a contract

Inputs Miller’s costs, triangular distributed Min Most likely Max Cost to prepare a bid Cost to complete project

Number of poten�al compe�tors Probability a given compe�tor bids

Min Most likely Max

Possible bids for Miller Miller's bid

Simula�on Miller’s cost to prepare a bid Miller’s cost to complete project Number of compe�ng bids Compe�tor index 4321 Compe�tors’ bids $12,818 Minimum compe�tor bid $12,818

Miller wins bid? (1 if yes, 0 if no) Miller’s profit

Parameters of triangular distribu�ons for each compe�tor’s bid (expressed as mul�ple of Miller’s most likely cost to complete project)

$15,000$14,500$14,000$13,500$13,000$12,500$12,000$11,500$11,000$10,500

0.9 1.3 1.8

4 0.5

$500$350$300 $9,000 $10,000 $15,000

$10,500

$459 $11,111

1 –$1,069

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16-2 Operations Models 7 8 3

cell, you select the cell and then click the Add Output button on @RISK’s ribbon. You can then label this output appropri- ately. We used the label Wins Bid.

6. Miller’s profit. If Miller submits a bid, the bid cost is lost for sure. Beyond that, the profit to Miller is the bid amount minus the cost of completing the project if the bid is won. Otherwise, Miller makes nothing. So enter the formula

5IF(B2651,B16-B20,0)2B19

in cell B27. Then designate this cell as an additional @RISK output cell. (We named it Profit.)

Running the Simulation Set the number of iterations to 1000, and set the number of simulations to 10 because there are 10 bid amounts Miller wants to test.

Discussion of the Simulation Results The summary results appear in Figure 16.3. For each simulation—that is, each bid amount—there are two outputs: 1 or 0 to indicate whether Miller wins the contract and Miller’s profit. The only interesting results for the 091 output are in the Mean column, which shows the fraction of iterations that resulted in 1’s. So you can see, for example, that if Miller bids $12,000 (simulation #4), the probability of winning the bid is estimated to be about 0.57. This probability clearly decreases as Miller’s bid increases.

Figure 16.3 Summary Results for Bidding Simulation

In terms of net profit, if you concentrate only on the Mean column, a bid amount of $13,500 (simulation #7) is the best. But as the other numbers in this figure indicate, the mean doesn’t tell the whole story. For example, if Miller bids $13,500, it could win the bid but still lose a considerable amount of money because of cost overruns. The histogram of profit in Figure 16.4 indicates this more clearly. It shows that in spite of the positive mean, many of the outcomes are negative.

So what should Miller do? If it doesn’t bid at all, its profit is a certain $0. If Miller is an expected profit maximizer, then the fact that several of the means in Figure 16.3 are positive indicates that bidding is better than not bidding, with a bid of $13,500 being the best bid. However, potential cost overruns and the corresponding losses are certainly a concern. Depending on Miller’s degree of risk aversion, the company might decide to (1) not bid at all, or (2) bid higher than $13,500 to minimize

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its worse loss. Still, we would caution Miller not to be too conservative. Rather than focusing on the Min (worst case) column in Figure 16.3, we would suggest focusing on the 5% column. This shows nearly how bad things could get (5% of the time it would be worse than this), and this 5th percentile remains fairly constant for higher bids.

Figure 16.4 Histogram of Profit with $13,500 Bid

16-2b Warranty Costs When you buy a new product, it usually carries a warranty. A typical warranty might state that if the product fails within a certain period such as one year, you will receive a new product at no cost, and it will carry the same warranty. However, if the product fails after the warranty period, you have to bear the cost of replacing the product. Due to random life- times of products, we need a way to estimate the warranty costs (to the manufacturer) of a product. The following example illustrates how this can be accomplished with simulation.

EXAMPLE

16.2 WARRANTY COSTS AT YAKKON Yakkon Company sells a popular camera for $400. This camera carries a warranty such that if the camera fails within 1.5 years, the company gives the customer a new camera for free. If the camera fails after 1.5 years, the warranty is no longer in effect. Every replacement camera carries exactly the same warranty as the original camera, and the cost to the company of sup- plying a new camera is always $225. Use simulation to estimate, for a given sale, the number of replacements under warranty and the NPV of profit from the sale, using a discount rate of 8%.

7 8 4 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

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16-2 Operations Models 7 8 5

Objective To use simulation to estimate the number of replacements under warranty and the total NPV of profit from a given sale.

Where Do the Numbers Come From? The warranty information is a policy decision made by the company. The hardest input to estimate is the probability distribu- tion of the lifetime of the product. We discuss this next.

Solution The only randomness in this problem concerns the time until failure of a new camera. Yakkon could estimate the distribution of time until failure from historical data. This would probably indicate a right-skewed distribution, as shown in Figure 16.5. If you look through the list of dis- tributions available in @RISK under Define Distributions, you will see several with this same basic shape. The one shown in Figure 16.5 is a commonly used distribution called the gamma distribution. We will use a gamma distribution in this example, although other choices such as the triangular are certainly possible.

The gamma distribution is a popular distribution when you want a right- skewed distribution of a positive quantity.

Figure 16.5 Right-Skewed Gamma Distribution

Selecting a Gamma Distribution The gamma distribution is characterized by two parameters, a and b. These determine its shape and location. It can be shown that the mean and standard deviation are m 5 ab and s 5 !ab. Alterna- tively, for any desired values of the mean and standard deviation, these equations can be solved for a and b, which leads to a 5 m2>s2 and b 5 s2>m. So, for example, if you want a gamma distribution with mean 2.5 and standard deviation 1 (which in this example would be based on camera lifetime data from the past), you should choose a 5 2.52>12 5 6.25 and b 5 12>2.5 5 0.4. These are the values shown in Figure 16.5 and the ones used for this example. The values in the figure (from @RISK) imply that the probability of failure before 1.5 years is about 0.15, so that the probability of failure out of warranty is about 0.85.

Developing the Simulation Model The variables for the model are shown in Figure 16.6, and the simulation model itself appears in Figure 16.7. (See the files Warranty Costs Big Picture.xlsx and Warranty Costs Finished.xlsx.) The particular random numbers in Figure 16.7 indi- cate an example (a rather unusual one) where there are two failures within warranty. However, because the lifetime of the

You can learn about distributions from @RISK’s Define Distribution window.

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second replacement (cell D17) is greater than 1.5, the company incurs only two replacement costs, as shown in cells B19 and C19. The model can be developed with the following steps.

1. Inputs. Enter the inputs in the blue cells. 2. Parameters of gamma distribution. As discussed previously, if you enter a desired mean and standard deviation (in cells

B5 and B6), you have to calculate the parameters of the gamma distribution. Do this by entering the formulas

5B5^2>B6^2

and

5B6^2>B5

in cells B7 and B8. 3. Lifetimes and times of failures. Generate at most five lifetimes and corresponding times of failures. (Why only five? You

could generate more, but it is extremely unlikely that this same customer would experience more than five failures within warranty, so five suffices.) As soon as a lifetime is greater than 1.5, the warranty period, no further lifetimes are required, in which case blanks can be recorded in row 17. With this in mind, enter the formulas

5RiskGamma(B7,B8)

5IF(B17*B10,RiskGamma(B7,B8),””)

and

5IF(C175””,””,IF(C17*$B$10,RiskGamma($B$7,$B$8),””))

in cells B17, C17, and C17, and copy the latter formula to cells E17 and F17. These formulas guarantee that once a blank is recorded in a cell, all cells to its right will also contain blanks. To get the actual times of failures, relative to time 0 when the customer originally purchases the camera, enter the formulas

5B17

and

5IF(C175””,””,B181C17)

in cells B18 and C18, and copy the latter across row 18. These values will be used for the NPV calculation because this requires the exact timing of cash flows.

Warranty Cost Model

Figure 16.6 Big Picture for Warranty Simulation Model

Failures within warranty

NPV of profit from customer

Warranty period Lifetime of camera

Time of failure

Discount rate Discounted cost

Cost to company Replacement cost (to company)

Cost of new camera (to customer)

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16-2 Operations Models 7 8 7

4. Costs and discounted costs. In row 19, enter the replacement cost of $225 or 0, depending on whether a failure occurs within warranty, and in row 20 discount these costs back to time 0, using the failure times in row 18. To do this, enter the formulas

5IF(B17*B10,B12,0)

and

5IF(C175””,0,IF(C17*$B$10,$B$12,0))

in cells B19 and C19, and copy this latter formula across row 19. Then enter the formula

5IF(B19+0,B19>(11$B$13)^B18,0)

in cell B20, and copy it across row 20. This formula uses the well-known fact that the present value of a cash flow at time t is the cash flow multiplied by 1>(1 1 r)t, where r is the discount rate.

5. Outputs. Calculate two outputs, the number of failures within warranty and the NPV of profit, with the formulas

5COUNTIF(B17:F17,”*”&$B$10)

and

5B112B122SUM(B20:F20)

in cells B22 and B23. Then designate these two cells as @RISK output cells. Note that the NPV is the margin from the sale (undiscounted) minus the sum of the discounted costs from replacements under warranty.

Figure 16.7 Warranty Simulation Model 1

2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23

FEDCBA Warranty costs for camera

Inputs Parameters of �me to failure distribu�on of any new camera (Gamma)

2.5Desired mean Desired stdev 1 Implied alpha 6.250 Implied beta 0.400

Warranty period 1.5 Cost of new camera (to customer) $400 Replacement cost (to company) $225 Discount rate 8%

Simula�on of new camera and its replacements (if any) 54321Camera

4.3331.4261.288Life�me 7.0472.7141.288Time of failure

000225225Cost to company 0.000.000.00182.59203.77Discounted cost

2.000Failures within warranty NPV of profit from customer ($211.36)

RiskGamma

To generate a random number from the gamma distribution, use the RiskGamma function in the form =RiskGamma(alpha,beta). The mean and standard deviation of this distribution are m 5 ab and s 5 !ab. Equivalently, a 5 m2>s2 and b 5 s2>m.

@RISK Function

Excel’s NPV function can be used only for cash flows that occur at the ends of the respective years. Otherwise, you have to discount cash flows manually.

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These results indicate that Yakkon is not suffering terribly from warranty costs. However, there are several ways the com- pany could decrease the effects of warranty costs. First, it could increase the price of the camera. Second, it could decrease the

Running the Simulation The @RISK setup is typical. Run 1000 iterations of a single simulation (because there is no RiskSimtable function).

Discussion of the Simulation Results The @RISK summary statistics and histograms for the two outputs appear in Figures 16.8 and 16.9. They show a fairly clear picture. About 85% of the time, there are no failures under warranty and the company makes a profit of $175, the margin from the camera sale. However, there is about a 12% chance of exactly one failure under warranty, in which case the company’s NPV of profit will be an approximate $50 loss (before discounting). Additionally, there is about a 3% chance that there will be even more failures under warranty, in which case the loss will be even greater. Note that in our 1000 iterations, the maximum number of failures under warranty was 4, and the maximum net loss was $515.94. On average, the NPV of profit was $139.75.

Figure 16.8 Histogram of Number of Failures

Figure 16.9 Histogram of NPV of Profit

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warranty period, say, from 1.5 years to 1 year. Third, it could change the terms of the warranty. For example, it could stipulate that if the camera fails within a year, the customer gets a new camera for free, whereas if the time to failure is between 1 and 1.5 years, the customer pays some pro rata share of the replacement cost. Finally, it could try to sell the customer an extended warranty—at a hefty price. We ask you to explore these possibilities in the problems.

16-2c Drug Production with Uncertain Yield In many manufacturing settings, products are produced in batches, and the usable yields from these batches are uncertain. This is particularly true in the drug industry. The fol- lowing example illustrates how a drug manufacturer can take this uncertainty into account when planning production.

EXAMPLE

16.3 TRYING TO MEET AN ORDER DUE DATE AT WOZAC Wozac Company is a drug manufacturer. Wozac has recently accepted an order from its best customer for 8000 ounces of a new miracle drug, and Wozac wants to plan its production schedule to meet the customer’s promised delivery date of Decem- ber 1. There are three sources of uncertainty that make planning difficult. First, the drug must be produced in batches, and there is uncertainty in the time required to produce a batch, which could be anywhere from 5 to 11 days. This uncertainty is described by the discrete distribution in Table 16.1. Second, the yield (usable quantity) from any batch is uncertain. Based on historical data, Wozac believes the yield can be modeled by a triangular distribution with minimum, most likely, and maximum values equal to 600, 1000, and 1100 ounces, respectively. Third, all batches must go through a rigorous inspection once they are completed. The probability that a typical batch passes inspection is only 0.8. With probability 0.2, the batch fails inspec- tion, and none of it can be used to help fill the order. Wozac wants to use simulation to help decide how many days prior to the due date it should begin production.

Table 16.1 Distribution of Days to Complete a Batch

Days Probability

5 0.05

6 0.10

7 0.20

8 0.30

9 0.20

10 0.10

11 0.05

Objective To use simulation to determine when Wozac should begin production for this order so that there is a high probability of com- pleting it by the due date.

Where Do the Numbers Come From? The important inputs here are the probability distributions of the time to produce a batch, the yield from a batch, and the inspection result. The probabilities we have assumed would undoubtedly be based on previous production data. For example, the company might have observed that about 80% of all batches in the past passed inspection. Of course, a discrete distribution is natural for the number of days to produce a batch, and a continuous distribution is appropriate for the yield from a batch.

Solution The variables for this model are shown in Figure 16.10. (See the file Drug Production Big Picture.xlsx.) The idea is to sim- ulate successive batches—their days to complete, their yields, and whether they pass inspection—and keep a running total of the usable ounces obtained so far. IF functions can then be used to check whether the order is complete or another batch is required. You need to simulate only as many as batches as are required to meet the order, and you should keep track of the days

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required to produce all of these batches. In this way, you can “back up” to see when production must begin to meet the due date. For example, if the simulation indicates that the order takes 96 days to complete, then production must begin on August 27, 96 days before the due date. (For simplicity, the model assumes that production occurs seven days a week.)

Figure 16.10 Big Picture for Drug Production Model

Days to produce batch

Batch passes inspec�on

Probability batch passes inspec�on

Due date

Yield of batch

Day to start

Has enough been produced?

Amount of drug required

Total days to complete

Batches required

Cumula�ve yield

Developing the Simulation Model The completed model appears in Figure 16.11. (See the file Drug Production Finished.xlsx.) It can be developed as follows.

1. Inputs. Enter the inputs in the blue cells. 2. Batch indexes. We don’t know ahead of time how many batches will be required to fill the order. There should be enough

rows in the simulation to cover the worst case that is likely to occur. After some experimentation, it is apparent that 25 batches are almost surely enough. Therefore, enter the batch indexes 1 through 25 in column A of the simulation section. (If 25 were not enough, you could always add more rows.) The idea, then, is to fill the entire range B16:F40 with formu- las. However, you can use appropriate IF functions in these formulas so that if enough has already been produced to fill the order, blanks are inserted in the remaining rows. For example, the scenario shown in Figure 16.11 is one where 13 batches were required, so blanks appear below row 28.

3. Days for batches. Simulate the days required for batches in column B. To do this, enter the formulas

5RiskDiscrete(G6:G12,H6:H12)

and

5IF(F16*+”No”,””,RiskDiscrete($G$6:$G$12,$H$6:$H$12))

in cell B16 and B17, and copy the latter formula down to cell B40. The IF function enters a blank in this cell if there is not a No in column F for the previous batch, that is, if the order was just completed in the previous batch or it has been completed for some time. Similar logic appears in later formulas.

4. Batch yields. Simulate the batch yields in column C. To do this, enter the formulas

5RiskTriang(B9,C9,D9)

and

5IF(F16*+”No”,””,RiskTriang($B$9,$C$9,$D$9))

in cells C16 and C17, and copy the latter formula down to cell C40. 5. Pass inspection? Check whether each batch passes inspection with the formulas

5IF(RAND()*Probability_pass,”Yes”,”No”)

and

5IF(F16*+”No”,””,IF(RAND()*Probability_pass,”Yes”,”No”))

Drug Production Model

You can use Excel’s RAND function inside an IF function to simulate whether some event occurs.

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16-2 Operations Models 7 9 1

in cells D16 and D17, and copy the latter formula down to cell D40. Note that you could use @RISK’s RiskUniform(0,1) function instead of RAND(), but there is no real advantage to doing so. They are essentially equivalent. (Besides, the aca- demic version of @RISK imposes an upper limit of 100 @RISK input functions per model, so it is often a good idea to substitute built-in Excel® functions when possible.)

6. Order filled? To keep track of the cumulative usable production and whether the order has been filled in columns E and F, first enter the formulas

5IF(D165”Yes”,C16,0)

and

5IF(E16+5Ounces_required,”Yes”,”Not yet”)

in cells E16 and F16 for batch 1. Then enter the general formulas

5IF(F16*+”No”,””,IF(D175”Yes”,C171E16,E16))

and

5IF(F16*+”No”,””,IF(E17+5Ounces_required,”Yes”,”Not yet”))

in cells E17 and F17, and copy them down to row 40. Note that the entry in column F is “Not yet” if the order is not yet complete. In the row that completes the order, it changes to “Yes,” and then it is blank in succeeding rows.

7. Summary measures. Calculate the batches and days required in cells I15 and I16 with the formulas

5INDEX(A16:A40,MATCH(”Yes”,F16:F40,0))

and

5SUM(B16:B40)

These are the two cells used as output cells for @RISK, so designate them as such. Also, calculate the day the order should be started to just meet the due date in cell I17 with the formula

5Due_date2I16

Planning produc�on of a drug A

Input sec�on

Simula�on model

Ounces required Due date

Distribu�on of days needed to produce a batch (discrete)

Distribu�on of yield (ounces) from each batch (triangular)

Prob of passing inspec�on

Batch 1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18

Enough? No No No No No No No No No No No No Yes

Outputs Batches required Days to complete Day to start

Sta�s�cal summary measures Max batches reqd

Avg days reqd Min days reqd Max days reqd 5th perc days reqd 95th perc days reqd

Range names used Batches_required Days_to_complete Due_date Probability _pass Ounces_required

93 62

161 71

121

B C D E F G H I J K L M

=Model!$l$15 =Model!$l$16 =Model!$B$5 =Model!$B$11 =Model!$B$4

Probability of mee�ng due date

30-Aug 30-Sep 23-Jun 21-Sep 2-Aug

Probability 0.990 0.955 0.836 0.503 0.127

Start date 15-Jul 1-Aug

15-Aug 1-Sep

15-Sep

8000 1-Dec

Min 600

0.8

Most likely 1000

Max 1100

Days 5 6 7 8 9

10 11

Probability 0.05 0.10 0.20 0.30 0.20 0.10 0.05

Days 10 6 9 8 9 9 9 10 8 9 9 8 7

CumYield 0.0

786.3 786.3

1630.1 1630.1 2690.4 3610.8 4324.5 5282.4 6315.9 7293.3 7293.3 8115.5

Yield 821.5 786.3

1012.2 843.7 754.6

1060.4 920.3 713.7 958.0

1033.5 977.4 843.5 822.2

Pass? No Yes No Yes No Yes Yes Yes Yes Yes Yes No Yes

13 111

12-Aug

Figure 16.11 Drug Production Simulation Model

Date subtraction in Excel allows you to calculate the number of days between two given dates.

