Homework
Obaid10
EGR202EconomicAnalysisLecture1.pptxEGR 202
Product Development
Engineering Economics and Decision Making
April 3
Engineering Economics and Decision Making
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The “dollars and cents” behind an engineering decision.
Helps to make decisions such as, “Should I invest in this project?”
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Source for much of the material in these slides:
Chan S. Park
Fundamentals of Engineering Economics
Source for today’s material.
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What you’ll be learning…
Basics of Time Value of Money.
The language of engineering economics:
P, i, n, N, F
How to solve problems using Compound Interest.
Two classic methods for evaluating economic feasibility of projects.
How to account for risk and uncertainty.
Overall, gain an understanding how time and uncertainty are defining aspects of any engineering economic decision and learn one means for dealing with these aspects.
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What you’ll be learning…
Basics of Time Value of Money.
The language of engineering economics:
P, i, n, N, F
How to solve problems using Compound Interest.
Two classic methods for evaluating economic feasibility of projects.
How to account for risk and uncertainty.
Overall, gain an understanding how time and uncertainty are defining aspects of any engineering economic decision and learn one means for dealing with these aspects.
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Time Value of Money
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Purchasing Power changes over time due to inflation (or deflation).
Earning Power changes because it can earn more money over time due to interest.
Time value of money is measured in terms of market interest rate.
Market interest rate includes gain in earning power and loss in purchasing power.
This is a two-edged sword whereby earnings grow (due to interest) but purchasing power decreases (due to inflation) as time goes on.
Start by thinking of ‘interest’ in general.
Money in a savings account yields a balance over time that is greater than sum of the deposits.
Borrowing money to buy a car means repaying an amount over time that is greater than the amount borrowed.
Interest is the cost of having
money available for use.
We are all familiar with this concept.
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Less familiar……
In the financial world,
money itself is a commodity (a marketable item or good) and,
like other goods that are bought and sold, money costs money.
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If money can be bought and sold, how much does money cost?
The market interest rate, quoted by financial institutions, establishes and measures the cost of money.
Takes into account the earning power, as well as the effect of purchasing power (inflation) perceived in the marketplace.
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What you’ll be learning…
Basics of Time Value of Money.
The language of engineering economics:
P, i, n, N, F
How to solve problems using Compound Interest.
Two classic methods for evaluating economic feasibility of projects.
How to account for risk and uncertainty.
Overall, gain an understanding how time and uncertainty are defining aspects of any engineering economic decision and learn one means for dealing with these aspects.
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Elements of Transactions….
Whether we are borrowing or lending money, there are some common elements in the transactions.
Common Elements form the
language of
‘Time Value of Money’
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Principal, P
An initial amount of money in a transaction at time zero.
Also referred to as present value (PV) or present worth (PW).
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Interest Rate, i
The interest rate per interest period.
Measures the cost (or price) of money.
Expressed as a % per period of time.
Usually quoted as annual percentage rate (APR)
Unless otherwise specified, i reflects both the earning and purchasing power (market interest rate)
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Interest Period, N
N = The total number of interest periods.
A specified length of time that marks the duration of the transaction – establishing the number of interest periods
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Future Amount of Money, F
F = A future sum of money at the end of the analysis period.
Results from the cumulative effects of the interest rate over a number of interest periods.
May be specified as (FN)
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Equal Amounts, A
A = A series of equal amounts, received or disbursed.
Example: monthly rental payments.
Often used to represent annual benefits or annual costs.
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Diagramming Equivalent
Free Body Diagrams
Engineering Mechanics
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i = 5%
0 1 2 3 4 5
$ 100
Years
F
N
P
Engineering Economics Cash Flow Diagram
n
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Cash Flow Diagram
Positive Flows
Receipts
Negative Flows
Disbursements
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i = 5%
0 1 2 3 4 5
$ 100
Years
F
N
P
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Arrows represents NET cash flows
$1,500
$500
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In this example, at Time 0, we paid out $1,500 and received $500. This is equivalently depicted as a net ‘disbursement’ of $1,000.
