5-1

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:

Download: https://testbankarea.com/download/775

or mirror link:

https://testbankarea.com/download/quantitative-analysis-management-12th-

edition-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-12th-

edition-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 ex-

pected 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 pro-

cess.

Teaching Suggestion 5.4: Management Input to the Exponential Smoothing Model.

5-2

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.

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 auto-

matically 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 interven-

tion.

5-3

ALTERNATIVE EXAMPLES

Alternative Example 5.1:

Moving averagedemand in previous periodsn

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 demand in period

weights

n n

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

Weights Applied Period

3 Last week

2 Two weeks ago

1 Three weeks ago

6 Sum of weights

A 3-week weighted moving average appears below.

Week Actual Bicycle Sales Three-Week Moving Average

1 8 -

2 10 -

3 9 -

4 11 [(3 9) + (2 10) + (1 8)]/6 = 9 16

5 10 [(3 11) + (2 9) + (1 10)]/6 = 10 16

6 13 [(3 10) + (2 11) + (1 9)]/6 = 10 16

7 — [(3 13) + (2 10) + (1 11)]/6 = 11 23

5-4

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 Janu-

ary was 22 batteries.

Absolute Absolute

Actual Forecast Deviation Forecast Deviation

Battery with With with with

Month Sales = 0.8 = 0.8 = 0.5 = 0.5

January 20 22 2 22 2

February 21 20.40 0.6 21 0

March 15 20.880 5.88 21 6

April 14 16.176 2.176 18 4

May 13 14.435 1.435 16 3

June 16 13.287 2.713 14.5 1.5

Sum of absolute deviations: 14.804 16.5

MAD: 2.467 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.

Time Sales

Year Period (Units) X2 XY

2007 1 100 1 100

2008 2 110 4 220

2009 3 122 9 366

2010 4 130 16 520

2011 5 139 25 695

2012 6 152 36 912

2013 7 164 49 1,148

X = 28 Y = 917 X2 = 140 XY= 3,961

5-5

2 22

28 9174 131

7 7

3,961 7 4 131 29310.464

28140 7 4

131 10.46 4 89.14

X YX Y

n n

XY nXYb

X nX

a Y bX

The trend equation is

^

0 1 89.14 10.464Y b b X 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

6 130 36 780

X = 21 Y = 730 X2 = 91 XY = 2600

1 2 22

0 1

21/ 6 3.5

730 / 6 121.667

2600 6(3.5)(121.667)2.57

91 6(3.5)

7302.57(3.5) 112.67

6

X

Y

XY nXYb

X nX

Yb b X

n

^

112.67 2.57Y X

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

=130.7

5-6

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.

Forecast errors 70MAD 11.7

6n

RSFE 24Tracking Signal 2.1 MADs

MAD 11.7

Table for Alternate Example 5.7

Forecast Actual Forecast Cumu-

lative

Track-

ing

Year Demand Demand Error RSFE Error Error MAD Signal

1 78 71 7 7 7 7 7.0 1.0

2 75 80 5 2 5 12 6.0 0.3

3 83 101 18 16 18 30 10.0 +1.6

4 84 84 0 16 0 30 7.5 +2.1

5 88 60 28 12 28 58 11.6 1.0

6 85 73 12 24 12 70 11.7 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

5-7

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 aver-

age of the absolute deviations. MAD is important because it can be used to help increase fore-

casting 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 in-

dex 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 ap-

propriate 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 ap-

propriate 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.

5-8

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.

5-9

5-16.

Three- Four-

Three- Month Four- Month

Actual Month Absolute Month Absolute

Month 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 10 26.75 8.75

Nov. 16 25.33 9.33 25.5 9.5

Dec. 14 20.67 6.67 23 9

58.33 62.25

Three-month MAD

58.336.48

9

Four-month MAD

62.257.78

8

The 3-month moving average appears to be more accurate. However, when weighted moving av-

erages were used, the MAD was 5.444.

5-10

5-17.

Year Demand 3-Year Mov-

ing Ave.

3-Year Wt. Moving

Ave.

