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5
C H A P T E R
Forecasting Models
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-thepants 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.
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.
ALTERNATIVE EXAMPLES
Alternative Example 5.1:
∑ demand in previous n periods
Moving average =
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.
52
Week
Actual
Bicycle Sales
Three-Week
Moving Average
1
2
3
4
5
6
7
8
10
9
11
10
13
—
(8 10 9)/3 9
(10 9 11)/3 10
(9 11 10)/3 10
(11 10 13)/3 11Z\c
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
Period
3
2
1
6
Last week
Two weeks ago
Three weeks ago
Sum of weights
A 3-week weighted moving average appears below.
Week
Actual
Bicycle
Sales
1
2
3
4
5
6
7
8
10
9
11
10
13
—
Three-Week Moving Average
[(3 9) (2 10) (1 8)]/6 9Z\n
[(3 11) (2 9) (1 10)]/6 10Z\n
[(3 10) (2 11) (1 9)]/6 10Z\n
[(3 13) (2 10) (1 11)]/6 11X\c
Alternative Example 5.3: A firm uses simple exponential
smoothing with a 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 ⫹ α(At ⫺ Ft)
⫽ 500 ⫹ 0.1(450 ⫺ 500) ⫽ 495 units
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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.
Actual
Battery
Sales
Month
January
February
March
April
May
June
20
21
15
14
13
16
Absolute
Deviation
with
␣ 0.8
Forecast
with
␣ 0.8
Absolute
Deviation
with
␣ 0.5
Forecast
with
␣ 0.5
22
2
20.40
0.6
20.880
5.88
16.176
2.176
14.435
1.435
13.287
2.713
Sum of absolute deviations: 15
22
21
21
18
16
14.5
2
0
6
4
3
31.5
16.5
MAD: 2.46
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
2000 sales.
Year
Sales (Units)
1993
1994
1995
1996
1997
1998
1999
100
110
122
130
139
152
164
1993
1994
1995
1996
1997
1998
1999
Alternative Example 5.6: The rated power capacity (in hours/
week) over the past 6 years has been:
Year
Rated Capacity
(hrs/wk)
1
2
3
4
5
6
115
120
118
124
123
130
Here is an alternative way to recode years which simplifies the
math since 兺X 0.
To minimize computations, transform the value of x (time) to simpler numbers. In this case, designate 1993 as year 1, 1994 as year
2, and so on.
Year
2.75
Time
Period
Sales
(Units)
1
2
3
4
5
6
17
兺x 28
100
110
122
130
139
152
164
兺y 917
x2
1
4
9
16
25
36
149
兺x 2 140
xy
100
220
366
520
695
912
1,148
兺 xy 3,961
∑ y 917
∑ x 28
y=
=
= 131
=
=4
n
7
n
7
∑ xy − nxy 3, 961 − (7)( 4 )(131) 293
=
= 10.464
b=
=
28
140 − (7)( 4 2 )
∑ x 2 − nx 2
x=
a = y − bx = 131 − 10.46( 4 ) = 89.14
Therefore, the least squares trend equation is,
yˆ = a + bx = 89.14 + 10.464 x
To project demand in 2000, we denote the year 2000 as x 8,
Sales in 2000 89.14 10.464(8) 172.85
Year
1
2
3
4
5
6
b=
a=
Renumbered
Year (x)
Capacity
(y)
x2
xy
2.5
1.5
.5
.5
1.5
2.5
兺X 0
115
120
118
124
123
130
兺Y 730
6.25
2.25
0.25
0.25
2.25
6.25
兺X2 17.5
287.5
180
59
62
184.5
325
兺XY 45
∑ XY
∑X2
=
45
= 2.57
17.5
∑ Y 730
=
= 121.67
n
6
y ⫽ 121.67 ⫹ 2.57X
Year 7 ⫽ 121.67 ⫹ (2.57)(3.5)
⫽131
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 70
MAD =
=
= 11.7
n
6
RSFE −24
Tracking Signal =
=
= −2.1 MADs
MAD 11.7
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Table for Alternate Example 5.7
Year
1
2
3
4
5
6
Forecast
Demand
Actual
Demand
Error
RSFE
Forecast
Error
Cumulative
Error
MAD
Tracking
Signal
78
75
83
84
88
85
71
80
101
84
60
73
7
5
18
0
28
12
7
2
16
16
12
24
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
MAD is important because it can be used to help increase forecasting accuracy.
