Write a 3.5-page to 4.5-page paper that highlights your reflections about relevant operations lessons from the Boeing tour and our course time studying Lean. The paper should address what you observed and learned at Boeing, and tie that to what you read i

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Home>Applied Sciences homework help>Write a 3.5-page to 4.5-page paper that highlights your reflections about relevant operations lessons from the Boeing tour and our course time studying Lean. The paper should address what you observed and learned at Boeing, and tie that to what you read i

Operations and Project Management ELCBUS 340 11 April 2018

1

Today’s Agenda

Class Administration – Littlefield Teams Verification

Forecasting

Solved Problems

Demand Patterns & Management

Forecast Error Measures

Forecasting Technique Parts I, II, and III

Criteria for Selecting a Time-Series Method

Discussion on Reading

2

50,000 Feet Overview of OPM

Managing Processes

Process Strategy

Process Performance & Quality

Constraint Management

Process Layout

Lean Systems

Process Analysis

Using Operations to Compete

Operations As a

Competitive Weapon

Operations Strategy

Project Management

Managing Value Chains

Supply Chain Strategy

Inventory Management

Location

Forecasting

Sales & Operations Planning

Scheduling

Resource Planning

Forecasting

3

Service vs. Manufacturing

Small vs. Medium vs. Large Firms

Industry type: hightech, healthcare, aerospace, IT, etc ….

A prediction of future events used for planning purposes.

What is a Forecast?

4

Finalize

and communicate

6

Review by Operating Committee

5

Revise forecasts

4

Consensus meetings and collaboration

3

Prepare initial forecasts

2

Adjust history file

1

Forecasting as a Process

5

SOME PRINCIPLES FOR THE FORECASTING PROCESS
Better processes yield better forecasts
Demand forecasting is being done in virtually every company, either formally or informally. The challenge is to do it well—better than the competition
Better forecasts result in better customer service and lower costs, as well as better relationships with suppliers and customers
The forecast can and must make sense based on the big picture, economic outlook, market share, and so on
The best way to improve forecast accuracy is to focus on reducing forecast error
Bias is the worst kind of forecast error; strive for zero bias
Whenever possible, forecast at more aggregate levels. Forecast in detail only where necessary
Far more can be gained by people collaborating and communicating well than by using the most advanced forecasting technique or model

Forecasting Principles

6

Collaborative planning, forecasting, and replenishment (CPFR)

A process for supply chain integration that allows a supplier and its customers to collaborate on making the forecast.

Adding Collaboration to the Process

Demand Patterns & Management

8

There are five basic time series patterns:

Horizontal

Trend

Seasonal

Cyclical

Random

A time series is the repeated observations of demand for a service or product in their order of occurrence. Time series plots are used to visualize Demand Patterns.

Demand Patterns

9

(a) Horizontal (b) Trend

(a) Seasonal (b) Cyclical

Demand Patterns

10

The process of changing demand patterns using one or more demand options.

Demand Management

Demand Management Options:

Complementary Products

Promotional Pricing

Prescheduled Appointments

Reservations

Revenue Management

Backlogs

Backorders and Stockouts

What?

How? (Forecasting Technique)

Key Decisions for Making Forecasts

Judgment methods

Causal methods

Time-series analysis

Trend projection using regression

Level of aggregation

Units of measurement

12

Forecast Error Measures

13

For any forecasting method, it is important to measure the accuracy of its forecasts.

Forecast error is simply the difference found by subtracting the forecast from actual demand for a given period, or

Where:

Et = forecast error for period t

Dt = actual demand in period t

Ft = forecast for period t

Et = Dt – Ft

Measures of Forecast Error

(|Et |/ Dt)(100)

n

MAPE =

CFE = Et

Et2

n

MSE =

|Et |

n

MAD =

Cumulative sum of forecast errors (Bias)

Average forecast error

Mean Squared Error

Mean Absolute Percent Error

Mean Absolute Deviation

Standard deviation

CFE

n

Ē=

 =

(Et – Ē)2

n – 1

Measures of Forecast Error

15

74

Month t Demand Dt Forecast Ft Error Et Error2 Et2 Absolute Error |Et| Absolute % Error (|Et|/Dt)(100)
1 200 225 –25
2 240 220 20
3 300 285 15
4 270 290 –20
5 230 250 –20 400 20 8.7
6 260 240 20 400 20 7.7
7 210 250 –40 1,600 40 19.0
8 275 240 35 1,225 35 12.7
Total –15 5,275 195 81.3%

Example

The following table shows the actual sales of upholstered chairs for a furniture manufacturer and the forecasts made for each of the last eight months.

