Chapter 4 - Forecasting Solution Manual.doc

CHAPTER 4 F O R E C A S T I N G

C H A P T E R

Forecasting

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

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DISCUSSION QUESTIONS 1. Qualitative models incorporate subjective factors into the forecasting model. Qualitative models are useful when subjective factors are important. When quantitative data are difficult to obtain, qualitative models may be appropriate. 2. Approaches are qualitative and quantitative. Qualitative is relatively subjective; quantitative uses numeric models. 3. Shortrange (under 3 months), mediumrange (3 months to 3 years), and longrange (over 3 years). 4. The steps that should be used to develop a forecasting system are: (a) Determine the purpose and use of the forecast (b) Select the item or quantities that are to be forecasted (c) Determine the time horizon of the forecast (d) Select the type of forecasting model to be used (e) Gather the necessary data (f) Validate forecasting model

the

(g) Make the forecast (h) Implement evaluate the results

and

5. Any three of: sales planning, production planning and budgeting, cash budgeting, analyzing various operating plans. 6. There is no mechanism for growth in these models; they are built exclusively from historical demand values. Such methods will always lag trends. 7. Exponential smoothing is a weighted moving average

4

where all previous values are weighted with a set of weights that decline exponentially. 8. MAD, MSE, and MAPE are common measures of forecast accuracy. To find the more accurate forecasting model, forecast with each tool for several periods where the demand outcome is known, and calculate MSE, MAPE, or MAD for each. The smaller error indicates the better forecast.

9. The Delphi technique involves: (a) Assembling a group of experts in such a manner as to preclude direct communication between identifiable members of the group (b) Assembling the responses of each expert to the questions or problems of interest (c) Summarizing these responses (d) Providing each expert with the summary of all responses (e) Asking each expert to study the summary of the responses and respond again to the questions or problems of interest. (f) Repeating steps (b) through (e) several times as necessary to obtain convergence in responses. If convergence has not been obtained by the end of the fourth cycle, the responses at that time should probably be accepted and the process terminated—little additional convergence is likely if the process is continued. 10. A time series model predicts on the basis of the

assumption that the future is a function of the past, whereas an associative model incorporates into the model the variables of factors that might influence the quantity being forecast. 11. A time series is a sequence of evenly spaced data points with the four components of trend, seasonality, cyclical, and random variation. 12. When the smoothing constant, , is large (close to 1.0), more weight is given to recent data; when  is low (close to 0.0), more weight is given to past data. 13. Seasonal patterns are of fixed duration and repeat regularly. Cycles vary in length and regularity. Seasonal indices allow “generic” forecasts to be made specific to the month, week, etc., of the application. 14. Exponential smoothing weighs all previous values with a set of weights that decline exponentially. It can place a full weight on the most recent period (with an alpha of 1.0). This, in effect, is the naïve approach, which places all its emphasis on last period’s actual demand.

18. Independent variable (x) is said to explain variations in the dependent variable (y). 19. Nearly every industry has seasonality. The seasonality must be filtered out for good mediumrange planning (of production and inventory) and performance evaluation. 20. There are many examples. Demand for raw materials and component parts such as steel or tires is a function of demand for goods such as automobiles. 21. Obviously, as we go farther into the future, it becomes more difficult to make forecasts, and we must diminish our reliance on the forecasts.

ETHICAL DILEMMA This exercise, derived from an actual situation, deals as much with ethics as with forecasting. Here are a few points to consider: 

15. Adaptive forecasting refers to computer monitoring of tracking signals and selfadjustment if a signal passes its present limit. 16. Tracking signals alert the user of a forecasting tool to periods in which the forecast was in significant error. 17. The correlation coefficient measures the degree to which the independent and dependent variables move together. A negative value would mean that as X increases, Y tends to fall. The variables move together, but move in opposite directions.






No one likes a system they don’t understand, and most college presidents would feel uncomfortable with this one. It does offer the advantage of depoliticizing the funds al location if used wisely and fairly. But to do so means all parties must have input to the process (such as smoothing constants) and all data need to be open to everyone. The smoothing constants could be selected by an agreed upon criteria (such as lowest MAD) or could be based on input from experts on the board as well as the college. Abuse of the system is tied to assigning

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alphas based on what results they yield, rather than what alphas make the most sense.

2. What happens to the graph when alpha equals one? The forecast follows the same pattern as the demand

4. At what level of alpha is the mean absolute deviation (MAD) minimized? alpha = .16

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No, they are not the same values. For example, an intercept of 57.81 with a slope of 9.44 yields a MAD of 7.17.

4.2 (a) No, the data appear to have no consistent pattern (see part d for graph).



Regression is open to abuse as well. Models can use many years of data yielding one result or few years yielding a totally different forecast. Selection of associative variables can have a major impact on results as well.

Active Model Exercises* ACTIVE MODEL 4.1: Moving Averages 1. What does the graph look like when n = 1? The forecast graph mirrors the data graph but one period later. 2. What happens to the graph as the number of periods in the moving average increases? The forecast graph becomes shorter and smoother. 3. What value for n minimizes the MAD for this data? n = 1 (a naïve forecast)

ACTIVE MODEL 4.2: Exponential Smoothing 1. What happens to the graph when alpha equals zero? The graph is a straight line. The forecast is the same in each period.

Year

1

2

3

4

5

6

7

Demand

7

9

5

9.0 7.0 6.4

13.0 7.7 7.8

8.0 9.0 11.0

12.0 10.0 9.6

(b) 3-year moving (c) 3-year weighted (except for the first forecast) but is offset by one period. This is a naïve forecast. 3. Generalize what happens to a forecast as alpha increases. As alpha increases the forecast is more sensitive to changes in demand. *Active Models 4.1, 4.2, 4.3, and 4.4 appear on our Web site, www.pearsonhighered.com/h eizer.

ACTIVE MODEL 4.3: Exponential Smoothing with Trend Adjustment

1. Scroll through different values for alpha and beta. Which smoothing constant appears to have the greater effect on the graph? alpha

8

9

10

11 Forecas t

13.0 9.0 11.0 7.0 11.0 11.0 11.3 11.0 10.9 12.2 10.5 10.6

END-OF-CHAPTER PROBLEMS 4.1 (a)

374 + 368 + 381  374.33 pints 3

(b)

Weight Week Pint ed of s Moving Use Averag d e August 360Error Forecasti Week of Pints Forecast 31ng  .20 Foreca Septe Error 389 st 381 mber 7 0 0 August 31 360 360 360 .1 = September 7 389 360 29 5.8 365.8 38.1 September 14 410 365.8 44.2 8.84 374.64 Septe 4101.272 375.91 368 September 21 381 374.64 6.36 mber 2  .3 14 = September 28 368 375.912 –7.912 –1.5824 374.32 96 110. 4 October 5 374 374.3296–.3296 –.06592 374.26 Septe 381 36 374 mber  .6 2. With beta set to zero, 21 = find the best alpha and 224. observe the MAD. Now find 4 the best beta. Observe the Septe 368 MAD. Does the addition of a mber 28 trend improve the forecast? Octobe 374 alpha = .11, MAD = r5 2.59; beta above .6 changes Forec the MAD (by a little) to ast 2.54. 372.9

ACTIVE MODEL 4.4: Trend Projections 1. What is the annual trend in the data? 10.54 2. Use the scrollbars for the slope and intercept to determine the values that minimize the MAD. Are these the same values that regression yields?


9.0 8.4

(c) The forecast is 374.26.

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(d) The threeyear moving average appears to give better results.

(c) The banking industry has a great deal of seasonality in its processing requirements 4.5 (a)

4.5 (c) Weighted 2 year M.A. with .6 weight for most recent year. Year

Mileage

1

(b)

4 5

The coming January = December = 23

Forecast

3,000 4,000 3,400 3,800 3,700

3,700 + 3,800 2 =3,750 miles 2 3

(b) [i] Naïve

[ii] 3month moving (20 + 21 + 23)/3 = 21.33

3,600 3,640 3,640

[iii] 6month weighted [(0.1  17) + (.1  18)

Forecast for year 6 is 3,740 miles.