5-1.
are:
5-9. 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.
The steps that are used to develop any forecasting system
1. Determine the use of the forecast.
2. Select the items or quantities that are to be forecasted.
5-10.
3. Determine the time horizon of the forecast.
4. Select the forecasting model.
5. Gather the necessary data.
Ft1 Ft 0(At Ft) Ft
This means that the forecast never changes.
If the smoothing constant equals 1, then
Ft1 Ft 1(At Ft) At
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 timeseries 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 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.
If the smoothing constant equals 0, then
This means that the forecast is always equal to the actual value in
the prior period.
5-11. 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.
5-12.
Month
Jan.
Feb.
Mar.
Apr.
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
Actual
Shed Sales
10
12
13
16
19
23
26
30
28
18
16
14
Four-Month Moving Average
(10 12 13 16)/4 51/4 12.75
(12 13 16 19)/4 60/4 15
(13 16 19 23)/4 70/4 17.75
(16 19 23 26)/4 84/4 21
(19 23 26 30)/4 98/4 24.5
(23 26 30 28)/4 107/4 26.75
(26 30 28 18)/4 102/4 25.5
(30 28 18 16)/4 92/4 23
The MAD 7.78
See solution to 5-13 for calculations.
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5-13.
Month
Actual
Shed Sales
ThreeMonth
Forecast
10
12
13
16
19
23
26
30
28
18
16
14
11.66
13.66
16
19.33
22.66
26.33
28
25.33
20.66
Jan.
Feb.
Mar.
Apr.
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
Three-month MAD =
58.35
= 6.48
9
Four-month MAD =
62.25
= 7.78
8
ThreeMonth
Absolute
Deviation
4.34
5.34
7
6.67
7.34
1.67
10
9.33
56.66
58.35
FourMonth
Forecast
12.75
15
17.75
21
24.5
26.75
25.5
23
FourMonth
Absolute
Deviation
6.25
8
8.25
9
3.5
8.75
9.5
69.25
62.25
The 3-month moving average appears to be more accurate. However, if weighted moving averages had been used, the results
might be different.
5-14.
1
2
3
4
5
6
7
8
9
10
11
Demand
4
6
4
5
10
8
7
9
12
14
15
Three-Year
Moving Averages
(4 6 4)/3
(6 4 5)/3
(4 5 10)/3
(5 10 8)/3
(10 8 7)/3
(8 7 9)/3
(7 9 12)/3
(9 12 14)/3
42⁄3
5
61⁄3
72⁄3
81⁄3
8
91⁄3
112⁄3
Weighted Three-Year
Moving Averages
sum of the weights
[(2 4) 6 4]/4 41⁄2
[(2 5) 4 6]/4 50
[(2 10) 5 4]/4 71⁄4
[(2 8) 10 5]/4 73⁄4
[(2 7) 8 10]/4 80
[(2 9) 7 8]/4 81⁄4
[(2 12) 9 7]/4 10
[(2 14) 12 9]/4 121⁄4
Total absolute deviations:
MAD for 3-year average 2.54
MAD for weighted 3-year average 2.32
The weighted moving average appears to be slightly more accurate
in its annual forecasts.
5-15.
Using Excel or QM for Windows, the trend line is
Y 2.22 1.05X
Where X time period (1, 2, . . .) Y demand
a
Year
Three-Year
Absolute Deviation
0.34
5.55
1.67
0.67
0.67
4.55
4.67
3.34
20.36
Three-Year Weighted
Absolute Deviation
0.55
5.55
0.75
0.75
1.55
3.75
4.55
2.75
18.5
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5-16. Using the forecasts in the previous problems we obtain the
absolute deviations given in the table below.