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Try it yourself….
Year 0 (today) Pay $1,000 for a machine
Year 4 Pay $750 for maintenance on machine
Year 4 Receive $500 in revenue from products made on machine
Year 7 Sell machine for $50
Assume an interest rate of 5%.
Construct the cash flow diagram for the following series of cash flows.
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And the cash flow diagram looks like……
0
1
2
3
4
5
6
7
$1,000
$250
$50
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i = 5%
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Think back to our double-edged…
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This means that a dollar today is worth more than a dollar tomorrow.
This worth is a function of market interest rate.
How does market interest rate affect worth?
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Methods of Calculating Interest
Simple Interest
Interest is earned on only the principal amount during each interest period.
Compound Interest (used in engineering economy)
Interest is earned on the principal amount plus the accumulated interest.
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Simple Interest
Simple interest is interest earned on only the principal amount during each interest period.
Suppose we deposit $1,000 in a bank at an interest rate of 9%. Using simple interest, how much will we have at the end of year 3?
P = $1,000
i = 10% (simple)
N = 3
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To solve, what should we do first?
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Cash Flow Diagram
0
1
2
3
P= $1,000
F= $ ?
i = 10% simple
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Simple Interest = (iP)
Interest earned the first year:
(interest rate) x Principal amount
= 10% x $1,000
= 0.10 x $1,000
= $100
Interest earned the second year:
(interest rate) x Principal amount
= 10% x $1,000
= 0.10 x $1,000
= $100
Interest earned the third year:
(interest rate) x Principal amount
= 10% x $1,000
= 0.10 x $1,000
= $100
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Cash Flow Diagram
0
1
2
3
P=$1,000
F=$1,300
i = 10% simple
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Ending Value (F)
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Let’s go back to our example….
Suppose we deposit $1000 in a bank at an interest rate of 10%. Using compound interest, how much will we have at the end of year 3?
P = $1,000
i = 10% (compound)
N = 3
Compound Interest - Interest is earned on the
principal amount plus the accumulated interest.
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Cash Flow Diagram
0
1
2
3
P = $1,000
F = $ ?
i = 10% compound
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Compound Interest – Earn interest on the principal and the interest
Interest earned the first year:
(interest rate) x Principal amount
= 10% x $1,000
= 0.10 x $1,000
= $100
Interest earned the second year:
(interest rate) x (Principal amount+ any interest earned)
= 10% x $1,100
= 0.10 x $1,100
= $110
Interest earned the third year:
(interest rate) x (Principal amount + any interest earned)
= 10% x $1,210
= 0.10 x $1,210
= $121
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Cash Flow Diagram
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0
10
$1,000
F=$1,331
i = 10% compound
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Future value = Principle + Interest Earned
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+
=
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Future value = Principle + Interest Earned
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+
=
Now that we see the pattern,
can we find general equation?
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Fundamental Law of Engineering Economics
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What you’ll be learning…
Basics of Time Value of Money.
The language of engineering economics:
P, i, n, N, F
How to solve problems using Compound Interest.
Two classic methods for evaluating economic feasibility of projects.
How to account for risk and uncertainty.
Overall, gain an understanding how time and uncertainty are defining aspects of any engineering economic decision and learn one means for dealing with these aspects.
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So now calculate using compound interest formula.
F = P(1+i)N
P = $10,000
i = 4% (compound)
N = 10
F = $10,000 (1 +.04)10 = $14,802
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Cash Flow Diagram
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0
10
$10,000
$14,802
i = 4% compound
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Practice Problem 1
If you deposit $1,000 now (n = 0) and $2,000 two years from now (n = 2) in a savings account that pays 3% compound interest, how much would you have at the end of year 10?
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Cash Flow Diagram
0 1 2 3 4 5 6 7 8 9 10
$1,000
$2,000
F
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i=3%
F=$3,877.46
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Practice Problem 2
If you deposit $12,000 now (n = 0) and $5,000 three years from now (n = 3) in a savings account that pays 5% compound interest, how much would you have at the end of year 15?