4.3

Year

1

2

3

Demand Naïve Exp. Smoothing

7

9.0 7.0 6.4

5.0 9.0 7.4

6

4

5

6

7

8

9

10

11

Forecast

9.0 13.0 8.0 12.0 13.0 9.0 11.0 7.0 5.0 9.0 13.0 8.0 12.0 13.0 9.0 11.0 6.5 7.5 9.7 9.0 10.2 11.3 10.4 10.6

Year

Mileage

1 2 3 4 5

3,000 4,000 3,400 3,800 3,700

MAD 

+

7.0 9.2

Two-Year � 420 �  Moving Average MAD  140 � � 3 � �

300  100. 3

4.5 (d) 3,500 3,700 Year 3,600 1 2 3 4 5

Mileage Forecast 3,000 4,000 3,400 3,800 3,700

Forecast Error

3,000 3,000 3,500 3,450 3,625 Total

(0.1  20) + (0.2  20) + (0.2  21) + (0.3  23)]/ 1.0

1,000

The forecast is 3,663 miles.

4.6 Y Sales January February March April May June July August September October November December

Naïve tracks the ups and downs best but lags the data by one period. Exponential smoothing is probably better because it smoothes the data and does not have as much variation.

Sum Average

TEACHING NOTE: Notice how well exponential smoothing forecasts the naïve.

= 20.6

X Period

20 21 15 14 13 16 17 18 20 20 21 23 218 18.2

(a)

4.4 (a) FJuly  FJune + 0.2(Forecasting error)  42 + 0.2(40 �42)  41.6 (b) FAugust  FJuly + 0.2(Forecasting error)  41.6 + 0.2(45 - 41.6)  42.3


[iv] Exponential smoothing with alpha = 0.3 FOct  18 + 0.3(20 - 18)  18.6

FNov  18.6 + 0.3(20 - 18.6)  19.0

FDec  19.02 + 0.3(21 - 19.02)  1

FJan  19.6 + 0.3(23 - 19.6)  20.6 10 11 12 78

[v] Trend � x  78, x  6.5, � y = 218, y  18. � xy - nx y � x 2 - nx 2 1474 - (12)(6.5)(18.2) 54.4 b   0.38 650 - 12(6.5)2 143 a  y - bx a  18.2 - 0.38(6.5)  15.73 b

Forecast = 15.73 + .38(1 3) = 20.67, where next January

CHAPTER 4 F O R E C A S T I N G 4.9

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(d) Table for Problem 4.9(d):  = .1

Month

Table for Problem 4.9 (a, b, c) Price per Forecast |Error| Chip

January February March April May June July August Septembe r October November December

$1.80 $1.80 1.67 1.80 Price per 1.70 1.79 Month Chip 1.85 1.78 January $1.80 1.90 1.79 February 1.67 1.87 1.80 March1.80 1.70 1.80 April 1.83 1.85 1.80 May 1.90 1.70 1.81 June 1.87 July 1.80 1.65 1.80 August 1.83 1.70 1.78 September 1.70 1.75 1.77 October 1.65 Totals November 1.70 December 1.75 MAD (total/12)

 = .3

 = .5

Forecast |Error| Forecast

$.00 $1.80 Two-Month .13 1.80 .09Moving 1.76 Average .07 1.74 .11 1.77 .07 1.81 .00 1.735 1.83 .03 1.685 1.82 1.775 .11 1.82 1.875 1.885 .15 1.79 1.835 .08 1.75 1.815 .02 1.765 1.73 $.8 1.675 6 1.675 $.072

Forecast

$.00 Three.13 Month Moving .06 Average .11 .13 .06 .03 1.723 .01 1.740 .12 1.817 1.873 .14 1.857 .05 1.833 .02 1.777 $.8 1.727 6 1.683 $.072 Totals

|Error| |Error|

$1.80 $.00 Two-Month 1.80 .13 1.74 Moving .04 1.72 Average .13 1.78 .12 1.84 .03 1.86 .035.06 1.83 .165.00 .125 1.83 .13 .005 .085 1.76 .11 .005 1.71 .01 .115 1.70 .115.05 $.81 .025 .075 . $.067 5

ThreeMonth Moving Average

 = .5 is preferable, using MAD, to  = .1 or  = .3. One could 4.10

Year Demand (a) 3-year moving (b) 3-year weighted

is the 13th month.

1

2

3

4

6

4

forecast becomes 56.3, or 56 patients.

4

5

6

5.0 10.0 8.0 4.7 5.0 6.3 4.5 5.0 7.3

4.8 (a)

7

8

9

10

11

Forecast

7.0 7.7 7.8

9.0 8.3 8.0

12.0 8.0 8.3

14.0 9.3 10.0

15.0 11.7 12.3

13.7 14.0

(96 + 88 + 90)  91.3 3

results appear in the figure below.

(c) Only trend provides an equation that can extend beyond one month 4.7 Present = Period (week) 6. ��1 � �1 � �1 � �1 � � a) So: F7  ��3 �A6 + � 4 �A5 + �4 �A4 + �6 �A3 � 1.0 � � � � � � � ��

where

�1 � �1 � �1 � �1 �  � �(52) + � �(63) + � �(48) + � �(70) = 56.76 patients, 3 4 4 � � � � � � �6 � or 57 patients

1 1 1 1 1.0 = � weights , , , 3 4 4 6

b) If the weights are 20, 15, 15, and 10, there will be no change in the forecast because these are the same relative weights as in part (a), i.e., 20/60, 15/60, 15/60, and 10/60. c) If the weights are 0.4, 0.3, 0.2, and 0.1, then the

MAD = 13.5/5 = 2.7

(d) MSE = 66.75/5 = 13.35 (e) MAPE = 14.94%/5 = 2.99% 4.9 (a, b) The computations for both the two and threemonth averages appear in the table; the


(c) MAD (twomonth moving average) = .750/10 = .075 MAD (threemonth moving average) = .793/9 = . 088 Therefore, the twomonth moving average seems to have performed better.

.127 .160 .053 .073 .027 .133 .127 .027 .067 .793

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4.11 (a) Year

(b) |Error| = |Actual – Forecast|

1 2 3 4 5 6 7 8 9 10 11 Forecast Year 1 2 3 4 5 6 7 8 9 10 11 MAD Demand 4 6.0 4.0 5.0 10.0 8.0 7.0 9.0 12.0 14.0 15.0 Exp. smoothing 5.1 4.6 2.4 Exp. Smoothing 5 4.7 5.1 14.8 1.3 4.8 1.1 6.4 0.2 6.9 5.2 6.9 1.6 7.5 0.1 8.9 2.1 10.4 4.5 11.8 These calculations were completed in Excel. Calculations are slightly different in Excel OM and POM for Windows due to rounding differences.

4.12

2 3 Actual 4 Demand 5

t

Day

1

Monday

88

2

Tuesday

72

3

Wednesday

68

4

Thursday

48

5

Friday

6

(c) The forecasts are about the same.

F2 = 88 + .25(88 – 88) = 88 + 0 = 88 F3 = 88 + .25(72 – 88) = 88 – 4 = 84 F4 = 84 + .25(68 – 84) = 84 – 4 = 80 F5 = 80 + .25(48 – 80) = 80 – 8 = 72 4.13 (a) Exponential smoothing,  = 0.6: Year 1 2 3 4 5 6

Demand 45 50 52 56 58 ?

(45 + 50 + 52)/3 = 4 (50 + 52 + 56)/3 = 52.7 (52 + 56 + 58)/3 = 55.3

?

 = 12.3 MAD = 6.2 (c) Trend projection:

Ft = Ft–1 + (At–1 – Ft–1) Let  = .25. Let Monday forecast demand = 88

50 52 56 58

Year

Demand

1 2 3 4 5 6

45 50 52 56 58 ?

Trend Projection 42.6 42.6 42.6 42.6 42.6 42.6

+ + + + + +

3.2 3.2 3.2 3.2 3.2 3.2

     

1 2 3 4 5 6

= = = = = =

45.8 49.0 52.2 55.4 58.6 61.8

 = 3.2 MAD = 0.64 Y  a + bX b

�XY �nXY �X 2 �nX 2

a  Y �bX

Exponential Smoothing X Y 41 1 41.0 + 0.6(45–41) = 45 43.4 2 5047.4 43.4 + 0.6(50–43.4) = 3 5250.2 47.4 + 0.6(52–47.4) = 4 5653.7 50.2 + 0.6(56–50.2) = 5 5856.3 53.7 + 0.6(58–53.7) =

XY

X2

45 100 156 224 290

1 4 9 16 25

Then: X = 15, Y = 261, 2 XY = 815, X = 55, X = 3, Y = 52.2 Therefore: Exponential smoothing,  = 815 �5 �3 �52.2 0.9: b  3.2 55 �5 �3 �3 Exponential Year Demand Smoothing a  52.20 �3.20 �3  42.6 Y6  42.6 + 3.2 �6  61.8 1 45 41  = 25.3 MAD = 5.06

2 3 4 5 6

50 52 56 58 ?