Year
Demand
3-Yr MA
|deviation|
3-Yr Wt. MA
|deviation|
Trend line
|deviation|
—
—
—
0.33
5.00
1.67
0.67
0.67
4.00
4.67
3.33
—
—
—
0.50
5.00
0.75
0.75
1.00
3.75
4.00
2.75
0.73
1.67
1.38
1.44
2.51
0.55
2.60
1.65
0.29
1.24
1.18
20.33
18.50
15.24
11
14
12
16
13
14
14
15
15
10
16
18
17
17
18
19
19
12
10
14
11
15
Total absolute
deviations ⫽
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
⫽ 5,000 ⫹ (0.3)(⫺ 1,000)
⫽ 4,700
The calculations are:
Year
Demand
New Forecast
2
3
4
5
6
7
8
9
10
11
6,000
4,000
5,000
10,000
8,000
7,000
9,000
12,000
14,000
15,000
4,700 5,000 (0.3)(4,000 5,000)
5,090 4,700 (0.3)(6,000 4,700)
4,763 5,090 (0.3)(4,000 5,090)
4,834 4,763 (0.3)(5,000 4,763)
6,384 4,834 (0.3)(10,000 4,834)
6,869 6,384 (0.3)(8,000 6,384)
6,908 6,869 (0.3)(7,000 6,869)
7,536 6,908 (0.3)(9,000 6,908)
8,875 7,536 (0.3)(12,000 7,536)
10,412 8,875 (0.3)(14,000 8,875)
The mean absolute deviation (MAD) can be used to determine
which forecasting method is more accurate.
Absolute
Deviation
4,500
5,000
7,250
7,750
8,000
8,250
10,000
12,250
Total:
Mean:
new forecast for year 2 ⫽ 5,000 ⫹ (0.3)(4,000 ⫺ 5,000)
⫽ 5,000 ⫺ 300
MAD (3-year moving average) 2.54
MAD (3-year weighted moving average) 2.31
MAD (trend line) 1.39
The trend line is best because the MAD is lowest.
Weighted
Moving
Average
5-17. 0.3. New forecast for year 2 is last period’s forecast (last period’s actual demand last period’s forecast):
500
5,000
750
750
1,000
3,750
4,000
12,750
18,500
2,312.5
Exp. Sm.
5,000
4,700
5,090
4,763
4,834
6,384
6,869
6,908
7,536
8,875
10,412
Absolute
Deviation
1,000
1,300
1,090
237
5,166
1,616
131
2,092
4,464
5,125
14,588
26,808
2,437
Thus, the 3-year weighted moving average model appears to be
more accurate.
5-18.
Year
1
2
3
4
5
6
Forecast
410.0
422.0
443.9
466.1
495.2
521.8
5-19.
Year
1
2
3
4
5
6
Sales
Forecast Using ␣ 0.6
450
495
518
563
584
?
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) 565.6
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
558.221 (0.9)(584 558.2) 581.4
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5-20.
Year
1
2
3
4
5
6
Actual
Sales
␣ 0.3
Forecast
Absolute
Deviation
␣ 0.6
Forecast
Absolute
Deviation
␣ 0.9
Forecast
Absolute
Deviation
410.0
434.0
470.6
499.0
537.4
565.8
40.0
61.0
47.4
64.0
46.6
—
259.0
410.0
446.0
490.1
515.2
558.2
581.4
40.0
49.0
27.9
47.8
25.8
—
190.5
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 372.8
MAD0.3 ⫽ 372.8/5 ⫽ 74.56
MAD0.6 ⫽ 259/5 ⫽ 51.8
MAD0.9 ⫽ 190.5/5 ⫽ 38.1
Because it has the lowest MAD, the smoothing constant 0.9
gives the most accurate forecast.
5-21.
Year
1
2
3
4
5
6
Sales
Three-Year Moving Average
450
495
518
563
584
?
(450 495 518)/3 487.667
(495 518 563)/3 525.333
(518 563 584)/3 555
5-22.
Year
Time
Period
X
1
2
3
4
5
1
2
3
4
5
Sales
Y
X2
XY
450
495
518
563
2,584
2,610
1
4
9
16
125
55
450
990
1554
2252
2920
8166
b ⫽ 33.6
a ⫽ 421.2
Y ⫽ 421.2 ⫹ 33.6X
Projected sales in year 6,
Y ⫽ 421.2 ⫹ (33.6)(6)
⫽ 622.8
5-23.
Year
1
2
3
4
5
6
Actual Sales
Three-Year Moving
Average Forecast
450
—
495
—
518
—
563
487.7
584
525.3
?