Calculate CFE, MSE, σ, MAD, and MAPE for this product.

16

The following table shows the actual sales of upholstered chairs for a furniture manufacturer and the forecasts made for each of the last eight months.

Calculate CFE, MSE, σ, MAD, and MAPE for this product.

625 25 12.5%
400 20 8.3
225 15 5.0
400 20 7.4
Month t Demand Dt Forecast Ft Error Et Error2 Et2 Absolute Error |Et| Absolute % Error (|Et|/Dt)(100)
1 200 225 –25
2 240 220 20
3 300 285 15
4 270 290 –20
5 230 250 –20 400 20 8.7
6 260 240 20 400 20 7.7
7 210 250 –40 1,600 40 19.0
8 275 240 35 1,225 35 12.7
Total –15 5,275 195 81.3%

Example

17

Using the formulas for the measures, we get:

CFE =

–15

Average forecast error (mean bias):

Mean squared error:

Cumulative forecast error (mean bias):

MSE =

Et2

n

5,275

8

=

659.4

=

CFE

n

–1.875

= =

15

8

Ē =

Example

18

Standard deviation:

Mean absolute deviation:

Mean absolute percent error:

S[Et – (–1.875)]2

n – 1

s =

S|Et |

n

MAD =

(S|Et |/ Dt)(100)

n

MAPE =

= 27.4

= = 24.4

195

8

= = 10.2%

81.3%

8

Example

19

A Cumulative Forecast Error (CFE) of –15 indicates that the forecast has a slight bias to overestimate demand.

The Mean Square Error (MSE), σ, and Mean Absolute Deviation (MAD) statistics provide measures of forecast error variability.

A MAD of 24.4 means that the average forecast error was 24.4 units in absolute value.

The value of σ, 27.4, indicates that the sample distribution of forecast errors has a standard deviation of 27.4 units.

A Mean Absolute Percent Error (MAPE) of 10.2 percent implies that, on average, the forecast error was about 10 percent of actual demand.

These measures become more reliable as the number of periods of data increases.

Example

20

Forecasting Techniques – Part I (Overview)

Judgment Methods

Casual Methods

Time-Series Analysis

21

Judgmental forecasts use contextual knowledge gained through experience and translate them into quantitative estimates.

Salesforce estimates

Executive opinion

Market research

Delphi method

Judgment Methods

Causal Methods are a quantitative forecasting method that uses historical data on independent variables, such as promotional campaigns, economic conditions, and competitors’ actions to predict demand.

Casual Methods

Time-Series Analysis is a statistical approach that relies heavily on historical demand data to project the future size of demand and recognizes trends and seasonal patterns.

Time-Series Analysis

Forecasting Techniques – Part II

Simple Moving Average

Weighted Moving Average

Exponential Smoothing

25

Naïve forecast

The forecast for the next period equals the demand for the current period (Forecast = Dt)

Horizontal Patterns: Estimating the average

Simple moving average

Weighted moving average

Exponential smoothing

Time Series Methods

Specifically, the forecast for period t + 1 can be calculated at the end of period t (after the actual demand for period t is known) as

Ft+1 = =

Sum of last n demands

n

Dt + Dt-1 + Dt-2 + … + Dt-n+1

n

where

Dt = actual demand in period t

n = total number of periods in the average

Ft+1 = forecast for period t + 1

Simple Moving Averages

a. Compute a three-week moving average forecast for the arrival of medical clinic patients in week 4. The numbers of arrivals for the past three weeks were as follows:

b. If the actual number of patient arrivals in week 4 is 415, what is the forecast error for week 4?

c. What is the forecast for week 5?