41.0 + 0.9(45–41) =(d) 44.6 Comparing the 44.6 + 0.9(50–44.6 ) = 49.5 results of the forecasting 49.5 + 0.9(52–49.5) = 51.8 methodologies for parts (a), 51.8 + 0.9(56–51.8) = 55.6 (b), and (c). 55.6 + 0.9(58–55.6) = 57.8 Forecast Methodology

 = 18.5 MAD = 3.7 (b) 3year average: Year 1

Demand

Exponential smoothing,  = 0.6 Exponential smoothing,  = 0.9 moving 3-year moving average Trend projection Three-Year Based on a mean absolute Moving Average

45


deviation criterion, the trend

MAD 5.06 3.7 6.2 0.64


projection is to be preferred over the exponential smoothing with  = 0.6, exponential smoothing with  = 0.9, or the 3year moving average forecast methodologies. 4.14 Method 1: MAD: (0.20 + 0.05 + 0.05 + 0.20)/4 = .125  better MSE : (0.04 + 0.0025 + 0.0025 + 0.04)/4 = . 021 Method 2: MAD: (0.1 + 0.20 + 0.10 + 0.11) / 4 = . 1275 MSE : (0.01 + 0.04 + 0.01 + 0.0121) / 4 = .018  better

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4.15 Year

Sales

2005 2006 2007 2008

450 495 518 563

2009

584

Forecast Three-Year Moving Average

(450 + 495 + 518)/3 = 487.7 (495 + 518 + 563)/3 = 525.3 (518 + 563 + 584)/3 = 555.0

2010

Forecast Expon Year

Sales

2005 2006 2007 2008

450 495 518 563

2009

584

2010

Smoothing  = 410.0 410 + 446 + 490.1 515.2 515.2 558.2 558.2 581.4

0.9(450 – 410) 0.9(495 – 446) + 0.9(518 – + 0.9(563 – + 0.9(584 –

4.16 Year

Time Period X

2005 2006 2007 2008 2009

1 2 3 4 5

Sales

2610

X  3, Y  522

b

�X 2 - nX 2

MAD 0.3  74.6 MAD 0.6  51.8 MAD 0.9  38.1

Y  a + bX �XY - nXY

(Refer to Solved Problem 4.1) For  = 0.3, absolute deviations for 2005–2009 are 40.0, 73.0, 74.1, 96.9, 88.8, respectively. So the MAD = 372.8/5 = 74.6.

8166 - (5)(3)(522) 336    33.6 55 - (5)(9) 10

a  Y - bX  522 - (33.6)(3)  421.2 y  421.2 + 33.6 x y  421.2 + 33.6 �6  622.8 Year

Sales

Forecast Trend

2005 2006 2007 2008 2009 2010

450 495 518 563 584

454.8 488.4 522.0 555.6 589.2 622.8

Because it gives the lowest MAD, the smoothing constant of  = 0.9 gives the most accurate forecast. 4.18 We need to find the smoothing constant . We know in general that Ft = Ft–1 + (At–1 – Ft–1); t = 2, 3, 4. Choose either t = 3 or t = 4 (t = 2 won’t let us find  because F2 = 50 = 50 + (50 – 50) holds for any ). Let’s pick t = 3. Then F3 = 48 = 50 + (42 – 50) or 48 = 50 + 42 – 50 or –2 = –8

4.17 Year

Sales

2005 2006 2007 2008

450 495 518 563

2009 2010

584

Forecast Exponential Smoothing 410.0 410 + 434 + 470.6 499.0 499 + 537.4 565.6

0.6(450 – 410) = 434.0 0.6(495 – 434) = 470.6 + 0.6(518 – 470.6) = 0.6(563 – 499) = 537.4 + 0.6(584 – 537.4) =


So, .25 =  Now we can find F5 : F5 = 50 + (46 – 50) F5 = 50 + 46 – 50 = 50 – 4 For  = .25, F5 = 50 – 4(.25) = 49 The forecast for time period 5 = 49 units.

34


4.19 Trend adjusted exponential smoothing:  = 0.1,  = 0.2

4.20 Trend

Unadjust ed Month Income Forecast Trend Februar y March April May June July August

70.0

65.0

0.0

68.5 64.8 71.7 71.3 72.8

65.5 65.9 65.92 66.62 67.31 68.16

0.1 0.16 0.13 0.25 0.33

MAD = 24.3/6 = 4.05, MSE = 113.2/6 = 18.87. Note that all numbers are rounded. Note: To use POM for Windows to solve this problem, a period 0, which contains the initial forecast and initial trend, must be added.

Month February March April May June July Totals Average August Forecast

Demand (y) 70.0 68.5 64.8 71.7 71.3 72.8 419.1

adjusted

Unadjusted Forecast

Trend

65.0 65.5 66.16 66.57 67.49 68.61

Adjusted Forecast

Error

|Error|

65.0 65.9 66.77 67.02 68.31 69.68

5.00 2.60 –1.97 4.68 2.99 3.12

5.0 2.6 1.97 4.68 2.99 3.12 20.3 6 3.39 (MAD)

0 0.4 0.61 0.45 0.82 1.06

16.42

69.85

2.74 (Bias)

71.30

Error2

25.00 6.76 3.87 21.89 8.91 9.76 76.19

12.70 (MSE)

Based upon the MSE criterion, the exponential smoothing with  = 0.1,  = 0.8 is to be preferred over the exponential smoothing with  = 0.1,  = 0.2. Its MSE of 12.70 is lower. Its MAD of 3.39 is also lower than that in Problem 4.19. FIT8  F8 + T8  27.14 + 2.45  29.59 exponential smoothing:  = 0.1,  = 0.8 F  A + ( 1 � ) ( F + T )  ( 0.2 ) ( 28) 9

8

4.21 F5   A4 + ( 1 � ) ( F4 + T4 )  ( 0.2 ) ( 19) + ( 0.8) ( 20.14 )

8

8

+ ( 0.8) ( 29

 3.8 + 16.11  19.91 T9   ( F9 �F8 ) + ( 1 �) T8  ( 0.4 ) ( 29.28 T5   ( F5 �F4 ) + ( 1 � ) T4  ( 0.4 ) ( 19.91 �17.82 ) + ( 0.6 ) ( 2.4 + ( 0.6 ) ( 2.32 )  0.4 ( 2.09 ) FIT9  F9 + T9  29.28 + 2.32  31.60 + 1.39  0.84 + 1.39  2.23

FIT5  F5 + T5  19.91 + 2.23  22.14 F6  A5 + ( 1 � ) ( F5 + T5 )  ( 0.2) ( 24) + ( 0.8) ( 22.14 )

 4.8 + 17.71  22.51

T6   ( F6 �F5 ) + ( 1 � ) T5  0.4 ( 22.51 �19.91) + 0.6 ( 2.23)

 0.4 ( 2.6 ) + 1.34

 1.04 + 1.34  2.38

FIT6  F6 + T6  22.51 + 2.38  24.89 4.22 F7  A6 + (1 �)( F6 + T6 )  (0.2)(21) + (0.8)(24.89)  4.2 + 19.91  24.11 T7  ( F7 �F6 ) + (1 �)T6  (0.4)(24.11 �22.51) + (0.6)(2.38)  2.07 FIT7  F7 + T7  24.11 + 2.07  26.18 F8  A7 + (1 � )( F7 + T7 )  (0.2)(31) + (0.8)(26.18)  27.14 T8   ( F8 �F7 ) + ( 1 � ) T7  0.4 ( 27.14 �24.11)

+ 0.6 ( 2.07)  2.45



Assume

4.23 Students must determine the naïve forecast for the four months. The naïve forecast for March is the February actual of 83, etc. (a)

Appearances X

Demand Y

3 4 7 6 8 5 9

3 6 7 5 10 7 ?

