Quantitative Analysis for Management 12th Edition Solutions Manual Render Stair Hanna Hale

Quantitative Analysis for Management 12th Edition Solutions Manual Render Stair Hanna Hale Completed download Comprehensive package: Solutions Manual, Answer key, Instructor Data, Excel Instructor for all chapters are included: https://testbankarea.com/download/quantitative-analysis-management-12thedition-solutions-manual-render-stair-hanna-hale/ Test Bank Quantitative Analysis for Management 12th Edition by Barry Render, Ralph M. Stair, Michael E. Hanna, Trevor S. Hale Completed download: https://testbankarea.com/download/quantitative-analysis-management-12thedition-test-bank-render-stair-hanna-hale/ CHAPTER 5

Forecasting TEACHING SUGGESTIONS Teaching Suggestion 5.1: Wide Use of Forecasting. Forecasting is one of the most important tools a student can master because every firm needs to conduct forecasts. It’s useful to motivate students with the idea that obscure sounding techniques such as exponential smoothing are actually widely used in business, and a good manager is expected to understand forecasting. Regression is 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 understand that qualitative analysis (judgmental modeling) plays an important role in predicting the future since not every factor can be quantified. 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 with too many variables. They also need to feel a part of the process. Teaching Suggestion 5.4: Management Input to the Exponential Smoothing Model. One of the strengths of exponential smoothing is that it allows decision makers to input constants that give weight to recent data. Most managers want to feel a part of the modeling process and appreciate the opportunity to provide input.

5-1

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

5-2

ALTERNATIVE EXAMPLES Alternative Example 5.1: Moving average 

 demand in previous n periods n

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

Three-Week

Week

Bicycle Sales

Moving Average

1

8

2

10

3

9

4

11

(8 + 10 + 9)/3 = 9

5

10

(10 + 9 + 11)/3 = 10

6

13

(9 + 11 + 10)/3 = 10

7

—

(11 + 10 + 13)/3 = 11 13

Alternative Example 5.2: Weighted moving average 

  weight for period n  demand in period n   weights

Bower’s Bikes decides to forecast bicycle sales by weighting the past 3 weeks as follows: Weights Applied 3 2 1 6

Period Last week Two weeks ago Three weeks ago Sum of weights

A 3-week weighted moving average appears below. Week 1 2 3 4 5 6 7

Actual Bicycle Sales 8 10 9 11 10 13 —

Three-Week Moving Average [(3  9) + (2  10) + (1  8)]/6 = 9 1 6 [(3  11) + (2  9) + (1  10)]/6 = 10 1 6 [(3  10) + (2  11) + (1  9)]/6 = 10 1 6 [(3  13) + (2  10) + (1  11)]/6 = 11 2 3

5-3

Alternative Example 5.3: A firm uses simple exponential smoothing with  = 0.1 to forecast demand. The forecast for the week of January 1 was 500 units, whereas actual demand turned out to be 450 units. The demand forecasted for the week of January 8 is calculated as follows. Ft+1 = Ft + (Yt – Ft) = 500 + 0.1(450 – 500) = 495 units Alternative Example 5.4: Exponential smoothing is used to forecast automobile battery sales. Two values of  are examined,  = 0.8 and  = 0.5. To evaluate the accuracy of each smoothing constant, we can compute the absolute deviations and MADs. Assume that the forecast for January was 22 batteries.

Month January February March April May June

Absolute Actual Forecast Deviation Battery with With Sales  = 0.8  = 0.8 20 22 2 21 20.40 0.6 15 20.880 5.88 14 16.176 2.176 13 14.435 1.435 16 13.287 2.713 Sum of absolute deviations: 14.804 MAD: 2.467

Forecast with  = 0.5 22 21 21 18 16 14.5

Absolute Deviation with  = 0.5 2 0 6 4 3 1.5 16.5 2.75

On the basis of this analysis, a smoothing constant of  = 0.8 is preferred to  = 0.5 because it has a smaller MAD. Alternative Example 5.5: Use the sales data given below to determine: (a) the least squares trend line, (b) the predicted value for 2014 sales.

