M05_REND6289_10_IM_C05

52

Actual Three-WeekWeek Bicycle Sales Moving Average

1 82 103 94 11 (8 � 10 � 9)/3 � 95 10 (10 � 9 � 11)/3 � 106 13 (9 � 11 � 10)/3 � 107 — (11 � 10 � 13)/3 � 11Z\c

Alternative Example 5.2: Weighted moving average

Bower’s Bikes decides to forecast bicycle sales by weighting thepast 3 weeks as follows:

Weights Applied Period

3 Last week2 Two weeks ago1 Three weeks ago6 Sum of weights

A 3-week weighted moving average appears below.

ActualBicycle

Week Sales Three-Week Moving Average

1 82 103 94 11 [(3 � 9) � (2 � 10) � (1 � 8)]/6 � 9Z\n

5 10 [(3 � 11) � (2 � 9) � (1 � 10)]/6 � 10Z\n

6 13 [(3 � 10) � (2 � 11) � (1 � 9)]/6 � 10Z\n

7 — [(3 � 13) � (2 � 10) � (1 � 11)]/6 � 11X\c

Alternative Example 5.3: A firm uses simple exponentialsmoothing with a � 0.1 to forecast demand. The forecast for theweek of January 1 was 500 units, whereas actual demand turnedout to be 450 units. The demand forecasted for the week of Janu-ary 8 is calculated as follows.

Ft�1 � Ft � α(At � Ft)

� 500 � 0.1(450 � 500) � 495 units

�∑ (weight for period )(demand in period n n))

∑ weights

TEACHING SUGGESTIONS

Teaching Suggestion 5.1: Wide Use of Forecasting.Forecasting is one of the most important tools a student can masterbecause every firm needs to conduct forecasts. It’s useful to moti-vate students with the idea that obscure sounding techniques suchas exponential smoothing are actually widely used in business, anda good manager is expected to understand forecasting. Regressionis commonly accepted as a tool in economic and legal cases.

Teaching Suggestion 5.2: Forecasting as an Art and a Science.Forecasting is as much an art as a science. Students should under-stand that qualitative analysis ( judgmental modeling) plays an im-portant role in predicting the future since not every factor can bequantified. Sometimes the best forecast is done by seat-of-the-pants methods.

Teaching Suggestion 5.3: Use of Simple Models.Many managers want to know what goes on behind the forecast.They may feel uncomfortable with complex statistical models withtoo many variables. They also need to feel a part of the process.

Teaching Suggestion 5.4: Management Input to the ExponentialSmoothing Model.One of the strengths of exponential smoothing is that it allows de-cision makers to input constants that give weight to recent data.Most managers want to feel a part of the modeling process andappreciate the opportunity to provide input.

Teaching Suggestion 5.5: Wide Use of Adaptive Models.With today’s dominant use of computers in forecasting, it ispossible for a program to constantly track the accuracy of amodel’s forecast. It’s important to understand that a programcan automatically select the best alpha and beta weights inexponential smoothing. Even if a firm has 10,000 products, theconstants can be selected very quickly and easily without humanintervention.

ALTERNATIVE EXAMPLES

Alternative Example 5.1:

Moving average

Bicycle sales at Bower’s Bikes are shown in the middle column of thefollowing table. A 3-week moving average appears on the right.

= ∑ demand in previous periodsn

n

5C H A P T E R

Forecasting Models

M05_REND6289_10_IM_C05.QXD 5/7/08 4:42 PM Page 52REVISED

CHAPTER 5 FORECAST ING MODELS 53

Alternative Example 5.4: Exponential smoothing is used toforecast automobile battery sales. Two values of � are examined,� � 0.8 and � � 0.5. To evaluate the accuracy of each smoothingconstant, we can compute the absolute deviations and MADs.Assume that the forecast for January was 22 batteries.

On the basis of this analysis, a smoothing constant of � � 0.8 ispreferred to � � 0.5 because it has a smaller MAD.

Alternative Example 5.5: Use the sales data given below to de-termine: (a) the least squares trend line, (b) the predicted value for2000 sales.

Year Sales (Units)

1993 1001994 1101995 1221996 1301997 1391998 1521999 164

To minimize computations, transform the value of x (time) to sim-pler numbers. In this case, designate 1993 as year 1, 1994 as year2, and so on.

Time SalesYear Period (Units) x2 xy

1993 1 100 1 1001994 2 110 4 2201995 3 122 9 3661996 4 130 16 5201997 5 139 25 6951998 6 152 36 9121999 17 164 149 1,148

�x � 28 �y � 917 �x2 � 140 �xy � 3,961

Absolute AbsoluteActual Forecast Deviation Forecast DeviationBattery with with with with

Month Sales � �� 0.8 � �� 0.8 � �� 0.5 � �� 0.5

January 20 22 2 22 2February 21 20.40 0.6 21 0March 15 20.880 5.88 21 6April 14 16.176 2.176 18 4May 13 14.435 1.435 16 3June 16 13.287 2.713 14.5 31.5

Sum of absolute deviations: 15 16.5

MAD: 2.46 2.75

Alternative Example 5.6: The rated power capacity (in hours/week) over the past 6 years has been:

Rated CapacityYear (hrs/wk)

1 1152 1203 1184 1245 1236 130

Here is an alternative way to recode years which simplifies themath since �X � 0.