Actual Forecas |Error| t March

101

120

April

96

114

May

89

110

Averages

y = 55

108

101

83

18

April May

96 89

101 96

June

108

89

19

The observations obviously do not form a straight line but do tend to cluster about a straight line over the range shown. (b) Leastsquares regression: Y  a + bX b

�XY - nXY �X 2 - nX 2

a  Y - bX

30 �4(2.5)

2



650 30

S Year1 Year2 Average Seaso Deman DemanYear1-Year2 D n d d Demand Fall Winter Spring Summe r

200 350 150 300

250 300 165 285

225.0 325.0 157.5 292.5

Sum of Ave Yr1 to Yr2 Demand 4 Average Yr1 to Yr2 Demand Seasonal index = Average Seasonal Demand New Annual Demand Yr3  �Seasonal index 4 1200  �Seasonal index 4

Average seasonal demand 

49  12.25 4 48.49% MAPE (for na� ve)   12.12%. 4 (c) If there are nine Naïve outperforms performances by Stone management. Temple Pilots, the estimated (c) MAD for the manager’s sales are: technique is 14.5, while Y9  .676 + 1.03 �9  .676 + 9.27  9.93 drums MAD for the naïve

4.24 (a) Graph of demand

650 �4(2.5)(55)

Average Yr1 to Yr2 � Yr1 Demand + Yr2 Demand � � � 2 Demand for season � �

MAD (for na� ve) 

forecast is only 12.25. MAPEs are 15.29% and 12.12%, respectively. So the naïve method is better.



A

b

March

2

4.26

Actual Naïve |Error|

(b)

2

The regression line is y = 5 + 20x. The forecast for May (x = 5) is y = 5 + 20(5) = 105.

X = 33, Y = 38, XY = 227, X2 = 199, X = 5.5, Y = 6.33. Therefore:

108

�xy - n x y

�x �nx 100   20 5 a  y �bx  55 �(20)(2.5) 5

227 �6 �5.5 �6.333  1.0286 199 �6 �5.5 �5.5 a  6.333 �1.0286 �5.5  .6762 58 Y  .676 + 1.03 x (rounded) MAD (for management)   14.5 4 The following figure shows 61.16% MAPE (for management)  both the data and the resulting  15.29% 4 equation: June

b

35

�10 drums

(d) R = .82 is the correlation coefficient, and R2 = .68 means 68% of the variation in sales can be explained by TV appearances.

20,000  1, 250 16 6,000 Average over spring:  1,500 4 1,500 Spring index:  1.2 1, 250 Average over all seasons:

�5,600 � Answer: � (1.2)  1,680 sailboats � 4 � � 4.27

4.25

Month January February March April Totals

Winter Number of Accidents (y) 30 40 60 90 220


2006 2007 2008 2009

1,400 1,200 1,000 900 4,500

Spring

Sum

1,500 1,400 1,600 1,500 6,000

1, 2, 2, 1, 7,

4.28

Avera

36


Quarte 2007 2008 r Winter Spring Summer Fall

73 104 168 74

65 82 124 52

2009

89 146 205 98

4.29 2011 is 25 years beyond 1986. Therefore, the 2011 quarter numbers are 101 through 104.

Then y = a + bx, where y = number sold, x = price, and

(1) Quarter

(2) Quarter Number

(3) Forecast (77 + . 43Q

Winter Spring Summer Fall

101 102 103 104

120.43 120.86 121.29 121.72

4.30 Given Y = 36 + 4.3X (a) Y = 36 + 4.3(70) = 337 (b) Y = 36 + 4.3(80) = 380 (c) Y = 36 + 4.3(90) = 423 4.31

Let x1, x2 , K , x6 be the prices and y1, y2 , K , y6 be the number sold. 4.33 (a) See the table below. 6 �xi X  i 12,880  Average price = 3.2583 - 5(3)(180) 2,880 - 2,700 (1) b6  55 - 45 55 - 5(3)2 6 yi �180 18 Y  i1  Average number sold = 550 (2) 610 6 a  180 - 3(18)  180 - 54  126 (3) �xi yi  9,783 i 1 y  126 + 18 x 6

2 �xi = 67.1925

i1

6

b

�xi yi - nx y

i 1 6

2 2 �x i - nx



(9, 783) - 6(3.25833)(550) 67.1925 - 6 (3.25833)2

i 1

-969.489   -277.6 3.49222 a  y - bx  550 - [(-277.6)(3.25)]  1, 454.6


(4)

x

4.32 (a)

y

xy

x2

16 330 5,280 12 270 3,240 18 380 6,840 14 300 CHAPTER 4 F O R E C A S4,200 TING 60 1,280 19,560

So at x = 2.80, y = 1,454.6 – 277.6($2.80) = 677.32. Now round to the nearest integer: Answer: 677 lattes.

the additional value. (c) Some other quantitative variables would 60 be age of the x  15 4 house, number of 1,280 bedrooms, size of y  320 the lot, and size of 4 �xy - nx y 19,560 - 4(15)(320) the garage, etc. 360 b    18 of 20 �x 2 - nx 2 920 - 4(15)2 (d) Coefficient determination = a  y - bx  320 - 18(15)  50 (0.63)2 = 0.397. Y  50 + 18 x This means that only about 39.7% (b) If the forecast is of the variability for 20 guests, the in the sales price bar sales forecast of a house is is 50 + 18(20) = explained by this $410. Each guest regression model accounts for an that only includes additional $18 in square footage as bar sales. the explanatory variable. (b) MSE = 160/5 = 32 4.36 (a) Given: Y = 90 + (c) MAPE = 13.23%/5 48.5X1 + 0.4X2 where: = 2.65%

4.34 Y = 7.5 + 3.5X1 + 4.5X2 + 2.5X3

Y  expected travel cost X1  number of days on the road

(a) 28 (b) 43 (c) 58 4.35 (a) Y� = 13,473 + 37.65(1860) = 83,502

X2  distance traveled, in miles

Therefore, the expected cost of the trip is $452.50. (b) The reimbursement request is much higher than predicted by the model. This request should probably be questioned by the accountant. (c) A number of other variables should be included, such as: 1. the type of travel (air or car) 2. conference fees, if any 3. costs entertaining customers

of

4. other transportation costs—cab, limousine, special tolls, or parking

In addition, the correlation r  0.68 (coefficient of correlation)coefficient of 0.68 is not exceptionally high. It If: indicates that the model Number of days on the explains approximately 46% road  X1 = 5 and of the overall variation in trip distance cost. This correlation traveled  X2 = 300 coefficient would suggest then: that the model is not a

(b) The predicted selling price is $83,502, but this is the average price for a Table for Problem 4.33 house of Year Transistors this size. (x) (y) There are 1 140 other

xy

x2

126 + 18x

Error

140

1

144

–4

4

162

–2

9

180

10

16

198

2

25

216

–6

factors 2 160 320 besides square 3 190 570 footage that will 4 200 800 impact 5 210 1,050 the selling 90 2,80 price of a Totals 1 5 0 0 house. If y = 180 x = 3 such a Y = 90 + 48.5  5 + house sold for 0.4  300 = 90 + 242.5 + 120 = $95,000, then 452.5 these other factors could be contributing to

5 5

particularly good one. 4.37 (a, b) Perio Demand Forecas


256 144 324 196 37 920

d 1 2 3 4 5 6 7 8 9 10

t 20 21 28 37 25 29 36 22 25 28

20 20 20.5 24.25 30.63 27.81 28.41 32.20 27.11 26.05

Cumulative error = 14.05; MAD = 5 Tracking = 14.05/5  2.82

4.38 (a) least squares equation: Y = –0.158 + 0.1308X (b) Y = – 0.158 + 0.1308(22) = 2.719 million (c) coefficient of correlation = r = 0.966 coefficient of determination = r2 = 0.934 4.39 Ye Pat X 2 ar ien X ts Y 1 36 1

Y2 X Y

1,29 36 6 2 33 4 1,08 66 9 3 40 9 1,60 12 0 0 4 41 16 1,68 16 Error 2 |% Error| 1 4 16 100 (4/140) = 5 40 25 2.86% 1,60 20 4 100 (2/160) = 0 0 6 55 36 1.25% 100 1003,02 (10/190) 33 = 5.26% 5 0 4 100 (2/200) = 7 60 49 1.00% 3,60 42 36 100 (6/210) = 0 0 8 54 64 2.86% 16 2,91 43 13.23 0 % 6 2 9 58 81 3,36 52 4 2 10 61 100 3,72 61 1 0

0.00 1.00 7.50 12.75 –5.63 1.19 7.59 –10.20 –2.10 1.95

38


55

478

385

Given: Y = a + bX where: b

 XY - nXY 2

 X - nX

2

a  Y - bX and X = 55, Y = 478, XY = 2900, X 2 = 385, Y 2 = 23892, X  5.5, Y  47.8, Then: 2900 - 10  5.5  47.8