Year 2007 2008 2009 2010 2011 2012 2013

Time Period 1 2 3 4 5 6 7 X = 28

Sales (Units) 100 110 122 130 139 152 164 Y = 917

5-4

X2 1 4 9 16 25 36 49 X2 = 140

XY 100 220 366 520 695 912 1,148 XY= 3,961

X

28  Y  917  131 4 Y n 7 n 7  XY  n X Y  3,961   7  4 131  293  10.464 b 2 28 140   7   42   X 2  nX X



a  Y  b X  131  10.46  4   89.14

The trend equation is ^

Y  b0  b1 X  89.14  10.464 X

To project demand in 2014, we denote the year 2014 as x = 8, Sales in 2000 = 89.14 + 10.464(8) = 172.85 Alternative Example 5.6: The rated power capacity (in hours/ week) over the past 6 years is shown in the table below.

Capacity Year

(Y)

X2

XY

1

115

1



2

120

4



3

118

9



4

124

16

496

5

123

25

615

6 X = 21

130 Y = 730

36 X = 91

780 XY = 2600

2

X  21/ 6  3.5 Y  730 / 6  121.667

 XY  n XY  2600  6(3.5)(121.667)  2.57 91  6(3.5)  X  nX  Y  b X  730  2.57(3.5)  112.67 

b1  b0

2

2

n

1

2

6

^

Y  112.67  2.57X

Forecast for year 7 = 112.67 + (2.57)(7) =130.7

5-5

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

MAD 

 Forecast errors  70  11.7 n

Tracking Signal 

6

RSFE 24   2.1 MADs MAD 11.7

Table for Alternate Example 5.7 Forecast

Actual

Year

Demand

Demand

1 2 3 4 5 6

78 75 83 84 88 85

71 80 101 84 60 73

Forecast Error

RSFE

7 5 18 0 28 12

7 2 16 16 12 24

Error

Cumulative Error

MAD

Tracking Signal

7 5 18 0 28 12

7 12 30 30 58 70

7.0 6.0 10.0 7.5 11.6 11.7

1.0 0.3 +1.6 +2.1 1.0 2.1

SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS 5-1. The steps that are used to develop any forecasting system are: 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 predict future trends. 5-3. The only difference between causal models and time-series models is that causal models take into account any factors that may influence the quantity being forecasted. Causal models use historical data as well. Time-series models use only historical data. 5-4. Qualitative models incorporate subjective factors into the forecasting model. Judgmental models are useful when subjective factors are important. When quantitative data are difficult to obtain, qualitative models are appropriate. 5-5. The disadvantages of the moving average forecasting model are that the averages always

5-6

stay within past levels, and the moving averages do not consider seasonal variations. 5-6. When the smoothing value, , is high, more weight is given to recent data. When  is low, more weight is given to past data. 5-7. The Delphi technique involves analyzing the predictions that a group of experts have made, then allowing the experts to review 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 a forecasting model by taking the average of the absolute deviations. MAD is important because it can be used to help increase forecasting accuracy. 5-9. The number of seasons depends on the number of time periods that occur before a pattern repeats itself. For example, monthly data would have 12 seasons because there are 12 months in a year. Quarterly data would have 4 seasons because there are 4 quarters in a year. Daily data would have 7 seasons because there are 7 days in a week. For daily data, it is common for many retail stores to have higher sales on Saturdays than on other days of the week, and a seasonal index would reflect that. 5-10. If a seasonal index equals 1, that season is just an average season. If the index is less than 1, that season tends to be lower than average. If the index is greater than 1, that season tends to be higher than average. 5-11. To remove the impact of seasonality in a time series, each observation is divided by the appropriate seasonal index. The resulting deseasonalized data is then used to develop a forecast. 5-12. The forecast based on the trend line (using the deseasonalized data) is multiplied by the appropriate seasonal index to adjust that forecast for the seasonal component. 5-13. If the smoothing constant equals 0, then Ft+1 = Ft + 0(Yt  Ft) = Ft This means that the forecast never changes. If the smoothing constant equals 1, then Ft+1 = Ft + 1(Yt  Ft) = Yt This means that the forecast is always equal to the actual value in the prior period. 5-14. A centered moving average (CMA) should be used if trend is present in data. If an overall average is used rather than a CMA, variations due to trend will be interpreted as variations due to seasonal factors. Thus, the seasonal indices will not be accurate.

5-7

5-15. Actual Month

Shed Sales

Four-Month Moving Average

Jan.

10

Feb.

12

Mar.

13

Apr.