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Cash Flow Diagram
0 1 2 3 4 5 15
$12,000
$5,000
F
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i=5%
F=$33,926.28
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Practice Problem 3
If you deposit $8,000 now (n = 0) and $5,000 two years from now (n = 2). Suppose you withdraw $3,000 four years from now. The money is in a savings account that pays 4% compound interest, how much would you have at the end of year 10?
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Cash Flow Diagram
0 1 2 3 4 5 6 7 8 9 10
$8,000
$5,000
F
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i=4%
$3,000
$14,888
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So how does this relate to your EGR202 project?
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You and/or an outside investor need(s) to determine WHETHER OR NOT TO INVEST in your project and time value of money is at the foundation of that decision.
You need to know the economic feasibility of your project.
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What you’ll be learning…
Basics of Time Value of Money.
The language of engineering economics:
P, i, n, N, F
How to solve problems using Compound Interest.
Two classic methods for evaluating economic feasibility of projects.
How to account for risk and uncertainty.
Overall, gain an understanding how time and uncertainty are defining aspects of any engineering economic decision and learn one means for dealing with these aspects.
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Economic Feasibility
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Two common approaches
Payback Period
Net Present Worth
Before we begin, we need another term.
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Minimum Acceptable Rate of Return or Minimally Attractive Rate of Return
Investment Pool
Minimal Attractive Rate of Return
(MARR)
If you are the investor, you need to decide whether to take money out of savings or borrow money to make the investment. You won’t do it unless it is likely to make a certain % in return.
Companies have ‘investment pools’, meaning that money taken from the
Investment Pool
needs to earn some established % in order for them to invest in the project.
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1) Initial Project Screening Method Payback Period
A measure of liquidity.
The time it will take to get back what it cost.
The time to recoup our investment.
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Payback Period
Conventional payback method (goes back to the 1950’s)
Ignores time value of money
Discounted payback method
Considers time value of money
Compare to an ‘acceptable period of time’
established by management policy
(We’ll use this one in class.)
In-Class Work
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| Year | A | B |
| 0 | -1000 | -3000 |
| 1 | 200 | 800 |
| 2 | 200 | 900 |
| 3 | 1200 | 1200 |
| 4 | 1200 | 1200 |
| 5 | 1200 | 5000 |
Assume Projects A and B have these expected cash flows over the next 5 years.
Find the Payback Period of Project A and B using the Discounted Method. Assume ‘cost of funds’ is 15%.
-1000 – 150 + 200 = -950
-950 – 142.5 + 200 = -892.5
-892.5 -133.875 +1200 = 173.625
Project A
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| Project A | |||
| Discounted Payback | |||
| Year | Cash Flow | Cost of Funds (15%) | Cum Cash Flow |
| 0 | -1000 | 0 | -1000 |
| 1 | 200 | (-1000 X 0.15)= -150 | -950 |
| 2 | 200 | (-950 X 0.15)= -142.5 | -892.5 |
| 3 | 1200 | (-892.5 X 0.15)= -133.875 | 173.625 |
| 4 | 1200 | (173.625 X 0.15)= 26.044 | 1399.67 |
| 5 | 1200 | (1399.67 X 0.15)= 209.95 | 2809.62 |
Our money is tied up, losing it’s MARR potential.
Cost of the funds used to support the project.
Assume in this case it is 15%
Project A
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| Project A | |||
| Discounted Payback | |||
| Year | Cash Flow | Cost of Funds (15%) | Cum Cash Flow |
| 0 | -1000 | 0 | -1000 |
| 1 | 200 | (-1000 X 0.15)= -150 | -950 |
| 2 | 200 | (-950 X 0.15)= -142.5 | -892.5 |
| 3 | 1200 | (-892.5 X 0.15)= -133.875 | 173.625 |
| 4 | 1200 | (173.625 X 0.15)= 26.044 | 1399.67 |
| 5 | 1200 | (1399.67 X 0.15)= 209.95 | 2809.62 |
Payback is between Years 2 and 3
Payback Period
0 1 2 3 4 5 -1000 -950 -892.5 173.625 1399.67 2809.62
Years
Cumulative Cash Flow ($)
Project B – work on your own.