555.0
Total absolute deviation
Absolute Deviation
—
—
—
75.3
58.7
—
134.0
Time-Series
Forecast
454.8
488.4
522.0
555.6
589.2
622.8
Absolute Deviation
4.8
6.6
4.0
7.4
5.2
—
28.0
57
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(see Problem 5-20)
MADmoving average ⫽ 134/2 ⫽ 67
MADregression ⫽ 28/5 ⫽ 5.6
Regression (trend line) is obviously the preferred method because
of its low MAD.
5-24. 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 average differences (MAD).
a, b.
Period
Month
Demand
Apr.
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
Jan.
Feb.
10
15
17
11
14
17
12
14
16
11
–
4
5
6
7
8
9
10
11
12
13
14
Average
Weighted Average
13.67
13.33
13.67
14
14.33
14
14
14.33
14.33
14
13.67
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 13.6667.
Weighted moving average forecast for February is 13.1667.
Because a three-period average forecasting method is used,
forecasts start for period 4. As can be seen, the MAD for the moving average is 2.2, and the MAD for the weighted moving average
is 2.7. 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.
5-25.
a.
Week
Actual
Miles
Forecast
(Ft)
Error
RSFE
Sum of
Absolute
Forecast
Errors
1
2
3
4
5
6
7
8
9
10
11
12
17
21
19
23
18
16
20
18
22
20
15
22
17.00
17.00
17.80
18.04
19.03
18.83
18.26
18.61
18.49
19.19
19.35
18.48
—
4.00
1.20
4.96
1.03
2.83
1.74
0.61
3.51
0.81
4.35
3.52
—
4.00
5.20
10.16
9.13
6.30
8.04
7.43
10.94
11.75
7.40
10.92
—
4.00
5.20
10.16
11.19
14.02
15.76
16.37
19.88
20.69
25.04
28.56
MAD
Track Signal
—
4.00
2.60
3.39
2.80
2.80
2.63
2.34
2.49
2.30
2.50
2.60
—
1
2
3
3.3
2.25
3.05
3.17
4.21
5.11
2.96
4.20
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b. The total MAD is 2.60.
c. RSFE is consistently positive. Tracking signal exceeds 5
MADs at week 10. This could indicate a problem.
5-26. 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,
At
Smoothed
Value,
Ft (␣ 0.1)
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
50
48
46
45
45
44
42
41
40
38
36
36
38
38
37
38
40
40
40
42
45
45
45
47
Forecast
Error
—
15
23
6
0
10
24
12
6
20
23
4
19
3
13
18
16
0
5
20
33
5
5
20
Smoothed
Value,
Ft (␣ 0.6)
50
41
31
37
42
38
27
29
32
25
19
32
46
39
31
45
51
44
39
51
66
56
46
58
Forecast
Error
—
15
16
8
9
7
18
3
6
12
10
21
23
11
14
24
10
12
10
21
23
16
16
18
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5-27.
Week
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
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FORECASTING MODELS
Using data from Problem 5-26, with 0.9
Actual
Value
At
Smoothed
Value
Ft
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
50
36
26
39
44
36
22
29
34
21
16
38
53
37
26
52
55
41
36
58
73
52
41
62
Forecast
Error
—
15
11
14
6
9
16
8
6
14
6
24
17
18
12
29
3
15
6
24
17
23
12
24
MAD 14.48
Note that in this problem, the initial forecast (for the first period) was
not used in computing the MAD. Either approach is considered valid.
5-28.
Exponential smoothing with 0.1
5-30. 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 trendline is Y ⫽ 237.7478 ⫹ 3.6658X
c. Quarter 1: Y ⫽ 237.7478 ⫹ 3.6658(17) ⫽ 300.0662
Quarter 2: Y ⫽ 237.7478 ⫹ 3.6658(18) ⫽ 303.7320
Quarter 3: Y ⫽ 237.7478 ⫹ 3.6658(19) ⫽ 307.3978
Quarter 4: Y ⫽ 237.7478 ⫹ 3.6658(20) ⫽ 311.0636
d. Quarter 1: 300.0662(0.8825) ⫽ 264.7938
Quarter 2: 303.7320(0.9816) ⫽ 298.1579
Quarter 3: 307.3978(0.9712) ⫽ 298.5336
Quarter 4: 311.0636(1.1569) ⫽ 359.8719
5-31. Letting
t ⫽ time period (1, 2, 3, . . . , 16)
Q1 ⫽ 1 if quarter 1, 0 otherwise
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
5-32. For a smoothing constant of 0.2, the forecast for year 11
is 6.489.