Renumbered CapacityYear Year (x) (y) x2 xy

1 �2.5 115 6.25 �287.52 �1.5 120 2.25 �1803 �.5 118 0.25 �594 �.5 124 0.25 �625 �1.5 123 2.25 �184.56 �2.5 130 6.25 �325

�X � 0 �Y � 730 �X2 � 17.5 �XY � 45

y � 121.67 � 2.57X

Year 7 � 121.67 � (2.57)(3.5)

�131

Alternative Example 5.7: The forecast demand and actual de-mand for 10-foot fishing boats are shown below. We compute thetracking signal and MAD.

Tracking SignalRSFE

MAD MADs= = − = −24

11 72 1

..

MAD Forecast errors= ∑ = =

n

70

611 7.

aY

n= ∑ = =730

6121 67.

bXY

X= ∑

∑= =

2

45

17 52 57

..

yy

n= ∑ = =917

7131

bxy nxy

x nx= ∑ −

∑ −= −

−2 2

3 961 7 4 131

140 7 4

, ( )( )( )

( )( 22

293

2810 464

).= =

a y bx= − = − =131 10 46 4 89 14. ( ) .

xx

n= ∑ = =28

74

Therefore, the least squares trend equation is,

To project demand in 2000, we denote the year 2000 as x � 8,

Sales in 2000 � 89.14 � 10.464(8) � 172.85

ˆ . .y a bx x= + = +89 14 10 464


54 CHAPTER 5 FORECAST ING MODELS

SOLUTIONS TO DISCUSSION QUESTIONS

AND PROBLEMS

5-1. The steps that are used to develop any forecasting systemare:

1. Determine the use of the forecast.

2. Select the items or quantities that are to be forecasted.

3. Determine the time horizon of the forecast.

4. Select the forecasting model.

5. Gather the necessary data.

6. Validate the forecasting model.

7. Make the forecast.

8. Implement the results.

5-2. A time-series forecasting model uses historical data to pre-dict future trends.

5-3. The only difference between causal models and time-series models is that causal models take into account any factorsthat may influence the quantity being forecasted. Causal modelsuse historical data as well. Time-series models use only historicaldata.

5-4. Qualitative models incorporate subjective factors into theforecasting model. Judgmental models are useful when subjectivefactors are important. When quantitative data are difficult to ob-tain, qualitative models are appropriate.

5-5. The disadvantages of the moving average forecastingmodel are that the averages always stay within past levels, and themoving averages do not consider seasonal variations.

5-6. When the smoothing value, �, is high, more weight is givento recent data. When � is low, more weight is given to past data.

5-7. The Delphi technique involves analyzing the predictionsthat a group of experts have made, then allowing the experts to re-view the data again. This process may be repeated several times.After the final analysis, the forecast is developed. The group of experts may be geographically dispersed.

5-8. MAD is a technique for determining the accuracy of aforecasting model by taking the average of the absolute deviations.

MAD is important because it can be used to help increase forecast-ing accuracy.

5-9. If a seasonal index equals 1, that season is just an averageseason. If the index is less than 1, that season tends to be lowerthan average. If the index is greater than 1, that season tends to behigher than average.

5-10. If the smoothing constant equals 0, then

Ft�1 � Ft � 0(At � Ft) � Ft

This means that the forecast never changes.If the smoothing constant equals 1, then

Ft�1 � Ft � 1(At � Ft) � At

This means that the forecast is always equal to the actual value inthe prior period.

5-11. A centered moving average (CMA) should be used iftrend is present in data. If an overall average is used rather than aCMA, variations due to trend will be interpreted as variations due toseasonal factors. Thus, the seasonal indices will not be accurate.

5-12.

ActualMonth Shed Sales Four-Month Moving Average

Jan. 10Feb. 12Mar. 13Apr. 16May 19 (10 � 12 � 13 � 16)/4 � 51/4 � 12.75June 23 (12 � 13 � 16 � 19)/4 � 60/4 � 15July 26 (13 � 16 � 19 � 23)/4 � 70/4 � 17.75Aug. 30 (16 � 19 � 23 � 26)/4 � 84/4 � 21Sept. 28 (19 � 23 � 26 � 30)/4 � 98/4 � 24.5Oct. 18 (23 � 26 � 30 � 28)/4 � 107/4 � 26.75Nov. 16 (26 � 30 � 28 � 18)/4 � 102/4 � 25.5Dec. 14 (30 � 28 � 18 � 16)/4 � 92/4 � 23

The MAD � 7.78

See solution to 5-13 for calculations.