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This formula uses date subtraction to find an elapsed time. (Again, the assumption is that production occurs every day of the week.)

This completes the simulation model development. The other entries in columns H through J will be explained shortly.

Dealing with Uncertain Timing

Many simulations that model a process over multiple time periods must deal with uncertain timing of events, such as when the manufacturing of an order will finish, which year sales of a new product will begin, and many others. Essentially, the spreadsheet model must generate random numbers that determine the timing and then play out the events. This can require tricky IF functions and possibly other functions. However, the hard work often involves getting the logic correct for only the first period or two. Then this logic can be copied down for the other periods. In other words, some time spent on developing the first row or two can result in a powerful model.

Fundamental Insight

Running the Simulation Set the number of iterations to 1000 and the number of simulations to 1, and then run the simulation as usual.

Discussion of the Simulation Results After running the simulation, you can obtain the distributions of the number of batches required and the number of days required in Figures 16.12 and 16.13.

Figure 16.12 Distribution of Batches Required

How should Wozac use this information? The key questions are how many batches will be required and (2) when produc- tion should start. To answer these questions, it is helpful to use several of @RISK’s statistical functions. Recall that these func- tions can be entered directly into the Excel model worksheet. (Also, recall that they provide useful information only after the simulation has been run.) These functions provide no new information you don’t already have from other @RISK windows, but they allow you to see (and manipulate) this information directly in the spreadsheet.

For the first question, enter the formula

5RiskMax(Batches_required)

in cell I20. (Refer to Figure 16.11.) It shows that the worst case from the 1000 iterations, in terms of batches required, is 20 batches. (If this maximum were 25, you would add more rows to the simulation model and run the simulation again.)

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You can answer the second question in two ways. First, you can calculate summary measures for days required and then back up from the due date. This is done in the range I22:J26. The formulas in column I are

5INT(RiskMean(Days_to_complete))

5RiskMin(Days_to_complete)

5RiskMax(Days_to_complete)

5RiskPercentile(Days_to_complete,0.05)

and

5RiskPercentile(Days_to_complete,0.95)

(The first uses the INT function to produce an integer.) You can then subtract each of these from the due date to obtain the potential starting dates in column J. Wozac should realize the pros and cons of these starting dates. For example, if the com- pany wants to be 95% sure of meeting the due date, it should start production on August 2. In contrast, if Wozac starts produc- tion on September 21, there is only a 5% chance of meeting the due date.

Alternatively, you can get a more direct answer to the question by using @RISK’s RiskTarget function. This allows you to find the probability of meeting the due date for any starting date, such as the trial dates in the range L22:L26. To do it, enter the formula

5RiskTarget(Days_to_complete,Due_date-L22)

in cell M22 and copy it down. This function returns the fraction of iterations where the (random) value in the first argument is less than or equal to the (fixed) value in the second argument. For example, you can see that 83.6% of the iterations have a value of days required less than or equal to 108, the number of days from August 15 to the due date.

What is our recommendation to Wozac? We suggest going with the 95th percen- tile—begin production on August 2. Then there is only a 5% chance of failing to meet the due date. But the table in the range L22:M26 also provides useful information. For each potential starting date, Wozac can see the probability of meeting the due date.

Figure 16.13 Distribution of Days Required

Using @RISK summary functions such as RiskMean, RiskPercen- tile, and others enables you to capture simulation results in the same work-sheet as the simulation model. These functions do not provide relevant results until the simulation is run.

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Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 1. In the Miller bidding model, the possible profits vary

from negative to positive for each of the 10 possible bids examined. a. For each of these, use @RISK’s RiskTarget function

to find the probability that Miller’s profit is positive. Do you believe these results should have any bearing on Miller’s choice of bid?

b. Use @RISK’s RiskPercentile function to find the 10th percentile for each of these bids. Can you explain why the percentiles have the values you obtain?

2. If the number of competitors in the Miller bidding model doubles, how does the optimal bid change?

3. Referring to the Miller bidding model, if the average bid for each competitor stays the same, but their bids exhibit less variability, does Miller’s optimal bid increase or decrease? To study this question, assume that each com- petitor’s bid, expressed as a multiple of Miller’s cost to complete the project, follows each of the following dis- tributions. a. Triangular with parameters 1.0, 1.3, and 2.4 b. Triangular with parameters 1.2, 1.3, and 2.2 c. Use @RISK’s Define Distributions window to check

that the distributions in parts a and b have the same mean as the original triangular distribution in the example, but smaller standard deviations. What is the common mean? Why is it not the same as the most likely value, 1.3?

4. In the Yakkon warranty model, the gamma distribution was used to model the skewness to the right of the lifetime distribution. Experiment to see whether the triangular dis- tribution could have been used instead. Let its minimum value be 0, and choose its most likely and maximum val- ues so that this triangular distribution has approximately the same mean and standard deviation as the gamma dis- tribution in the example. (Use @RISK’s Define Distribu- tions window and trial and error to do this.) Then run the simulation and comment on similarities or differences between your outputs and the outputs in the example.

5. See how sensitive the results in the Yakkon warranty model are to the following changes. For each part, make the change indicated, run the simulation, and comment on any differences between your outputs and the outputs in the example. a. The cost of a new camera is increased to $450. b. The warranty period is decreased to one year. c. The terms of the warranty are changed. If the camera

fails within one year, the customer gets a new camera for free. However, if the camera fails between 1 year and 1.5 years, the customer pays a pro rata share of the new camera, increasing linearly from 0 to full price. For example, if it fails at 1.2 years, which is 40% of the way from 1 to 1.5, the customer pays 40% of the full price.

d. The customer pays $50 up front for an extended war- ranty. This extends the warranty to three years. This extended warranty is just like the original, so that if the camera fails within three years, the customer gets a new camera for free.

6. In the Wozac drug production model, we commented on the 95th percentile on days required and the correspond- ing date. If the company begins production on this date, then it is 95% sure to complete the order by the due date. We found this date to be August 2. Do you always get this answer? Find out by (1) running the simulation 10 more times, each with 1000 iterations, and finding the 95th percentile and corresponding date in each, and (2) running the simulation once more, but with 10,000 iterations. Comment on the difference between sim- ulations (1) and (2) in terms of accuracy. Given these results, when would you recommend that production should begin?

7. In the Wozac drug production model, suppose you want to run five simulations, where the probability of pass- ing inspection is varied from 0.6 to 1.0 in increments of 0.1. Use the RiskSimtable function appropriately to do this. Comment on the effect of this parameter on the key outputs. In particular, does the probability of passing inspection have a large effect on when production should start? (Note: When this probability is low, it might be necessary to produce more than 25 batches, the maxi- mum built into the model. Check whether this maximum should be increased.)

16-3 Financial Models There are many financial applications where simulation can be applied. Future cash flows, future stock prices, and future interest rates are some of the many uncertain variables financial analysts must deal with. In every direction they turn, they see uncertainty. In this section, we analyze a few typical financial applications that can benefit from simulation modeling.

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16-3 Financial Models 7 9 5

16-3a Financial Planning Models Many companies, such as GM, Eli Lilly, Procter & Gamble, and Pfizer, use simulation in their capital budgeting and financial planning processes. Simulation can be used to model the uncertainty associated with future cash flows. In particular, simulation can be used to answer questions such as the following:

• What are the mean and variance of a project’s net present value (NPV)? • What is the probability that a project will have a negative NPV? • What are the mean and variance of a company’s profit during the next fiscal year? • What is the probability that a company will have to borrow more than $2 million

during the next year?

The following example illustrates how simulation can be used to evaluate the financial success of a new car.

EXAMPLE

16.4 DEVELOPING A NEW CAR AT GF AUTO General Ford (GF) Auto Corporation is developing a new model of compact car. This car is assumed to generate sales for the next five years. GF has gathered information about the following quantities through focus groups with the marketing and engi- neering departments.

• Fixed cost of developing car. This cost is assumed to be $600 million. The fixed cost is incurred at the beginning of year 1, before any sales are recorded.

• Margin per car. This is the unit selling price minus the variable cost of producing a car. GF assumes that in year 1, the mar- gin will be $4000. Every other year, GF assumes the margin will decrease by 4%.1

• Sales. The demand for the car is the uncertain quantity. In its first year, GF assumes sales—number of cars sold—will be triangularly distributed with parameters 30,000, 55,000, and 65,000. Every year after that, the company assumes that sales will decrease by some percentage, where this percentage is triangularly distributed with parameters 5%, 8%, and 10%. GF also assumes that the percentage decreases in successive years are independent of one another.

• Depreciation and taxes. The company will depreciate its development cost on a straight-line basis over the lifetime of the car. The corporate tax rate is 21%.

• Discount rate. GF figures its cost of capital at 7%.

Given these assumptions, GF wants to develop a simulation model that will evaluate its NPV of after-tax cash flows for this new car over the five-year time horizon.

Objective To simulate the cash flows from the new car model, from the development time to the end of its life cycle, so that GF can esti- mate the NPV of after-tax cash flows from this car.

Where Do the Numbers Come From? There are many inputs to this problem. As we indicated, they are probably obtained from experts within the company and from focus groups of potential customers.

Solution The variables for the model are shown in Figure 16.14. (See the file New Car Development Big Picture.xlsx.) This model is like most financial multiyear spreadsheet models. The completed model extends several years to the right, but most of the work is for the first year or two. From that point, you can copy to the other years to complete the model.

1 The margin decreases because the company assumes variable costs tend to increase through time, whereas selling prices tend to remain fairly constant through time.

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Developing the Simulation Model The simulation model for GF appears in Figure 16.15. (See the file New Car Development Finished .xlsx.) It can be formed as follows.

1. Inputs. Enter the given inputs in the blue cells. 2. Unit sales. Generate first-year sales in cell B12 with the formula

5RiskTriang(E5,F5,G5)

Then generate the reduced sales in later years by entering the formula

5B12*(12RiskTriang($E$6,$F$6,$G$6))

in cell C12 and copying it across row 12. Note that each sales figure is a random fraction of the previous sales figure.

Figure 16.14 Big Picture for GF Auto Simulation Model

NPV of cash flows

Unit sales Year 1 contribu�on

Unit contribu�on

Deprecia�on

Before tax profit

A�er tax profitTax rate

Cash flow

Discount rate

Annual decrease in contribu�on

Revenue minus variable cost

Fixed development cost

New Car Development Model

New car simula�on A

Inputs

Simula�on

Fixed development cost Year 1 contribu�on Annual decrease in contribu�on Tax rate Discount rate

Parameters of triangular distribu�ons

Year 1 sales Annual decay rate

End of year Unit sales Unit contribu�on Revenue minus variable cost Deprecia�on Before tax profit A�er tax profit Cash flow

NPV of cash flows

B C D E F G

$600,000,000 $4,000

4% 21%

Min 30000

Most likely 55000

Max 65000

10%

1 56529

$4,000 $226,116,773 $120,000,000 $106,116,773

$83,832,251 $203,832,251

$89,240,670

2 51303

$3,840 $197,004,495 $120,000,000

$77,004,495 $60,833,551

$180,833,551

3 46936

$3,686 $173,026,079 $120,000,000

$53,026,079 $41,890,602

$161,890,602

4 43585

$3,539 $154,246,551 $120,000,000

$34,246,551 $27,054,776

$147,054,776

5 41017

$3,397 $139,350,833 $120,000,000

$19,350,833 $15,287,158

$135,287,158

Figure 16.15 GF Auto Simulation Model

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16-3 Financial Models 7 9 7

3. Contributions. Calculate the unit contributions in row 13 by entering the formulas

5B5

and

5B13*(12$B$6)

in cells B13 and C13, and copying the latter across. Then calculate the contributions in row 14 as the product of the cor- responding values in rows 12 and 13.

4. Depreciation. Calculate the depreciation each year in row 15 as the development cost in cell B4 divided by 5. This is exactly what “straight-line depreciation” means.

5. Before-tax and after-tax profits. To calculate the before-tax profit in any year, subtract the depreciation from total contribution, so each value in row 16 is the difference between the corresponding values in rows 14 and 15. The reason is that depreciation isn’t taxed. To calculate the after-tax profits in row 17, multiply each before-tax profit by one minus the tax rate in cell B7. Finally, each cash flow in row 18 is the sum of the corresponding values in rows 15 and 17. Here depreciation is added back to get the cash flow.

6. NPV. Calculate the NPV of cash flows in cell B20 with the formula

5 2B41NPV(B8,B18:F18)

and designate it as an @RISK output cell (the only output cell). Here, we are assuming that the development cost is incurred right now, so that it isn’t discounted, and that all other cash flows occur at the ends of the respective years. This allows the NPV function to be used directly.

Running the Simulation Set the number of iterations to 1000 and the number of simulations to 1, and then run the simulation as usual.

Discussion of the Simulation Results After running @RISK, you obtain the histogram in Figure 16.16. These results are somewhat comforting, but also a cause of concern for GF. On the bright side, the mean NPV is slightly more than $28 million, and there is some chance that the NPV could go well above that figure, even above $177 million. However, there is also a downside, as shown by the two sliders in the histogram. One slider has been placed over an NPV of 0. As the histogram indicates, there is a 65.9% chance of a positive NPV, but there is a 34.1% chance of it being negative. The second slider has been positioned at its default 5th percentile set- ting. Financial analysts often call this percentile the value at risk at the 5% level, or VaR 5%, because it indicates nearly the worst possible outcome. From this simulation, you can see that GF’s VaR 5% is approximately a $111.77 million loss.

Depreciation is subtracted to get before-tax profit, but it is then added back after taxes have been deducted.

Figure 16.16 Histogram of NPV

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What is most responsible for this huge variability in NPV, the variability in first-year sales or the variability in annual sales decreases? This can be answered with @RISK’s tornado graph, which lets you see which random inputs have the most effect on a specified output. (See Figure 16.17.) To get this graph, click the tornado button below the histogram in Figure 16.16 and select the Change in Output Mean option. This graph answers the question emphatically. Variability in first-year sales is by far the largest influence on NPV. It is very highly correlated with NPV. The annual decreases in sales are important, but they have much less effect on NPV. If GF wants to get a more favorable NPV distribution, it should do all it can to boost first-year sales—and make the first-year sales distribution less variable.

The value at risk at the 5% level, or VaR 5%, is the 5th percentile of a distribu- tion, and it is often used in financial models. It indicates nearly the worst possible outcome.

Financial analysts typically look at VaR 5% to see how bad—or more precisely, almost how bad—things could get.

Figure 16.17 Tornado Graph for NPV

2 It turns out that the NPV in this model is linear in the two random inputs. When an output is linear in the inputs, the deterministic model using means of inputs always gives the correct mean output (aside from some randomness), so that the flaw of averages in the form from the previous chapter does not occur. Even so, a deterministic model still provides no indication of how bad or how good things could get.

Before finishing this example, we revisit the flaw of averages. What if GF used a deterministic model to estimate NPV? Would the results match those from the sim- ulation? We tried this two ways, once by entering the most likely values of the inputs instead of the random numbers, and once by entering the means instead of the random numbers. The results (not shown here) are available in the last sheet of the finished ver- sion of the file. As you can check, the difference between the two NPVs is huge. In this case, the NPV by using means is very close to the mean NPV from the simulation, about $28 million. But if the company used most likely values for the inputs in its determin- istic model, which certainly seems sensible, the NPV would be about $77 million, off by a factor of more than two, another variation of the flaw of averages. Besides this problem, neither deterministic model provides even a hint that the company has about a 34.1% chance of a negative NPV.2

If you create a deterministic model using the most likely values of the uncertain inputs, you can possibly get an output value that is nowhere near the mean of that output.

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16-3 Financial Models 7 9 9

The Mean Isn’t Everything

Many discussions of simulation focus on the mean of an output variable. This makes sense, given the importance of EMV for decision making, as discussed in Chapter 6. After all, EMV is just the mean of a monetary output. However, ana- lysts in many areas, including finance, are often at least as interested in the extreme values of an output distribution. For example, the VaR 5% discussed in this exam- ple indicates nearly how bad things could get if unlucky outcomes occur. If large amounts of money are at stake, particularly potential losses, companies might not want to focus only on the mean. They should be aware of potential disasters as well. Of course, simulation also shows the bright side, the extremes on the right that could occur if lucky outcomes occur. Managers shouldn’t be so conservative that they focus only on the negative outcomes and ignore the upside potential.

Fundamental Insight

16-3b Cash Balance Models All companies track their cash balance over time. As specific payments come due, com- panies sometimes need to take out short-term loans to keep a minimal cash balance. The following example illustrates one such application.

EXAMPLE

16.5 MAINTAINING A MINIMAL CASH BALANCE AT ENTSON The Entson Company believes that its monthly sales during the period from November of the current year to July of next year are normally distributed with the means and standard deviations given in Table 16.2. Each month Entson incurs fixed costs of $250,000. In March taxes of $150,000 and in June taxes of $50,000 must be paid. Dividends of $50,000 must also be paid in June. Entson estimates that its receipts in a given month are a weighted sum of sales from the current month, the previous month, and two months ago, with weights 0.2, 0.6, and 0.2. In symbols, if Rt and St represent receipts and sales in month t, then

Rt 5 0.2St 2 2 1 0.6St 2 1 1 0.2St (16.1)

The materials and labor needed to produce a month’s sales must be purchased one month in advance, and the cost of these averages to 80% of the product’s sales. For example, if sales in February are $1,500,000, then the February materials and labor costs are $1,200,000, but these must be paid in January.

Table 16.2 Monthly Sales (in Thousands of Dollars) for Entson

Nov. Dec. Jan. Feb. Mar. Apr. May Jun. Jul.