Week Patient Arrivals
1 400
2 380
3 411

Simple Moving Averages - Example

28

a. The moving average forecast at the end of week 3 is:

Week Patient Arrivals
1 400
2 380
3 411

b. The forecast error for week 4 is

F4 =

= 397.0

411 + 380 + 400

3

E4 = D4 – F4

= 415 – 397 = 18

c. The forecast for week 5 requires the actual arrivals from weeks 2 through 4, the three most recent weeks of data

F5 =

= 402.0

415 + 411 + 380

3

Simple Moving Averages - Example

29

Estimating with Simple Moving Average using the following customer-arrival data:

Month Customer arrival
1 800
2 740
3 810
4 790

Use a three-month moving average to forecast customer arrivals for month 5

F5 =

= 780

D4 + D3 + D2

3

790 + 810 + 740

3

=

Forecast for month 5 is 780 customer arrivals

Simple Moving Averages – Example 2

If the actual number of arrivals in month 5 is 805, what is the forecast for month 6?

F6 =

= 801.667

D5 + D4 + D3

3

805 + 790 + 810

3

=

Forecast for month 6 is 802 customer arrivals

Month Customer arrival
1 800
2 740
3 810
4 790

Simple Moving Averages – Example 2

Forecast error is simply the difference found by subtracting the forecast from actual demand for a given period, or

Given the three-month moving average forecast for month 5, and the number of patients that actually arrived (805), what is the forecast error?

Forecast error for month 5 is 25

Et = Dt – Ft

E5 =

805 – 780

= 25

Simple Moving Averages – Example 2

In the weighted moving average method, each historical demand in the average can have its own weight, provided that the sum of the weights equals 1.0.

The average is obtained by multiplying the weight of each period by the actual demand for that period, and then adding the products together

Ft+1 = W1D1 + W2D2 + … + WnDt-n+1

Weighted Moving Averages

33

Using the customer arrival data in Example 2, let

W1 = 0.50, W2 = 0.30, and W3 = 0.20. Use the weighted moving average method to forecast arrivals for month 5.

= 0.50(790) + 0.30(810) + 0.20(740)

F5 = W1D4 + W2D3 + W3D2

= 786

Forecast for month 5 is 786 customer arrivals.

Given the number of customers that actually arrived (805),

what is the forecast error?

Forecast error for month 5 is 19.

E5 =

805 – 786

= 19

Weighted Moving Averages - Example

34

If the actual number of arrivals in month 5 is 805, compute the forecast for month 6:

= 0.50(805) + 0.30(790) + 0.20(810)

F6 = W1D5 + W2D4 + W3D3

= 801.5

Forecast for month 6 is 802 customer arrivals.

Weighted Moving Averages - Example

35

A sophisticated weighted moving average that calculates the average of a time series by implicitly giving recent demands more weight than earlier demands

Requires only three items of data

The last period’s forecast

The demand for this period

A smoothing parameter, alpha (α), where 0 ≤ α ≤ 1.0

The equation for the forecast is

Ft+1 = α(Demand this period) + (1 – α)(Forecast calculated last period)

= αDt + (1 – α)Ft

Exponential Smoothing

36

The emphasis given to the most recent demand levels can be adjusted by changing the smoothing parameter.

Larger α values emphasize recent levels of demand and result in forecasts more responsive to changes in the underlying average.

Smaller α values treat past demand more uniformly and result in more stable forecasts.

Exponential Smoothing

37

Reconsider the patient arrival data. It is now the end of week 3 so the actual arrivals is known to be 411 patients. Using α = 0.10, calculate the exponential smoothing forecast for week 4.

What was the forecast error for week 4 if the actual demand turned out to be 415?

What is the forecast for week 5?

Exponential Smoothing - Example

Week Patient Arrivals
1 400
2 380
3 411

38

a. To obtain the forecast for week 4, using exponential smoothing with and the initial forecast of 390*, we calculate the average at the end of week 3 as:

F4 =

Thus, the forecast for week 4 would be 392 patients.