Table for Alternate Example 5.7

Forecast Actual Forecast Cumulative TrackingYear Demand Demand Error RSFE Error Error MAD Signal

1 78 71 �7 �7 7 7 7.0 �1.02 75 80 5 �2 5 12 6.0 �0.33 83 101 18 16 18 30 10.0 �1.64 84 84 0 16 0 30 7.5 �2.15 88 60 �28 �12 28 58 11.6 �1.06 85 73 �12 �24 12 70 11.7 �2.1



5-14.

Three-Year Weighted Three-Year Three-Year Three-Year WeightedYear Demand Moving Averages Moving Averages Absolute Deviation Absolute Deviation

1 42 63 4 sum of the weights4 5 (4 � 6 � 4)/3 � 42⁄3 [(2 � 4) � 6 � 4]/4 � 41⁄2 0.34 0.555 10 (6 � 4 � 5)/3 � 5 [(2 � 5) � 4 � 6]/4 � 50 5.55 5.556 8 (4 � 5 � 10)/3 � 61⁄3 [(2 � 10) � 5 � 4]/4 � 71⁄4 1.67 0.757 7 (5 � 10 � 8)/3 � 72⁄3 [(2 � 8) � 10 � 5]/4 � 73⁄4 0.67 0.758 9 (10 � 8 � 7)/3 � 81⁄3 [(2 � 7) � 8 � 10]/4 � 80 0.67 1.559 12 (8 � 7 � 9)/3 � 8 [(2 � 9) � 7 � 8]/4 � 81⁄4 4.55 3.75

10 14 (7 � 9 � 12)/3 � 91⁄3 [(2 � 12) � 9 � 7]/4 � 10 4.67 4.5511 15 (9 � 12 � 14)/3 � 112⁄3 [(2 � 14) � 12 � 9]/4 � 121⁄4 3.34 2.75

Total absolute deviations: 20.36 18.5

MAD for 3-year average � 2.54

MAD for weighted 3-year average � 2.32

The weighted moving average appears to be slightly more accuratein its annual forecasts.

5-15. Using Excel or QM for Windows, the trend line is

Y � 2.22 � 1.05X

Where X � time period (1, 2, . . .) Y � demand

The 3-month moving average appears to be more accurate. How-ever, if weighted moving averages had been used, the resultsmight be different.

Three- Four-Three- Month Four- Month

Actual Month Absolute Month AbsoluteMonth Shed Sales Forecast Deviation Forecast Deviation

Jan. 10Feb. 12Mar. 13Apr. 16 11.66 4.34May 19 13.66 5.34 12.75 6.25June 23 16 7 15 8July 26 19.33 6.67 17.75 8.25Aug. 30 22.66 7.34 21 9Sept. 28 26.33 1.67 24.5 3.5Oct. 18 28 10 26.75 8.75Nov. 16 25.33 9.33 25.5 9.5Dec. 14 20.66 56.66 23 69.25

58.35 62.25

5-13.

Three-month MAD

Four-month MAD = =62 25

87 78

..

= =58 35

96 48

..

a



5-17. � � 0.3. New forecast for year 2 is last period’s forecast ��(last period’s actual demand � last period’s forecast):

new forecast for year 2 � 5,000 � (0.3)(4,000 � 5,000)

� 5,000 � (0.3)(� 1,000)

� 5,000 � 300

� 4,700The calculations are:

Year Demand New Forecast

2 6,000 4,700 � 5,000 � (0.3)(4,000 � 5,000)3 4,000 5,090 � 4,700 � (0.3)(6,000 � 4,700)4 5,000 4,763 � 5,090 � (0.3)(4,000 � 5,090)5 10,000 4,834 � 4,763 � (0.3)(5,000 � 4,763)6 8,000 6,384 � 4,834 � (0.3)(10,000 � 4,834)7 7,000 6,869 � 6,384 � (0.3)(8,000 � 6,384)8 9,000 6,908 � 6,869 � (0.3)(7,000 � 6,869)9 12,000 7,536 � 6,908 � (0.3)(9,000 � 6,908)

10 14,000 8,875 � 7,536 � (0.3)(12,000 � 7,536)11 15,000 10,412 � 8,875 � (0.3)(14,000 � 8,875)

The mean absolute deviation (MAD) can be used to determinewhich forecasting method is more accurate.

WeightedMoving Absolute Absolute

Year Demand Average Deviation Exp. Sm. Deviation

1 4,000 5,000 1,0002 6,000 4,700 1,3003 4,000 5,090 1,0904 5,000 4,500 500 4,763 2375 10,000 5,000 5,000 4,834 5,1666 8,000 7,250 750 6,384 1,6167 7,000 7,750 750 6,869 1318 9,000 8,000 1,000 6,908 2,0929 12,000 8,250 3,750 7,536 4,464

10 14,000 10,000 4,000 8,875 5,12511 15,000 12,250 12,750 10,412 14,588

Total: 18,500 26,808Mean: 2,312.5 2,437

5-18.

5-19.