16

May

19

(10 + 12 + 13 + 16)/4 = 51/4 = 12.75

June

23

(12 + 13 + 16 + 19)/4 = 60/4 = 15

July

26

(13 + 16 + 19 + 23)/4 = 70/4 = 17.75

Aug.

30

(16 + 19 + 23 + 26)/4 = 84/4 = 21

Sept.

28

(19 + 23 + 26 + 30)/4 = 98/4 = 24.5

Oct.

18

(23 + 26 + 30 + 28)/4 = 107/4 = 26.75

Nov.

16

(26 + 30 + 28 + 18)/4 = 102/4 = 25.5

Dec.

14

(30 + 28 + 18 + 16)/4 = 92/4 = 23

The MAD = 7.78 See solution to 5-16 for calculations.

5-8

5-16. Three-

Month

Four-

Three-

Month

Four-

Month

Actual

Month

Absolute

Month

Absolute

Shed Sales

Forecast

Deviation

Forecast

Deviation

Jan.

10

Feb.

12

Mar.

13

Apr.

16

11.67

4.33

May

19

13.67

5.33

12.75

6.25

June

23

16

7

15

8

July

26

19.33

6.67

17.75

8.25

Aug.

30

22.67

7.33

21

9

Sept.

28

26.33

1.67

24.5

3.5

Oct.

18

28

26.75

8.75

Nov.

16

25.33

9.33

25.5

9.5

Dec.

14

20.67

6.67

23

9

10

58.33

 Three-month MAD

 Four-month MAD

62.25

58.33  6.48 9

62.25  7.78 8

The 3-month moving average appears to be more accurate. However, when weighted moving averages were used, the MAD was 5.444.

5-9

5-17.

Year

Demand

1

4

2

6

3

4

4

5

3-Year Moving Ave.

3-Year Wt. Moving Ave.

(4 + 6 + 4)/3

[(2  4) + 6 + 4]/4 = 4 12

0.34

0.55

[(2  5) + 4 + 6]/4 =5

5

5

[(2  10) + 5 +4]/4 = 7 14

1.67

0.75

[(2  8) + 10 +5]/4 = 7 34

0.67

0.75

[(2  7) + 8 +10]/4 =8

0.67

1

4

3.75

4.67

4

3.34

2.75

20.36

18.55

= 4 23 5

10

(6 + 4 + 5)/3 =5

6

8

(4 + 5 + 10)/3 = 6 13

7

7

(5 + 10 + 8)/3 = 7 23

8

9

(10 + 8 + 7)/3 = 8 13

9

12

(8 + 7 + 9)/3 =8

10

14

(7 + 9 + 12)/3 = 9 13

11

15 (9 + 12 + 14)/3 =11 2 3

[(2  9) + 7 + 8]/4 = 8 14 [(2  12) + 9 +7]/4 = 10 [(2  14) + 12+9]/4 = 12 1 4

Total absolute deviations:

3-Year Abs. Deviation

3-Year Wt. Abs. Deviation

MAD for 3-year average = 20.36/8 = 2.55 MAD for weighted 3-year average = 18.55/8 = 2.32 The weighted moving average appears to be slightly more accurate in its annual forecasts.

5-10

5-18. Using Excel or QM for Windows, the trend line is Y = 2.22 +1.05X Where X = time period (1, 2, . . ., 11) Y = demand 5-19. Using the forecasts in the previous problem we obtain the absolute deviations given in the table below.

Year

Demand

3-Yr MA

3-Yr Wt. MA

Trend line

|deviation|

|deviation|

|deviation|

1

4

—

—

0.73

2

6

—

—

1.67

3

4

—

—

1.38

4

5

0.34

0.55

1.44

5

10

5.00

5.00

2.51

6

8

1.67

0.75

0.55

7

7

0.67

0.75

2.60

8

9

0.67

1.00

1.65

9

12

4.00

3.75

0.29

10

14

4.67

4.00

1.24

11

15

3.34

2.75

1.18

20.36

18.55

15.24

Total absolute deviations =

MAD (3-year moving average) = 2.55 MAD (3-year weighted moving average) = 2.32 MAD (trend line) = 1.39 The trend line is best because the MAD for that method is lowest. 5-20.  = 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 = 4,700

5-11

The 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)

12

11,789 = 10,412 + (0.3)(15,000-10,412)

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

Year 1 2 3 4 5 6 7 8 9 10 11

Demand 4,000 6,000 4,000 5,000 10,000 8,000 7,000 9,000 12,000 14,000 15,000

Weighted Moving Average

Absolute Deviation

Exp. Sm. 5,000 4,700 5,090 4,763 4,834 6,384 6,869 6,908 7,536 8,875 10,412

4,500 500 5,000 5,000 7,250 750 7,750 750 8,000 1,000 8,250 3,750 10,000 4,000 12,250 2,750 Total: 18,500 Mean: 2,312.5 Thus, the 3-year weighted moving average model appears to be more accurate.