3-Year Abs.

Deviation

3-Year Wt. Abs.

Deviation

1 4

2 6

3 4

4 5 (4 + 6 + 4)/3

= 4 23

[(2 4) + 6 + 4]/4

= 4 12

0.34 0.55

5 10 (6 + 4 + 5)/3

= 5

[(2 5) + 4 + 6]/4

= 5

5 5

6 8 (4 + 5 + 10)/3

= 6 13

[(2 10) + 5 +4]/4

= 7 14

1.67 0.75

7 7 (5 + 10 + 8)/3

= 7 23

[(2 8) + 10 +5]/4

= 7 34

0.67 0.75

8 9 (10 + 8 + 7)/3

= 8 13

[(2 7) + 8 +10]/4

= 8

0.67 1

9 12 (8 + 7 + 9)/3

= 8

[(2 9) + 7 + 8]/4

= 8 14

4 3.75

10 14 (7 + 9 + 12)/3

= 9 13

[(2 12) + 9 +7]/4

= 10

4.67 4

11 15 (9 + 12 + 14)/3

=11 23

[(2 14) + 12+9]/4

= 12 14

3.34 2.75

Total absolute deviations: 20.36 18.55

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.

5-11

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.

3-Yr MA 3-Yr Wt.

MA

Trend line

Year Demand |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

Total absolute deviations = 20.36 18.55 15.24

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

5-12

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

12

15,000

10,412 = 8,875 + (0.3)(14,000 8,875)

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.

Weighted

Moving Absolute Absolute

Year Demand Average Deviation Exp. Sm. Deviation

1 4,000 5,000 1,000

2 6,000 4,700 1,300

3 4,000 5,090 1,090

4 5,000 4,500 500 4,763 237

5 10,000 5,000 5,000 4,834 5,166

6 8,000 7,250 750 6,384 1,616

7 7,000 7,750 750 6,869 131

8 9,000 8,000 1,000 6,908 2,092

9 12,000 8,250 3,750 7,536 4,464

10 14,000 10,000 4,000 8,875 5,125

11 15,000 12,250 2,750 10,412 4,588

Total: 18,500 26,808

Mean: 2,312.5 2,437

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

5-13

5-21. = 0.30

Year 1 2 3 4 5 6

Forecast 410.0 422.0 443.9 466.1 495.2 521.8

5-22.

Year Sales Forecast Using = 0.6 Forecast Using = 0.9

1 450 410

2 495 410 + (0.6)(450 410) = 434 410 + (0.9)(450 410) = 446

3 518 434 + (0.6)(495 434) = 470.6 446 + (0.9)(495 446) = 490.1

4 563 470.6 + (0.6)(518 470.6) = 499.0 490.1 + (0.9)(518 490.1) = 515.21

5 584 499 + (0.6)(563 499) = 537.4 515.21 + (0.9)(563 515.21) = 558.2

6 ? 537.4 + (0.6)(584 537.4) = 565.4 558.2 + (0.9)(584 558.2) = 581.4

5-23.

Actual = 0.3 Absolute = 0.6 Absolute = 0.9 Absolute

Year Sales Forecast Devia-

tion

Forecast Devia-

tion

Forecast Devia-

tion

1 450 410.0 40.0 410.0 40.0 410.0 40.0

2 495 422.0 73.0 434.0 61.0 446.0 49.0

3 518 443.9 74.1 470.6 47.4 490.1 27.9

4 563 466.1 96.9 499.0 64.0 515.2 47.8

5 584 495.2 88.8 537.4 46.6 558.2 25.8

6 ? 521.8 — 565.4 — 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 = 38.1

Because it has the lowest MAD, the smoothing constant = 0.9 gives the most accurate forecast.

5-24.

Year Sales Three-Year Moving Average

1 450

2 495

3 518

4 563 (450 + 495 + 518)/3 = 487.67

5 584 (495 + 518 + 563)/3 = 525.3

6 ? (518 + 563 + 584)/3 = 555

5-14

5-25.