0.10(411) + 0.90(390) = 392.1

* Here the initial forecast of 390 is the average of the first two weeks of demand. POM for Windows and OM Explorer, on the other hand, simply use the actual demand for the first week as the default setting for the initial forecast for period 1, and do not begin tracking forecast errors until the second period.

Exponential Smoothing - Example

39

b. The forecast error for week 4 is

c. The new forecast for week 5 would be

E4 =

F5 =

or 394 patients.

415 – 392 = 23

0.10(415) + 0.90(392.1) = 394.4

Exponential Smoothing - Example

Week Patient Arrivals
2 380
3 411
4 415

40

Forecasting Techniques – Part III

Trend Patterns Using Linear Regression

Multiplicative Seasonal Method

41

A dependent variable is related to one or more independent variables by a linear equation

The independent variables are assumed to “cause” the results observed in the past

Simple linear regression model is a straight line

Y = a + bX

where

Y = dependent variable

X = independent variable

a = Y-intercept of the line

b = slope of the line

Linear Regression

Dependent variable

Independent variable

X

Y

Estimate of

Y from

regression

equation

Regression

equation:

Y = a + bX

Actual

value

of Y

Value of X used

to estimate Y

Deviation,

or error

Linear Regression

The sample correlation coefficient, r

Measures the direction and strength of the relationship between the independent variable and the dependent variable.

The value of r can range from –1.00 ≤ r ≤ 1.00

The sample coefficient of determination, r2

Measures the amount of variation in the dependent variable about its mean that is explained by the regression line

The values of r2 range from 0.00 ≤ r2 ≤ 1.00

The standard error of the estimate, syx

Measures how closely the data on the dependent variable cluster around the regression line

Linear Regression

A trend in a time series is a systematic increase or decrease in the average of the series over time.

The forecast can be improved by calculating an estimate of the trend.

Trend Projection with Regression accounts for the trend with simple regression analysis.

Trend Patterns Using Regression

45

Medanalysis, Inc., provides medical laboratory services

Managers are interested in forecasting the number of blood analysis requests per week

There has been a national increase in requests for standard blood tests.

The arrivals over the next 16 weeks are given in the Table on the next slide.

What is the forecasted demand for the next three periods?

Trend Patterns – Regression Example

46

Week Arrivals Week Arrivals
1 28 9 61
2 27 10 39
3 44 11 55
4 37 12 54
5 35 13 52
6 53 14 60
7 38 15 60
8 57 16 75

Arrivals at Medanalysis, Inc.

Trend Patterns – Regression Example

Trend Patterns – Regression Example

Trend Patterns – Regression Example

Use OM Explorer to project the following weekly demand data using trend projection with regression.

What is the forecasted demand for periods 11-14?

Week Demand Week Demand
1 2 3 4 5 24 34 29 27 39 6 7 8 9 10 42 39 56 45 43

Trend Patterns – Regression Example 2

Trend Patterns – Regression Example 2

Trend Patterns – Regression Example 2

Multiplicative seasonal method

A method whereby seasonal factors are multiplied by an estimate of average demand to arrive at a seasonal forecast.

Additive seasonal method

A method in which seasonal forecasts are generated by adding a constant to the estimate of average demand per season.

Seasonal Factors

53

For each year, calculate the average demand for each season by dividing annual demand by the number of seasons per year.

For each year, divide the actual demand for each season by the average demand per season, resulting in a seasonal factor for each season.

Calculate the average seasonal factor for each season using the results from Step 2.

Calculate each season’s forecast for next year.

Multiplicative seasonal method

Multiplicative Seasonal Method

54

The manager wants to forecast customer demand for each quarter of year 5, based on an estimate of total year 5 demand of 2,600 customers.

The manager of the Stanley Steemer carpet cleaning company needs a quarterly forecast of the number of customers expected next year. The carpet cleaning business is seasonal, with a peak in the third quarter and a trough in the first quarter. The quarterly demand data from the past 4 years are on the next two slides.