Year 1 2 3 4 5 6

Forecast 410.0 422.0 443.9 466.1 495.2 521.8

Year Sales Forecast Using � �� 0.6 Forecast Using � �� 0.9

1 4502 495 410 � (0.6) (450 � 410) � 434 410 � (0.9)(450 � 410) � 4463 518 434 � (0.6) (495 � 434) � 470.6 446 � (0.9)(495 � 446) � 490.14 563 470.6 � (0.6)(518 � 470.6) � 499.0 490.1 � (0.9)(518 � 490.1) � 515.215 584 499 � (0.6) (563 � 499) � 537.4 515.21 � (0.9)(563 � 515.21) � 558.26 ? 537.4 � (0.6)(584 � 537) � 565.6 558.221 � (0.9)(584 � 558.2) � 581.4

Thus, the 3-year weighted moving average model appears to bemore accurate.

5-16. Using the forecasts in the previous problems we obtain theabsolute deviations given in the table below.

3-Yr MA 3-Yr Wt. MA Trend line Year Demand |deviation| |deviation| |deviation|

11 14 — — 0.7312 16 — — 1.6713 14 — — 1.3814 15 0.33 0.50 1.4415 10 5.00 5.00 2.5116 18 1.67 0.75 0.5517 17 0.67 0.75 2.6018 19 0.67 1.00 1.6519 12 4.00 3.75 0.2910 14 4.67 4.00 1.2411 15 3.33 2.75 1.18

Total absolutedeviations � 20.33 18.50 15.24

MAD (3-year moving average) � 2.54MAD (3-year weighted moving average) � 2.31MAD (trend line) � 1.39The trend line is best because the MAD is lowest.



5-20.

Actual � �� 0.3 Absolute � � 0.6 Absolute � � 0.9 AbsoluteYear Sales Forecast Deviation Forecast Deviation Forecast Deviation

1 450 410.0 40.0 410.0 40.0 410.0 40.02 495 422.0 73.0 434.0 61.0 446.0 49.03 518 443.9 74.1 470.6 47.4 490.1 27.94 563 466.1 96.9 499.0 64.0 515.2 47.85 584 495.2 88.8 537.4 46.6 558.2 25.86 ? 521.8 — 565.8 — 581.4 —

Total absolute deviation 372.8 259.0 190.5

MAD��0.3 � 372.8/5 � 74.56

MAD��0.6 � 259/5 � 51.8

MAD��0.9 � 190.5/5 � 38.1Because it has the lowest MAD, the smoothing constant � � 0.9gives the most accurate forecast.

5-21.

Year Sales Three-Year Moving Average

1 4502 4953 5184 563 (450 � 495 � 518)/3 � 487.6675 584 (495 � 518 � 563)/3 � 525.3336 ? (518 � 563 � 584)/3 � 555

5-22.

TimePeriod Sales

Year X Y X2 XY

1 1 450 1 4502 2 495 4 9903 3 518 9 15544 4 563 16 22525 5 2,584 125 2920

2,610 55 8166

b � 33.6

a � 421.2

Y � 421.2 � 33.6X

Projected sales in year 6,

Y � 421.2 � (33.6)(6)

� 622.8

5-23.

Three-Year Moving Time-SeriesYear Actual Sales Average Forecast Absolute Deviation Forecast Absolute Deviation

1 450 — — 454.8 4.82 495 — — 488.4 6.63 518 — — 522.0 4.04 563 487.7 75.3 555.6 7.45 584 525.3 58.7 589.2 5.26 ? 555.0 — 622.8 —

Total absolute deviation 134.0 28.0



MAD��0.3 � 74.56 (see Problem 5-20)

MADmoving average � 134/2 � 67

MADregression � 28/5 � 5.6Regression (trend line) is obviously the preferred method becauseof its low MAD.

5-24. To answer the discussion questions, two forecasting mod-els are required: a three-period moving average and a three-periodweighted moving average. Once the actual forecasts have beenmade, their accuracy can be compared using the mean average dif-ferences (MAD).

a, b.

Period Month Demand Average Weighted Average

4 Apr. 10 13.67 14.55 May 15 13.33 12.676 June 17 13.67 13.57 July 11 14 15.178 Aug. 14 14.33 13.679 Sept. 17 14 13.50

10 Oct. 12 14 1511 Nov. 14 14.33 1412 Dec. 16 14.33 13.8313 Jan. 11 14 14.6714 Feb. – 13.67 13.17

c. MAD for moving average is 2.2. MAD for weighted aver-age is 2.72. Moving average forecast for February is 13.6667.Weighted moving average forecast for February is 13.1667.

Because a three-period average forecasting method is used,forecasts start for period 4. As can be seen, the MAD for the mov-ing average is 2.2, and the MAD for the weighted moving averageis 2.7. Thus, based on this analysis, the moving average appears tobe more accurate. The forecast for February is about 14.

d. There are many other factors to consider, including sea-sonality and any underlying causal variables such as advertis-ing budget.

5-25. a.