5-12

Absolute Deviation 1,000 1,300 1,090 237 5,166 1,616 131 2,092 4,464 5,125 4,588 26,808 2,437

5-21.  = 0.30 Year Forecast 5-22. Year 1 2 3 4 5 6

Sales 450 495 518 563 584 ?

1

2

3

4

5

6

410.0

422.0

443.9

466.1

495.2

521.8

Forecast Using  = 0.6 410 410 + (0.6)(450  410) = 434 434 + (0.6)(495  434) = 470.6 470.6 + (0.6)(518  470.6) = 499.0 499 + (0.6)(563  499) = 537.4 537.4 + (0.6)(584  537.4) = 565.4

Forecast Using  = 0.9 410 + (0.9)(450  410) = 446 446 + (0.9)(495  446) = 490.1 490.1 + (0.9)(518  490.1) = 515.21 515.21 + (0.9)(563  515.21) = 558.2 558.2 + (0.9)(584  558.2) = 581.4

5-23. Year 1 2 3 4 5 6

Actual Sales

 = 0.3

Absolute Forecast Deviation 450 410.0 40.0 495 422.0 73.0 518 443.9 74.1 563 466.1 96.9 584 495.2 88.8 ? 521.8 — Total absolute deviation

 = 0.6 Forecast 410.0 434.0 470.6 499.0 537.4 565.4

Absolute Deviation 40.0 61.0 47.4 64.0 46.6 —

372.8

259.0

 = 0.9 Forecast 410.0 446.0 490.1 515.2 558.2 581.4

Absolute Deviation 40.0 49.0 27.9 47.8 25.8 — 190.5

MAD=0.3 = 372.8/5 = 74.56 MAD=0.6 = 259/5 = 51.8 MAD=0.9 = 190.5/5 = 38.1 Because it has the lowest MAD, the smoothing constant  = 0.9 gives the most accurate forecast. 5-24. Year 1 2 3 4 5 6

Sales 450 495 518 563 584 ?

Three-Year Moving Average

(450 + 495 + 518)/3 = 487.67 (495 + 518 + 563)/3 = 525.3 (518 + 563 + 584)/3 = 555

5-13

5-25. Time Period X 1 2 3 4 5

Year 1 2 3 4 5

Sales Y 450 495 518 563 584 2,610

X2 1 4 9 16 25 55

XY 450 990 1554 2252 2920 8166

b1 = 33.6 b0 = 421.2 Y = 421.2 + 33.6X Projected sales in year 6, Y = 421.2 + (33.6)(6) = 622.8 5-26.

Year 1 2 3 4 5 6

Three-Year Moving Average Forecast

Actual Sales 450 — 495 — 518 — 563 487.7 584 525.3 ? 555.0 Total absolute deviation

MAD=0.3 = 74.56

Time-Series Absolute Deviation — — — 75.3 58.7 — 134.0

Forecast 454.8 488.4 522.0 555.6 589.2 622.8

(see Problem 5-23)

MADmoving average = 134/2 = 67 MADregression = 28/5 = 5.6 Regression (trend line) is obviously the preferred method because of its low MAD.

5-14

Absolute Deviation 4.8 6.6 4.0 7.4 5.2 — 28.0

5-27. To answer the discussion questions, two forecasting models are required: a three-period moving average and a three-period weighted moving average. Once the actual forecasts have been made, their accuracy can be compared using the mean absolute differences (MAD). a., b. Because a three-period average forecasting method is used, forecasts start for period 4. Period 4 5 6 7 8 9 10 11 12 13 14

Month Apr. May June July Aug. Sept. Oct. Nov. Dec. Jan. Feb.