Mean 1500 1600 1800 1500 1900 2600 2400 1900 1300

Standard Deviation 70 75 80 80 100 125 120 90 70

At the beginning of January, Entson has $250,000 in cash. The company wants to ensure that each month’s ending cash balance never falls below $250,000. This means that Entson might have to take out short-term (one-month) loans. For exam- ple, if the ending cash balance at the end of March is $200,000, Entson will take out a loan for $50,000, which it will then pay back (with interest) one month later. The interest rate on a short-term loan is 1% per month. At the beginning of each month, Entson earns interest of 0.5% on its cash balance. The company wants to use simulation to estimate the maximum loan it will need to take out to meet its desired minimum cash balance. Entson also wants to analyze how its loans will vary over time, and it wants to estimate the total interest paid on these loans.

Objective To simulate Entson’s cash flows and the loans the company must take out to meet a minimum cash balance.

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8 0 0 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

Where Do the Numbers Come From? Although there are many monetary inputs in the problem statement, they should all be easily accessible. Of course, Entson chooses the minimum cash balance of $250,000 as a matter of company policy.

Solution The variables for this model appear in Figure 16.18. (See the file Cash Balance Big Picture.xlsx.) Clearly, there is a consid- erable amount of bookkeeping in this simulation, so it is a good idea to list the events in chronological order that occur each month. We assume the following:

• Entson observes its beginning cash balance.

• Entson receives interest on its beginning cash balance.

• Receipts arrive and expenses are paid (including payback of the previous month’s loan, if any, with interest).

• If necessary, Entson takes out a short-term loan.

• The final cash balance is observed, which becomes next month’s beginning cash balance.

Total interest on loans

Maximum loan

Final cash balance

Cash balance before loan

Interest on cash balance

Beginning cash balance

Loan payback (principal + interest)

Receipts Interest rate

on cash Ini�al cash in

January

Tax, dividend expenses

Interest rate on loan

Minimum cash balance

Sales

Fixed costs

Timing of receipts

Material, labor cost as % of sales

Material, labor costs

Loan amount (if any)

Figure 16.18 Big Picture for Cash Balance Simulation Model

Developing the Simulation Model The completed simulation model appears in Figure 16.19. (See the file Cash Balance Finished.xlsx.) It requires the following steps.

1. Inputs. Enter the inputs in the blue cells. Note that loans are simulated (in row 42) only for the period from January to June of next year. However, sales figures are required (in row 28) in November and December of the current year to generate receipts for January and February. Also, July sales are required for next year to generate the material and labor costs paid in June.

2. Actual sales. Generate the sales in row 28 by entering the formula

5RiskNormal(B6,B7)

in cell B28 and copying across. 3. Beginning cash balance. For January of next year, enter the cash balance with the formula

5B19

in cell D31. Then for the other months, enter the formula

5D43

in cell E31 and copy it across row 31. This reflects that the beginning cash balance for one month is the final cash balance from the previous month.

Cash Balance Model

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4. Incomes. Entson’s incomes (interest on cash balance and receipts) are entered in rows 32 and 33. To calculate these, enter the formulas

5$B$24*D31

and

5SUMPRODUCT($B$14:$D$14,B28:D28)

in cells D32 and D33, and copy them across rows 32 and 33. This latter formula, which is based on Equation (16.1), mul- tiplies the fixed weights in row 14 by the relevant sales and adds these products to calculate receipts.

5. Expenses. Entson’s expenses (fixed costs, taxes and dividends, material and labor costs, and payback of the previous month’s loan) are entered in rows 35 through 39. Calculate these by entering the formulas

5D9

5D10

5$B$17*E28

5D42

Figure 16.19 Cash Balance Simulation Model

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46

A B C D E F G H I J Entson cash balance simula�on

Inputs Distribu�on of monthly sales (normal)

Nov Dec Jan Feb Mar Apr May Jun Jul 130000912400260019001500180016001500Mean

St Dev 700912012510080807570

Monthly fixed cost 250250250250250250 Tax, dividend expenses 0010005100

Receipts in any month are of form: A*(sales from 2 months ago)+B*(previous month’s sales)+C*(current month’s sales), where: A B C

0.2 0.6 0.2

Cost of materials and labor for next month, spent this month, is a percentage of product’s sales from next month, where the percentage is: 80%

Ini�al cash in January Minimum cash balance

250 250

Monthly interest rates Interest rate on loan 1.0% Interest rate on cash 0.5%

Simula�on Nov Dec Jan Feb Mar Apr May Jun Jul

Actual sales 1516.008 1518.567 1809.091 1498.674 1952.505 2382.129 2391.557 1966.894 1443.801

Cash, receipts Beginning cash balance 250.000

250 250 250 250

Interest on cash balance 1.892 1.250 1.2501.892 378.471 257.263 250.000 250.000

1.250 250.000

1.250 1688.903 1651.523 1947.664 2298.090 2304.7391576.160Receipts

Costs Fixed costs Tax, dividend expenses 0

250 10000 150

250 0

Material, labor expenses 1198.939 1562.004 1905.704 1913.245 1573.515 1155.041 399.258

3.993 866.419

8.664 645.631

6.456 Loan payback (principal) 0.0000.000000.0 Loan payback (interest) 0.0000.000000.0

Cash balance before loan 257.263 645.631

257.263

378.471

–395.631

250.000 250.000 250.000 647.697 866.419

–616.419 –149.258 399.258 0.000

647.697 Loan amount (if any) 0.0000.000 Final cash balance

Maximum loan Total interest on loans

J F b Mb

All monetary values are in $1000s.

One user suggested that the formula for total interest in cell B46 should sum only through June, i.e., not include July. You could argue it either way, so feel free to change the formula in cell B46 if you like.

866.419 19.113

16-3 Financial Models 8 0 1

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8 0 2 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

and

5D42*$B$23

in cells D35, D36, D37, E38, and E39, respectively, and copying these across rows 35 through 39. (For the loan payback, we are assuming that no loan payback is due in January.)

6. Cash balance before loan. Calculate the cash balance before the loan (if any) by entering the formula

5SUM(D31:D33)2SUM(D35:D39)

in cell D41 and copying it across row 41. 7. Amount of loan. If the value in row 41 is below the minimum cash balance ($250,000), Entson must borrow enough to

bring the cash balance up to this minimum. Otherwise, no loan is necessary. Therefore, enter the formula

5MAX($B$202D41,0)

in cell D42 and copy it across row 42. (You could use an IF function, rather the MAX function, to accomplish the same result.)

8. Final cash balance. Calculate the final cash balance by entering the formula

5D411D42

in cell D43 and copying it across row 43. 9. Maximum loan, total interest. Calculate the maximum loan from January to June in cell B45 with the formula

5MAX(D42:I42)

Then calculate the total interest paid on all loans in cell B46 with the formula

5SUM(E39:J39)

10. Output range. In the usual way, designate cells B45 and B46 as output cells. Also, designate the entire range of loans, D42:I42, as an output range. To do this, select this range and click the @RISK Add Output button. It will ask you for a name of the output. We suggest “Loans.” Then a typical formula in this range, such as the formula for cell E42, will be

5RiskOutput(”Loans”,2)1MAX($B$202E41,0)

This indicates that cell E42 is the second cell in the Loans output range.

Running the Simulation Set the number of iterations to 1000 and the number of simulations to 1. Then run the simulation in the usual way.

Discussion of the Simulation Results The summary results from the simulation are shown in Figure 16.20. They indicate that the maximum loan varies considerably, from a low of about $470,000 to a high of almost $1.5 million. The average is about $952,000. You can also see that Entson is spending close to $20,000 on average in interest on the loans, although the actual amounts vary considerably from one itera- tion to another.

You can also gain insights from the summary trend graph of the series of loans, shown in Figure 16.21. To obtain this graph, click the fourth button at the bottom of the Results Summary window in Figure 16.20. (This button is also available in any distribution graph window.) This graph clearly shows how the loans vary through time. The middle line is the expected loan amount. The inner bands extend to one standard deviation on either side of the mean, and the outer bands extend to the 5th and 95th percentiles. You can see that the largest loans are required in March and April.

Is it intuitively clear why the required loans peak in March and April? After all, why should Entson need money in months when its sales tend to be relatively high? There are two factors working here. First, Entson has to pay its costs early. For exam- ple, it has to pay 80% of its April sales for labor and material expenses in March. Second, most of its receipts arrive late. For example, 80% of its receipts from sales in March are not received until after March. Therefore, the answer to the question is that the timing and amounts of loans are fairly complex. Of course, this is why Entson builds a simulation model in the first place.

The loan amounts are determined by the random cash inflows and outflows and the fact that Entson’s policy is to maintain a minimum cash balance.

An @RISK output range, as opposed to a single output cell, allows you to obtain a summary graph that shows the whole simulated range at once. This range is typically a time series.

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Figure 16.21 Summary Trend Graph of Loans Through Time

Figure 16.20 Summary Measures for Cash Balance Simulation

16-3c Investment Models Individual investors typically want to choose investment strategies that meet some pre- specified goal. The following example is typical. Here, a person wants to meet a retire- ment goal, starting at an early age.

16-3 Financial Models 8 0 3

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8 0 4 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

EXAMPLE

16.6 INVESTING FOR RETIREMENT Attorney Sally Evans has just begun her career. At age 25, she has 40 years until retirement, but she realizes that now is the time to start investing. She plans to invest $1000 at the beginning of each of the next 40 years. Each year, she plans to put fixed percentages—the same each year—of this $1000 into stocks, Treasury bonds (T-bonds), and Treasury bills (T-bills). However, she is not sure which percentages to use. (We call these percentages investment weights.) She has historical annual returns from stocks, T-bonds, and T-bills from 1946 to 2007. These are listed in the file Retirement Planning.xlsx. This file also includes inflation rates for these years. For example, for 1993, the annual returns for stocks, T-bonds, and T-bills were 9.99%, 18.24%, and 2.90%, respectively, and the inflation rate was 2.75%. Sally would like to use simulation to help decide what investment weights to use, with the objective of achieving a large investment value, in today’s dollars, at the end of 40 years.

Objective To use simulation to estimate the value of Sally’s future investments, in today’s dollars, from several investment strategies in T-bills, T-bonds, and stocks.

Where Do the Numbers Come From? Historical returns and inflation rates, such as those quoted here, are available on the Web. In fact, you might want to find more recent data than we have provided here.

Solution The variables for this model appear in Figure 16.22. (See the file Retirement Planning Big Picture.xlsx.) The most difficult modeling aspect is settling on a way to use historical returns and inflation factors to generate future values of these quanti- ties. We suggest using a scenario approach. You can think of each historical year as a possible scenario, where the scenario specifies the returns and inflation factor for that year. Then for any future year, you randomly choose one of these scenarios. It seems intuitive that more recent scenarios ought to have a greater chance of being chosen. To implement this idea, you can give a weight (not to be confused with the investment weights) to each scenario, starting with weight 1 for the most recent year, 2007. Then the weight for any year is a damping factor mul- tiplied by the weight from the next year. For example, the weight for 1996 is the damping factor multiplied by the weight for 1997. To change these weights to probabilities, you can divide each weight by the sum of all the weights. The damping factor illustrated here is 0.98. Others could be used instead, and it is not clear which produces the most realistic results.

You can simulate future scenarios by randomly choosing past scenarios, giving higher probabilities to more recent scenarios.

Figure 16.22 Big Picture for Retirement Planning Simulation Model

Deflator for inflation

Scenario returns, inflation

Investment weights Ending cash

Amount to invest each year

Historical returns and inflation

Damping factor for probabilities

Probabilities of scenarios

Beginning cash Future scenarios

Final cash (today’s dollars)

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16-3 Financial Models 8 0 5

The other difficult part of the solution is choosing “good” investment weights. This is really an optimization problem: find three weights that add to 1 and pro- duce the largest mean final cash. (@RISK contains a powerful tool, RiskOptimizer, that solves this type of optimization–simulation problem, but we will not discuss it here.) This example illustrates several sets of weights, where some percentage is put into stocks and the remainder is split evenly between T-bonds and T-bills, and see which does best. You can try other sets if you like.

Developing the Simulation Model The historical data and the simulation model (each with some rows hidden) appear in Figures 16.23 and 16.24. (Again, see the file Retirement Planning Finished.xlsx.) It can be developed as follows.

1. Inputs. Enter the data in the blue regions of Figures 16.23 and 16.24.

Without a tool like RiskOptimizer, you cannot find the “best” set of investment weights, but the simulation model lets you experiment with various sets of weights.

Investing for Retirement Model

Figure 16.23 Historical Data, Inputs, and Probabilities

1 2 3 4 5 6 7 8 9

58 59 60 61 62 63 64 65 66 67

A B C D E F G

Historical data and probabili�es

Planning for re�rement

Year 1946 1947 1948 1949 1950 1999 2000 2001 2002 2003 2004 2005 2006 2007

ProbWts 0.2916 0.2976 0.3036 0.3098 0.3161 0.8508 0.8681 0.8858 0.9039 0.9224 0.9412 0.9604 0.9800 1.0000

35.7115

Probability 0.0082 0.0083 0.0085 0.0087 0.0089 0.0238 0.0243 0.0248 0.0253 0.0258 0.0264 0.0269 0.0274 0.0280 1.0000

T-Bonds –0.0010 –0.0263

0.0340 0.0645 0.0006

–0.0825 0.1666 0.0557 0.1512 0.0038 0.0449 0.0287 0.0196 0.0488

Stocks –0.0807

0.0571 0.0550 0.1879 0.3171 0.2089

–0.0903 –0.1185 –0.2198

0.2841 0.1070 0.0485 0.1563 0.1021

Infla�on 0.1817 0.0901 0.0271

–0.0180 0.0579 0.0270 0.0340 0.0160 0.0159 0.0227 0.0268 0.0339 0.0324 0.0285

Sums—>

T-Bills 0.0035 0.0050 0.0081 0.0110 0.0120 0.0439 0.0537 0.0573 0.0180 0.0180 0.0218 0.0431 0.0488 0.0548

2. Weights. The investment weights used for the model are in rows 10 through 12 of Figure 16.23. (For example, the first set puts 80% in stocks and 10% in each of T-bonds and T-bills.) You can simulate all three sets of weights simultaneously with a RiskSimtable and VLOOKUP combination as follows. First, enter the formula

5RiskSimtable(51,2,36) in cell I16. Then enter the formula

5VLOOKUP($I$16,LTable1,2)

in cell J16 and copy it to cells K16 and L16. Then modify the formulas in these latter two cells, changing the last argument of the VLOOKUP to 3 and 4, respectively. For example, the formula in cell L16 should end up as

5VLOOKUP($I$16,LTable1,4)

The effect is that you can run three simulations, one for each set of weights in rows 10 through 12. 3. Probabilities. Enter value 1 in cell F66. Then enter the formula

5$J$4*F66

in cell F65 and copy it up to cell F5. Sum these values with the SUM function in cell F67. Then to convert them to proba- bilities (numbers that add to 1), enter the formula

5F5>$F$67

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8 0 6 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

Figure 16.24 Retirement Simulation Model

3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 56 57 58 59 60 61 62 63

I J K L M N O P Q Inputs Damping factor 0.98 Range names used Yearly investment =Model!$I$10:$L$12LTable1$1,000 Planning horizon =Model!$A$5:$E$66LTable2years40

Weights =Model!$J$16:$L$16 Alterna�ve sets of weights to test

Index T-Bills T-Bonds Stocks 1 0.10 0.10 0.80 2 0.20 0.20 0.60 3 0.30 0.30 0.40

Weights used Index T-Bills T-Bonds Stocks

1 0.10 0.10 0.80

Output from simula�on below Final cash (today’s dollars) $47,921

Simula�on model Future year

1 2 3 4 5 6

33 34 35 36 37 38 39 40

Beginning cash $1,000 $1,889 $3,026 $4,165 $5,555 $6,428

$102,516 $135,562 $182,967 $181,887 $222,579 $244,116 $291,427 $305,431

T-Bills 1.0693 1.0351 1.0431 1.0521 1.1471 1.0521 1.0580 1.0154 1.0781 1.0213 1.0218 1.0149 1.0547 1.0653

T-Bonds 0.9889 1.0805 1.0287 0.9974 1.0185 0.9974 1.0919 0.9390 1.0618 1.0097 1.0449 0.9606 0.9731 1.1210

Stocks 0.8534 1.0767 1.0485 1.1106 0.9509 1.1106 1.3720 1.4336 0.9683 1.2689 1.1070 1.2402 1.0523 1.0401

Ending cash $889

$2,026 $3,165 $4,555 $5,428 $7,029

$134,562 $181,967 $180,887 $221,579 $243,116 $290,427 $304,431 $320,919

Deflator 0.919 0.893 0.864 0.825 0.757 0.723 0.194 0.191 0.180 0.179 0.174 0.164 0.158 0.149

Infla�on 1.0880 1.0290 1.0339 1.0472 1.0894 1.0472 1.0701 1.0176 1.0611 1.0067 1.0268 1.0587 1.0441 1.0549

Scenario 1973 1992 2005 1968 1981 1968 1975 1958 1990 1961 2004 1951 1987 1970

2 3 4 5 Column offset for lookup2

in cell G5 and copy it down to cell G66. Note how the probabilities for more recent years are considerably larger. When scenarios are selected randomly, recent years will have a greater chance of being chosen. (The SUM formula in cell G67 confirms that the probabilities sum to 1.)

4. Scenarios. Moving to the model in Figure 16.24, the goal is to simulate 40 scenarios in columns K through O, one for each year of Sally’s investing. To do this, enter the formulas

5RiskDiscrete($A$5:$A$66,$G$5:$G$66)

and

511VLOOKUP($K24,LTable2,L$22)

in cells K24 and L24, and copy this latter formula to the range M24:O24. Then copy all of these formulas down to row 63. Make sure you understand how the RiskDiscrete and VLOOKUP functions combine to achieve the goal. (Also, check the list of range names used at the top of Figure 16.24.) The RiskDiscrete function randomly generates a year from column A, using the probabilities in column G. Then the VLOOKUP function captures the data from this year. (You add 1 to the VLOOKUP to get a value such as 1.08, rather than 0.08.) This is the key to the simulation. (By the way, do you see why Excel’s RANDBETWEEN function isn’t used to generate the years in column K? The reason is that this function makes all possible years equally likely, and the goal is to make more recent years more likely.)

5. Beginning, ending cash. The bookkeeping part is straightforward. Begin by entering the formula

5J5

in cell J24 for the initial investment. Then enter the formulas

5J24*SUMPRODUCT(Weights,L24:N24)

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16-3 Financial Models 8 0 7

and

5$J$51P24

in cells P24 and J25 for ending cash in the first year and beginning cash in the second year. The former shows how the beginning cash grows in a given year. You should think it through carefully. The latter implies that Sally reinvests her pre- vious money, plus she invests an additional $1000. Copy these formulas down columns J and P.