2900 - 2629 271 b    3.28 385 - 302.5 82.5 385 - 10  5.52 a  47.8 - 3.28  5.5  29.76 and Y = 29.76 + 3.28X. For: X  11: Y  29.76 + 3.28 � 11  65.8 X  12: Y  29.76 + 3.28 � 12  69.1

Year Patients X Y 1

36

2

33

3

40

4

41

5

40

6

55

7

60

8

54

9

58

10

61

Therefore:

Trend Forecast 8

94.8

54

8,987.0

9

103.3

58

10,670.9

10

116.2

61

13,502.4

Column Totals

854.0

478

76,129. 9

29.8 + 3.28 33.1 29.8 + 3.28 36.3 29.8 + 3.28 39.6 29.8 + 3.28 42.9 29.8 + 3.28 46.2 29.8 + 3.28 49.4 29.8 + 3.28 52.7 29.8 + 3.28 56.1 29.8 + 3.28 59.3 29.8 + 3.28 62.6

Given: Y = a + bX where b

Year 12  69.1 patients

r





n  XY -  X  Y n  X 2 - (  X ) 2  n  Y 2 - (  Y ) 2      10  2900 - 55  478 10  385 - 552  10  23892 - 4782     29000 - 26290

3850 - 3025  238920 - 228484 2710 2710    0.924 825  10436 2934.3

and X = 854, Y = 478, XY = 42558.6, X2 = 76129.9, Y 2 = 23892, X = 85.4, Y = 47.8. Then: b

42558.6 - 10  85.4  47.8 2

76129.9 - 10  85.4 1737.4   0.543 3197.3 a  47.8 - 0.543  85.4  1.43



42558.6 76129.

The MAD is 3.26—this is approximately 7% of the and Y = 1.43 + 0.543X average number of patients For: and 10% of the minimum X  131.2 : Y  1.43 + 0.543(131.2)  72 number of patients. We also X  90.6 : Y  1.43 + 0.543(90.6)  50.6 see absolute deviations, for years 5, 6, and 7 in the range Therefore: 5.6–7.3. The comparison of Crime rate = 131.2 the MAD with the average  72.7 patients and minimum number of Crime rate = 90.6 patients and the  50.6 patients comparatively large Note that rounding deviations during the middle differences occur when Also note that a crime rate of 131.2 is outside the range of the years indicate that the forecast solving with Excel. data set used to determine the regression equations, so caution is model is not exceptionally 4.41 (a) It appears from the advised. accurate. It is more useful for following graph predicting general trends than that the points do the actual number of patients scatter around a to be seen in a specific year. straight line. 4.40

Year

Crime Patient s Rate X Y

r  0.853

1

58.3

36

The coefficient of determination of 0.853 is quite respectable—indicating our original judgment of a “good” fit was appropriate.

2

61.1

33

3

73.4

40

4

75.7

41

5

81.1

40

6

89.0

55

7

101.1

60 (Summe

2

 X 2 - nX 2

a  Y - bX

Year 11  65.8 patients The model “seems” to fit the data pretty well. One should, however, be more precise in judging the adequacy of the model. Two possible approaches are computation of (a) the correlation coefficient, or (b) the mean absolute deviation. The correlation coefficient:

 XY - nXY


(b) Developing the regression relationship, we have: Tourists

Ridership


r months) (Millions) (1,000,000s ) Year (X) (Y) 1 2 3 4 5 6 7 8 9 10 11 12

7 2 6 4 14 15 16 12 14 20 15 7

1.5 1.0 1.3 1.5 2.5 2.7 2.4 2.0 2.7 4.4 3.4 1.7

Syx  

 Y 2 - a  Y - b  XY n-2 71.59 - 0.511  27.1 - 0.159  352.9 12 - 2

71.59 - 13.85 - 56.11  .163 10  .404 (rounded to .407 in POM for Windows software) 

(f) The correlation coefficient and the coefficient of determination are given by:


 XY - nXY  X 2 - nX 2

a  Y - bX and X = 132, Y = 27.1, XY = 352.9, X2 = 1796, Y 2 = 71.59, X = 11, Y = 2.26. Then: 352.9 - 298.3 54.6    0.159 1796 - 1452 344 1796 - 12  112 a  2.26 - 0.159  11  0.511 b

and

352.9 - 12  11  2.26

Y = 0.511 + 0.159X (c) Given a tourist population of 10,000,000, the model predicts a ridership of:

Y = 0.511 + 0.159  10 = 2.101, or 2,101,000 persons. (d) If there are no tourists at all, the model predicts a ridership of 0.511, or 511,000 persons. One would not place much confidence in this forecast, however, because the number of tourists (zero) is outside the range of data used to develop the model. (e) The standard error of the estimate is given by:

39

the data. Moving averages smooth out seasonality. Exponential smoothing can forecast January next year, but not farther. Because seasonality is strong, a naïve model that students create on their own might be best. (b) One model might be: Ft+1 = At–11 That is forecastnext period = actualone year earlier to account for

seasonality. But this ignores the trend. n  XY -  X One very good approach Y r be to 2 calculate the  n  X 2 - (  X ) 2  would n Y 2 - ( Y )    increase from each month last  year to each month this year, 12  352.9 - 132  27.1  sum all 12 increases, and 12  1796 - 1322  12  71.59 - 27.12    divide by 12. The forecast for   next year would equal the 4234.8 - 3577.2  value for the same month this - 734.41 year plus the average increase 21552 - 17424  859.08 over the 12 months of last 657.6 657.6   year.  0.917 4128  124.67 64.25  11.166 (c) Using this model, the January forecast for next year and r 2  0.840 becomes: 4.42 (a) This problem 148 gives students a F25  17 +  17 + 12  29 12 chance to tackle a realistic problem where 148 = total monthly in business, i.e., increases from last year to not enough data to this year. make a good The forecasts for each of forecast. As can be the months of next year seen in the then become: accompanying Jan. 29 figure, the data Feb. 26 contains both Mar. 32 seasonal and trend Apr. 35 factors. May. 42 Jun.

50

Both history and forecast for the next year are shown in the accompanying figure:

Averaging methods are not appropriate with trend, seasonal, or other patterns in Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

July. Aug. Sep. Oct. Nov. Dec.

40


standard error of the estimate and the MAD, the 0.2 constant is better. However, other smoothing constants need to be examined.

4.43 (a) and (b) See the following table: Actual Smoothed Week

Value

Value

t

A(t)

Ft ( = 0.2)

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

50 35 25 40 45 35 20 30 35 20 15 40 55 35 25 55 55 40 35 60 75 50 40 65

+50.0 +50.0 +47.0 +42.6 +42.1 +42.7 +41.1 +36.9 +35.5 +35.4 +32.3 +28.9 +31.1 +35.9 +36.7 +33.6 +37.8 +41.3 +41.0 +39.8 +43.9 +50.1 +50.1 +48.1 +51.4

Forecast

MAD

(c) Students should note how stable the smoothed values are for  = 0.2. When compared to actual week 25 calls of 85, the smoothing constant,  = 0.6, appears to do a slightly better job. On the basis of the



4.44 Week t 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

Smoothed 87Trend Estimate So: MAD:  10.875 Value 8 Ft ( = 0.3) Tt (  = 0.2) 39

Actual Value At

50.000 35.000 25.000 40.000 45.000 35.000 20.000 30.000 35.000 20.000 15.000 40.000 55.000 35.000 25.000 55.000 55.000 40.000 35.000 60.000 75.000 50.000 40.000 65.000

50.00010.875 50.000 45.500 38.720 37.651 38.543 36.555 30.571 28.747 29.046 25.112 20.552 24.526 32.737 33.820 31.649 38.731 44.664 44.937 43.332 49.209 58.470 58.445 54.920 59.058

 3.586

0.000 0.000 –0.900 –2.076 –1.875 –1.321 –1.455 –2.361 –2.253 –1.743 –2.181 –2.657 –1.331 0.578 0.679 0.109 1.503 2.389 1.966 1.252 2.177 3.594 2.870 1.591 2.100

Forecast FITt 50.000 50.000 44.600 36.644 35.776 37.222 35.101 28.210 26.494 27.303 22.931 17.895 23.196 33.315 34.499 31.758 40.234 47.053 46.903 44.584 51.386 62.064 61.315 56.511 61.158