Demand 10 15 17 11 14 17 12 14 16 11 –

Average 13.67 13.33 13.67 14 14.33 14 14 14.33 14.33 14 13.67

Weighted Average 14.5 12.67 13.5 15.17 13.67 13.50 15 14 13.83 14.67 13.17

c. MAD for moving average is 2.2. MAD for weighted average is 2.72. Moving average forecast for February is 13.67. Weighted moving average forecast for February is 13.17. Thus, based on this analysis, the moving average appears to be more accurate. The forecast for February is about 14. d. There are many other factors to consider, including seasonality and any underlying causal variables such as advertising budget. 5-28. a.  = 0.20

Week

Actual Miles

1 2 3 4 5 6 7 8 9 10 11 12

17 21 19 23 18 16 20 18 22 20 15 22

Forecast (Ft) 17.00 17.00 17.80 18.04 19.03 18.83 18.26 18.61 18.49 19.19 19.35 18.48

Error — +4.00 +1.20 +4.96 1.03 2.83 +1.74 0.61 +3.51 +0.81 4.35 +3.52

RSFE — +4.00 +5.20 +10.16 +9.13 +6.30 +8.04 +7.43 +10.94 +11.75 +7.40 +10.92

b. The total MAD is 2.60.

5-15

Sum of Absolute Forecast Errors

MAD

— 4.00 5.20 10.16 11.19 14.02 15.76 16.37 19.88 20.69 25.04 28.56

— 4.00 2.60 3.39 2.80 2.80 2.63 2.34 2.49 2.30 2.50 2.60

Tracking Signal — 1 2 3 3.3 2.25 3.05 3.17 4.21 5.11 2.96 4.20

c. RSFE is consistently positive. Tracking signal exceeds 2 MADs at week 10. This could indicate a problem. 5-29. a., b. See the accompanying table for a comparison of the calculations for the exponentially smoothed forecasts using constants of 0.1 and 0.6. c. Students should note how stable the smoothed values for the 0.1 smoothing constant are. When compared to actual week 25 calls of 85, the 0.6 smoothing constant appears to do a better job. On the basis of the forecast error, the 0.6 constant is better also. However, other smoothing constants need to be examined.

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

Actual Value, Yt 50 35 25 40 45 35 20 30 35 20 15 40 55 35 25 55 55 40 35 60 75 50 40 65

Smoothed Value, Ft( = 0.1) 50 50.00 48.50 46.15 45.54 45.48 44.43 41.99 40.79 40.21 38.19 35.87 36.28 38.16 37.84 36.56 38.40 40.06 40.05 39.55 41.59 44.93 45.44 44.90 46.91

Forecast Error — -15.00 -23.50 -6.15 -0.54 -10.48 -24.43 -11.99 -5.79 -20.21 -23.19 4.13 18.72 -3.16 -12.84 18.44 16.60 -0.06 -5.05 20.45 33.41 5.07 -5.44 20.10

5-16

Smoothed Value, Ft( = 0.6) 50 50.00 41.00 31.40 36.56 41.62 37.65 27.06 28.82 32.53 25.01 19.00 31.60 45.64 39.26 30.70 45.28 51.11 44.45 38.78 51.51 65.60 56.24 46.50 57.60

Forecast Error — -15.00 -16.00 8.60 8.44 -6.62 -17.65 2.94 6.18 -12.53 -10.01 21.00 23.40 -10.64 -14.26 24.30 9.72 -11.11 -9.45 21.22 23.49 -15.60 -16.24 18.50

5-30. If the initial forecast is 40, the forecast for time period 25 is 46.11 when  = 0.1 and 57.60 when  = 0.6. If the initial forecast is 60, the forecast for time period 25 is 47.71 when  = 0.1 and 57.60 when  = 0.6. Note that when  = 0.6, the forecast for time period 25 is 57.60 for an initial forecast of 40, 50, and 60. This illustrates how little impact the initial forecast has on forecasts many periods into the future when the smoothing constant is higher. 5-31. Exponential smoothing with  = 0.1 Month Feb. March April May June July Aug.

Income 70.0 68.5 64.8 71.7 71.3 72.8

Forecast 65.0 65.0 + 0.1(70  65) = 65.5 65.5 + 0.1(68.5  65.5) = 65.8 65.8 + 0.1(64.8  65.8) = 65.7 65.7 + 0.1(71.7  65.7) = 66.3 66.3 + 0.1(71.3  66.3) = 66.8 66.8 + 0.1(72.8  66.8) = 67.4 MAD =

Error — 3.0 1.0 6.0 5.0 6.0 4.20

Note that in this problem, the initial forecast (for the first period) was not used in computing the MAD. Either approach is considered valid. 5-32. Exponential smoothing with  = 0.3 Month Feb. March April May June July Aug.