Month
Income
Forecast
Error
Year
Rate
Feb.
March
April
May
June
July
Aug.
70.0
68.5
64.8
71.7
71.3
72.8
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
—
3.0
1.0
6.0
5.0
6.0
1
2
3
4
5
6
7
8
9
10
11
7.2
7
6.2
5.5
5.3
5.5
6.7
7.4
6.8
6.1
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.
5-29.
Exponential smoothing with 0.3
Forecast
|Error|
7.2
7.2
7.16
6.968
6.674
6.400
6.220
6.316
6.533
6.586
6.489
0
0.2
0.96
1.468
1.374
0.900
0.480
1.084
0.267
0.486
MAD = 0.722
For a smoothing constant of 0.4, the forecast for year 11 is 6.458.
Month
Income
Forecast
Error
Feb.
March
April
May
June
July
Aug.
70.0
68.5
64.8
71.7
71.3
72.8
65.0
66.5
67.1
66.4
68.0
69.0
70.1
—
2.0
2.3
5.3
3.3
3.8
MAD 3.34
Based on MAD, 0.3 produces a better forecast than 0.1
(of Problem 5-28).
Note that in this problem, the initial forecast (for the first period) was
not used in computing the MAD. Either approach is considered valid.
Year
Rate
Forecast
|Error|
1
2
3
4
5
6
7
8
9
10
11
7.2
7
6.2
5.5
5.3
5.5
6.7
7.4
6.8
6.1
7.2
7.2
7.12
6.752
6.251
5.871
5.722
6.113
6.628
6.697
6.458
0
0.2
0.92
1.252
0.951
0.371
0.978
1.287
0.172
0.597
MAD = 0.673
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For a smoothing constant of 0.6, the forecast for year 11 is 6.401.
Year
Rate
Forecast
|Error|
1
2
3
4
5
6
7
8
9
10
11
7.2
7
6.2
5.5
5.3
5.5
6.7
7.4
6.8
6.1
7.2
7.2
7.08
6.552
5.921
5.548
5.519
6.228
6.931
6.852
6.401
0
0.2
0.88
1.052
0.621
0.048
1.181
1.172
0.131
0.752
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FORECASTING MODELS
5-33. 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
5-34.
(Average demand (year 1 demand) + (year 2 demand)
⫽
for season)
2
MAD = 0.604
For a smoothing constant of 0.8, the forecast for year 11 is 6.256.
Year
1
2
3
4
5
6
7
8
9
10
11
Rate
Forecast
|Error|
7.2
7
6.2
5.5
5.3
5.5
6.7
7.4
6.8
6.1
7.2
7.2
7.04
6.368
5.674
5.375
5.475
6.455
7.211
6.882
6.256
0
0.2
0.84
0.868
0.374
0.125
1.225
0.945
0.411
0.782
Overall average (sum of all values)
=
demand
8
Season index =
(average for season)
overall average demand
new annual demand
4
1, 200
=
× season index
4
Year 3 demand =
MAD = 0.577
The lowest MAD is 0.577 for a smoothing constant of 0.8.
Solution Table for Problem 5-34
Season
Year 1
Demand
Year 2
Demand
(Average Year 1Year 2 Demand)
Average
Season
Demand
Season
Index
Year 3
Demand
Fall
Winter
Spring
Summer
200
350
150
300
250
300
165
285
225.0
325.0
157.5
292.5
250
250
250
250
0.90
1.30
0.63
1.17
270
390
189
351
5-35. Using Excel, the trend equation is Y ⫽ 1582.61 ⫹ 612.37X.
For 2008, X ⫽ 19; Y ⫽ 1582.61 ⫹ 612.37(19) ⫽ 13217.6
For 2009, X ⫽ 20; Y ⫽ 1582.61 ⫹ 612.37(20) ⫽ 13830.0
For 2010, X ⫽ 21; Y ⫽ 1582.61 ⫹ 612.37(21) ⫽ 14442.4
The MSE from the Excel output is 1654334.7.