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Payback between Year 4 and Year 5
| Project B | |||
| Discounted Payback | |||
| Year | Cash Flow | Cost of Funds (15%) | Cum Cash Flow |
| 0 | -3000 | 0 | -3000 |
| 1 | 800 | (-3000 X 0.15) = -450 | -2650 |
| 2 | 900 | (-2650 X 0.15) = -397.5 | -2147.5 |
| 3 | 1200 | (-2147.5 X 0.15) = -322.125 | -1269.625 |
| 4 | 1200 | (-1269.625 X 0.15) = -190.444 | -260.07 |
| 5 | 5000 | (-260.07 X 0.15) = -39.01 | 4700.92 |
Payback Period
Cumulative Cash Flow
0 1 2 3 4 5 -3000 -2650 -2147.5 -1269.625 -260.07 4700.92
Years
Cumulative CAsh Flow ($)
How does management select an ‘acceptable payback period’?
Consider the type of investment…..
Product with a long life cycle
vs.
High tech product with short life cycle
In the case of a high tech product, the product may be obsolete in a short time period. In that case, management would require a short payback period.
| Project B | ||
| Discounted Payback | ||
| Year | Cash Flow | Cum Cash Flow |
| 0 | -3000 | -3000 |
| 1 | 800 | -2650 |
| 2 | 900 | -2147.5 |
| 3 | 1200 | -1269.625 |
| 4 | 1200 | -260.07 |
| 5 | 5000 | 4700.92 |
What would we decide based on Payback….
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| Project A | ||
| Discounted Payback | ||
| Year | Cash Flow | Cum Cash Flow |
| 0 | -1000 | -1000 |
| 1 | 200 | -950 |
| 2 | 200 | -892.5 |
| 3 | 1200 | 173.625 |
| 4 | 1200 | 1399.67 |
| 5 | 1200 | 2809.62 |
Payback
Between 2 and 3 years
Payback
Between 4 and 5 years
Based solely on Payback, we would choose Project A
But clearly that overlooks all money earned or lost AFTER payback period
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Final comments on payback period
Good screening tool for liquidity.
Lacks a measure of investment worth for profitabili
So where do we go from here?
In addition to Payback period, we need a method that considers all cash flow elements involved in our project.
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2) Net Present Worth Analysis
Present value of all the cash inflows
minus the
Present value of all the cash outflows
Net Present Worth or Net Present Value
What criterion should we use?
If NPW(i) > 0, then accept project.
But why NPW(i) > 0 ?
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Because of the i used.
(i ) = interest rate the firm wants to earn on its investments.
Represents the rate that the firm can invest the money in its investment pool.
We call that i, the MARR (the minimum attractive rate of return)
NPW
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In essence,
Determining whether the anticipated cash inflows from a proposed project are sufficiently attractive to invest funds in the projects.
Our process….
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Determine our MARR
Estimate the service life of the project
Estimate the cash inflow for each period over the service life
Estimate the cash outflow for each period over the service life
Find the present worth of the cash inflow and cash outflow
Compare to NPW criterion (NPW > 0)
PW(i)=NPW(i)
If PW(i) > 0, accept the investment
If PW(i) = 0, remain indifferent
If PW(i) < 0, reject the investment
Example: Make and Sell Widgets.
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Determine our MARR
Many companies use a MARR of 12%.
This varies between investors, but we’ll use 12% for this analysis.
Estimate the service life of the project
Assume we plan to make widgets for 8 years.
You need to estimate the service life for your project.
Example: Make and Sell Widgets
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3. Estimate the cash inflow for each period over the service life.
Cash inflow is the money we will make from selling our product.