Sum ofAbsolute

Actual Forecast ForecastWeek Miles (Ft) Error RSFE Errors MAD Track Signal

1 17 17.00 — — — — —2 21 17.00 �4.00 �4.00 4.00 4.00 13 19 17.80 �1.20 �5.20 5.20 2.60 24 23 18.04 �4.96 �10.16 10.16 3.39 35 18 19.03 �1.03 �9.13 11.19 2.80 3.36 16 18.83 �2.83 �6.30 14.02 2.80 2.257 20 18.26 �1.74 �8.04 15.76 2.63 3.058 18 18.61 �0.61 �7.43 16.37 2.34 3.179 22 18.49 �3.51 �10.94 19.88 2.49 4.21

10 20 19.19 �0.81 �11.75 20.69 2.30 5.1111 15 19.35 �4.35 �7.40 25.04 2.50 2.9612 22 18.48 �3.52 �10.92 28.56 2.60 4.20



b. The total MAD is 2.60.c. RSFE is consistently positive. Tracking signal exceeds 5MADs at week 10. This could indicate a problem.

5-26. a, b. See the accompanying table for a comparison ofthe calculations for the exponentially smoothed forecasts usingconstants of 0.1 and 0.6.

c. Students should note how stable the smoothed values forthe 0.1 smoothing constant are. When compared to actualweek 25 calls of 85, the 0.6 smoothing constant appears to doa better job. On the basis of the forecast error, the 0.6 con-stant is better also. However, other smoothing constants needto be examined.

Actual Smoothed SmoothedWeek, Value, Value, Forecast Value, Forecast

t At Ft (� �� 0.1) Error Ft (� �� 0.6) Error

1 50 50 — —2 35 50 �15 50 �153 25 48 �23 41 �164 40 46 �6 31 �85 45 45 0 37 �96 35 45 �10 42 �77 20 44 �24 38 �188 30 42 �12 27 �39 35 41 �6 29 �6

10 20 40 �20 32 �1211 15 38 �23 25 �1012 40 36 �4 19 �2113 55 36 �19 32 �2314 35 38 �3 46 �1115 25 38 �13 39 �1416 55 37 �18 31 �2417 55 38 �16 45 �1018 40 40 0 51 �1219 35 40 �5 44 �1020 60 40 �20 39 �2121 75 42 �33 51 �2322 50 45 �5 66 �1623 40 45 �5 56 �1624 65 45 �20 46 �1825 47 58



5-27. Using data from Problem 5-26, with � � 0.9

Actual SmoothedValue Value Forecast

Week At Ft Error

1 50 50 —2 35 50 �153 25 36 �114 40 26 145 45 39 66 35 44 �97 20 36 �168 30 22 89 35 29 6

10 20 34 �1411 15 21 �612 40 16 2413 55 38 1714 35 53 �1815 25 37 �1216 55 26 2917 55 52 318 40 55 �1519 35 41 �620 60 36 2421 75 58 1722 50 73 �2323 40 52 �1224 65 41 2425 62

MAD � 14.48

Note that in this problem, the initial forecast (for the first period) wasnot used in computing the MAD. Either approach is considered valid.

5-28. Exponential smoothing with � � 0.1

Month Income Forecast Error

Feb. 70.0 65.0 —March 68.5 65.0 � 0.1 (70 � 65) � 65.5 3.0April 64.8 65.5 � 0.1(68.5 � 65.5) � 65.8 �1.0May 71.7 65.8 � 0.1(64.8 � 65.8) � 65.7 6.0June 71.3 65.7 � 0.1(71.7 � 65.7) � 66.3 5.0July 72.8 66.3 � 0.1(71.3 � 66.3) � 66.8 6.0Aug. 66.8 � 0.1(72.8 � 66.8) � 67.4

MAD � 4.20


5-29. Exponential smoothing with � � 0.3

Month Income Forecast Error

Feb. 70.0 65.0 —March 68.5 66.5 2.0April 64.8 67.1 �2.3May 71.7 66.4 5.3June 71.3 68.0 3.3July 72.8 69.0 3.8Aug. 70.1

MAD � 3.34

Based on MAD, � � 0.3 produces a better forecast than � � 0.1(of Problem 5-28).


5-30. Using QM for Windows, we select Forecasting - TimeSeries and multiplicative decomposition. Then specify CenteredMoving Average and we have the following results:

a. Quarter 1 index � 0.8825; Quarter 2 index � 0.9816;Quarter 3 index � 0.9712; Quarter 4 index � 1.1569

b. The trendline is Y � 237.7478 � 3.6658X

c. Quarter 1: Y � 237.7478 � 3.6658(17) � 300.0662

Quarter 2: Y � 237.7478 � 3.6658(18) � 303.7320

Quarter 3: Y � 237.7478 � 3.6658(19) � 307.3978

Quarter 4: Y � 237.7478 � 3.6658(20) � 311.0636

d. Quarter 1: 300.0662(0.8825) � 264.7938

Quarter 2: 303.7320(0.9816) � 298.1579

Quarter 3: 307.3978(0.9712) � 298.5336

Quarter 4: 311.0636(1.1569) � 359.8719

5-31. Letting

t � time period (1, 2, 3, . . . , 16)

Q1 � 1 if quarter 1, 0 otherwise



Note: if Q1 � Q2 � Q3 � 0, then it is quarter 4.