6. Deflators. You eventually need to deflate future dollars to today’s dollars. The proper way to do this is to calculate defla- tors (also called deflation factors). Do this by entering the formula

51>O24

in cell Q24. Then enter the formula

5Q24>O25

in cell Q25 and copy it down. The effect is that the deflator for future year 20, say, in cell Q43, is 1 divided by the product of all 20 inflation factors up through that year. (This is similar to discounting for the time value of money, but the relevant discount rate, now the inflation rate, varies from year to year.)

7. Final cash. Calculate the final value in today’s dollars in cell K19 with the formula

5P63*Q63

Then designate this cell as an @RISK output cell.

Running the Simulation Set the number of iterations to 1000 and the number of simulations to 3 (one for each set of investment weights to be tested). Then run the simulation as usual.

Discussion of the Simulation Results Summary results appear in Figure 16.25. The first simulation, which invests the most heavily in stocks, is easily the winner. Its mean final cash, slightly more than $155,000 in today’s dollars, is much greater than the means for the other two sets of weights. The first simulation also has a much larger upside potential (its 95th percentile is close to $368,000), and even its downside is slightly better than the others: Its 5th percentile is the best, and its minimum is only slightly worse than the mini- mum for the other sets of weights.

Figure 16.25 Summary Results for Retirement Simulation

Nevertheless, the histogram for simulation 1 (put 80% in stocks), shown in Figure 16.26, indicates a lot of variability—and skewness—in the distribution of final cash. As in Example 16.4, the concept of value at risk (VaR) is useful. Recall that VaR 5% is defined as the 5th percentile of a distribution and is often the value investors worry about. Perhaps Sally should rerun the simula- tion with different investment weights, with an eye on the weights that increase her VaR 5%. Right now it is about $36,500—not too good considering that she invests $40,000 total. She might not like the prospect of a 5% chance of ending up with no more than this. We also encourage you to try running this simulation with other investment weights, both for the 40-year horizon and (after modifying the spreadsheet model slightly) for shorter time horizons such as 10 or 15 years. Even though the stock strategy appears to be best for a long horizon, it is not necessarily guaranteed to dominate for a shorter time horizon.

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Figure 16.26 Histogram of Final Cash with 80% in Stocks

11. In the Entson cash balance model, is the $250,000 min- imum cash balance requirement really “costing” the company very much? Answer this by rerunning the simulation with minimum required cash balances of $50,000, $100,000, $150,000, and $200,000. Use the RiskSimtable function to run all simulations at once. Comment on the outputs from these simulations. In par- ticular, comment on whether the company appears to be better off with a lower minimum cash balance.

12. Run the investing for retirement model with a damp- ing factor of 1.0 (instead of 0.98), again using the same three sets of investment weights. Explain in words what it means, in terms of the simulation, to have a damping factor of 1. Then comment on the differences, if any, between your simulation results and those in the example.

13. The simulation output from the investing for retirement model indicates that an investment heavy in stocks pro- duces the best results. Would it be better to invest entirely in stocks? Answer this by rerunning the simulation. Is there any apparent downside to this strategy?

14. Modify the investing for retirement model so that you use only the years 1975 to 2007 of historical data. Run the simulation for the same three sets of investment weights. Comment on whether your results differ in any important way from those in the example.

15. Rerun the investing for retirement model with a planning horizon of 10 years; 15 years; 25 years. For each, which

Problems

Level A 8. Rerun the GF Auto new car simulation model but

now introduce uncertainty into the fixed development cost. Let it be triangularly distributed with parameters $500 million, $550 million, and $750 million. (You can check that the mean of this distribution is $600 million, the same as the cost given in the example.) Comment on the differences between your output and those in the example. Would you say these differences are important for the company?

9. Rerun the GF Auto new car simulation model but now use the RiskSimtable function appropriately to simulate discount rates of 5%, 7.5%, 10%, and 12.5%. Comment on how the output changes as the discount rate increases.

10. In the Entson cash balance model, the timing is such that some receipts are delayed by one or two months, and the payments for materials and labor must be made a month in advance. Change the model so that all receipts are received immediately, and payments made this month for materials and labor are 80% of sales this month (not next month). The period of interest is again January through June. Rerun the simulation, and comment on any differences between your outputs and those from the example.

8 0 8 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

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16-3 Financial Models 8 0 9

set of investment weights maximizes the VaR 5% (the 5th percentile) of final cash in today’s dollars? Does it appear that a portfolio heavy in stocks is better for long horizons but not for shorter horizons?

Level B 16. Change the GF Auto new car simulation model as fol-

lows. It is the same as before for years 1 through 5, including depreciation through year 5. However, the car might sell through year 10. Each year after year 5, the company examines sales. If fewer than 35,000 cars were sold that year, there is a 50% chance the car won’t be sold after that year. Modify the model and run the sim- ulation. Keep track of two outputs: NPV (through year 10) and the number of years of sales.

17. Based on Kelly (1956). You currently have $100. Each week you can invest any amount of money you cur- rently have in a risky investment. With probability 0.4, the amount you invest is tripled (e.g., if you invest $100, you increase your asset position by $300), and, with probability 0.6, the amount you invest is lost. Consider the following investment strategies:

• Each week, invest 10% of your money. • Each week, invest 30% of your money. • Each week, invest 50% of your money.

Use @RISK to simulate 100 weeks of each strategy 1000 times. Which strategy appears to be best in terms of the maximum growth rate? (In general, if you can multiply your investment by M with probability p and lose your investment with probability q 5 1 2 p, you should invest a fraction 3p(M 2 1) 2 q4 >(M 2 1) of your money each week. This strategy maximizes the expected growth rate of your fortune and is known as the Kelly criterion.) (Hint: If an initial wealth of I dol- lars grows to F dollars in 100 weeks, the weekly growth rate, labeled r , satisfies F 5 (1 1 r)100*I , so that r 5 (F>I)1>100 2 1.)

18. Amanda has 30 years to save for her retirement. At the beginning of each year, she puts $5000 into her retire- ment account. At any point in time, all of Amanda’s retirement funds are tied up in the stock market. Suppose the annual return on stocks follows a normal distribution with mean 12% and standard deviation 25%. What is the probability that at the end of 30 years, Amanda will have reached her goal of having $1,000,000 for retire- ment? Assume that if Amanda reaches her goal before 30 years, she will stop investing. (Hint: Each year you should keep track of Amanda’s beginning cash posi- tion—for year 1, this is $5000—and Amanda’s ending cash position. Of course, Amanda’s ending cash posi- tion for a given year is a function of her beginning cash position and the return on stocks for that year. To esti- mate the probability that Amanda meets her goal, use an

IF statement that returns 1 if she meets her goal and 0 otherwise.)

19. In the financial world, there are many types of com- plex instruments called derivatives that derive their value from the value of an underlying asset. Consider the following simple derivative. A stock’s current price is $80 per share. You purchase a derivative whose value to you becomes known a month from now. Specifi- cally, let P be the price of the stock in a month. If P is between $75 and $85, the derivative is worth noth- ing to you. If P is less than $75, the derivative results in a loss of 100*(75 2 P) dollars to you. (The factor of 100 is because many derivatives involve 100 shares.) If P is greater than $85, the derivative results in a gain of 100*(P 2 85) dollars to you. Assume that the distri- bution of the change in the stock price from now to a month from now is normally distributed with mean $1 and standard deviation $8. Let E be the expected gain/ loss from this derivative. It is a weighted average of all the possible losses and gains, weighted by their likeli- hoods. (Of course, any loss should be expressed as a negative number. For example, a loss of $1500 should be expressed as 2$1500.) Unfortunately, this is a dif- ficult probability calculation, but E can be estimated by an @RISK simulation. Perform this simulation with at least 1000 iterations. What is your best estimate of E?

20. Suppose you currently have a portfolio of three stocks, A, B, and C. You own 500 shares of A, 300 of B, and 1000 of C. The current share prices are $42.76, $81.33, and $58.22, respectively. You plan to hold this portfo- lio for at least a year. During the coming year, econo- mists have predicted that the national economy will be awful, stable, or great with probabilities 0.2, 0.5, and 0.3, respectively. Given the state of the economy, the returns (one-year percentage changes) of the three stocks are independent and normally distributed. How- ever, the means and standard deviations of these returns depend on the state of the economy, as indicated in the file P16_20.xlsx. a. Use @RISK to simulate the value of the portfolio and

the portfolio return in the next year. How likely is it that you will have a negative return? How likely is it that you will have a return of at least 25%?

b. Suppose you had a crystal ball where you could predict the state of the economy with certainty. The stock returns would still be uncertain, but you would know whether your means and standard deviations come from row 6, 7, or 8 of the file P16_20.xlsx. If you learn, with certainty, that the economy is going to be great in the next year, run the appropriate sim- ulation to answer the same questions as in part a. Repeat this if you learn that the economy is going to be awful. How do these results compare with those in part a?

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8 1 0 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

16-4 Marketing Models There are many opportunities for marketing departments to use simulation. They face uncer- tainty in the brand-switching behavior of customers, the entry of new brands into the mar- ket, customer preferences for different attributes of products, the effects of advertising on sales, and so on. We examine several marketing applications of simulation in this section.

16-4a Customer Loyalty Models What is a loyal customer worth to a company? This is an extremely important question for companies. Companies know that if customers become dissatisfied with the company’s product, they are likely to switch and never return. Marketers refer to this customer loss as churn. The loss in profit from churn can be large, particularly because long-standing customers tend to be more profitable in any given year than new customers. The following example uses a reasonable model of customer loyalty and simulation to estimate the worth of a customer to a company. It is based on the excellent discussion of customer loyalty in Reichheld (1996).

EXAMPLE

16.7 LONG-TERM VALUE OF A CUSTOMER AT CCAMERICA CCAmerica is a credit card company that does its best to gain customers and keep their business in a highly competitive indus- try. The first year a customer signs up for service typically results in a loss to the company because of various administrative expenses. However, after the first year, the profit from a customer is typically positive, and this profit tends to increase through the years. The company has estimated the mean profit from a typical customer to be as shown in column B of Figure 16.28 below. (See the file Customer Loyalty.xlsx.) For example, the company expects to lose $40 in the customer’s first year but to gain $87 in the fifth year—provided that the customer stays loyal that long. For modeling purposes, we assume that the actual profit from a customer in a customer’s given year of service is normally distributed with mean shown in Figure 16.28 and standard deviation equal to 10% of the mean. At the end of each year, the customer leaves the company, never to return, with probability 0.15, the churn rate. Alternatively, the customer stays with probability 0.85, the retention rate. The company wants to estimate the NPV of the net profit from any such customer who has just signed up for service at the beginning of year 1, at a discount rate of 8%, assuming that the cash flow occurs in the middle of the year.3 It also wants to see how sensitive this NPV is to the retention rate.

Objective To use simulation to find the NPV of a customer, and to see how this varies with the retention rate.

Where Do the Numbers Come From? The numbers in column B of Figure 16.28 are undoubtedly averages, based on the historical records of many customers. To build in randomness for any particular customer, we need a probability distribution around the numbers in this figure. We arbitrarily chose a normal distribution centered on the historical average and a standard deviation of 10% of the average. These are educated guesses. Finally, the churn rate is a number very familiar to marketing people, and it can also be estimated from historical customer data.

Solution The variables for this model appear in Figure 16.27. (See the file Customer Loyalty Big Picture .xlsx.) The idea is to keep simulating profits (or a loss in the first year) for the customer until the customer churns. We simulate 30 years of potential profits, but this could be varied.

Developing the Simulation Model The simulation model appears in Figure 16.28. (See the file Customer Loyalty Finished.xlsx.) It can be developed with the following steps.

3 This assumption makes the NPV calculation slightly more complex, but it is probably more realistic than the usual assumption that cash flows occur at the ends of the years.

As usual, Excel’s RAND function can be used inside an IF statement to determine whether a given event occurs.

Customer Loyalty Model

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16-4 Marketing Models 8 1 1

Figure 16.27 Big Picture for Customer Loyalty Simulation Model

NPV of annual profits

Discounted profitDiscount rate

Actual profit for year

Standard deviation of profit (% of mean)

Quits after this year?

Mean profit from customer Retention rate

Years loyal

Customer loyalty model in the credit card industry A

Inputs Reten�on rates to try Reten�on rate Discount rate Stdev % of mean

Es�mated means Simula�on Outputs Year

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Discounted profit �39.93

59.55 58.58 67.16 65.76 71.99 60.74 52.10 42.86 49.42 48.83 52.89 36.61 48.24 35.96 34.80 32.92 36.39 37.60 35.89 30.14 35.81 27.01 24.95 24.23

0.00 0.00 0.00 0.00 0.00

NPV Years loyal

Means Simula�on

1 2 3 4 5

Reten�on rate 0.75 0.80 0.85 0.90 0.95

NPV $149.70 $184.67 $264.03 $388.32 $663.44

Years loyal 4.26 4.92 6.51 9.25

16.13

B C D E F G H I J

0.75 0.80 0.85 0.90 0.950.75 0.08 10%

Mean Profit(if s�ll here) �40.00

66.00 72.00 79.00 87.00 92.00 96.00 99.00

103.00 106.00 111.00 116.00 120.00 124.00 130.00 137.00 142.00 148.00 155.00 161.00 161.00 161.00 161.00 161.00 161.00 161.00 161.00 161.00 161.00 161.00

Quits at end of year? No No No No No No No No No No No No No No No No No No No No No No No No Yes

Actual profit �41.50

66.84 71.01 87.93 92.98

109.92 100.17

92.80 82.43

102.66 109.56 128.17

95.80 136.34 109.77 114.73 117.20 139.92 156.15 160.96 145.97 187.33 152.59 152.22 159.68

0.00 0.00 0.00 0.00 0.00

$1,030.50 25

800 20.00 15.00 10.00 5.00 0.00

600 400 200

0.70 0.75 0.80 0.85 Reten�on rate

Sensi�vity to Reten�on Rate

0.90 0.95 1.00 0

NPV Years loyal

Figure 16.28 Customer Loyalty Model

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8 1 2 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

1. Inputs. Enter the given inputs in the blue cells. 2. Retention rate. Although an 85% retention rate was given in the statement of the problem, it is useful to investigate reten-

tion rates from 75% to 95%, as shown in row 4. To run a separate simulation for each of these, enter the formula

5RiskSimtable(D4:H4)

in cell B4. 3. Timing of churn. In column C, use simulation to discover when the customer churns. This column will contain a sequence

of No values, followed by a Yes, and then a sequence of blanks (or all No values if the customer never churns). To generate these, enter the formulas

5IF(RAND()*12B4,”Yes”,”No”)

and

5IF(C11*+”No”,””,IF(RAND()*12$B$4,”Yes”,”No”))

in cells C11 and C12, and copy the latter formula down column C. Study these formulas carefully to see how the logic works. Note that they do not rely on @RISK functions. Excel’s RAND function can be used any time you want to simu- late whether or not an event occurs.

4. Actual and discounted profits. Profits (or a loss in the first year) occur as long as there is not a blank in column C. Therefore, simulate the actual profits by entering the formula

5IF(C11*+””,RiskNormal(B11,$B$6*ABS(B11)),0)

in cell D11 and copying it down. (The absolute value function, ABS, is required in case any of the cash flows are negative. A normal distribution cannot have a negative standard deviation.) Then discount these appropriately in column E by entering the formula

5D11>(11$B$5)^(A1120.5)

in cell E11 and copying it down. Note how the exponent of the denominator accounts for the cash flow in the middle of the year.

5. Outputs. Keep track of two outputs, the total NPV and the number of years the customer stays with the company. Calcu- late the NPV in cell H10 by summing the discounted values in column E. (They have already been discounted, so the NPV function is not needed.) To find the number of years the customer is loyal, count the number of No values plus the number of Yes values, that is, all nonblanks. Calculate this in cell H11 with the formula

5COUNTIF(C11:C40,”No”)1COUNTIF(C11:C40,”Yes”)

Finally, designate both of cells H10 and H11 as @RISK output cells.

Running the Simulation Set the number of iterations to 1000 and the number of simulations to 5 (one for each potential retention rate). Then run the simulation as usual.

Discussion of the Simulation Results Summary results for all five retention rates appear in Figure 16.29. You can see that the mean NPV and the mean number of years loyal are quite sensitive to the retention rate.

To follow up on this observation, you can use the RiskMean function to capture the means in columns I and J of the model sheet and then create line charts of them as a function of the reten- tion rate. (See Figure 16.28, where years loyal is shown on a secondary axis.) These line charts show the rather dramatic effect the retention rate can have on the value of a customer. For example, if it increases from the current 85% to 90%, the mean NPV increases by about 47%. If it increases from 85% to 95%, the mean NPV increases by about 151%. In the other direction, if the retention rate decreases from 85% to 80%, the mean NPV decreases by about 30%. This is why credit card companies are so anxious to keep their customers.

Careful discounting is required if cash flows occur in the middle of a year.

Varying the retention rate can have a large impact on the value of a customer.

09953_ch16_ptg01_779-836.indd 812 04/03/19 1:20 PM

The following example is a variation of the customer loyalty example. We now inves- tigate the effect of offering a customer an incentive to remain loyal.

Figure 16.29 Summary Results for Customer Loyalty Model

EXAMPLE

16.8 THE VALUE OF A FREE MAINTENANCE AGREEMENT AT JAMESONS

Companies value loyal customers, and they sometimes go to great lengths to keep their customers loyal. This example inves- tigates whether one such plan is worth its cost. We consider a nationwide company called Jamesons, which sells electronic appliances. Specifically, we will focus on sales of smart phones. To attract customers, the company is considering giving customers a free maintenance agreement with each purchase of a smart phone. The unit profit without free maintenance is cur- rently $75. The company believes this will decrease to $60 with free maintenance. Their thinking is that about 5% of custom- ers will actually use the free maintenance, and for each such customer, the company will lose about $300. Hence the average decrease in profit per purchaser is about $15.