This is a function of demand (units sold) X selling price.
Demand is forecasted (usually by marketing and sales functions)
Selling price (can be estimated by similar products in the marketplace)
Forecasted Demand
Year 0
Year 1 100 widgets
Year 2 Double Year 1 = 200 widgets
Year 3` Double Year 2 = 400 widgets
Years 4 – 8 Mass production mode = 1500 widgets per year
Estimated Selling price
Year 0 and Year 1 $300 per widget
Years 2 through 8 $250 per widget
You need to estimate demand and selling price of your product over the life of your project.
Example: Make and Sell Widgets
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4. Estimate the cash outflow for each period over the service life
Year 0 (today) $100,000 for initial product development effort.
Year 1 $50,000 more for development effort during Year 1.
Year 1 $25,000 for equipment and tooling
Year 3 $45,000 for equipment and tooling
Year 4 $10,000 for equipment and tooling
Year 1 – Material cost is $1.25 per widget / labor cost $1.00 per widget
Year 2 – Material cost is $0.80 per widget / labor cost $0.90 per widget
Years 3 to 8 – Material cost is $0.64 per widget / labor cost is $0.15 per widget
Use forecasting and estimation techniques for material and labor costs.
Another option – use online costing sites (next slide)
Example: Make and Sell Widgets
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4. Estimate the cash outflow for each period over the service life
Many online sources for costing parts.
Example: Make and Sell Widgets
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5. Find the present worth of the cash inflow and cash outflow
Construct a cash flow diagram as shown on next slide.
We need a cash flow diagram of anticipated cash flows for life of project.
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| Year 0 (Today) | Year 1 | Year 2 | Year 3 | Year 4 | Year 5 | Year 6 | Year 7 | Year 8 | |
| Sales Revenue | |||||||||
| Sales Volume, (units) | 100 | 200 | 400 | 1500 | 1500 | 1500 | 1500 | 1500 | |
| Unit Sales Price ($/unit) | $300 | $300 | $250 | $250 | $250 | $250 | $250 | $250 | |
| Total Revenue | $30,000 | $60,000 | $100,000 | $375,000 | $375,000 | $375,000 | $375,000 | $375,000 | |
| Product Development | $100,000 | $75,000 | $50,000 | ||||||
| Equipment and Tooling | $25,000 | $0 | $45,000 | $10,000 | |||||
| Marketing / Sales | $10,000 | $10,000 | $10,000 | $10,000 | $10,000 | $10,000 | $10,000 | $10,000 | |
| Cost of Goods Sold (Variable Costs) | |||||||||
| Material ($/unit) | $1.25 | $0.80 | $0.64 | $0.64 | $0.64 | $0.64 | $0.64 | $0.64 | |
| Labor ($/unit) | $1.00 | $0.90 | $0.15 | $0.15 | $0.15 | $0.15 | $0.15 | $0.15 | |
| Cost of Goods Sold (Variable Costs) | $225 | $340 | $316 | $1,186 | $1,186 | $1,186 | $1,186 | $1,186 | |
| Total Costs | $100,000 | $110,225 | $60,340 | $55,316 | $11,186 | $21,186 | $11,186 | $11,186 | $11,186 |
| Period Cash Flow | -$100,000 | -$80,225 | -$340 | $44,684 | $363,814 | $353,814 | $363,814 | $363,814 | $363,814 |
0 1 2
$100,000
$80,225
$340
$44,684
$363,814
$353,814
$363,814
$363,814
$363,814
EXAMPLE
3 4 5 6 7 8
Example: Make and Sell Widgets
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5. Find the present worth of the cash inflow and cash outflow
Using the fundamental law of engineering economics, solve for P.
P is the present value of a future amount.