Time

Period Sales

Year X Y X2 XY

1 1 450 1 450

2 2 495 4 990

3 3 518 9 1554

4 4 563 16 2252

5 5 584 25 2920

2,610 55 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.

Three-Year

Moving

Time-Series

Year Actual

Sales

Average Forecast Absolute De-

viation

Forecast Absolute

Deviation

1 450 — — 454.8 4.8

2 495 — — 488.4 6.6

3 518 — — 522.0 4.0

4 563 487.7 75.3 555.6 7.4

5 584 525.3 58.7 589.2 5.2

6 ? 555.0 — 622.8 —

Total absolute deviation 134.0 28.0

MAD=0.3 = 74.56 (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.

5-15

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 Month Demand Average Weighted Average

4 Apr. 10 13.67 14.5

5 May 15 13.33 12.67

6 June 17 13.67 13.5

7 July 11 14 15.17

8 Aug. 14 14.33 13.67

9 Sept. 17 14 13.50

10 Oct. 12 14 15

11 Nov. 14 14.33 14

12 Dec. 16 14.33 13.83

13 Jan. 11 14 14.67

14 Feb. – 13.67 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

Sum of

Absolute

Actual Forecast Forecast

Week Miles (Ft) Error RSFE Errors MAD Tracking

Signal

1 17 17.00 — — — — —

2 21 17.00 +4.00 +4.00 4.00 4.00 1

3 19 17.80 +1.20 +5.20 5.20 2.60 2

4 23 18.04 +4.96 +10.16 10.16 3.39 3

5 18 19.03 1.03 +9.13 11.19 2.80 3.3

6 16 18.83 2.83 +6.30 14.02 2.80 2.25

7 20 18.26 +1.74 +8.04 15.76 2.63 3.05

8 18 18.61 0.61 +7.43 16.37 2.34 3.17

9 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 5.11

11 15 19.35 4.35 +7.40 25.04 2.50 2.96

12 22 18.48 +3.52 +10.92 28.56 2.60 4.20

b. The total MAD is 2.60.

5-16

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 exponen-

tially 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.

Actual Smoothed Smoothed

Week, Value, Value, Forecast Value, Forecast

t Yt Ft( = 0.1) Error Ft( = 0.6) Error

1 50 50 — 50 —

2 35 50.00 -15.00 50.00 -15.00

3 25 48.50 -23.50 41.00 -16.00

4 40 46.15 -6.15 31.40 8.60

5 45 45.54 -0.54 36.56 8.44

6 35 45.48 -10.48 41.62 -6.62

7 20 44.43 -24.43 37.65 -17.65

8 30 41.99 -11.99 27.06 2.94

9 35 40.79 -5.79 28.82 6.18

10 20 40.21 -20.21 32.53 -12.53

11 15 38.19 -23.19 25.01 -10.01

12 40 35.87 4.13 19.00 21.00

13 55 36.28 18.72 31.60 23.40

14 35 38.16 -3.16 45.64 -10.64

15 25 37.84 -12.84 39.26 -14.26

16 55 36.56 18.44 30.70 24.30

17 55 38.40 16.60 45.28 9.72

18 40 40.06 -0.06 51.11 -11.11

19 35 40.05 -5.05 44.45 -9.45

20 60 39.55 20.45 38.78 21.22

21 75 41.59 33.41 51.51 23.49

22 50 44.93 5.07 65.60 -15.60

23 40 45.44 -5.44 56.24 -16.24

24 65 44.90 20.10 46.50 18.50

25 46.91 57.60

5-17

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 fore-

casts many periods into the future when the smoothing constant is higher.