Using computer software we get

Y � 281.6 � 3.7t � 75.7Q1 � 48.9Q2 � 52.1Q3

The forecasts for the next 4 quarters are:

Y � 281.6 � 3.7(17) � 75.7(1) � 48.9(0) � 52.1(0) � 268.7

Y � 281.6 � 3.7(18) � 75.7(0) � 48.9(1) � 52.1(0) � 299.2

Y � 281.6 � 3.7(19) � 75.7(0) � 48.9(0) � 52.1(1) � 299.7

Y � 281.6 � 3.7(20) � 75.7(0) � 48.9(0) � 52.1(0) � 355.4

5-32. For a smoothing constant of 0.2, the forecast for year 11is 6.489.

Year Rate Forecast |Error|

1 7.2 7.2 02 7 7.2 0.23 6.2 7.16 0.964 5.5 6.968 1.4685 5.3 6.674 1.3746 5.5 6.400 0.9007 6.7 6.220 0.4808 7.4 6.316 1.0849 6.8 6.533 0.267

10 6.1 6.586 0.48611 6.489

MAD = 0.722For a smoothing constant of 0.4, the forecast for year 11 is 6.458.


1 7.2 7.2 02 7 7.2 0.23 6.2 7.12 0.924 5.5 6.752 1.2525 5.3 6.251 0.9516 5.5 5.871 0.3717 6.7 5.722 0.9788 7.4 6.113 1.2879 6.8 6.628 0.172

10 6.1 6.697 0.59711 6.458

MAD = 0.673



For a smoothing constant of 0.6, the forecast for year 11 is 6.401.


1 7.2 7.2 02 7 7.2 0.23 6.2 7.08 0.884 5.5 6.552 1.0525 5.3 5.921 0.6216 5.5 5.548 0.0487 6.7 5.519 1.1818 7.4 6.228 1.1729 6.8 6.931 0.131

10 6.1 6.852 0.75211 6.401

MAD = 0.604For a smoothing constant of 0.8, the forecast for year 11 is 6.256.


1 7.2 7.2 02 7 7.2 0.23 6.2 7.04 0.844 5.5 6.368 0.8685 5.3 5.674 0.3746 5.5 5.375 0.1257 6.7 5.475 1.2258 7.4 6.455 0.9459 6.8 7.211 0.411

10 6.1 6.882 0.78211 6.256

MAD = 0.577The lowest MAD is 0.577 for a smoothing constant of 0.8.

5-33. To compute a seasonalized or adjusted sales forecast, wejust multiply each seasonal index by the appropriate trend forecast.

Y � seasonal index � Ytrend forecast

Hence for:

Quarter I: YI � (1.30)($100,000) � $130,000

Quarter II: YII � (0.90)($120,000) � $108,000

Quarter III: YIII � (0.70)($140,000) � $98,000

Quarter IV: YIV � (1.10)($160,000) � $176,000

5-34.

(Average demandfor season)

Overall averagedemand

= ×1 200

4

,season index

Year 3 demandnew annual demand

4=

Season index(average for season)

overall av=

eerage demand

sum of all values= ( )

8

(year 1 demand) +�

((year 2 demand)

2

Solution Table for Problem 5-34

AverageYear 1 Year 2 (Average Year 1- Season Season Year 3

Season Demand Demand Year 2 Demand) Demand Index Demand

Fall 200 250 225.0 250 0.90 270Winter 350 300 325.0 250 1.30 390Spring 150 165 157.5 250 0.63 189Summer 300 285 292.5 250 1.17 351

5-35. Using Excel, the trend equation is Y � 1582.61 � 612.37X.

For 2008, X � 19; Y � 1582.61 � 612.37(19) � 13217.6

For 2009, X � 20; Y � 1582.61 � 612.37(20) � 13830.0

For 2010, X � 21; Y � 1582.61 � 612.37(21) � 14442.4

The MSE from the Excel output is 1654334.7.

5-36. a. With a smoothing constant of 0.3, the forecast for 2008is 11211.2 with MSE � 3246841.

b. Using QM for Windows, the best smoothing constant is1.0. This gives the lowest MSE of 1443842.

5-37. Using Excel, the trend equation is Y � 1.1940 � 0.0095X.

For January of 2007, X � 13; Y � 1.1940 � 0.0095(13) � 1.318.

For February of 2007, X � 14; Y � 1.1940 � 0.0095(14) � 1.327.

5-38. The forecast for January 2007 would be 1.286.

The MSE with the trend equation is 0.0003. The MSE with thisexponential smoothing model is 0.0010.