Multiplicative Seasonal Method - Example

55

60

YEAR 1 YEAR 2
     
Q Demand Seasonal Factor (1) Demand Seasonal Factor (2)
1 45 45/250 = 0.18 70 70/300 = 0.23
2 335 335/250 = 1.34 370 370/300 = 1.23
3 520 520/250 = 2.08 590 590/300 = 1.97
4 100 100/250 = 0.40 170 170/300 = 0.57
Total 1,000 1,200  
Average 1,000/4 = 250   1,200/4 = 300  

Multiplicative Seasonal Method - Example

YEAR 3 YEAR 4
     
Q Demand Seasonal Factor (3) Demand Seasonal Factor (4)
1 100 100/450 = 0.22 100 100/550 = 0.18
2 585 585/450 = 1.30 725 725/550 = 1.32
3 830 830/450 = 1.84 1160 1160/550 = 2.11
4 285 285/450 = 0.63 215 215/550 = 0.39
Total 1,800 2,200  
Average 1,800/4 = 450   2,200/4 = 550  

Multiplicative Seasonal Method - Example

Quarterly Forecasts

Quarter Forecast
1 650 x 0.2043 = 132.795
2 650 x 1.2979 = 843.635
3 650 x 2.001 = 1,300.06
4 650 x 0.4977 = 323.505
Quarter Average Seasonal Factor
1 0.2043
2 1.2979
3 2.0001
4 0.4977

Average Seasonal Factor

Multiplicative Seasonal Method - Example

58

60

Figure 8.7

Multiplicative Seasonal Method - Example

59

Suppose the multiplicative seasonal method is being used to forecast customer demand. The actual demand and seasonal indices are shown below.

Year 1 Year 2 Average Index
Quarter Demand Index Demand Index
1 100 0.40 192 0.64 0.52
2 400 1.60 408 1.36 1.48
3 300 1.20 384 1.28 1.24
4 200 0.80 216 0.72 0.76
Average 250 300

Multiplicative Seasonal Method – Example 2

1400 units ÷ 4 quarters = 350 units

Quarter Average Index
1 0.52
2 1.48
3 1.24
4 0.76

Forecast for Quarter 1 =

Forecast for Quarter 2 =

Forecast for Quarter 3 =

Forecast for Quarter 4 =

0.52(350) ≈ 182 units

1.48(350) ≈ 518 units

1.24(350) ≈ 434 units

0.76(350) ≈ 266 units

Multiplicative Seasonal Method – Example 2

1320 units ÷ 4 quarters = 330 units

Quarter Average Index
1 0.52
2 1.48
3 1.24
4 0.76

If the projected demand for Year 3 is 1320 units, what is the forecast for each quarter of that year?

Forecast for Quarter 1 =

Forecast for Quarter 2 =

Forecast for Quarter 3 =

Forecast for Quarter 4 =

0.52(330) ≈ 172 units

1.48(330) ≈ 488 units

1.24(330) ≈ 409 units

0.76(330) ≈ 251 units

Multiplicative Seasonal Method – Example 2

Time Series-Method Selection

63

Criteria:

Minimizing bias (CFE)

Minimizing MAPE, MAD, or MSE

Maximizing r2 for trend projections using regression

Using a holdout sample analysis

Using a tracking signal

Meeting managerial expectations of changes in the components of demand.

Minimizing the forecast errors in recent periods.

Time-Series Method Selection Criteria

64

Using Statistical Criteria:

For more stable demand patterns, use lower a values or larger n values to emphasize historical experience.

For more dynamic demand patters, use higher a values or smaller n values.

Choosing a Time-Series Method

Holdout sample

Actual demands from the more recent time periods in the time series that are set aside to test different models developed from the earlier time periods.

Choosing a Time-Series Method

A measure that indicates whether a method of forecasting is accurately predicting actual changes in demand.

Tracking signal =

CFE

MAD

Each period, the CFE and MAD are updated to reflect current error, and the tracking signal is compared to some predetermined limits.