Prior to this year, 50,000 customers were loyal to Jamesons and 100,000 customers were loyal to their competitors. (Loyalty is defined in terms of where the customer bought his or her last smart phone.) There are a number of uncertain quan- tities, and we assume they are all triangularly distributed. Their parameters (minimum, most likely, and maximum) are as follows. (1) The percentage of the 150,000 customers who purchase a smart phone in any given year has parameters 20%, 25%, and 40%. (2) The annual percentage change in unit profit has parameters 3%, 5%, and 6%. (3) In any year, the percentage of Jamesons’ loyal customers who remain loyal has parameters 56%, 60%, and 66% if there is no free maintenance, and they increase to 60%, 64%, and 70% with free maintenance. (4) Similarly, the percentage of the competitors’ loyal customers who switch to Jamesons has parameters 27%, 30%, and 34% if there is no free maintenance, and they increase to 32%, 35%, and 39% with free maintenance. These inputs are shown in the top section of Figure 16.31 below.

Jamesons is hoping that the decrease in unit profit from the free maintenance agreement will be more than offset by the higher loyalty percentages. Using a 15-year planning horizon, does the NPV of profits with a 8% discount rate confirm the company’s hopes?

16-4 Marketing Models 8 1 3

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8 1 4 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

Objective To use simulation to see whether it makes sense for Jamesons to give a free maintenance agreement to DVD player purchasers.

Where Do the Numbers Come From? In the previous example, we discussed the switching rates, which would be estimated from extensive customer data. The other data in the problem statement are straightforward to obtain.

Solution The variables for this model are shown in Figure 16.30. (See the file Free Maintenance Big Picture.xlsx.) The solution strategy is to compare two simulations, one without free maintenance and one with it. Because they are so similar, you can use RiskSimtable to run both simulations. We make one assumption that is common in marketing but might not be intuitive. We assume that only purchasers in a given year have any chance of switching loyalty in the next year. For example, if a customer is loyal to Jamesons and doesn’t purchase a smart phone in a given year, this customer is automatically loyal to Jamesons in the next year.

Figure 16.30 Big Picture for Free Maintenance Simulation Model

Profit contribution

Initial # loyal to them

Initial # loyal to us

Initial unit profit

Customers loyal to them

Customers loyal to us

% change in unit profit

% loyal to us who purchase

% loyal to them who purchase

% who stay loyal to us

% who switch loyalty to us

Offer free maintenance?

Unit profit

Purchases of our product

Discount rate NPV of our profits

Developing the Simulation Model The completed simulation model appears in Figure 16.31. (See the file Free Maintenance 1 Finished.xlsx.) It can be devel- oped with the following steps.

1. Inputs. Enter the given inputs in the blue cells. 2. Maintenance decision. The current “no free maintenance” policy is labeled simulation #1 and

the proposed “free maintenance” policy is labeled simulation #2, so enter the formula

5RiskSimtable(51,26) in cell B21.

Free Maintenance Agreement Model

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16-4 Marketing Models 8 1 5

3. Percentages who purchase. We assume that each year a random percentage of Jamesons’ loyal customers and a random percentage of the competitors’ loyal customers purchase a DVD player. Each of these is generated from the triangular distribution in rows 9–11, so enter the formula

5RiskTriang($B$9,$B$10,$B$11)

in cell B24 and copy it to the range B24:Q25. 4. Percentage who stay or become loyal. Each year a random percentage of the customers previously loyal to Jamesons

remain loyal, and a random percentage of the competitors’ previously loyal customers switch loyalty to Jamesons. Also, the distributions of these random percentages depend on the company’s maintenance policy. Therefore, enter the formula

5IF($B$2151,RiskTriang($G$8,$G$9,$G$10),RiskTriang($H$8,$H$9,$H$10))

in cell C26, enter the formula

5IF($B$2151,RiskTriang($G$13,$G$14,$G$15),RiskTriang($H$13,$H$14,$H$15))

in cell C27, and copy these across their rows. 5. Numbers of loyal customers. Create links to cells B5 and B6 in cells B28 and B29. Then, remembering that only pur-

chasers in a given year can switch loyalty, calculate the number of customers loyal to Jamesons in year 1 with the formula

5B28*((1-B24)1B24*C26)1B29*B25*C27

in cell C28 and copy it across row 28. Similarly, calculate the number of customers loyal to the competitors in year 1 with the formula

5B29*((1-B25)1B25*(1-C27))1B28*B24*(1-C26)

in cell C29 and copy it across row 29. These are basic bookkeeping formulas. Jamesons’ loyal customers are those who (1) were loyal and didn’t purchase; (2) were loyal, purchased, and stayed loyal; and (3) weren’t loyal, purchased, and switched loyalty. Similar logic holds for the competitors’ loyal customers.

6. Purchasers at Jamesons. Calculate the number of purchasers at Jamesons in year 1 with the formula

5C24*C28

in cell C30 and copy it across row 30.

Figure 16.31 Free Maintenance Simulation Model

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

Free maintenance agreement - is it worth it? A

Common inputs Inputs that depend on policy Loyal customers in previous year To our brand To their brand

% of poten�al customers who purchase in any year (triangular distribu�on) Minimum Most likely Maximum

Annual % growth in profit contribu�on (triangular distribu�on) Minimum Most likely Maximum

Discount rate

Simula�on Index of simula�on

Year % loyal to us who purchase % loyal to them who purchase % who stay loyal to us % who switch loyalty to us Customers loyal to us Customers loyal to them Purchases of our product % change in unit profit Unit profit Profit contribu�on

NPV

B C D E F G H I J K L M N O P Q

Unit profit

% of our loyal customers who remain loyal (triangular distribu�on) Minimum Most likely Maximum

% of their loyal customers who switch to us (triangular distribu�on) Minimum Most likely Maximum

50000 100000

Not free $75

56% 60% 66%

27% 30% 34%

Free $60

60% 64% 70%

32% 35% 39%

20% 25% 40%

1 27.4% 26.6% 59.5% 30.5% 54424 95576 14895

$75.00 $1,117,157

2 23.2% 27.4% 59.4% 28.2% 55550 94450 12862 5.37%

$79.03 $1,016,407

3 32.9% 24.1% 58.7% 31.1% 58295 91705 19165 3.89%

$82.10 $1,573,477

4 34.0% 25.4% 59.5% 28.7% 56886 93114 19327 4.95%

$86.17 $1,665,302

5 30.7% 32.5% 57.9% 30.7% 56023 93977 17197 5.61%

$91.00 $1,564,988

6 39.3% 36.2% 63.0% 31.1% 59166 90834 23277 5.41%

$95.93 $2,232,861

7 25.9% 22.8% 60.2% 29.1% 59474 90526 15375 3.93%

$99.70 $1,532,883

8 33.0% 24.7% 62.0% 32.1% 60250 89750 19874 5.63%

$105.31 $2,093,028

9 30.9% 33.5% 59.9% 28.6% 58626 91374 18137 5.05%

$110.63 $2,006,493

10 26.9% 24.8% 57.7% 27.9% 59494 90506 16032 5.12%

$116.29 $1,864,397

11 24.3% 30.7% 62.3% 30.8% 60366 89634 14674 5.83%

$123.07 $1,805,893

12 32.7% 36.2% 61.9% 27.2% 62275 87725 20356 3.26%

$127.08 $2,586,811

13 28.9% 31.5% 56.5% 30.7% 63164 86836 18236 4.17%

$132.38 $2,414,209

14 25.1% 32.9% 65.3% 30.7% 65235 84765 16365 5.12%

$139.16 $2,277,431

15 33.0% 28.6% 62.3% 31.1% 67753 82247 22369 4.69%

$145.69 $3,258,973

In this version, different random percentages are generated for each year. (See the formulas in the green cells below). Also, note how the index in cell B21, along with VLOOKUPs and RiskSimtable func�ons), permit you to run two simula�ons, one without free maintenance and one with it.

3% 5% 6%

0 21.3% 28.7%

50000 100000

$15,235,876

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8 1 6 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

7. Monetary outcomes. These are straightforward. Start by entering the formula

5IF($B$21=1,G5,H5)

for unit profit in year 1 in cell C32. Then enter the formulas

5RiskTriang($B$14,$B$15,$B$16)

5C32*(11D31)

and

5C32*C30

in cells D31, D32, and C33, respectively, and copy them across their rows. Finally, calculate the NPV with the formula

5NPV(B18,C33:Q33)

in cell B35, and designate this as an @RISK output cell.

Running the Simulation Set up @RISK to run 1000 iterations and 2 simulations, one for each maintenance decision to be tested. Then run the simula- tion as usual.

Discussion of the Simulation Results The summary measures for the two simulations appear in Figure 16.32. Using the current inputs, the free maintenance initia- tive does not look good. Every measure, except possibly the standard deviation, is worse with the free maintenance agreement (simulation 1) than without it (simulation 2). Evidently, the increase in loyal customers does not compensate for the decrease in unit profit. If Jamesons is reasonably confident about the inputs for this model, it should scrap the free maintenance idea. However, it might want to perform some sensitivity analysis on the decrease in unit profit or the increase in loyalty percent- ages (or both) to see when the free maintenance agreement starts looking attractive. We tried two possibilities. First, if the decrease in unit profit is only $7.50, not $15, and everything else remains the same, the two mean NPVs are very close, so the free maintenance agreement might be worth trying. Second, if the decrease in unit profit remains at $15, but all of the input percentages in the ranges H8:H10 and H13:H15 increase by five percentage points, the mean NPV with the free maintenance agreement is still slightly lower than the mean NPV without it. Evidently, the company can’t take this big a hit in its profit margin unless it can convince a lot more customers to stay or become loyal.

Figure 16.32 Summary Measures for Comparing Two Decisions

There is an interesting modeling issue in this example. For each of the random quantities, we have generated a new random value each year. Would it be better to generate one random number from each triangular distribution and use it for each year? Would it make a difference in the results? The modified simulation appears in the file Free Maintenance 2 Finished.xlsx. The summary measures from this simulation appear in Figure 16.33. If we are interested in comparing the mean NPV with no free maintenance versus free maintenance, we get about the same comparison in either model. The main difference between Figures 16.32 and 16.33 is the variability. Are you surprised that the models with more random numbers in Figure 16.32 have much smaller standard deviations than those in Figure 16.33? Evidently, there is an averaging effect. When different random numbers are used for each year, the highs and lows tend to cancel out, resulting in lower variability in NPV.

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16-4b Marketing and Sales Models We conclude this marketing section with a model of selling condos. The main issue is the timing of sales, and we demonstrate how a deterministic model of this timing can provide very misleading results.

Figure 16.33 Summary Results for Modified Model

EXAMPLE

16.9 SELLING CONDOS AT BLACKSTONE DEVELOPMENT Blackstone Development Company has just finished building 120 high-end condos, each priced at $300,000. Blackstone has hired another company, Pletcher Marketing, to market and sell these condos. Pletcher will incur all of the marketing and maintenance costs, assumed to be $800 per unsold condo per month, and it will receive a 10% commission ($30,000) from Blackstone at the time of each condo sale. Because Blackstone wants these condos to be sold in a timely manner, it has offered Pletcher a $200,000 bonus at the end of the first year if at least half of the condos have been sold, and an extra $500,000 bonus at the end of the second year if all of the condos have been sold. Pletcher estimates that it can sell five condos per month on average, so that it should be able to collect the bonuses. However, Pletcher also realizes that there is uncertainty about the number of sales per month. How should this uncertainty be modeled, and will the resulting simulation model give different qualitative results than a deterministic model where exactly five condos are sold per month?

Objective To develop a simulation model that allows us to see how the uncertain timing affects the monetary outcomes for Pletcher, and to compare this simulation model to a deterministic model with no uncertainty about the timing of sales.

Where Do the Numbers Come From? The inputs are straightforward from Blackstone’s agreement with Pletcher. The only difficulty is determining an appropriate probability model for the timing of sales, which we discuss next.

Solution To make a fair comparison between a deterministic model with five sales per month and a simulation model with uncertainty in the timing of sales, we need a discrete distribution for monthly sales that has mean 5. One attractive possibility is to use the Poisson distribution discussed briefly in Chapter 5. It is discrete, and it has only one parameter, the mean. The Poisson distribution has one theoretical drawback in that it allows all nonnegative integers to occur, but this has no practical effect. As shown in Figure 16.34, the Poisson distribution with mean 5 has virtually no probability of values larger than, say, 15.

Developing the Simulation Model The variables for the model appear in Figure 16.35. (See the file Selling Condos Big Picture.xlsx.) The deterministic model is straightforward and is not shown here. By selling a sure five condos per Selling Condos Model

16-4 Marketing Models 8 1 7

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8 1 8 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

Figure 16.34 Poisson Distribution with Mean 5

Figure 16.35 Big Picture for Condo Selling Simulation Model

Poten�al bonuses

Condos sold

Condos to sell

Discount rate

Condos le to be sold

Commission from sales

Unit marke�ng, maintenance cost

Commission per condo sale

Maintenance cost

Cash flow

Mean monthly demand

Demand

Bonuses received

NPV of all cash flows

Months to sell out

month, Pletcher sells all condos by the end of year 2, receives both bonuses, and realizes an NPV (including bonuses) of $2,824,333. However, this is not very realistic. The steps for creating a more realistic simulation model follow. (See Fig- ure 16.36, with several hidden columns, and the file Selling Condos Finished.xlsx.) Because of the uncertain timing of sales, we cannot say when all 120 condos will be sold. It could be before 24 months or well after 24 months. Therefore, we model it through 40 months. (By experimenting, we found that all 120 condos will almost surely be sold in 40 months.)

1. Inputs. Enter the given inputs in the blue ranges. 2. Random demands. Generate the random demands for condos (the number of people who would like to buy) by entering

the formula

5IF(B16+0,RiskPoisson($B$12),””)

in cell B15 and copying across to month 40. The IF function checks whether there are still any condos available in that month. If there aren’t, a blank is recorded. Similar logic appears in many of the other formulas.

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16-4 Marketing Models 8 1 9

3. Number remaining and sold. In cell B16, enter a link to cell B3. In cell B17, find the number sold as the minimum of supply and demand with the formula

5IF(B16+0,MIN(B16,B15),””)

In cell C16, find the number remaining to be sold with the formula

5IF(B16+0,B16-B17,0)

Then copy the formulas in cells C16 and B17 across. Note that a 0, not a blank, is recorded in row 16 after all condos have been sold. This makes all the other IF functions work correctly.

4. Monetary values. Enter the formulas

5IF(B16+0,$B$4*(B16-B17),””)

5IF(B16+0,$B$5*B17,””)

and

5IF(B16+0,SUM(B19:B21)-B18,””)

in cells B18, B19, and B22, respectively, and copy these across. For the bonuses, enter the formulas

5IF(SUM(B17:M17)+5B3>2,B6,0)

and

5IF(SUM(B17:Y17)5B3,B7,0)

in cells M20 and Y21. These capture the all-or-nothing nature of the bonuses. 5. Outputs. Three interesting outputs are the number of months required to sell out, the total bonus earned, and the NPV of

the cash flows, including bonuses. Calculate these in cells B24–B26 with the formulas

5COUNTIF(B16:AO16,”+0”)

5M201Y21

and

5NPV(B8,B22:AO22)

Then designate them as @RISK output cells.

Running the Simulation Set @RISK to run 1000 iterations for a single simulation. Then run the simulation in the usual way.

Figure 16.36 Condo Selling Simulation Model

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

AOANAMAFAEADACABCBA

Condos to sell 120 Unit marketing, maintenance cost $800 Commission per condo sale $30,000 Bonus if at least half sold in year 1 $200,000 Extra bonus if all sold in 2 years $500,000 Discount rate (monthly) 0.8%

Mean demand per month 5

403938313029282721Month Demand 4 5 7 1 6

0 0 0 0 0Condos left to be sold 120 116 10 3 2 Condos sold this month 4 5 7 1 2 Maintenance cost $0$1,600$2,400$88,800$92,800 Commission from sales $60,000$30,000$210,000$150,000$120,000 Bonus at end of year 1 Bonus at end of year 2 Cash flow $60,000$28,400$207,600$27,200 $61,200

Months to sell out 29 Total bonus received NPV of cash flows

$0 $1,991,40126

Marketing and selling condos

Simulation model Distribution of demand for condos each month (Poisson distributed)

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Discussion of the Simulation Results Recall that the deterministic model sells out in 24 months, receives both bonuses, and achieves an NPV of about $2.82 million. As you might guess, the simulation model doesn’t do this well. The main problem is that there is a fairly good chance that one or both bonuses will not be received. Distributions of the three outputs appear in Figures 16.37 through 16.39. Figure 16.37 shows that although 24 months is (nearly) the most likely number of months to sell out, there was at least one scenario where it took only 18 months and another where it took 32 months. Figure 16.38 shows the four possibilities for bonuses: receive neither, receive one or the other, or receive both. Unfortunately for Pletcher, the first three possibilities are fairly likely; the probability of receiving both bonuses is only about 0.37. Finally, the shape of the NPV distribution in Figure 16.39, with three separate peaks, is influenced heavily by the bonuses or lack of them. On average, the NPV is only about $2.38 million, much less than estimated by the deterministic model. This is still one more example—a dramatic one—of the flaw of averages.

Figure 16.37 Distribution of Months to Sell Out

Figure 16.38 Distribution of Total Bonus Received

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Figure 16.39 Distribution of NPV

Problems

Level A 21. Suppose that Coke and Pepsi are fighting for the cola

market. Each week each person in the market buys one case of Coke or Pepsi. If the person’s last purchase was Coke, there is a 0.90 probability that this person’s next purchase will be Coke; otherwise, it will be Pepsi. (You can assume that there are only two brands in the mar- ket.) Similarly, if the person’s last purchase was Pepsi, there is a 0.80 probability that this person’s next pur- chase will be Pepsi; otherwise, it will be Coke. Currently half of all people purchase Coke, and the other half pur- chase Pepsi. Simulate one year (52 weeks) of sales in the cola market and estimate each company’s average weekly market share and each company’s ending mar- ket share in week 52. Do this by assuming that the total market size is fixed at 100,000 customers. (Hint: Use the RiskBinomial function. However, if your model requires more RiskBinomial functions than the number allowed in the academic version of @RISK, remember that you can instead use the BINOM.INV function to generate binomially distributed random numbers. This takes the form =BINOM.INV(n,p,RAND()).)

22. Seas Beginning sells clothing by mail order. An import- ant question is when to strike a customer from the com- pany’s mailing list. At present, the company strikes a customer from its mailing list if a customer fails to order from six consecutive catalogs. The company wants to know whether striking a customer from its list after a

customer fails to order from four consecutive catalogs results in a higher profit per customer. The following data are available:

• If a customer placed an order the last time she received a catalog, then there is a 20% chance she will order from the next catalog.

• If a customer last placed an order one catalog ago, there is a 16% chance she will order from the next catalog she receives.

• If a customer last placed an order two catalogs ago, there is a 12% chance she will order from the next catalog she receives.