5-31. 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 3.0

April 64.8 65.5 + 0.1(68.5 65.5) = 65.8 1.0

May 71.7 65.8 + 0.1(64.8 65.8) = 65.7 6.0

June 71.3 65.7 + 0.1(71.7 65.7) = 66.3 5.0

July 72.8 66.3 + 0.1(71.3 66.3) = 66.8 6.0

Aug. 66.8 + 0.1(72.8 66.8) = 67.4

MAD = 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 Income Forecast Error

Feb. 70.0 65.0 —

March 68.5 66.5 2.0

April 64.8 67.1 2.3

May 71.7 66.4 5.3

June 71.3 68.0 3.3

July 72.8 69.0 3.8

Aug. 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 decompo-

sition. 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

5-18

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 statisti-

cally significant and r2 = 0.001

b) Using QM for Windows for the multiplicative decomposition method with 4 seasons and us-

ing a centered moving average, the seasonal indices are 1.47, 0.96, 0.70, and 0.87 for quarters 1-

4 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 pre-

sent.

5-19

5-36. For a smoothing constant of 0.2, the forecast for year 11 is 6.489.

Year Rate Forecast |Error|

1 7.2 7.2 0

2 7 7.2 0.2

3 6.2 7.16 0.96

4 5.5 6.968 1.468

5 5.3 6.674 1.374

6 5.5 6.400 0.900

7 6.7 6.220 0.480

8 7.4 6.316 1.084

9 6.8 6.533 0.267

10 6.1 6.586 0.486

11 6.489

MAD = 0.722

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

Year Rate Forecast |Error|

1 7.2 7.2 0

2 7 7.2 0.2

3 6.2 7.12 0.92

4 5.5 6.752 1.252

5 5.3 6.251 0.951

6 5.5 5.871 0.371

7 6.7 5.722 0.978

8 7.4 6.113 1.287

9 6.8 6.628 0.172

10 6.1 6.697 0.597

11 6.458

MAD = 0.673

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

Year Rate Forecast |Error|

1 7.2 7.2 0

2 7 7.2 0.2

3 6.2 7.08 0.88

4 5.5 6.552 1.052

5 5.3 5.921 0.621

6 5.5 5.548 0.048

7 6.7 5.519 1.181

8 7.4 6.228 1.172

9 6.8 6.931 0.131

10 6.1 6.852 0.752

11 6.401

MAD = 0.604

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

5-20

Year Rate Forecast |Error|

1 7.2 7.2 0

2 7 7.2 0.2

3 6.2 7.04 0.84

4 5.5 6.368 0.868

5 5.3 5.674 0.374

6 5.5 5.375 0.125

7 6.7 5.475 1.225

8 7.4 6.455 0.945

9 6.8 7.211 0.411

10 6.1 6.882 0.782

11 6.256

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 Year 1 Year 2 Average Seasonal Index

Fall 200 250 225.0 225/250 = 0.90

Winter 350 300 325.0 325/250 = 1.30

Spring 150 165 157.5 257.5/250 = 0.63

Summer 300 285 292.5 292.5/250 = 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.

5-21

5-39. Using Excel with X = 1, 2, 3, …, 20 for the years 1994-2013 respectively, the trend equa-

tion 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.(ig-noring 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.

5-22

5-47. The trend line found using Excel is: Crime Rate = 51.98 + 6.09(time). Note these coeffi-

cients 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 co-

efficients 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 statisti-

cally significant, and the coefficient of determination is 0.84.

5-23

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% in-

crease) 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 leav-

ing the end zone prices more reasonable.

5-24

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 evi-

denced 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 Feb-

ruary 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 Unadjusted forecast Seasonal index Adjusted forecast

January 325.852 1.447 471.5

February 326.711 1.393 455.1

March 327.57 1.379 451.7

April 328.429 1.074 352.7

May 329.288 1.039 342.1

June 330.147 0.797 263.1

July 331.006 0.813 269.1

August 331.865 0.720 238.9

September 332.724 0.667 221.9

October 333.583 0.747 249.2

November 334.442 0.891 298.0

December 335.301 1.033 346.4

5-25

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 in-

dependent 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.

More download links:

quantitative analysis for management 12th edition answer key

quantitative analysis for management 12th edition test bank

quantitative analysis for management 12th edition solutions pdf

quantitative analysis for management solutions manual

quantitative analysis for management solutions pdf

quantitative analysis for management 11th edition solutions manual free

quantitative analysis for management 11th edition solutions free

5-26

quantitative analysis for management 12th edition solution manual pdf