5-39. With a � 0.4, forecast for 2004 � 10,339 and MAD �837. With a � 0.6, forecast for 2004 � 10,698 and MAD � 612. 5-40. Using Excel, the trend line is: GDP � 6142.7 �

441.4(time). For 2004 (time � 12) the forecast is GDP � 6142.7 �441.4(12) � 11,439.5.5-41. The trend line found using Excel is: Patients � 29.73 �

3.28(time). Note these coefficients are rounded. For the next3 years (time � 11, 12, and 13) the forecasts for the number ofpatients are:

Patients � 29.73 � 3.28(11) � 65.8 Patients � 29.73 � 3.28(12) � 69.1 Patients � 29.73 � 3.28(13) � 72.4

The coefficient of determination is 0.85, so the model is a fairmodel.

SOLUTIONS TO INTERNET HOMEWORK PROBLEMS



Deposits and GSP over Time

0

20

40

60

80

100

1950 1960 1970 1980 1990 2000 2010

Time

DEPOSITS

GSP

5-42. The trend line found using Excel is: Crime Rate � 51.98� 6.09(time). Note these coefficients are rounded. For thenext 3 years (time � 11, 12, and 13) the forecasts for the crimerates are:

Crime Rate � 51.98 � 6.09(11) � 118.97Crime Rate � 51.98 � 6.09(12) � 125.06Crime Rate � 51.98 � 6.09(13) � 131.15

The coefficient of determination is 0.96, so this is a very goodmodel.

5-43. The regression equation (from Excel) is: Patients � 1.23 �0.54(crime rate). Note these coefficients are rounded. If the crimerate is 131.2, the forecast number of patients is:

Patients � 1.23 � 0.54(131.2) � 72.1

If the crime rate is 90.6, the forecast number of patients is:

Patients � 1.23 � 0.54(90.6) � 50.2

The coefficient of determination is 0.90, so this is a good model.

5-44. With a � 0.6, forecast for 2003 � 86.2 and MAD �3.42. With a � 0.2, forecast for 2003 � 63.87 and MAD � 7.23.The model with a � 0.6 is better since it has a lower MAD.

5-45. With a � 0.6, forecast for 2003 � 4.86 and MAD �0.23. With a � 0.2, forecast for 2003 � 4.52 and MAD � 0.48.The model with a � 0.6 is better since it has a lower MAD.

5-46. The trend line (coefficients from Excel are rounded) fordeposits is:

Deposits � �18.968 � 1.638(time)For 2003, 2004, and 2005, time � 45, 46, and 47 respectively. Theforecasts are:

Deposits � �18.968 � 1.638(45) � 54.7Deposits � �18.968 � 1.638(46) � 56.4Deposits � �18.968 � 1.638(47) � 58.0

The trend line (coefficients from Excel are rounded) for GSP is:GSP � 0.090 � 0.112(time). The forecasts are:GSP � 0.090 � 0.112(45) � 5.1GSP � 0.090 � 0.112(46) � 5.2GSP � 0.090 � 0.112(47) � 5.4

5-47. The regression equation from Excel is Deposits � �17.64 � 13.59(GSP)

In the scatterplot of this data that follows, the pattern appears tochange around 1985. There are definitely different relationshipsbefore 1985 and after 1985, so perhaps the model should be devel-oped with 1985 as the first year of data.



50000

SWU Football Attendance

40000

30000

20000

10000

0

2001

Att

end

ance

2003 2005

Year

2007

FORECASTING ATTENDANCE AT SWU FOOTBALL GAMES

1. Because we are interested in annual attendance and thereare six years of data, we find the average attendance in eachyear shown in the table below. A graph of this indicates a lin-ear trend in the data. Using Trend Analysis in the forecastingmodule of QM for Windows we find the equation:

Y � 31,660 � 2,305.714X

Where Y is attendance and X is the time period (X � 1 for2002, 2 for 2003, etc.).For this model, r2 � 0.98 which indicates this model is veryaccurate.

Attendance in 2008 is projected to beY � 31,660 � 2,305.714(7) � 47,800

Attendance in 2009 is projected to be

Y � 31,660 � 2,305.714(8) � 50,105At this rate, the stadium, with a capacity of 54,000, will be“maxed out” (filled to capacity) in 2011.

Year 2002 2003 2004 2005 2006 2007

Attendance 34840 35380 38520 40500 43320 45820

2. Based upon the projected attendance and tickets prices of$20 in 2008 and $21 (a 5% increase) in 2009, the projectedrevenues are:

47,800(20) � $958,000 in 2008 and50,105(21) � $1,052,205 in 2009.

3. The school might consider another expansion of the sta-dium, or raise the ticket prices more than 5% per year. An-other possibility is to raise the prices of the best seats whileleaving the end zone prices more reasonable.

SOLUTION TO INTERNET CASES

SOLUTION TO AKRON ZOOLOGICAL PARK CASE

1. The instructor can use this question to have the student calcu-late a simple linear regression, using real-world data. The idea isthat attendance is a linear function of expected admission fees.Also, the instructor can broaden this question to include severalother forecast techniques. For example, exponential smoothing,last-period demand, or n-period moving averages can be assigned.It can be explained that mean absolute deviation (MAD) is one ofbut a few methods by which analysts can select the more appropri-ate forecast technique and outcome.