• If a customer last placed an order three catalogs ago, there is an 8% chance she will order from the next catalog she receives.

• If a customer last placed an order four catalogs ago, there is a 4% chance she will order from the next catalog she receives.

• If a customer last placed an order five catalogs ago, there is a 2% chance she will order from the next catalog she receives.

It costs $2 to send a catalog, and the average profit per order is $30. Assume a customer has just placed an order. To maximize expected profit per customer, would Seas Beginning make more money canceling such a cus- tomer after six nonorders or four nonorders?

23. Based on Babich (1992). Suppose that each week each of 300 families buys a gallon of orange juice from com- pany A, B, or C. Let pA denote the probability that a gallon produced by company A is of unsatisfactory quality, and define pB and pC similarly for companies B

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8 2 2 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

and C. If the last gallon of juice purchased by a family is satisfactory, the next week they will purchase a gallon of juice from the same company. If the last gallon of juice purchased by a family is not satisfactory, the family will purchase a gallon from a competitor. Consider a week in which A families have purchased juice A, B families have purchased juice B, and C families have purchased juice C. Assume that families that switch brands during a period are allocated to the remaining brands in a man- ner that is proportional to the current market shares of the other brands. For example, if a customer switches from brand A, there is probability B>(B 1 C) that he will switch to brand B and probability C>(B 1 C) that he will switch to brand C. Suppose that the market is currently divided equally: 10,000 families for each of the three brands. a. After a year, what will the market share for each firm

be? Assume pA 5 0.10, pB 5 0.15, and pC 5 0.20. (Hint: You will need to use the RiskBinomial func- tion to see how many people switch from A and then use the RiskBinomial function again to see how many switch from A to B and from A to C. However, if your model requires more RiskBinomial func- tions than the number allowed in the academic ver- sion of @RISK, remember that you can instead use the BINOM.INV function to generate binomially distributed random numbers. This takes the form =BINOM.INV(n,p,RAND()).)

b. Suppose a 1% increase in market share is worth $10,000 per week to company A. Company A believes that for a cost of $1 million per year it can cut the percentage of unsatisfactory juice cartons in half. Is this worthwhile? (Use the same values of pA, pB, and pC as in part a.)

Level B 24. The customer loyalty model in Example 16.7 assumes

that once a customer leaves (becomes disloyal), that customer never becomes loyal again. Assume instead that there are two probabilities that drive the model, the retention rate and the rejoin rate, with values 0.75 and 0.15, respectively. The simulation should follow a cus- tomer who starts as a loyal customer in year 1. From then on, at the end of any year when the customer was loyal, this customer remains loyal for the next year with probability equal to the retention rate. But at the end of any year the customer is disloyal, this customer becomes loyal the next year with probability equal to the rejoin rate. During the customer’s nth loyal year with the com- pany, the company’s mean profit from this customer is the nth value in the mean profit list in column B. Keep track of the same two outputs as in the example, and also keep track of the number of times the customer rejoins.

25. We are all aware of the fierce competition by mobile phone service companies to get our business. For

example, AT&T is always trying to attract Verizon’s customers, and vice versa. Some even give away prizes to entice us to sign up for a guaranteed length of time. This example is based on one such offer. We assume that a mobile provider named Syncit is willing to give a customer a free laptop computer, at a cost of $300 to Syncit, if the customer signs up for a guaranteed two years of service. During that time, the cost of service to the customer is a constant $60 per month, or $720 annually. After two years, we assume the cost of service increases by 2% annually. We assume that in any year after the guaranteed two years, the probability is 0.7 that the customer will stay with Syncit. This probability is the retention rate. We also assume that if a customer has switched to another mobile service, there is always a probability of 0.1 that the customer will (without any free laptop offer) willingly rejoin Syncit. The company wants to see whether this offer makes financial sense in terms of NPV, using a 7% discount rate. It also wants to see how the NPV varies with the retention rate. Simulate a 15-year time horizon, both with and without the free offer, to estimate the difference. (For the situation with- out the free offer, assume the customer has probability 0.5 of signing up with Syncit during year 1.)

26. Suppose that GLC earns a $2000 profit each time a per- son buys a car. We want to determine how the expected profit earned from a customer depends on the qual- ity of GLC’s cars. We assume a typical customer will purchase 10 cars during her lifetime. She will purchase a car now (year 1) and then purchase a car every five years—during year 6, year 11, and so on. For simplicity, we assume that Hundo is GLC’s only competitor. We also assume that if the consumer is satisfied with the car she purchases, she will buy her next car from the same company, but if she is not satisfied, she will buy her next car from the other company. Hundo produces cars that satisfy 80% of its customers. Currently, GLC produces cars that also satisfy 80% of its customers. Consider a customer whose first car is a GLC car. If profits are discounted at 7% annually, use simulation to estimate the value of this customer to GLC. Also estimate the value of a customer to GLC if it can raise its customer satisfaction rating to 85%, to 90%, or to 95%. You can interpret the satisfaction value as the probability that a customer will not switch companies.

27. Mutron Company is thinking of marketing a new drug used to make pigs healthier. At the beginning of the current year, there are 1,000,000 pigs that could use the product. Each pig will use Mutron’s drug or a com- petitor’s drug once a year. The number of pigs is fore- cast to grow by an average of 5% per year. However, this growth rate is not a sure thing. Mutron assumes that each year’s growth rate is an independent draw from a normal distribution, with probability 0.95 that the growth rate will be between 3% and 7%. Assuming it

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16-5 Simulating Games of Chance 8 2 3

enters the market, Mutron is not sure what its share of the market will be during year 1, so it models this with a triangular distribution. Its worst-case share is 20%, its most likely share is 40%, and its best-case share is 70%. In the absence of any new competitors entering this mar- ket (in addition to itself), Mutron believes its market share will remain the same in succeeding years. How- ever, there are three potential entrants (in addition to Mutron). At the beginning of each year, each entrant that has not already entered the market has a 40% chance of entering the market. The year after a competitor enters, Mutron’s market share will drop by 20% for each new competitor who entered. For example, if two competitors

enter the market in year 1, Mutron’s market share in year 2 will be reduced by 40% from what it would have been with no entrants. Note that if all three entrants have entered, there will be no more entrants. Each unit of the drug sells for $2.20 and incurs a variable cost of $0.40. Profits are discounted by 8% annually. a. Assuming that Mutron enters the market, use simula-

tion to find its NPV for the next 10 years from the drug.

b. Again assuming that Mutron enters the market, it can be 95% certain that its actual NPV from the drug is between what two values?

16-5 Simulating Games of Chance We realize that this is a book about business applications. However, it is instructive (and fun) to see how simulation can be used to analyze games of chance, including sports con- tests. Indeed, many analysts refer to Monte Carlo simulation, and you can guess where that name comes from—the gambling casinos of Monte Carlo.

16-5a Simulating the Game of Craps Most games of chance are great candidates for simulation because they are, by their very nature, driven by randomness. In this section, we examine one such game that is extremely popular in the gambling casinos: the game of craps. In its most basic form, craps is played as follows. A player rolls two dice and observes the sum of the two sides turned up. If this sum is 7 or 11, the player wins immediately. If the sum is 2, 3, or 12, the player loses immediately. Otherwise, if this sum is any other number (4, 5, 6, 8, 9, or 10), that num- ber becomes the player’s point. Then the dice are thrown repeatedly until the sum is the player’s point or 7. In case the player’s point occurs before a 7, the player wins. But if a 7 occurs before the point, the player loses. The following example uses simulation to deter- mine the properties of this game.

EXAMPLE

16.10 ESTIMATING THE PROBABILITY OF WINNING AT CRAPS

Joe Gamble loves to play craps at the casinos. He suspects that his chances of winning are less than fifty-fifty, but he wants to find the probability that he wins a single game of craps.

Objective To use simulation to find the probability of winning a single game of craps.

Where Do the Numbers Come From? There are no input numbers here, only the rules of the game.

Solution The simulation is of a single game. By running this simulation for many iterations, you can find the probability that Joe wins a single game of craps. If his intuition is correct (and surely it must be, or the casino could not stay in business), this probability is less than 0.5.

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Developing the Simulation Model The simulation model is for a single game. (See Figure 16.40 and the file Craps Finished.xlsx. No “big picture” is provided for this model.) There is a subtle problem here: The number of tosses of the dice nec- essary to determine the outcome of a single game is unknown. Theoretically, the game could continue forever, with the player waiting for his point or a 7.However, it is extremely unlikely that more than, say, 40 tosses are necessary in a single game. (This can be shown by a probability argument not presented here.) Therefore, you can simulate 40 tosses and use only those that are necessary to determine the outcome of a single game. The steps required are as follows.

Game of Craps

Figure 16.40 Simulation of Craps Game

1 2 3 4 5 6 7 8 9

10 11 12 13 14

A B C D E F G H I J Craps Simula�on

Simulated tosses Toss Die 1 Die 2 Sum Win on this toss? Lose on this toss? Con�nue? Summary results from simula�on

1 0 0 0 0 0 1

0 0 0 0 0 0

Yes Yes Yes Yes Yes No

Win? (1 if yes, 0 if no) 2 Number of tosses 3 4 Pr(winning) 0.495 5 Expected number of tosses 3.344 6 7 8 9

10 9 5 5

11 10 4 8 5 9

6 4 1 1 6 4 2 6 4 3

4 5 4 4 5 6 2 2 1 6

1 6

1. Simulate tosses. Simulate the results of 40 tosses in the range B5:D44 by entering the formula

5RANDBETWEEN(1,6)

in cells B5 and C5 and the formula

5SUM(B5:C5)

in cell D5. Then copy these to the range B6:D44. (Recall that the RANDBETWEEN function generates a random integer between the two specified values such that all values are equally likely, so it is perfect for tossing a die. You could also use @RISK’s RiskIntUniform function, which works exactly like RANDBETWEEN.)

As in many spreadsheet simulation models, the concepts in this model are simple. The key is careful bookkeeping.

RiskIntUniform

The @RISK function RiskIntUniform in the form 5RiskIntUniform(N1,N2) works exactly like Excel’s RANDBETWEEN function.

@RISK Function

2. First toss outcome. Determine the outcome of the first toss with the formulas

5IF(OR(D557,D5511),1,0)

5IF(OR(D552,D553,D5512),1,0)

and

5IF(AND(E550,F550),”Yes”,”No”)

in cells E5, F5, and G5, respectively. Note that the OR condition checks whether Joe wins right away (in which case a 1 is recorded in cell E5). Similarly, the OR condition in cell F5 checks whether he loses right away. In cell G5, the AND condition checks whether both cells E5 and F5 are 0, in which case the game continues. Otherwise, the game is over.

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3. Outcomes of other tosses. Assuming the game continues beyond the first toss, Joe’s point is the value in cell D5. Then he is waiting for a toss to have the value in cell D5 or 7, whichever occurs first. To implement this logic, enter the formulas

5IF(OR(G55”No”,G55””),””,IF(D65$D$5,1,0))

5IF(OR(G55”No”,G55””),””,IF(D657,1,0))

and

5IF(OR(G55”No”,G55””),””,IF(AND(E650,F650),”Yes”,”No”))

in cells E6, F6, and G6, respectively, and copy these to the range E7:G44. The OR condition in each formula checks whether the game just ended on the previous toss or has been over for some time, in which case blanks are entered. Other- wise, the first two formulas check whether Joe wins or loses on this toss. If both of these return 0, the third formula returns Yes (and the game continues). Otherwise, it returns No (and the game has just ended).

4. Game outcomes. Keep track of two aspects of the game in @RISK output cells: whether Joe wins or loses and how many tosses are required. To find these, enter the formulas

5SUM(E5:E44)

and

5COUNT(E5:E44)

in cells J5 and J6, and designate each of these as an @RISK output cell. Note that both functions, SUM and COUNT, ignore blank cells.

5. Simulation summary. Although you can get summary measures in the various @RISK results windows after you run the simulation, it is useful to see some key summary measures right on the model sheet. To obtain these, enter the formula

5RiskMean(J5)

in cell J8 and copy it to cell J9. As the labels indicate, the RiskMean in cell J8, being an aver- age of 0’s and 1’s, is just the fraction of iterations where Joe wins. The average in cell J9 is the average number of tosses until the game’s outcome is determined.

Running the Simulation Set the number of iterations to 10,000 (partly for variety and partly to obtain a more accurate result) and the number of simula- tions to 1. Then run the simulation as usual.

Discussion of the Simulation Results After running @RISK, the summary results in cells J8 and J9 of Figure 16.40 (among others) are available. Our main interest is in the average in cell J8. It represents the best estimate of the proba- bility of winning, 0.495. (It can be shown with a probability argument that the exact probability of winning in craps is 0.493.) You can also see that the average number of tosses needed to determine the outcome of a game was about 3.34. (The maximum number of tosses ever needed on these 10,000 iterations was 25.)

Recall that the mean (or average) of a sequence of 0’s and 1’s is the fraction of 1’s in the sequence. This can typically be interpreted as a probability.

Perhaps surprisingly, the probability of winning in craps is 0.493, only slightly less than 0.5.

16-5b Simulating the NCAA Basketball Tournament Each year the suspense reaches new levels as “March Madness” approaches, the time of the NCAA Basketball Tournament. Which of the 68 teams in the tournament will reach the “Sweet Sixteen,” which will go on to the prestigious “Final Four,” and which team will be crowned champion? The excitement at Indiana University is particularly high, given the strong basketball tradition here, so it has become a yearly tradition at IU (at least for the authors) to simulate the NCAA Tournament right after the brackets have been announced. We share that simulation in the following example.

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EXAMPLE

16.11 MARCH MADNESS At the time this example was written, the most recent NCAA Basketball Tournament was the 2018 tournament, won by Villanova. Of course, on the Sunday evening when the 68-team field was announced, we did not know which team would win. All we knew were the pairings (which teams would play which other teams) and the team ratings, based on Jeff Sagarin’s nationally syndicated rating system. We now show how to simulate the tournament and keep a tally of the winners.

Objective To simulate the NCAA basketball tournament and keep a tally on the number of times each team wins the tournament.

Where Do the Numbers Come From? As soon as you learn the pairings for the next NCAA tournament, you can perform a Web search for “Sagarin ratings” to find the latest ratings.

Solution We need to make one probabilistic assumption. From that point, it is a matter of “playing out” the games and doing the required bookkeeping. To understand this probabilistic assumption, suppose team A plays team B and Sagarin’s ratings for these teams are, say, 85 and 78. Then Sagarin pre- dicts that the actual point differential in the game (team A’s score minus team B’s score) will be the difference between the ratings, or 7.4 We take this one step further. We assume that the actual point differential is normally distributed with mean equal to Sagarin’s prediction, 7, and standard devi- ation 10. (Why 10? This is an estimate based on an extensive analysis of historical data. However, the spreadsheet model is set up so that you can change the standard deviation to a different value if you prefer.) Then if the actual point differential is positive, team A wins. If it is negative, team B wins.

Developing the Simulation Model We provide only an outline of the simulation model. You can see the full details in the file March Madness 2018.xlsx. (This file includes the data for the 2018 tournament, but you can easily mod- ify it for future tournaments.) The entire simulation is on a single Model sheet. Columns A through C list team indexes, team names, and Sagarin ratings. If two teams are paired in the first round, they are placed next to one another in the list. Also, all teams in a given region are listed together. (The regions are color-coded.) Columns H through N contain the simulation. The first-round results are at the top, the sec- ond-round results are below these, and so on. Winners from one round are automatically carried over to the next round with appropriate formulas. Selected portions of the model appear in Figure 16.41. This figure shows one possible scenario where Villanova did indeed win. We now describe the essential features of the model. (We didn’t use @RISK in this model. We instead used Excel’s built-in functions and a data table to replicate the results.)

1. Simulate rounds. Jumping ahead to the fourth-round simulation in Figure 16.41, the winners from the previous round 3 are captured, and then the games in round 4 are simulated. The key formulas are in columns H and I. For example, the formulas in cells K137 and J137 are

5VLOOKUP(F137,LTable,3)2VLOOKUP(F138,LTable,3)

and

=NORM.INV(RAND(),K137,$F$3)

The first of these looks up the ratings of the two teams involved and subtracts to get the predicted point spread. The second formula simulates a point spread with the predicted point spread as its mean and the value in cell F3, 10, as its standard deviation. The rest of the formulas do the appropriate bookkeeping. You can view the details in the file.

2. Outputs. The model uses a data table to replicate the following outputs for each team: (1) whether the team wins the tournament, (2) whether the team reaches the final game, and (3) whether the team reaches the final four (the semi-finals). Then it shows tallies of these in tabular and graphical form. For example, after one run of 1000 replications, Villanova won the tournament 189 times, reached the final game 284 times, and reached the final four 437 times. They were clearly the favorite entering the tournament, and they backed it up by winning.

We model the point spread as normally distributed, with mean equal to the difference between the Sagarin ratings and standard deviation 10.

March Madness Model

4 In general, there is also a home-court advantage, but we assume all games in the tournament are on “neutral” courts, so that there is no advantage to either team.

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Figure 16.41 NCAA Basketball Simulation Model (Last Three Rounds Only)

135

136

137

138

139

140

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

Results of Round 4 H

Semifinals

Teams Kentucky

Tennessee Xavier

North Carolina Villanova Arkansas

Auburn Duke

Teams Kentucky

Xavier Villanova

Duke

Teams Kentucky Villanova

Finals

Winner

I KJ L M N O

Predicted 1.42

�2.4

9.89

-8.59

Predicted �0.46

1.38

Predicted �6.57

Simulated 7.13

0.69

14.44

–17.26

Simulated 5.48

4.06

Simulated –11.58

Winner Kentucky

Xavier

Villanova

Duke

Lowest seed to win

Winner Kentucky

Villanova

Lowest seed to win

Winner Villanova

11 17 32 34 48 58 67

17 34 67

Index of winner Seed

Indexes

IndexesGame

Game

Index of winner

Seed

Problems

Level A 28. The game of Chuck-a-Luck is played as follows: You

pick a number between 1 and 6 and toss three dice. If your number does not appear, you lose $1. If your num- ber appears x times, you win $x. On the average, use simulation to find the average amount of money you will win or lose on each play of the game.

29. A martingale betting strategy works as follows. You begin with a certain amount of money and repeatedly play a game in which you have a 40% chance of winning any bet. In the first game, you bet $1. From then on, every time you win a bet, you bet $1 the next time. Each time you lose, you dou- ble your previous bet. Currently you have $63. Assuming you have unlimited credit, so that you can bet more money than you have, use simulation to estimate the profit or loss you will have after playing the game 50 times.