Forecas t Error

Y

377 2.93 585 3.00 690 3.45 608 3.66 390 2.88 415 2.15 481 2.53 729 3.22

n

Tracking signal 

X

421 2.90

To evaluate the trend adjusted exponential smoothing model, actual week 25 calls are compared to the forecasted value. The model appears to be producing a forecast approximately midrange between that given by simple exponential smoothing using  = 0.2 and  = 0.6. Trend adjustment does not appear to give any significant improvement. �( At - Ft )

501 1.99

t 1

Month

At

Ft

May June July August Septemb er October Novembe r Decembe r

100 80 110 115 105

100 104 99 101 104

110 125

104 105

120

109

MAD

613 2.75 709 3.90 366 1.60 Column totals

 XY - nXY  X 2 - nX 2

a  Y - bX

0.000 –15.000 –19.600 3.356 9.224 –2.222 –15.101 1.790 8.506 –7.303 –7.931 22.105 31.804 1.685 –9.499 23.242 14.766 –7.053 –11.903 15.416 23.614 –12.064 –21.315 8.489

4.46 (a)

4.45

b

6885 36.96

Given: Y = a + bX where:


41

42


and X = 6885, Y = 36.96, XY = 20299.5, X 2 = 3857893, Y 2 = 110.26, X = 529.6, Y = 2.843, Then:

(b, c) Amit wants to forecast by exponential smoothing (setting February’s forecast equal to January’s 20299.5 - 13  529.6  2.843 20299.5 - 19573.5 sales) with alpha  b  0.1. Barbara wants 3857893 - 3646190 3857893 - 13  529.62 to use a 3period 726   0.0034 moving average. 211703 a  2.84 - 0.0034  529.6  1.03

Sales

Amit

400 380 410 375 405

— 400 398 399.2 396.8

and Y = 1.03 + 0.0034X As an indication of the usefulness of this relationship, we can calculate the correlation coefficient: r



 

January February March April May

n  XY -  X  Y 2

 n  X 2 - (  X )  n  Y 2 - ((d) Y )     forecast observations, while Barbara’s moving 13  20299.5 - 6885  36.96 average does not start 13  3857893 - 68852  13  110.26 - 36.962  until month 4. Also note    that the MAD for Amit 263893.5 - 254469.6 is an average of 4 50152609 - 47403225 1433.4 - 1366.0  numbers, while Barbara’s is only 2. 9423.9

2749384  67.0 9423.9   0.692 1658.13  8.21 r 2  0.479 A correlation coefficient of 0.692 is not particularly high. The coefficient of 2 determination, r , indicates that the model explains only 47.9% of the overall variation. Therefore, while the model does provide an estimate of GPA, there is considerable variation in GPA, which is as yet unexplained. For

2 Note that Amit has more

Amit’s MAD for exponential smoothing (16.11) is lower than that of Barbara’s moving average (19.17). So his forecast seems to be better.

(b) X  350: Y  1.03 + 0.0034 � 350  2.22 (c) X  800: Y  1.03 + 0.0034 � 800  3.75 Note: When solving this problem, care must be taken to interpret significant digits. Also note that X = 800 is outside the range of the data set used to determine the regression relationship, so caution is advised. 4.47 (a) There is not a strong linear trend in sales over time.

4.48 (a)

27 28 Quarter Contracts Sales Y 29 X 1 2 3 4 5 6 7 8 Barba Totals

153 172 197 178 185 199 205 226 1,51 5 189.375

Average

8 10 15 9 12 13 12 16 95

7.70 10.10 15.20

5.497 6.818 8.787

(Continued)

11.875

396.67 b = (18384 – 8  189.375  388.33 11.875)/(290,413 – 8  MAD

189.375  189.375) = 0.1121 a = 11.875 – 0.1121  189.375 = –9.3495 Sales ( y) = –9.349 + 0.1121 (Contracts) (b) r  (8 � 18384 - 1515 � 95)

((8 � 290,413 - 15152 )(8 � 1183 - 952 ))

 0.8963 Sxy  1183 - (-9.3495 � 95) - (0.112 � 18384 / 6)  1.3408 r 2  .8034 4.49 (a)

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

Method  Exponential Smoothing 0.6 = Deposits Forecast (Y ) 0.25 0.24 0.24 0.26 0.25 0.30 0.31 0.32 0.24 0.26 0.25 0.33 0.50 0.95 1.70 2.30 2.80 2.80 2.70 3.90 4.90 5.30 6.20 4.10 4.50 6.10

0.25 0.25 0.244 0.241 0.252 0.251 0.280 0.298 0.311 0.268 0.263 0.255 0.300 0.420 0.738 1.315 1.906 2.442 2.656 2.682 3.413 4.305 4.90 5.680 4.732 4.592


2.20 3.28 6.41


4.49 (a) (Continued)

Year 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 TOTALS AVERAGE

appropriate choice. Measures of error and goodnessoffit are really irrelevant. Exponential smoothing provides a forecast only of deposits for the next year— and thus does not

Method  Exponential Smoothing 0.6 = Deposits Forecast (Y ) 18.10 24.10 25.60 30.30 36.00 31.10 31.70 38.50 47.90 49.10 55.80 70.10 70.90 79.10 94.00 787.30

12.6350 15.9140 20.8256 23.69 27.6561 32.6624 31.72 31.71 35.784 43.0536 46.6814 52.1526 62.9210 67.7084 74.5434

17.8932

Forecasting Summary Table

8 9 10 11 12

0.32 0.24 0.26 0.25 0.33

13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

0.50 0.95 1.70 2.30 2.80 2.80 2.70 3.90 4.90 5.30 6.20 4.10 4.50 6.10 7.70 10.10 15.20 18.10

Exponential Smoothing

Method used:

Linear Regression (Trend Analysis)

Linear Regression

Y = –18.968 + 1.638  YEAR MAD MSE Standard error using n – 2 in denominator Correlation coefficient

31 31 32 32 Next period forecast = 86.2173 33 33 34 34 35 35 36 (Trend36 Method  Linear Regression Analysis) 37 Year Period (X ) Deposits 37 38 38 1 1 0.25 39 39 2 2 0.24 40 40 3 3 0.24 41 41 4 4 0.26 42 42 5 5 0.25 43 43 6 6 0.30 44 44 7 7 0.31 8 9 10 11 12

43

24.10 25.60 30.30 36.00 31.10 31.70 38.50 47.90 49.10 55.80 70.10 70.90 79.10 94.00

3.416 34.39 6.075

13 14 15 16 17

1.20 1.20 1.20 1.60 1.50

10.587 171.817 13.416

Y = –17.636 + 13.59364  GSP 10.255 204.919 14.651

0.846

0.813

0.50 0.95 1.70 2.30 2.80

18 1.60 2.80 19 1.70 2.70 20 1.90 3.90 21 1.90 4.90 22 2.30 5.30 23 2.50 6.20 24 2.80 4.10 25 2.90 4.50 26 3.40 6.10 27 3.80 7.70 TOTALS 990.0 787.30 28 4.10 10.10 0 29 4.00 15.20 AVERAGE 22.50 17.893 30 4.00 18.10 31 3.90 24.10 32 3.80 25.60 33 3.80 30.30 Method  Least squares–Simple Regression on 34 3.70 36.00 35 4.10 31.10 a b 36 4.10 31.70 –17.636 13.5936 37 4.00 38.50 Coefficients GSP Deposit 38 4.50 47.90 : s 39 4.60 49.10 Year (X) (Y ) Forecast 40 4.50 55.80 4.60 70.10 1 0.40 0.25 –12 41 4.60 70.90 2 0.40 0.24 –12 42 4.70 79.10 3 0.50 0.24 –10 43 5.00 94.00 4 0.70 0.26 –8 44 5 0.90 0.25 –5 TOTALS 6 1.00 0.30 –4 AVERAGE 7 1.40 0.31 1 8 1.70 0.32 5 Given that one 9 1.30 0.24 wishes to develop 0.036086 a fiveyear 10 1.20 0.26 –1 forecast, trend 11 1.10 0.25 –2 analysis is the 12 0.90 0.33 –5

–1 –1 –1 4 2 4 5 8 8 13 16 20 21 28 34 38 36 36 35 34 34 32 38 38 36 43 44 43 44 44 46 50


address the five year forecast problem. In order to use the regression model based upon GSP, one must first develop a model to forecast GSP, and then use the forecast of GSP in the model to forecast deposits. This requires the development of two models—one of which (the model for GSP) must be based solely on time as the independent variable (time is the only other variable we are given). (b) One could make a case for exclusion of the older data. Were we to exclude data from roughly the first 25 years, the forecasts for the later years would likely be considerably more accurate. Our argument would be

44


4.51

that a change that caused an increase in the rate of growth appears to have taken place at the end of that period. Exclusion of this data, however, would not change our choice of forecasting model because we still need to forecast deposits for a future fiveyear period.