First, we perform a linear regression with time as the inde-pendent variable. The model that results is

admissions � 44,352 � 9,197 � year(where year is coded as 1 � 1989, 2 � 1990, etc.)

r � 0.88

MAD � 9,662

MSE � 201,655,824So the forecasts for 1999 and 2000 are 145,519 and 154,716, re-spectively. Using a weighted average of $2.875 to represent gatereceipts per person, revenues for 1999 and 2000 are $418,367 and$444,808, respectively.

To complicate the situation further, students may legitimatelyuse a regression model to forecast admission fees for each of thethree categories, or for the weighted average fee. This numberwould then replace $2.875.

Here is the result of a linear regression using weighted aver-age admission fees as the predicting (independent) variable.Weights are obtained each year by taking 35% of adult fees, plus50% of children’s fees, plus 15% of group fees. The weighted feeseach year (1989–1998) are $0.975, $0.975, $0.975, $0.975,$1.275, $1.775, $1.775, $2.275, $2.20, $2.875.

Gate admissions � 31,451 � 39,614 � (average fee in given year)

r � 0.847

MAD � 13,212

MSE � 254,434,912

If we assume that admission fees are not raised in 1999 and 2000, expected gate admissions � 145,341 in each year and



revenues � $417,856. Comparing the earlier time-series model tothis second regression, we note that the r is higher and MAD andMSE are lower in the time-series approach.2. The student should respond that the other factors are the vari-ability of the weather, the special events, the competition, and therole of advertising.

Kwik Lube1. The relationship between Kwik Lube sales (y), average indus-try sales (x), and year (t with t � 1 corresponding to 1972) isshown in the table below. The x and y values are in thousands ofdollars. One could try a multiple regression analysis but the corre-lation of y with just x is 0.998, leading one to use the simple linearregression equation: y � 2.99x � 1.42.

t x y

1 22 682 25 753 24 754 26 785 33 996 35 1047 39 1208 44 133

The year 1971 was excluded since the Kwik Lube revenues werenot for an entire year. 1979 (t � 8) was the last year of Kwik Lubeoperation without the competition from Speedy Lube. The fore-casted sales for 1980 would be estimated using the average indus-try sales of $47,000 for x:

y � 2.99(47) � 1.42 � 141.95

and the forecasted sales for 1981 would use the industry sales of$52,000:

y � 2.99(52) � 1.42 � 156.90

The estimated lost sales is the difference between the forecastedand actual sales: (141,950 � $156,900) � ($111,000 � $111,000) �$76,850.

A 95% prediction interval for 1980 is 141.95 � 5.20 and for1981 is 156.90 � 5.80. Thus, despite the danger of extrapolation,the results of a regression outside the range of the data, one can bereasonably certain that the lost sales were at least $65,850.2. Without the questionnaire study, the best estimate of lost saleswould be from the regression of y on t:

y � 9.38t � 51.8

with a somewhat lower correlation. The estimated lost sales wouldbe $59,820, about $20,000 less than the estimate based on averageindustry sales. Even recovering as little as 10 percent of this dif-ference would pay for the study.

3. The lawsuit filed by Dick Johnson should discuss two basicareas which will build a sound case for damages being awarded inhis favor.

The first factor involves the concept behind setting up a fran-chise. Franchises are designed so that independent owners canstart a business with a well-known name (and consequently, withan already-captured market). This, coupled with proven strategiesand expertise given to a franchise purchaser by the franchiseseller, reduces the usually high probability of a new businessgoing under in its infancy stage. The franchise fee is the cost paidfor the reduced risk of a new enterprise.

Naturally, the franchising firm will protect itself against compe-tition in a franchise contract. A franchise holder who violates suchclauses has, in essence, gained free proven strategies and has capital-ized on them. Thus, the franchising firm has been damaged by thefact that a competitor has gained information without paying for it.

This is the case with Kwik Lube. A franchise owner, T. A.Williams, has benefited from Johnson’s expertise more than is justified by the monetary gains earned from franchise fees. This isnot simply an economic issue, however, for such a situation was



thought of before by Johnson. He had sought to protect himselfwith a noncompetition clause in his franchised contract. Thus,Williams is legally in the wrong for his breach of contract.

What this first area of discussion in the lawsuit does is to deter-mine that there, in fact, has been damage done to the plaintiff, John-son. The second area to be discussed in the lawsuit should deal withhow those damages can be mitigated by the defendant, Williams.

Usually in lawsuits, there is a problem with measuring thedamage done. Johnson, however, can measure his loss by forecast-ing sales and then comparing actual sales to predicted sales.

In summary, the lawsuit should discuss how damage was in-curred to plaintiff, Johnson, and how said damage should and/orcould be mitigated. A well-presented lawsuit or petition to the courtshould result in a favorable judgment for the owner of Kwik Lube.


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