30. You have $5 and your opponent has $10. You flip a fair coin and if heads comes up, your opponent pays you $1. If tails comes up, you pay your opponent $1. The game is finished when one player has all the money or after 100 tosses, whichever comes first. Use simulation to estimate the probability that you end up with all the money and the probability that neither of you goes broke in 100 tosses.

Level B 31. Assume a very good NBA team has a 70% chance of

winning in each game it plays. During an 82-game sea- son what is the average length of the team’s longest win- ning streak? What is the probability that the team has a winning streak of at least 16 games? Use simulation to answer these questions, where each iteration of the sim- ulation generates the outcomes of all 82 games.

32. You are going to play the Wheel of Misfortune Game against the house. The wheel has 10 equally likely num- bers: 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50. The goal is to get a total as close as possible to 50 points without exceeding 50. You go first and spin the wheel. Based on your first spin, you can decide whether you want to spin again. (You can spin no more than twice.) After you are done, it is the house’s turn. If your total is more than 50, the house doesn’t need a turn; it wins automatically. Oth- erwise, the house spins the wheel. After its first spin, it can spin the wheel again if it wants. (The house can also spin no more than twice.) Then the winner is determined, where a tie goes to you. Use simulation to estimate your probability of winning the game if you and the house both use best strategies. What are the best strategies?

33. Consider the following card game. The player and dealer each receive a card from a 52-card deck. At the end of the game the player with the highest card wins; a tie goes to

16-5 Simulating Games of Chance 8 2 7

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8 2 8 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

• Pull your goalie if you are behind at any point in the last minute of the game; put him back in if you tie the score.

Which strategy maximizes your probability of winning or tying the game? Simulate the game using 10- second increments of time. Use the RiskBinomial function to determine whether a team scores a goal in a given 10-second segment. This is reasonable because the probability of scoring two or more goals in a 10-second period is near zero.

35. You are playing tennis against a top tennis pro, and you have a 42% chance of winning each point. (You are good!) a. Use simulation to estimate the probability you will

win a particular game. Note that the first player to score at least four points and have at least two more points than his or her opponent wins the game.

b. Use simulation to determine your probability of win- ning a set. Assume that the first player to win six games wins the set if he or she is at least two games ahead; otherwise, the first player to win seven games wins the set. (We substitute a single game for the usual tiebreaker.)

c. Use simulation to determine your probability of win- ning a match. Assume that the first player to win three sets wins the match.

the dealer. (You can assume that Aces count 1, Jacks 11, Queens 12, and Kings 13.) After the player receives his card, he keeps the card if it is 7 or higher. If the player does not keep the card, the player and dealer swap cards. Then the dealer keeps his current card (which might be the player’s original card) if it is 9 or higher. If the dealer does not keep his card, he draws another card. Use simulation with at least 1000 iterations to estimate the probability that the player wins. (Hint: See the file P16_33.xlsx to see a clever way of simulating cards from a deck so that the same card is never dealt more than once.)

34. Based on Morrison and Wheat (1984). When his team is behind late in the game, a hockey coach usually waits until there is one minute left before pulling the goalie out of the game. Using simulation, it is possible to show that coaches should pull their goalies much sooner. Suppose that if both teams are at full strength, each team scores an average of 0.05 goal per minute. Also, suppose that if you pull your goalie you score an average of 0.08 goal per minute and your opponent scores an average of 0.12 goal per minute. Suppose you are one goal behind with five minutes left in the game. Consider the following two strategies:

• Pull your goalie if you are behind at any point in the last five minutes of the game; put him back in if you tie the score.

16-6 Conclusion We claimed in the previous chapter that spreadsheet simulation, especially together with an add-in like @RISK, is a powerful tool. After seeing the examples in this chapter, you should now appreciate how powerful and flexible simulation is. Unlike Solver optimization models, where you often make simplifying assumptions to achieve linearity, for example, you can allow virtually anything in simulation models. All you need to do is relate output cells to input cells with appropriate formulas, where any of the input cells can contain probability distributions to reflect uncertainty. The results of the simulation then show the distribution of any particular output. It is no wonder that companies such as GM, Eli Lilly, and many others are increas- ingly relying on simulation models to analyze their corporate operations.

Summary of Key Terms TERM EXPLANATION EXCEL PAGE

Gamma distribution Right-skewed distribution of nonnegative values useful for many quantities such as the lifetime of an appliance

836

Value at risk at the 5% level (VaR 5%)

Fifth percentile of distribution of some output, usually a monetary output; indicates nearly the worst possible outcome

850

Churn When customers stop buying a product or service and switch to a competitor’s offering

864

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16-6 Conclusion 8 2 9

Problems

Conceptual Questions C.1. We have separated the examples in this chapter into

operations, finance, marketing, and sports categories. List at least one other problem in each of these catego- ries that could be attacked with simulation. For each, identify the random inputs, possible probability distri- butions for them, and any outputs of interest.

C.2. Suppose you are an HR (human resources) manager at a big university, and you sense that the university is becoming too top-heavy with full professors. That is, there do not seem to be as many younger professors at the assistant and associate levels as there ought to be. How could you study this problem with a simulation model, using current and/or proposed promotions, hiring, firing, and retirement policies?

C.3. You are an avid basketball fan, and you would like to build a simulation model of an entire game so that you could compare two different strategies, such as man- to-man versus zone defense. Is this possible? What might make this simulation model difficult to build?

C.4. Suppose you are a financial analyst and your com- pany runs many simulation models to estimate the profitability of its projects. If you had to choose just two measures of the distribution of any important out- put such as net profit to report, which two would you choose? Why? What information would be missing if you reported only these two measures? How could they be misleading?

C.5. Software development is an inherently risky and uncer- tain process. For example, there are many examples of software that couldn’t be “finished” by the sched- uled release date—bugs still remained and features weren’t ready. (Many people believe this was the case with Office 2007.) How might you simulate the devel- opment of a software product? What random inputs would be required? Which outputs would be of inter- est? Which measures of the probability distributions of these outputs would be most important?

C.6. Health care is continually in the news. Can (or should) simulation be used to help solve, or at least study, some of the difficult problems associated with health care? Provide at least two examples where simulation might be useful.

Level A 36. You now have $3000. You will toss a fair coin four

times. Before each toss you can bet any amount of your money (including none) on the outcome of the toss. If heads comes up, you win the amount you bet. If tails comes up, you lose the amount you bet. Your goal is to reach $6000. It turns out that you can maximize your chance of reaching $6000 by betting either the money you have on hand or $6000 minus the money you

have on hand, whichever is smaller. Use simulation to estimate the probability that you will reach your goal with this betting strategy.

37. You now have $10,000, all of which is invested in a sports team. Each year there is a 60% chance that the value of the team will increase by 60% and a 40% chance that the value of the team will decrease by 60%. Estimate the mean and median value of your investment after 50 years. Explain the large difference between the estimated mean and median.

38. Suppose you have invested 25% of your portfolio in four different stocks. The mean and standard deviation of the annual return on each stock are shown in the file P16_38.xlsx. The correlations between the annual returns on the four stocks are also shown in this file. a. What is the probability that your portfolio’s annual

return will exceed 20%? b. What is the probability that your portfolio will lose

money during the year? 39. A ticket from Indianapolis to Orlando on Deleast

Airlines sells for $150. The plane can hold 100 people. It costs Deleast $8000 to fly an empty plane. Each per- son on the plane incurs variable costs of $30 (for food and fuel). If the flight is overbooked, anyone who cannot get a seat receives $300 in compensation. On average, 95% of all people who have a reservation show up for the flight. To maximize expected profit, how many res- ervations for the flight should Deleast book? (Hint: The function RiskBinomial can be used to simulate the num- ber who show up. It takes two arguments: the number of reservations booked and the probability that any ticketed person shows up.)

40. Based on Marcus (1990). The Balboa mutual fund has beaten the Standard and Poor’s 500 during 11 of the last 13 years. People use this as an argument that you can beat the market. Here is another way to look at it that shows that Balboa’s beating the market 11 out of 13 times is not unusual. Consider 50 mutual funds, each of which has a 50% chance of beating the market during a given year. Use simulation to estimate the probabil- ity that over a 13-year period the best of the 50 mutual funds will beat the market for at least 11 out of 13 years. This probability turns out to exceed 40%, which means that the best mutual fund beating the market 11 out of 13 years is not an unusual occurrence after all.

41. You have been asked to simulate the cash inflows to a toy company for the next year. Monthly sales are inde- pendent random variables. Mean sales for the months January through March and October through December are $80,000, and mean sales for the months April through September are $120,000. The standard devia- tion of each month’s sales is 20% of the month’s mean sales. Model the method used to collect monthly sales as follows: • During each month a certain fraction of new sales

will be collected. All new sales not collected become one month overdue.

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8 3 0 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

45. The annual demand for Prizdol, a prescription drug manufactured and marketed by the NuFeel Company, is normally distributed with mean 50,000 and standard deviation 12,000. Assume that demand during each of the next 10 years is an independent random number from this distribution. NuFeel needs to determine how large a Prizdol plant to build to maximize its expected profit over the next 10 years. If the company builds a plant that can produce x units of Prizdol per year, it will cost $16 for each of these x units. NuFeel will pro- duce only the amount demanded each year, and each unit of Prizdol produced will sell for $3.70. Each unit of Prizdol produced incurs a variable production cost of $0.20. It costs $0.40 per year to operate a unit of capacity. a. Among the capacity levels of 30,000, 35,000, 40,000,

45,000, 50,000, 55,000, and 60,000 units per year, which level maximizes expected profit? Use simulation to answer this question.

b. Using the capacity from your answer to part a, NuFeel can be 95% certain that actual profit for the 10-year period will be between what two values?

46. A company is trying to determine the proper capac- ity level for its new electric car. A unit of capacity provides the potential to produce one car per year. It costs $10,000 to build a unit of capacity and the cost is charged equally over the next five years. It also costs $400 per year to maintain a unit of capacity (whether or not it is used). Each car sells for $14,000 and incurs a variable production cost of $10,000. The annual demand for the electric car during each of the next five years is believed to be normally distributed with mean 50,000 and standard deviation 10,000. The demands during different years are assumed to be independent. Profits are discounted at a 7.5% annual interest rate. The company is working with a five-year planning horizon. Capacity levels of 30,000, 40,000, 50,000, 60,000, and 70,000 are under consideration. You can assume that the company never produces more than demand, so there is never any inventory to carry over from year to year. a. Assuming that the company is risk neutral, use simu-

lation to find the optimal capacity level. b. Using the answer to part a, there is a 5% chance that

the actual discounted profit will exceed what value, and there is a 5% chance that the actual discounted profit will be less than what value?

c. If the company is risk averse, how might the optimal capacity level change?

47. The DC Cisco office is trying to predict the revenue it will generate next week. Ten deals may close next week. The probability of each deal closing and data on the possible size of each deal (in millions of dollars) are listed in the file P16_47.xlsx. Use simulation to estimate total reve- nue. Based on the simulation, the company can be 95% certain that its total revenue will be between what two numbers?

• During each month a certain fraction of one-month overdue sales is collected. The remainder becomes two months overdue.

• During each month a certain fraction of two-month overdue sales is collected. The remainder is written off as bad debt.

You are given the information in the file P16_41.xlsx from past months. Using this information, build a sim- ulation model that generates the total cash inflow for each month. Develop a simple forecasting model and build the error of your forecasting model into the simu- lation. Assuming that there are $120,000 of one-month- old sales outstanding and $140,000 of two-month-old sales outstanding during January, you are 95% sure that total cash inflow for the year will be between what two values?

42. Consider a device that requires two batteries to function. If either of these batteries dies, the device will not work. Currently there are two new batteries in the device, and there are three extra new batteries. Each battery, once it is placed in the device, lasts a random amount of time that is triangularly distributed with parameters 15, 18, and 25 (all expressed in hours). When any of the batter- ies in the device dies, it is immediately replaced by an extra if an extra is still available. Use @RISK to simu- late the time the device can last with the batteries cur- rently available.

43. Consider a drill press containing three drill bits. The cur- rent policy (called individual replacement) is to replace a drill bit when it fails. The firm is considering chang- ing to a block replacement policy in which all three drill bits are replaced whenever a single drill bit fails. Each time the drill press is shut down, the cost is $100. A drill bit costs $50, and the variable cost of replacing a drill bit is $10. Assume that the time to replace a drill bit is negligible. Also, assume that the time until fail- ure for a drill bit follows an exponential distribution with a mean of 100 hours. Determine which replace- ment policy (block or individual replacement) should be implemented. (Hint: To generate an exponentially dis- tributed time to failure, you can use 5RiskExpon(100) or 52100*LN(RAND()). The latter uses only Excel functions, so it can be used to get around @RISK’s 100- function limit in the academic version.)

44. Appliances Unlimited (AU) sells refrigerators. Any refrigerator that fails before it is three years old is replaced for free. Of all refrigerators, 3% fail during their first year of operation; 5% of all one-year-old refrigerators fail during their second year of operation; and 7% of all two-year-old refrigerators fail during their third year of operation. a. Use simulation to estimate the fraction of all refriger-

ators that will have to be replaced. b. It costs $500 to replace a refrigerator, and AU sells

10,000 refrigerators per year. If the warranty period were reduced to two years, how much per year in replacement costs would be saved?

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16-6 Conclusion 8 3 1

the RiskUniform function to model the actual value of the field and the competitors’ bids.)

50. Suppose you begin year 1 with $5000. At the begin- ning of each year, you put half of your money under a mattress and invest the other half in Whitewater stock. During each year, there is a 50% chance that the White- water stock will double, and there is a 50% chance that you will lose half of your investment. To illustrate, if the stock doubles during the first year, you will have $3750 under the mattress and $3750 invested in White- water during year 2. You want to estimate your annual return over a 30-year period. If you end with F dollars, your annual return is (F>5000)1>30 2 1. For exam- ple, if you end with $100,000, your annual return is 201>30 2 1 5 0.105, or 10.5%. Run 1000 replications of an appropriate simulation. Based on the results, you can be 95% certain that your annual return will be between which two values?

51. Mary Higgins is a freelance writer with enough spare time on her hands to play the stock market fairly seri- ously. Each morning she observes the change in stock price of a particular stock and decides whether to buy or sell, and if so, how many shares to buy or sell. Assume that on day 1, she has $100,000 cash to invest and that she spends part of this to buy her first 500 shares of the stock at the current price of $50 per share. From that point on, she follows a fairly simple “buy low, sell high” strategy. Specifically, if the price has increased three days in a row, she sells 25% of her shares of the stock. If the price has increased two days in a row (but not three), she sells 10% of her shares. In the other direction, if the price has decreased three days in a row, she buys up to 25% more shares, whereas if the price has decreased only two days in a row, she buys up to 10% more shares. The reason for the “up to” proviso is that she cannot buy more than she has cash to pay for. Assume a fairly sim- ple model of stock price changes, as described in the file P16_51.xlsx. Each day the price can change by as much as $2 in either direction, and the probabilities depend on the previous price change: decrease, increase, or no change. Build a simulation model of this strategy for a period of 75 trading days. (You can assume that the stock price on each of the previous two days was $49.) Choose interesting @RISK output cells, and then run @ RISK for at least 1000 iterations and report your find- ings.

52. You are considering a 10-year investment project. At present, the expected cash flow each year is $10,000. Suppose, however, that each year’s cash flow is nor- mally distributed with mean equal to last year’s actual cash flow and standard deviation $1000. For example, suppose that the actual cash flow in year 1 is $12,000. Then year 2 cash flow is normal with mean $12,000 and standard deviation $1000. Also, at the end of year 1, your best guess is that each later year’s expected cash flow will be $12,000.

Level B 48. A common decision is whether a company should buy

equipment and produce a product in house or outsource production to another company. If sales volume is high enough, then by producing in house, the savings on unit costs will cover the fixed cost of the equipment. Suppose a company must make such a decision for a four-year time horizon, given the following data. Use simulation to estimate the probability that producing in house is better than outsourcing.

• If the company outsources production, it will have to purchase the product from the manufacturer for $18 per unit. This unit cost will remain constant for the next four years.

• The company will sell the product for $40 per unit. This price will remain constant for the next four years.

• If the company produces the product in house, it must buy a $400,000 machine that is depreciated on a straight-line basis over four years, and its cost of pro- duction will be $7 per unit. This unit cost will remain constant for the next four years.

• The demand in year 1 has a worst case of 10,000 units, a most likely case of 14,000 units, and a best case of 16,000 units.

• The average annual growth in demand for years 2–4 has a worst case of 10%, a most likely case of 20%, and a best case of 26%. Whatever this annual growth is, it will be the same in each of the years.

• The tax rate is 21%. • Cash flows are discounted at 7% per year.

49. Consider an oil company that bids for the rights to drill in offshore areas. The value of the right to drill in a given offshore area is highly uncertain, as are the bids of the competitors. This problem demonstrates the “winner’s curse.” The winner’s curse states that the optimal bid- ding strategy entails bidding a substantial amount below the company’s assumed value of the product for which it is bidding. The idea is that if the company does not bid under its assumed value, its uncertainty about the actual value of the product will often lead it to win bids for products on which it loses money (after paying its high bid). Suppose Royal Conch Oil (RCO) is trying to deter- mine a profit-maximizing bid for the right to drill on an offshore oil site. The actual value of the right to drill is unknown, but it is equally likely to be any value between $10 million and $110 million. Seven competitors will bid against RCO. Each bidder’s (including RCO’s) esti- mate of the value of the drilling rights is equally likely to be any number between 50% and 150% of the actual value. Based on past history, RCO believes that each competitor is equally likely to bid between 40% and 60% of its value estimate. Given this information, what fraction (within 0.05) of RCO’s estimated value should it bid to maximize its expected profit? (Hint: You can use

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8 3 2 C H A P T E R 1 6 S i m u l a t i o n M o d e l s

Suppose that at the end of year t 2 1, n competitors are present (including Play Things). Then during year t, a fraction 0.9 9 0.1n of the company’s loyal cus- tomers (last year’s purchasers) will buy a doll from Play Things this year, and a fraction 0.2 9 0.04n of customers currently in the market who did not pur- chase a doll last year will purchase a doll from Play Things this year. Adding these two provides the mean sales for this year. Then the actual sales this year is normally distributed with this mean and standard de- viation equal to 7.5% of the mean.

a. Use @RISK to estimate the expected NPV of this project. Use a discount rate of 7%.

b. Use the percentiles in @RISK’s output to find an interval such that you are 95% certain that the compa- ny’s actual NPV will be within this interval.