Demand

Exponentially Smoothed Forecast

1 2 3 4 5 6 7

7 9 5 9 13 8 Forecast

5 5 + 0.2 5.4 + 0.2 6.12 + 0.2 5.90 + 0.2 6.52 + 0.2 7.82 + 0.2

4.52 Actual

Forecast

95 108 123 130

ADDITIONAL HOMEWORK PROBLEMS

100 110 120 130

(b) 3month weighted moving average

MAD = 10/4 = 2.5, MSE = 38/4 = 9.5 4.53 (a) 3month moving average:

These problems, which appear on www.myomlab.com, provide an additional 13 problems that you may wish to assign.

Three-Month Month

4.50

(a) (b) (c)

Period

Week

1

2

3

4

Registration Naïve 2-week moving 4-week moving

22

21 22

25 21 21.5

27 25 23

January 6 7 February 35 March 29 33 27 April 35 29 26 31 32 23.7 May 27 29 5 5

Sales 8 11 14 3716 3310 31 3115

9

Moving Average 10

Forecast

41 37 37 (11 41+ 14 37 + 16)/3 = 35 39 39 33.5(14 35+ 16 37 + 10)/3 =

June

17

(16 + 10 + 15)/3 =

July

11

(10 + 15 + 17)/3 =

August

14

(15 + 17 + 11)/3 =

September

17

(17 + 11 + 14)/3 =

October

12

(11 + 14 + 17)/3 =

November

14

(14 + 17 + 12)/3 =

December

16

(17 + 12 + 14)/3 =

January

11

(12 + 14 + 16)/3 =

February

0 0 M A D = 2 . 2 0

(14 + 16 + 11)/3 =

 = 2 2 . Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.


 9.17 Average4.56 To Absolute compute Three-Month yMoving seasonalized or adjusted sales x  3.5 Moving Deviation

(c) Based on a Sales Mean Month

Absolute Deviation January 11 criterion, 14the February 3month moving March 16 average 10 with April MAD = 2.2 is May 15 to be preferred 17 over June July 11 the 3month August weighted moving 14 September average 17 with October MAD = 2.72. 12

forecast, we just multiply each seasonalized index by the Period 7 forecast = appropriate trend forecast. 13.07 (1  11 + 2  14 + 3  16)/6 = 14.50 4.50 Y� Y�  Index Trend forecast Period 12+ forecast = = 12.67Seasonal (1  14 + 2  16 3  10)/6 2.33 18.64, but this is far Hence, for (1  16 + 2  10 + 3  15)/6 = 13.50 3.50 outside the range of = 15.17Quarter I: Y� (1  10 + 2  15 +3  17)/6 4.17 I  1.25  120,000  150,000 valid data. (1  15 + 2  17 + 3  11)/6 = 13.67 0.33 � Quarter II: Y3.50 (1  17 + 2  11 + 3  14)/6 = 13.50 II  0.90  140,000  126,000 (1  11 + 2  14 + 3  17)/6 = 15.00 3.00 � Quarter III: Y  0.75  160,000  120,000 (1 (1 (1 (1

   

14 17 12 14

complex model are interest rates and cycle or seasonal factors. 4.54 (a) Actua Cumulativ l Wee Miles Foreca Error Error k st 1 2 3 4 5

17 21 19 23 18

17.00 17.00 17.80 18.04 19.03

6

16

18.83

7 8

20 18

18.26 18.61

9 10 11

22 20 15

18.49 19.19 19.35

12

22

18.48

0.00 –4.00 –1.20 –4.96 +1.0 3 +2.8 3 –1.74 +0.6 1 –3.51 –0.81 +4.3 5 –3.52

(b) The MAD 28.56/12 = 2.38

=

(c) The cumulative error and tracking signals appear to be consistently negative, and at week 10, the tracking signal exceeds 5 MADs. 4.55

y

x

x2

7 9 5 11 10 13 55

1 2 3 4 5 6 21

1 4 9 16 25 36 91

(b) Correlation coefficient:

y  5.27 + 1.11x

November 14 (d) Other factors 16 that December January might be included 11 February in a more

–

+ + + +

2 2 2 2

   

17 12 14 16

+ + + +

3 3 3 3

   

Week 1 Week 2 Week 3 Week 4 Averages

III 12)/6 = 14.00 0.00 � Quarter IV: Y2.17 14)/6 = 13.83 IV  1.10  180,000  198,000 16)/6 = 14.67 3.67 4.57 11)/6 = 13.17 Mon Tue Wed Thu Fri Sat

210 215 220 225 217. 5

 = 27.17

178 250 MAD = 2.72 180 250 176 260 178 260 17 255 8

215 213 220 225 218. 3

160 165 175 176 16 9

(a) Seasonal indices: 1.066 (Mon) 0.873 (Tue) 1.25 (Wed) 1.07 (Thu) 0.828 (Fri) 0.913 (Sat)

(b) To calculate for Monday of Week 5 = 201.74 + 0.18(25)  1.066 = 219.9 rounded to 220 – –

–

45

180 185 190 190 186.3

r



 

n  XY -  X  Y

n  X 2 - (  X ) 2  n  Y 2    5  70 - 15  20

5  55 - 152  5  130 - 2   350 - 300  275 - 225  650 - 400  50  0.45 111.80

Overall average = 204

The correlation coefficient indicates that there is a positive correlation between bank deposits and consumer price indices in

Forecast 220 (Mon) 180 (Tue) 258 (Wed) 221 (Thu) 171 (Fri) 189 (Sat)

4.58 (a) 4000 0.20(15,000) = 7,000

+

(b) 4000 + 0.20(25,000) = 9,000 4.59 (a) 35 + 20(80) + 50(3.0) = 1,785 (b) 35 + 20(70) + 50(2.5) = 1,560 4.60 Given: X = 15, Y = 2 2 20, XY = 70, X = 55, Y = 130, X = 3, Y = 4  XY - nXY (a) b   X 2 - nX 2

Birmingham, Alabama—i.e., as one variable tends to increase (or decrease), the other tends to increase (or decrease).

a  Y - bX error of 70 - 5  3  4 70 4.60 (c) Standard - 60 10 b    1 the estimate: 55 - 45 10 55 - 5  32 a  4 - 1 3  4 - 3  1 Y  1 + 1X Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

46


Syx  

 Y 2 - a  Y - b  XY 130 - 1  20 - 1  70  n-2 3 130 - 20 - 70  3

40  13.3  3.65 3

4.61

Column Totals

X

Y

2 1 4 5 3

4 1 4 6 5

15

20

days rain.

CASE STUDIES

 XY - nXY  X 2 - nX 2

55 - 5  32 a  4 - 1  3  1.0







1. A plot of the data indicates a linear trend (least squares) model might be appropriate for forecasting. Using linear trend you obtain the following: x

This is the second in a series of integrated case studies that run throughout the text. 1. One way to address the case is with separate forecasting models for each game. Clearly, the homecoming game (week 2) and the fourth game (craft festival) are unique attendance situations.

70 - 60  1.0 55 - 45

and Y = 1.0 + 1.0X. The correlation coefficient: r



SOUTHWESTERN UNIVERSITY: B

1

and X = 15, Y = 20, XY = 2 2 70, X = 55, Y = 94, X = 3, Y = 4. Then: 70 - 5  4  3

Selection of an appropriate time series forecasting model based upon a plot of 94 the data. - 1  20 - 1  70



The 5 - 2importance of combining a qualitative 94 - 20 - 70   1.333  1.15 model with a quantitative 3 model in situations where technological change is occurring. 4.62 Using software, the Games lost = 6.41 + 0.533 

a  Y - bX

b

 Y 2 - a  Y - b  XY  n-2

Syx 

regression equation is:


Standard error of the estimate:

n  XY -  X  Y

Game

 n  X 2 - (  X ) 2  n  Y 2 - (  Y ) 2      5  70 - 15  20 5  55 - 152  5  94 - 202     50 50   0.845 50  70 59.16



350 - 300 275 - 225 470 - 400 

1 2 3 4 5 Total

Model y y y y y

= = = = =

30,713 37,640 36,940 22,567 30,440

+ + + + +

2,534x 2,146x 1,560x 2,143x 3,146x

(where y = attendance and x = time) 2. Revenue in 2010 = (239,000) ($50/ticket) = $11,950,000 Revenue in 2011 = (250,530) ($52.50/ticket) = $13,152,825 3. In games 2 and 5, the forecast for 2011 exceeds stadium capacity. With this appearing to be a continuing trend, the time has come for a new or expanded stadium. 2

DIGITAL CELL PHONE, INC.