Income 70.0 68.5 64.8 71.7 71.3 72.8

Forecast Error 65.0 — 66.5 2.0 67.1 2.3 66.4 5.3 68.0 3.3 69.0 3.8 70.1 MAD = 3.34

Based on MAD,  = 0.3 produces a better forecast than  = 0.1 (of Problem 5-29). Note that in this problem, the initial forecast (for the first period) was not used in computing the MAD. Either approach is considered valid. 5-33. Using QM for Windows, we select Forecasting - Time Series and multiplicative decomposition. Then specify Centered Moving 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 trend line is Y = 237.7478 + 3.6658X

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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 5-34. Letting t = time period (1, 2, 3, . . . , 16) Q1 = 1 if quarter 1, 0 otherwise Q2 = 1 if quarter 2, 0 otherwise Q3 = 1 if quarter 3, 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 5-35 a. Using computer software we get Y = 197.5 – 0.34X where X = time period. The slope is -0.34 which indicates a small negative trend. Note that the results are not statistically significant and r2 = 0.001 b) Using QM for Windows for the multiplicative decomposition method with 4 seasons and using a centered moving average, the seasonal indices are 1.47, 0.96, 0.70, and 0.87 for quarters 14 respectively. The trend equation found with the deseasonalized data is Y = 176.63+ 2.20X. The slope of 2.20 indicates a positive trend of 2.20 per time period. The results are statistically significant. c) The negative slope of the trend line in part (a) was found when the seasonality was ignored. The first quarter has a high seasonal index, so the first observation was very large relative to the last observation. Thus, by looking at the raw data, which was used for the trend line in part (a), it appeared that there was a negative trend but in reality this was due to the seasonal variations and not due to trend. The decomposition method is better to use when there is a seasonal pattern present.

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5-36. For a smoothing constant of 0.2, the forecast for year 11 is 6.489. Year 1 2 3 4 5 6 7 8 9 10 11

Rate 7.2 7 6.2 5.5 5.3 5.5 6.7 7.4 6.8 6.1

Forecast |Error| 7.2 0 7.2 0.2 7.16 0.96 6.968 1.468 6.674 1.374 6.400 0.900 6.220 0.480 6.316 1.084 6.533 0.267 6.586 0.486 6.489 MAD = 0.722

For a smoothing constant of 0.4, the forecast for year 11 is 6.458. Year 1 2 3 4 5 6 7 8 9 10 11

Rate 7.2 7 6.2 5.5 5.3 5.5 6.7 7.4 6.8 6.1

Forecast 7.2 7.2 7.12 6.752 6.251 5.871 5.722 6.113 6.628 6.697 6.458

|Error| 0 0.2 0.92 1.252 0.951 0.371 0.978 1.287 0.172 0.597 MAD = 0.673

For a smoothing constant of 0.6, the forecast for year 11 is 6.401. Year 1 2 3 4 5 6 7 8 9 10 11

Rate 7.2 7 6.2 5.5 5.3 5.5 6.7 7.4 6.8 6.1

Forecast 7.2 7.2 7.08 6.552 5.921 5.548 5.519 6.228 6.931 6.852 6.401

|Error| 0 0.2 0.88 1.052 0.621 0.048 1.181 1.172 0.131 0.752 MAD = 0.604

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

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Year 1 2 3 4 5 6 7 8 9 10 11

Rate 7.2 7 6.2 5.5 5.3 5.5 6.7 7.4 6.8 6.1

Forecast 7.2 7.2 7.04 6.368 5.674 5.375 5.475 6.455 7.211 6.882 6.256

|Error| 0 0.2 0.84 0.868 0.374 0.125 1.225 0.945 0.411 0.782 MAD = 0.577

The lowest MAD is 0.577 for a smoothing constant of 0.8. 5-37. To compute a seasonalized or adjusted sales forecast, we just multiply each seasonal index by the appropriate trend forecast. Ŷ = seasonal index  Ŷtrend forecast Hence for: Quarter I: ŶI = (1.30)($100,000) = $130,000 Quarter II: ŶII = (0.90)($120,000) = $108,000 Quarter III: ŶIII = (0.70)($140,000) = $98,000 Quarter IV: ŶIV = (1.10)($160,000) = $176,000 5-38. The total sales in year 1 is 1,000, and the total sales in year 2 is 1,000. Thus, there is no trend and the seasonal indices can be calculated using an overall average. Overall average = (200 + 350 + 150 + 300 + 250 + 300 + 165 + 285)/8 = 250 The table below give the average for each season and the seasonal indices.