54. An automobile manufacturer is considering whether to introduce a new model called the Racer. The profitabil- ity of the Racer depends on the following factors:

• The fixed cost of developing the Racer is triangularly distributed with parameters $3, $4, and $5, all in billions.

• Year 1 sales are normally distributed with mean 200,000 and standard deviation 50,000. Year 2 sales are normally distributed with mean equal to actual year 1 sales and standard deviation 50,000. Year 3 sales are normally distributed with mean equal to actual year 2 sales and standard deviation 50,000.

• The selling price in year 1 is $25,000. The year 2 selling price will be 1.053year 1 price 1 $50 (% diff1)4 where % diff1 is the number of percentage points by which actual year 1 sales differ from expected year 1 sales. The 1.05 factor accounts for inflation. For example, if the year 1 sales figure is 180,000, which is 10 percentage points below the expect- ed year 1 sales, then the year 2 price will be 1.05325,000 1 50( 2 10)4 5 $25,725. Similarly, the year 3 price will be 1.053year 2 price 1 $50(% diff2)4 where % diff2 is the percentage by which actual year 2 sales differ from expected year 2 sales.

• The variable cost in year 1 is triangularly distributed with parameters $10,000, $12,000, and $15,000, and it is assumed to increase by 5% each year.

Your goal is to estimate the NPV of the new car during its first three years. Assume that the company is able to produce exactly as many cars as it can sell. Also, assume that cash flows are discounted at 7.5%. Simulate 1000 trials to estimate the mean and standard deviation of the NPV for the first three years of sales. Also, determine an interval such that you are 95% certain that the NPV of the Racer during its first three years of operation will be within this interval.

55. It costs a pharmaceutical company $40,000 to produce a 1000-pound batch of a drug. The average yield from a batch is unknown but the best case is 90% yield (that is, 900 pounds of good drug will be produced), the most

a. Estimate the mean and standard deviation of the NPV of this project. Assume that cash flows are discounted at a rate of 7% per year.

b. Now assume that the project has an abandon- ment option. At the end of each year you can abandon the project for the value given in the file P16_52.xlsx. For example, suppose that year 1 cash flow is $4000. Then at the end of year 1, you expect cash flow for each remaining year to be $4000. This has an NPV of less than $62,000, so you should abandon the project and collect $62,000 at the end of year 1. Estimate the mean and standard deviation of the project with the abandonment option. How much would you pay for the abandonment option? (Hint: You can abandon a project at most once. So in year 5, for example, you abandon only if the sum of future expected NPVs is less than the year 5 abandonment value and the project has not yet been abandoned. Also, once you abandon the project, the actual cash flows for future years are zero. So in this case the future cash flows after abandonment should be zero in your model.)

53. Play Things is developing a new Miley Cyrus doll. The company has made the following assumptions:

• The doll will sell for a random number of years from 1 to 10. Each of these 10 possibilities is equally likely.

• At the beginning of year 1, the potential market for the doll is one million. The potential market grows by an average of 5% per year. The company is 95% sure that the growth in the potential market during any year will be between 3% and 7%. It uses a normal distribution to model this.

• The company believes its share of the potential mar- ket during year 1 will be at worst 20%, most likely 40%, and at best 50%. It uses a triangular distribution to model this.

• The variable cost of producing a doll during year 1 has a triangular distribution with parameters $8, $10, and $12.

• The current selling price is $20. • Each year, the variable cost of producing the doll will

increase by an amount that is triangularly distributed with parameters 4.5%, 5%, and 6.5%. You can assume that once this change is generated, it will be the same for each year. You can also assume that the company will change its selling price by the same percentage each year.

• The fixed cost of developing the doll (which is incurred right away, at time 0) has a triangular distri- bution with parameters $4, $6, and $12 million.

• There is currently one competitor in the market. Dur- ing each year that begins with four or fewer competi- tors, there is a 20% chance that a new competitor will enter the market.

• Year t sales (for t 7 1) are determined as follows.

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16-6 Conclusion 8 3 3

and b 5 0.5. (This implies a mean of one year.) Use @ RISK to simulate the six-year period. Include as outputs (1) your total cost, (2) the number of failures during the warranty period, and (3) the number of devices you own during the six-year period.

58. Work the previous problem for a case in which the one- year warranty requires you to pay for the new device even if failure occurs during the warranty period. Spe- cifically, if the device fails at time t, measured relative to the time it went into use, you must pay $300t for a new device. For example, if the device goes into use at the beginning of April and fails nine months later, at the beginning of January, you must pay $225. The reasoning is that you got 9>12 of the warranty period for use, so you should pay that fraction of the total cost for the next device. As before, however, if the device fails outside the warranty period, you must pay the full $300 cost for a new device.

59. Based on Hoppensteadt and Peskin (1992). The following model (the Reed–Frost model) is often used to model the spread of an infectious disease. Suppose that at the begin- ning of period 1, the population consists of five diseased people (called infectives) and 95 healthy people (called susceptibles). During any period there is a 0.05 probabil- ity that a given infective person will encounter a particu- lar susceptible. If an infective encounters a susceptible, there is a 0.5 probability that the susceptible will contract the disease. An infective lives for an average of 10 peri- ods with the disease. To model this, assume that there is a 0.10 probability that an infective dies during any given period. Use @RISK to model the evolution of the popu- lation over 100 periods. Use your results to answer the following questions. [Hint: During any period there is probability 0.05(0.50) 5 0.025 that an infective will infect a particular susceptible. Therefore, the probability that a particular susceptible is not infected during a period is (1 2 0.025)n, where n is the number of infectives pres- ent at the end of the previous period.] a. What is the probability that the population will

die out? b. What is the probability that the disease will die out? c. On the average, what percentage of the population is

infected by the end of period 100? d. Suppose that people use infection “protection” during

encounters. The use of protection reduces the prob- ability that a susceptible will contract the disease during a single encounter with an infective from 0.50 to 0.10. Now answer parts a through c under the assumption that everyone uses protection.

60. Chemcon has taken over the production of Nasacure from a rival drug company. Chemcon must build a plant to produce Nasacure by the beginning of 2010. Once the plant is built, the plant’s capacity cannot be changed. Each unit sold brings in $10 in revenue. The fixed cost (in dollars) of producing a plant that can produce x units per year of the drug is 5,000,000 1 10x. This cost

likely case is 85% yield, and the worst case is 70% yield. The annual demand for the drug is unknown, with the best case being 22,000 pounds, the most likely case 18,000 pounds, and the worst case 12,000 pounds. The drug sells for $60 per pound and leftover amounts of the drug can be sold for $8 per pound. To maximize annual expected profit, how many batches of the drug should the company produce? You can assume that it will pro- duce the batches only once, before demand for the drug is known.

56. A truck manufacturer produces the Off Road truck. The company wants to gain information about the discounted profits earned during the next three years. During a given year, the total number of trucks sold in the United States is 500,000 1 50,000G 2 40,000I, where G is the number of percentage points increase in gross domestic product during the year and I is the number of percent- age points increase in the consumer price index during the year. During the next three years, Value Line has made the predictions listed in the file P16_56.xlsx. In the past, 95% of Value Line’s G predictions have been accurate within 6%, and 95% of Value Line’s I predic- tions have been accurate within 5%. You can assume that the actual G and I values are normally distributed each year.

At the beginning of each year, a number of competitors might enter the trucking business. The probability distri- bution of the number of competitors that will enter the trucking business is also given in the same file. Before competitors join the industry at the beginning of year 1, there are two competitors. During a year that begins with n competitors (after competitors have entered the business, but before any have left, and not counting Off Road), Off Road will have a market share given by 0.5(0.9)n. At the end of each year, there is a 20% chance that any competitor will leave the industry. The selling price of the truck and the production cost per truck are also given in the file. Simulate 1000 replications of the company’s profit for the next three years. Estimate the mean and standard deviation of the discounted three- year profits, using a discount rate of 8% and Excel’s NPV function. Do the same if the probability that any competitor leaves the industry during any year increases to 50%.

57. Suppose you buy an electronic device that you operate continuously. The device costs you $300 and carries a one-year warranty. The warranty states that if the device fails during its first year of use, you get a new device for no cost, and this new device carries exactly the same war- ranty. However, if it fails after the first year of use, the warranty is of no value. You plan to use this device for the next six years. Therefore, any time the device fails out- side its warranty period, you will pay $300 for another device of the same kind. (We assume the price does not increase during the six-year period.) The time until failure for a device is gamma distributed with parameters a 5 2

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62. You are unemployed, 21 years old, and searching for a job. Until you accept a job offer, the following situation occurs. At the beginning of each year, you receive a job offer. The annual salary associated with the job offer is equally likely to be any number between $20,000 and $100,000. You must immediately choose whether to accept the job offer. If you accept an offer with salary $x, you receive $x per year while you work (assume you retire at age 70), including the current year. Assume that cash flows are discounted so that a cash flow received one year from now has a present value of 0.93. You decide to accept the first job offer that exceeds w dollars. a. Use simulation to determine the value of w (within

$10,000) that maximizes the expected NPV of earn- ings you will receive the rest of your working life.

b. Repeat part a, but now assume that you get a 3% raise in salary every year after the first year you accept the job.

63. A popular restaurant in Indianapolis does a brisk busi- ness, filling virtually all of its seats from 6 p.m. until 9 p.m. Tuesday through Sunday. Its current annual revenue is $2.34 million. However, it does not currently accept credit cards, and it is thinking of doing so. If it does, the bank will charge 4% on all receipts during the first year. (To keep it simple, you can ignore taxes and tips and focus only on the receipts from food and liquor.) Depending on receipts in year 1, the bank might then reduce its fee in succeeding years, as indicated in the file P16_63.xlsx. (This would be a one-time reduction, at the end of year 1 only.) This file also contains parameters of the two uncertain quantities, credit card usage (percent- age of customers who will pay with credit cards) and increased spending (percentage increase in spending by credit card users, presumably on liquor but maybe also on more expensive food). The restaurant wants to sim- ulate a five-year horizon. Its base case is not to accept credit cards at all, in which case it expects to earn $2.34 million in revenue each year. It wants to use simula- tion to explore other options, where it will accept credit cards in year 1 and then continue them in years 2–5 if the bank fee is less than or equal to some cutoff value. For example, one possibility is to accept credit cards in year 1 and then discontinue them only if the bank fee is less than or equal to 3%. You should explore the cutoffs 2% to 4% in increments of 0.5%. Which policy provides with the largest mean increase in revenue over the five- year horizon, relative to never using credit cards?

64. The Ryder Cup is a three-day golf tournament played every other year with 12 of the best U.S. golfers against 12 of the best European golfers. They play 16 team matches (each match has two U.S. golfers against two European golfers) on Friday and Saturday, and they play 12 singles matches (each match has a single U.S. golfer against a European golfer) on Sunday. Each match is either won or tied. A win yields 1 point for the winning team and 0 points for the losing team. A tie yields 0.5

is assumed to be incurred at the end of 2010. In fact, you can assume that all cost and sales cash flows are incurred at the ends of the respective years. If a plant of capacity x is built, the variable cost of producing a unit of Nasacure is 6 2 0.1(x 2 1,000,000)>100,000. For example, a plant capacity of 1,100,000 units has a vari- able cost of $5.90. Each year a plant operating cost of $1 per unit of capacity is also incurred. Based on a fore- casting sales model from the previous 10 years, Chem- con forecasts that demand in year t, Dt, is related to the demand in the previous year, Dt 2 1, by the equation Dt 5 67,430 1 0.985Dt 2 1 1 et, where et is normally distributed with mean 0 and standard deviation 29,320. The demand in 2009 was 1,011,000 units. If demand for a year exceeds production capacity, all demand in excess of plant capacity is lost. If demand is less than capacity, the extra capacity is simply not used. Chemcon wants to determine a capacity level that maximizes expected discounted profits (using a discount rate of 7.5%) for the time period 2010 through 2019. Use simulation to help it do so.

61. Tinkan Company produces one-pound cans for the Canadian salmon industry. Each year the salmon spawn during a 24-hour period and must be canned immedi- ately. Tinkan has the following agreement with the salmon industry. The company can deliver as many cans as it chooses. Then the salmon are caught. For each can by which Tinkan falls short of the salmon industry’s needs, the company pays the industry a $2 penalty. Cans cost Tinkan $1 to produce and are sold by Tinkan for $2 per can. If any cans are left over, they are returned to Tinkan and the company reimburses the industry $2 for each extra can. These extra cans are put in storage for next year. Each year a can is held in storage, a car- rying cost equal to 20% of the can’s production cost is incurred. It is well known that the number of salmon harvested during a year is strongly related to the number of salmon harvested the previous year. In fact, using past data, Tinkan estimates that the harvest size in year t, Ht (measured in the number of cans required), is related to the harvest size in the previous year, Ht 2 1, by the equa- tion Ht 5 Ht 2 1et, where et is normally distributed with mean 1.02 and standard deviation 0.10.

Tinkan plans to use the following production strategy. For some value of x, it produces enough cans at the beginning of year t to bring its inventory up to x 1 Ĥt, where Ĥt is the predicted harvest size in year t. Then it delivers these cans to the salmon industry. For exam- ple, if it uses x 5 100,000, the predicted harvest size is 500,000 cans, and 80,000 cans are already in inventory, then Tinkan produces and delivers 520,000 cans. Given that the harvest size for the previous year was 550,000 cans, use simulation to help Tinkan develop a produc- tion strategy that maximizes its expected profit over the next 20 years. Assume that the company begins year 1 with an initial inventory of 300,000 cans.

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probability 0.475 and won with probability 0.525. (These are the historical fractions of holes that have been tied and won in singles matches in the past few Ryder Cups.) Note that each match is “match play,” so the only thing that counts on each hole is whether a golfer has fewer strokes than the other golfer— winning a hole by one stroke is equivalent to winning the hole by two or more strokes in match play. The player win- ning the most holes wins the match, unless they tie.

b. Run another simulation, using the estimated prob- ability h as an input, to estimate the probability that the U.S. will score at least 8.5 points in the 12 singles matches.

point for each team. A team needs 14.5 points to win the Cup. If each team gets 14 points, the tournament is a tie, but the preceding winner gets to keep the Cup. In 1999, the U.S. was behind 10 points to 6 after the team matches. To win the Cup, the U.S. needed at least 8.5 points on Sunday, a very unlikely outcome, but they pulled off the miracle and won. Use simulation to estimate the probability of the U.S. scoring at least 8.5 points in the 12 singles matches, assuming all golfers in the tournament are essentially equal. Proceed as follows. a. Use simulation to estimate the probability, call it h

(for half), that a given match ends in a tie. To do this, you can assume that any of the 18 holes is tied with

CASE 16.1 College Fund Investment Your next-door neighbor, Scott Jansen, has a 12-year-old daughter, and he intends to pay the tuition for her first year of college six years from now. The tuition for the first year will be $22,500. Scott has gone through his budget and finds that he can invest $3000 per year for the next six years. Scott has opened accounts at two mutual funds. The first fund fol- lows an investment strategy designed to match the return of the S&P 500. The second fund invests in 3-month Treasury bills. Both funds have very low fees.

Scott has decided to follow a strategy in which he contributes a fixed fraction of the $3000 to each fund. An adviser from the first fund suggested that in each year he should invest 80% of the $3000 in the S&P 500 fund and the other 20% in the T-bill fund. The adviser explained that the S&P 500 has averaged much larger returns than the T-bill fund. Even though stock returns are risky investments in the short run, the risk should be fairly minimal over the longer six-year period. An adviser from the second fund recom- mended just the opposite: invest 20% in the S&P 500 fund and 80% in T-bills, because treasury bills are backed by the United States government. If you follow this allocation, he said, your average return will be lower, but at least you will have enough to reach your $22,500 target in six years.

Not knowing which adviser to believe, Scott has come to you for help.

Questions 1. The file C16_01.xlsx contains annual returns of the

S&P 500 and 3-month Treasury bills from 1960. Sup- pose that in each of the next 72 months (six years), it is equally likely that any of the historical returns will occur. Develop a spreadsheet model to simulate the two suggested investment strategies over the six-year period. Plot the value of each strategy over time for a single iter- ation of the simulation. What is the total value of each strategy after six years? Do either of the strategies reach the target?

2. Simulate 1000 iterations of the two strategies over the six-year period. Create a histogram of the final fund val- ues. Based on your simulation results, which of the two strategies would you recommend? Why?

3. Suppose that Scott needs to have $25,000 to pay for the first year’s tuition. Based on the same simulation results, which of the two strategies would you recommend now? Why?

4. What other real-world factors might be important to con- sider in designing the simulation and making a recom- mendation?

CASE 16.2 Bond Investment Strategy An investor is considering the purchase of zero-coupon U.S. Treasury bonds. A 30-year zero-coupon bond yielding 8% can be purchased today for $9.94. At the end of 30 years, the owner of the bond will receive $100. The yield of the bond is related to its price by the following equation:

P 5 100

(1 1 y)t

Here, P is the price of the bond, y is the yield of the bond, and t is the maturity of the bond measured in years. Evalu- ating this equation for t 5 30 and y 5 0.08 gives P 5 9.94.

The investor is planning to purchase a bond today and sell it one year from now. The investor is interested in eval- uating the return on the investment in the bond. Suppose, for example, that the yield of the bond one year from now is 8.5%. Then the price of the bond one year later will be

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$9.39 [5100/(1 1 0.085)29]. The time remaining to maturity is t 5 29 because one year has passed. The return for the year is 25.54% [= (9.39 2 9.94)/9.944.

In addition to the 30-year-maturity zero-coupon bond, the investor is considering the purchase of zero-coupon bonds with maturities of 2, 5, 10, or 20 years. All of the bonds are currently yielding 8.0%. (Bond investors describe this as a flat yield curve.) The investor cannot predict the future yields of the bonds with certainty. However, the investor believes that the yield of each bond one year from now can be modeled by a normal distribution with mean 8% and standard deviation 1%.

Questions 1. Suppose that the yields of the five zero-coupon bonds

are all 8.5% one year from today. What are the returns of each bond over the period?

2. Using a simulation with 1000 iterations, estimate the expected return of each bond over the year. Estimate the standard deviations of the returns.

3. Comment on the following statement: “The expected yield of the 30-year bond one year from today is 8%. At that yield, its price would be $10.73. The return for the year would be 8% [= (10.73 2 9.94)/9.94]. Therefore, the average return for the bond should be 8% as well. A simulation isn’t really necessary. Any difference between 8% and the answer in Question 2 must be due to simulation error.”

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