=
Example: Make and Sell Widgets
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5. Find the present worth of the cash inflow and cash outflow
0 1 2
$100,000
$80,225
$340
$44,684
$363,814
$353,814
$363,814
$363,814
$363,814
3 4 5 6 7 8
=
$787,707
Example: Make and Sell Widgets
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6. Compare to NPW criterion (NPW > 0)
PW(i)=NPW(i)
If PW(i) > 0, accept the investment
If PW(i) = 0, remain indifferent
If PW(i) < 0, reject the investment
$787,707 > 0, so accept the investment
But remember these numbers are based on estimates and forecasts.
There is risk and uncertainty in this decision.
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What you’ll be learning…
Basics of Time Value of Money.
The language of engineering economics:
P, i, n, N, F
How to solve problems using Compound Interest.
Two classic methods for evaluating economic feasibility of projects.
How to account for risk and uncertainty.
Overall, gain an understanding how time and uncertainty are defining aspects of any engineering economic decision and learn one means for dealing with these aspects.
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Engineering Decisions vs. Economic Decisions
Engineering Design Decisions
Much of what we use in our decisions is known and doesn’t change with time
Known physical properties
Principles of physics, chemistry
Engineering Economic Decisions
Much of what we use in our decisions is unknown and changes with time
Forecasts of demand (sales)
Forecasts of sales prices
Forecasts of material costs
Known and
Doesn’t change with time
Unknown and
Changes with time
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Engineering Decisions vs. Economic Decisions
Engineering Design Decisions
Much of what we use in our decisions is known and doesn’t change with time
Known physical properties
Principles of physics, chemistry
Engineering Economic Decisions
Much of what we use in our decisions is unknown and changes with time
Forecasts of demand (sales)
Forecasts of sales prices
Forecasts of material costs
Known and
Doesn’t change with time
Unknown and
Changes with time
Risk and Uncertainty affect our Decisions
What can we do to minimize this risk?
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Example: Make and Sell Widgets
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Forecasted Demand
Year 0
Year 1 100 widgets
Year 2 Double Year 1 = 200 widgets
Year 3`Double Year 2 = 400 widgets
Years 4 – 8 Mass production mode = 1500 widgets per year
One area of ‘uncertainty’ is the demand we forecasted.
Let’s look at this in more detail. Let’s assume the following.
| Most Likely 50% chance | Worst Case 25 % chance | Best Case 25% Chance | |
| Year 1 | 100 | 50 | 100 |
| Year 2 | 200 | 50 | 500 |
| Year 3 | 400 | 100 | 1000 |
| Years 4 - 8 | 1500 | 100 | 3000 |
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Example: Make and Sell Widgets
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| Most Likely 50% chance | Worst Case 25 % chance | Best Case 25% Chance | |
| Year 1 | 100 | 50 | 100 |
| Year 2 | 200 | 50 | 500 |
| Year 3 | 400 | 100 | 1000 |
| Years 4 - 8 | 1500 | 100 | 3000 |
| NPW | $787,707 | -$209,663 | $1,924,611 |
Expected NPW = (0.5)$787,707 + (0.25)(-$209,663)+(0.25)($1,924,611)
= $822,590
In this situation, decision remains the same, invest in project
Example: Make and Sell Widgets Calculate Payback Period
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| Year | Cash Flow | Cost of Funds | Cumulative Cash Flow |
| 0 | -$100,000 | -$100,000 | |
| 1 | -$80,225 | -$12,000 | -$192,225 |
| 2 | -$340 | -$23,067 | -$215,632 |
| 3 | $44,684 | -$25,876 | -$196,824 |
| 4 | $363,814 | -$23,619 | $143,370 |
| 5 | $353,814 | $17,204 | $514,389 |
| 6 | $363,814 | $61,727 | $939,929 |
| 7 | $363,814 | $112,792 | $1,416,535 |
| 8 | $363,814 | $169,984 | $1,950,332 |
Discounted Payback Period
Between 3 and 4 years.
Is this Payback Period acceptable for us and/or our investors?
Payback Period
Cumulative Cash Flow
0 1 2 3 4 5 6 7 8 -100000 -192225 -215632 -196824 143370 514388 939929 1416534 1950332
Year
Cumulative Cash Flow