Season Fall Winter Spring Summer

Year 1 200 350 150 300

Year 2 250 300 165 285

Average 225.0 325.0 157.5 292.5

Seasonal Index 225/250 = 0.90 325/250 = 1.30 257.5/250 = 0.63 292.5/250 = 1.17

The total sales for year 3 are predicted to be 1200, which is an average of 300 for each of the 4 quarters. When these are adjusted for seasonality using the seasonal indices shown above, the forecasts are: Fall 300(0.90) = 270; Winter 300(1.30) = 390; Spring 300(0.63) = 189; Summer 300(1.17) = 351.

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5-39. Using Excel with X = 1, 2, 3, …, 20 for the years 1994-2013 respectively, the trend equation is Ŷ = 5371.8 + 397.37X. For 2014, X = 21; Ŷ = 5371.85 + 397.37(21) = 13,716.62 For 2015, X = 22; Ŷ = 5371.85 + 397.37(22) = 14,113.99 For 2016, X = 23; Ŷ = 5371.85 + 397.37(23) = 14,511.36 The MSE from the Excel output is 2,626,267. 5-40. Using QM for Windows, the forecast is 13,401.800 and the MSE = 2,867,351(ignoring the first error). This MSE is higher than the MSE found using a trend line, so the trend line provides better forecasts. However, other values for the two smoothing constants might result in better forecasts and a lower MSE. 5-41. a. With a smoothing constant of 0.4, the forecast for 2014 is 12174 with MSE =3,773,916.(ignoring the first error) b. Using QM for Windows, the best smoothing constant is 0.99. This gives the lowest MSE of 2,634,898. 5-42. Using Excel, the trend equation is Ŷ = 1.299 – 0.002X. For January of 2010, X = 13; Ŷ = 1.299 – 0.002(13) = 1.273. For February of 2010, X = 14; Ŷ = 1.299 – 0.002(14)= 1.271. The MSE = 0.00084 5-43. The forecast for January 2010 would be 1.288. The MSE with the trend equation is 0.00084. The MSE (with time period 1 included) with this exponential smoothing model is 0.00096. If time period 1 is omitted, this is 0.00105. SOLUTIONS TO INTERNET HOMEWORK PROBLEMS 5-44. With  = 0.4, forecast for 2011 = 10338 and MAD = 836 (including first error). With  = 0.6, forecast for 2011 = 10697 and MAD = 612. 5-45. Using Excel, the trend line is: GDP = 6,142.7 + 441.4(time). For 2011 (time = 12) the forecast is GDP = 6,142.7 + 441.4 (12) = 11,4389.5. 5-46. The trend line found using Excel is: Patients = 29.73 + 3.28(time). Note these coefficients are rounded. For the next 3 years (time = 11, 12, and 13) the forecasts for the number of patients 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 fair model.

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5-47. The trend line found using Excel is: Crime Rate = 51.98 + 6.09(time). Note these coefficients are rounded. For the next 3 years (time = 11, 12, and 13) the forecasts for the crime rates are: Crime Rate = 51.98 + 6.09(11) = 118.97 Crime Rate = 51.98 + 6.09(12) = 125.06 Crime Rate = 51.98 + 6.09(13) = 131.15 The coefficient of determination is 0.96, so this is a very good model. 5-48. The regression equation (from Excel) is: Patients = 1.23 + 0.54(crime rate). Note these coefficients are rounded. If the crime rate 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. 5-49. With  = 0.6, forecast for year 11 = 86.22 and MAD = 10.21. With  = 0.2, forecast for year 11 = 64.68 and MAD = 18.76. The model with  = 0.6 is better since it has a lower MAD. 5-50. With  = 0.6, forecast for year 11 = 4.86 and MAD = 0.163. With  = 0.2, forecast for year 11 = 4.57 and MAD = 0.285. The model with  = 0.6 is better since it has a lower MAD. 5-51. The trend line (coefficients from Excel are rounded) for deposits is: Deposits = 19.047 + 6.868X where X = time period For years 11, 12, and 13, the forecasts are: Deposits = 19.047 + 6.868(11) = 95.49 Deposits = 19.047 + 6.868(12) = 101.46Deposits = 19.047 + 6.868(13) = 108.33. The trend line (coefficients from Excel are rounded) for GSP is: GSP = 3.953 + 0.094X. The forecasts are: GSP = 3.953 + 0.094(11) = 4.99 GSP = 3.953 + 0.094(12) = 5.08 GSP = 3.953 + 0.094(13) = 5.18 5-52. The regression equation from Excel is Deposits = -219.731 + 61.868X where X = GSP The model is useful because the p-value for the F-test is 0.002 which means the model is statistically significant, and the coefficient of determination is 0.84.