CFE

MADt

or

Tracking Signals

The MAD can be calculated as the simple average of all absolute errors or as a weighted average determined by the exponential smoothing method

MADt = α|Et| + (1 – α)MADt-1

If forecast errors are normally distributed with a mean of 0, the relationship between σ and MAD is simple

σ = (  /2)(MAD)  1.25(MAD)

MAD = 0.7978σ  0.8σ where p = 3.1416

Tracking Signals

+2.0 –

+1.5 –

+1.0 –

+0.5 –

0 –

–0.5 –

–1.0 –

–1.5 –

| | | | |

0 5 10 15 20 25

Observation number

Tracking signal

Out of control

Control limit

Control limit

Figure 8.8

Tracking Signals

69

86

This slide completes Figure 13.9 and shows the use of a tracking signal.

Solved Problems

70

Chicken Palace periodically offers carryout five-piece chicken dinners at special prices. Let Y be the number of dinners sold and X be the price. Based on the historical observations and calculations in the following table, determine the regression equation, correlation coefficient, and coefficient of determination. How many dinners can Chicken Palace expect to sell at $3.00 each?

Observation Price (X) Dinners Sold (Y)
1 $2.70 760
2 $3.50 510
3 $2.00 980
4 $4.20 250
5 $3.10 320
6 $4.05 480
Total $19.55 3,300
Average $ 3.26 550

Solved Problem 1

We use the computer to calculate the best values of a, b, the correlation coefficient, and the coefficient of determination

r 2 = 0.71

r = –0.84

b = –277.63

a = 1,454.60

The regression line is

Y = a + bX =

1,454.60 – 277.63X

For an estimated sales price of $3.00 per dinner

Y = a + bX =

1,454.60 – 277.63(3.00)

= 621.71 or 622 dinners

Solved Problem 1

The Polish General’s Pizza Parlor is a small restaurant catering to patrons with a taste for European pizza. One of its specialties is Polish Prize pizza. The manager must forecast weekly demand for these special pizzas so that he can order pizza shells weekly. Recently, demand has been as follows:

Week Pizzas Week Pizzas
June 2 50 June 23 56
June 9 65 June 30 55
June 16 52 July 7 60

a. Forecast the demand for pizza for June 23 to July 14 by using the simple moving average method with n = 3 then using the weighted moving average method with and weights of 0.50, 0.30, and 0.20, with 0.50.

b. Calculate the MAD for each method.

Solved Problem 2

a. The simple moving average method and the weighted moving average method give the following results:

Current Week Simple Moving Average Forecast for Next Week Weighted Moving Average Forecast for Next Week
June 16
June 23
June 30
July 7

= 55.7 or 56

52 + 65 + 50

3

[(0.5  52) + (0.3  65) + (0.2  50)] = 55.5 or 56

= 57.7 or 58

56 + 52 + 65

3

= 54.3 or 54

55 + 56 + 52

3

[(0.5  56) + (0.3  52) + (0.2  65)] = 56.6 or 57

[(0.5  55) + (0.3  56) + (0.2  52)] = 54.7 or 55

= 57.0 or 57

60 + 55 + 56

3

[(0.5  60) + (0.3  55) + (0.2  56)] = 57.7 or 58

Solved Problem 2

b. The mean absolute deviation is calculated as follows:

Simple Moving Average Weighted Moving Average
Week Actual Demand Forecast for This Week Absolute Errors |Et| Forecast for This Week Absolute Errors |Et|
June 23 56 56 56
June 30 55 58 57
July 7 60 54 55

|56 – 56| = 0

|55 – 58| = 3

|60 – 54| = 6

MAD = = 3

0 + 3 + 6

3

MAD = = 2.3

0 + 2 + 2

3

|56 – 56| = 0

|55 – 57| = 2

|60 – 55| = 5

For this limited set of data, the weighted moving average method resulted in a slightly lower mean absolute deviation. However, final conclusions can be made only after analyzing much more data.