Objectives:


Totals

y

x2

xy

48 87 1,44 1,79 2,29 2,93 3,48 3,44 3,99 4,96 5,35 6,30 7,47 7,37 8,10 8,03 8,63 10,31 9,65 9,96 10,18 11,57 12,69 14,08 15,20 15,52 16,52 16,88 18,21 18,15 19,43 18,49 19,30 19,75 22,12 23,61

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

480 436 482 448 458 489 498 430 444 496 487 525 575 527 540 502 508 573 508 498 485 526 552 587 608 597 612 603 628 605 627 578 585 581 632 656

1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 625 676 729 784 841 900 961 1,024 1,089 1,156 1,225 1,296

666

19,366

16,206

Average 18.5

537.9

450.2

378,6

10,5


378,661 - 36 �18.5Analysis for Management, �537.9 20390.0   5.25 10th 3885.0 Hall �x - nx 16,206 - (36 �18.52 ) ed., Prentice Publishing. a  y - bx  537.9 - 5.25 �18.5  440.85 Cases = 432.28 + 5.73 n �xy - �x �y (time), r2 = .93, MAD = r [ n �x 2 - ( �x )2 ][ n �y 2 - ( �y)212.84 ]

b

�xy - nx y

2

2



(36)(378,661) - c) Additive seasonal model, (666)(19,366) Cases = 444.29 + 5.06 2 [(36) �(16,206) - (666) ][(36)(10,558,246) - (19,366)2 ] (time), r2 = .86, MAD = 13,631,796 - 12,897,756 20.02  [(583,416) - (443,556)][380,096,856) - (375,041,956)] d) Additive seasonal model, with centered moving 737,040 734,040   averages. [139,860][5,054,900] 706,978,314,000 Cases = 431.31 + 5.72 734,040 (time), r2 = .94, MAD =   .873 840,820 12.28 

r 2  .76 y  440.85 + 5.25 (time) r = 0.873 indicating a reasonably good fit The student should report the linear trend results, but deflate the forecast obtained based upon qualitative information about industry and technology trends. Because there is limited seasonality in the data, the linear trend analysis above provides a good r2 of .76. However, a more precise forecast can be developed addressing the seasonality issue, which is done below. Methods a and c yield r2 of . 85 and .86, respectively, and methods b and d, which also center the seasonal adjustment, yield r2 of .93 and .94, respectively. 2. Four approaches to decomposition of The Digital Cell Phone data can address seasonality, as follows: a) Multiplicative seasonal model, Cases = 443.87 + 5.08 (time), r2 = .85, MAD = 20.89 b) Multiplicative Seasonal Model, with centered moving averages (CMA), which is not covered in our text but can be seen in Render, Stair, and Hanna’s Quantitative

The two methods that use the average of all data have very similar results, and the two CMA methods also look quite close. As suggested with this analysis, CMA is typically the better technique.

VIDEO CASE STUDY FORECASTING AT HARD ROCK CAFE There is a short video (8 minutes) available from Prentice Hall and filmed specifically for this text that supplements this case. 1. Hard Rock uses forecasting for (1) sales (guest counts) at cafes, (2) retail sales, (3) banquet sales, (4) concert sales, (5) evaluating managers, and (6) menu planning. They could also employ these techniques to forecast: retail store sales of individual (SKU) product demands; sales of each entrée; sales at each work station, etc. 2. The POS system captures all the basic sales data needed to drive individual cafe’s scheduling/ordering. It also is aggregated at corporate HQ. Each entrée sold is counted as one guest at a Hard Rock Cafe. 3. The weighting system is subjective, but is reasonable. More weight is given to each of the past 2 years than to 3

years ago. This system actually protects managers from large sales variations outside their control. One could also justify a 50%– 30%–20% model or some other variation.

2003 2004 2005 2006 2007 2008 2009

4. Other predictors of cafe sales could include season of year (weather); hotel occupancy; spring break from colleges; beef prices; promotional budget; etc.

47

51.80 54.92 69.70 68.90 63.72 84.73 78.74

Southeast Airline D Year

Airframe Cost per Aircraft

2003 13.29 2004 25.15 5. Y  a + bx 2005 32.18 2006 31.78 Month Advertising Guest Count 2007 25.34 X Y 2008 32.78 1 14 21 2009 35.56 2 17 24 3 25 27 Utilizing the software 4 25 32 package provided with this 5 35 29 text, we can develop the 6

35

37

7

45

43

8

50

43

following regression equations for the variables of interest:

Northern Airlines— Airframe Maintenance Cost:  Cost = 36.10 + 0.0026 9 60 54  Airframe age  Coefficient of 10 60 66 determination = 0.7695 Totals 36 37  Coefficient of 6 6 correlation = 0.8772 Averag 36.6 37.6 e Northern Airlines—Engine Maintenance Cost:  Cost = 20.57 + 0.0026  Airframe age 15,772 - 10 �36.6 �37.6  Coefficient of b  0.7996 �.8 15,910 - 10 �36.62 determination = 0.6124 of a  37.6 - 0.7996 �36.6  8.3363 �8.3  Coefficient correlation = 0.7825 Y  8.3363 + 0.7996 X At $65,000; y  8.3 + .8 (65)  8.3 + 52 = 60.3, or 60,300 guests.

For the instructor who asks other questions than this one: r2  0.8869 Std. error  5.062

Southeast Airlines— Airframe Maintenance Cost:  Cost = 4.60 + 0.0032  Airframe age  Coefficient of determination = 0.3905  Coefficient of correlation = 0.6249

Southeast Airlines—Engine Maintenance Cost;  Cost = –0.67 + 0.0041  Airframe age THE NORTHSOUTH  Coefficient of AIRLINE determination = 0.4600  Coefficient of Northern Airline Data correlation = 0.6782

ADDITIONAL CASE STUDY*

Year

43.4 38.5 51.4 58.7 45.4 50.2 79.6

Airframe Cost per Aircraft


Engine per Airc

18.8 31.5 40.4 22.1 19.6 32.5 38.0

48


The following graphs portray both the actual data and the regression lines for airframe and engine maintenance costs for both airlines.

*This case study appears on our companion Web site, at www.pearsonhighered.com/heize r, and at www.myomlab.com.

Note that the two graphs have been drawn to the same scale to facilitate comparisons between the two airlines. Comparison: 

Northern Airlines: There seem to be modest correlations between maintenance costs and airframe age

for Northern Airlines. There is certainly reason to conclude, however, that airframe age is not the only important factor.



Southeast Airlines: The relationships between maintenance costs and airframe age for Southeast Airlines are much less well defined. It is even more obvious that airframe age is not the only important factor —perhaps not even the most important factor.

Overall, it would seem that: 

Northern Airlines has the smallest variance in maintenance costs —indicating that its daytoday management of maintenance is working pretty well.



Maintenance costs seem to be more a function of airline than of airframe age.



The airframe and engine maintenance costs for Southeast Airlines are not only lower, but more nearly similar than those for Northern Airlines. From the graphs, at least, they appear to be rising more sharply with age.



From an overall perspective, it appears that Southeast Airlines may perform more efficiently on sporadic or emergency repairs, and Northern Airlines may place more emphasis on preventive maintenance.

Ms. Young’s report should conclude that: 

There is evidence to suggest that maintenance costs could be made to be a function of airframe age by implementing more effective


management practices. 



The difference between maintenance procedures of the two airlines should be investigated. The data with which she is currently working does not provide conclusive results.

Concluding Comment: The question always arises, with this case, as to whether the data should be merged for the two airlines, resulting in two regressions instead of four. The solution provided is that of the consultant who was hired to analyze the data. The airline’s own internal analysts also conducted regressions, but did merge the data sets. This shows how statisticians can take different views of the same data.

Chapter 4 - Forecasting Solution Manual.doc

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