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CASE STUDIES FORECASTING ATTENDANCE AT SWU FOOTBALL GAMES 1. Because we are interested in annual attendance and there are six years of data, we find the average attendance in each year shown in the table below. A graph of this indicates a linear trend in the data. Using Trend Analysis in the forecasting module 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 for 2008, 2 for 2009, etc.). For this model, r2 = 0.98 which indicates this model is very accurate. Attendance in 2014 is projected to be Y = 31,660 + 2,305.714(7) = 47,800 Attendance in 2015 is projected to be Y = 31,660 + 2,305.714(8) = 50,105 At this rate, the stadium, with a capacity of 54,000, will be “maxed out” (filled to capacity) in 2017. Year

2008

2009

2010

2011

2012

2013

Average Attendance

34840

35380

38520

40500

43320

45820

2. Based upon the projected attendance and tickets prices of $20 in 2014 and $21 (a 5% increase) in 2015, the projected revenues are: 47,800(20) = $956,000 in 2014 and 50,105(21) = $1,052,205 in 2015. 3. The school might consider another expansion of the stadium, or raise the ticket prices more than 5% per year. Another possibility is to raise the prices of the best seats while leaving the end zone prices more reasonable.

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FORECASTING MONTHLY SALES 1.

The scatter plot of the data shows a definite seasonal pattern with higher sales in the winter months and lower sales in the summer and fall months. There is a slight upward trend as evidenced by the fact that for each month, the sales increased from the first year to the second, and again form the second year to the third. 2. A trend line based on the raw data is found to be: Y = 330.889 – 1.162X The slope of the trend line is negative which would indicate that sales are declining over time. However, as previously noted, sales are increasing. The high seasonal index in January and February causes the trend line on the unadjusted data to appear to have a negative slope. 3. There is a definite seasonal pattern and a definite trend in the data. Using the decomposition method in QM for Windows, the trend equation (based on the deseasonalized data) is Y = 294.069 + 0.859X The table below gives the seasonal indices, the unadjusted forecasts found using the trend line, and the final (adjusted) forecasts for the next year. Month January February March April May June July August September October November December

Unadjusted forecast 325.852 326.711 327.57 328.429 329.288 330.147 331.006 331.865 332.724 333.583 334.442 335.301 5-24

Seasonal index 1.447 1.393 1.379 1.074 1.039 0.797 0.813 0.720 0.667 0.747 0.891 1.033

Adjusted forecast 471.5 455.1 451.7 352.7 342.1 263.1 269.1 238.9 221.9 249.2 298.0 346.4

SOLUTION TO INTERNET CASE SOLUTION TO AKRON ZOOLOGICAL PARK CASE 1. The instructor can use this question to have the student calculate a simple linear regression, using real-world data. The attendance would be the dependent variable and time would be the independent variable. From the attendance, the expected revenues could be determined. Also, the instructor can broaden this question to include several other 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 of but a few methods by which analysts can select the more appropriate forecast technique and outcome. First, we perform a linear regression with time as the independent variable. The model that results is admissions = 44,352 + 9,197  year (where year is coded as 1 = first year, 2 = second year, etc.) r = 0.88 MAD = 9,662 MSE = 201,655,824 So the forecasts for the next two years are 145,519 and 154,716, respectively. Using a weighted average of $2.875 to represent gate receipts per person, revenues for 1999 and 2000 are $418,367 and $444,808, respectively. Students could also consider the impact of the increasing fees to see if the increase had an impact on attendance. 2. The student should respond that the other factors are the variability of the weather, the special events, the competition, and the role of advertising.

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