Solved Problem 2

The monthly demand for units manufactured by the Acme Rocket Company has been as follows:

Month Units Month Units
May 100 September 105
June 80 October 110
July 110 November 125
August 115 December 120

a. Use the exponential smoothing method to forecast June to January. The initial forecast for May was 105 units; α = 0.2.

b. Calculate the absolute percentage error for each month from June through December and the MAD and MAPE of forecast error as of the end of December.

c. Calculate the tracking signal as of the end of December. What can you say about the performance of your forecasting method?

Solved Problem 3

a.

Current Month, t Calculating Forecast for Next Month Ft+1 = αDt + (1 – α)Ft Forecast for Month t + 1
May June
June July
July August
August September
September October
October November
November December
December January

0.2(100) + 0.8(105)

= 104.0 or 104

0.2(80) + 0.8(104.0)

0.2(110) + 0.8(99.2)

= 99.2 or 99

= 101.4 or 101

0.2(115) + 0.8(101.4)

0.2(105) + 0.8(104.1)

0.2(110) + 0.8(104.3)

0.2(125) + 0.8(105.4)

0.2(120) + 0.8(109.3)

= 104.1 or 104

= 104.3 or 104

= 105.4 or 105

= 109.3 or 109

= 111.4 or 111

Solved Problem 3

b.

–24
24
30.0%
11
11
10.0
Month, t Actual Demand, Dt Forecast, Ft Error, Et = Dt – Ft Absolute Error, |Et| Absolute Percent Error, (|Et|/Dt)(100)
June 80 104
July 110 99
August 115 101
September 105 104
October 110 104
November 125 105
December 120 109
Total 765
14 14 12.0
1 1 1.0
6 6 5.5
20 20 16.0
11 11 9.2
39 87 83.7%

|Et |

n

MAD =

(|Et |/Dt)(100)

n

MAPE =

= = 11.96%

83.7%

7

= = 12.4

87

7

Solved Problem 3

c. As of the end of December, the cumulative sum of forecast errors (CFE) is 39. Using the mean absolute deviation calculated in part (b), we calculate the tracking signal:

The probability that a tracking signal value of 3.14 could be generated completely by chance is small. Consequently, we should revise our approach. The long string of forecasts lower than actual demand suggests use of a trend method.

Tracking signal =

CFE

MAD

= = 3.14

39

12.4

Solved Problem 3

The Northville Post Office experiences a seasonal pattern of daily mail volume every week. The following data for two representative weeks are expressed in thousands of pieces of mail:

Day Week 1 Week 2
Sunday 5 8
Monday 20 15
Tuesday 30 32
Wednesday 35 30
Thursday 49 45
Friday 70 70
Saturday 15 10
Total 224 210

a. Calculate a seasonal factor for each day of the week.

b. If the postmaster estimates 230,000 pieces of mail to be sorted next week, forecast the volume for each day.

Solved Problem 4

Week 1 Week 2
Day Mail Volume Seasonal Factor (1) Mail Volume Seasonal Factor (2) Average Seasonal Factor [(1) + (2)]/2
Sunday 5 8
Monday 20 15
Tuesday 30 32
Wednesday 35 30
Thursday 49 45
Friday 70 70
Saturday 15 10
Total 224 210
Average 224/7 = 32 210/7 = 30

5/32 = 0.15625

20/32 = 0.62500

30/32 = 0.93750

8/30 = 0.26667

15/30 = 0.50000

32/30 = 1.06667

0.21146

0.56250

1.00209

35/32 = 1.09375

49/32 = 1.53125

70/32 = 2.18750

15/32 = 0.46875

30/30 = 1.00000

45/30 = 1.50000

70/30 = 2.33333

10/30 = 0.33333

1.04688

1.51563

2.26042

0.40104

Solved Problem 4

b. The average daily mail volume is expected to be 230,000/7 = 32,857 pieces of mail. Using the average seasonal factors calculated in part (a), we obtain the following forecasts:

6,948

18,482

32,926

0.21146(32,857) =

0.56250(32,857) =

1.00209(32,857) =

34,397

49,799

74,271

13,177

230,000

1.04688(32,857) =

1.51563(32,857) =

2.26042(32,857) =

0.40104(32,857) =

Day Calculations Forecast
Sunday
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Total

Solved Problem 4