5-36. a. With a smoothing constant of 0.3, the forecast for 2008
is 11211.2 with MSE ⫽ 3246841.
b. Using QM for Windows, the best smoothing constant is
1.0. This gives the lowest MSE of 1443842.
5-37.
Using Excel, the trend equation is Y ⫽ 1.1940 ⫹ 0.0095X.
For January of 2007, X ⫽ 13; Y ⫽ 1.1940 ⫹ 0.0095(13) ⫽ 1.318.
For February of 2007, X ⫽ 14; Y ⫽ 1.1940 ⫹ 0.0095(14) ⫽ 1.327.
5-38.
The forecast for January 2007 would be 1.286.
The MSE with the trend equation is 0.0003. The MSE with this
exponential smoothing model is 0.0010.
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS
5-39. With a 0.4, forecast for 2004 10,339 and MAD 837. With a 0.6, forecast for 2004 10,698 and MAD 612.
5-40. Using Excel, the trend line is: GDP 6142.7 ⫹
441.4(time). For 2004 (time 12) the forecast is GDP 6142.7 ⫹
441.4(12) 11,439.5.
5-41. The trend line found using Excel is: Patients 29.73 ⫹
3.28(time). Note these coefficients are rounded. For the 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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5-42. 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.
5-43. 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.
5-44. With a 0.6, forecast for 2003 86.2 and MAD 3.42. With a 0.2, forecast for 2003 63.87 and MAD 7.23.
The model with a 0.6 is better since it has a lower MAD.
5-46. The trend line (coefficients from Excel are rounded) for
deposits is:
Deposits ⫺18.968 ⫹ 1.638(time)
For 2003, 2004, and 2005, time 45, 46, and 47 respectively. The
forecasts are:
Deposits ⫺18.968 ⫹ 1.638(45) 54.7
Deposits ⫺18.968 ⫹ 1.638(46) 56.4
Deposits ⫺18.968 ⫹ 1.638(47) 58.0
The trend line (coefficients from Excel are rounded) for GSP is:
GSP 0.090 ⫹ 0.112(time). The forecasts are:
GSP 0.090 ⫹ 0.112(45) 5.1
GSP 0.090 ⫹ 0.112(46) 5.2
GSP 0.090 ⫹ 0.112(47) 5.4
5-47. The regression equation from Excel is
Deposits ⫺17.64 ⫹ 13.59(GSP)
In the scatterplot of this data that follows, the pattern appears to
change around 1985. There are definitely different relationships
before 1985 and after 1985, so perhaps the model should be developed with 1985 as the first year of data.
5-45. With a 0.6, forecast for 2003 4.86 and MAD 0.23. With a 0.2, forecast for 2003 4.52 and MAD 0.48.
The model with a 0.6 is better since it has a lower MAD.
Deposits and GSP over Time
100
80
60
DEPOSITS
40
GSP
20
0
1950
1960
1970
1980
Time
1990
2000
2010
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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
2002, 2 for 2003, etc.).
For this model, r2 ⫽ 0.98 which indicates this model is very
accurate.
SWU Football Attendance
Attendance
50000
40000
30000
20000
10000
0
2001
2003
2005
2007
Year
Attendance in 2008 is projected to be
Y ⫽ 31,660 ⫹ 2,305.714(7) ⫽ 47,800
Attendance in 2009 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 2011.
Year
2002
Attendance 34840
2003
2004
35380 38520
2005
2006
2007
40500
43320
45820
2. Based upon the projected attendance and tickets prices of
$20 in 2008 and $21 (a 5% increase) in 2009, the projected
revenues are:
47,800(20) ⫽ $958,000 in 2008 and
50,105(21) ⫽ $1,052,205 in 2009.
FORECASTING MODELS
63
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.
SOLUTION TO INTERNET CASES
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 idea is
that attendance is a linear function of expected admission fees.
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 1989, 2 1990, etc.)
r ⫽ 0.88
MAD ⫽ 9,662
MSE ⫽ 201,655,824
So the forecasts for 1999 and 2000 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.
To complicate the situation further, students may legitimately
use a regression model to forecast admission fees for each of the
three categories, or for the weighted average fee. This number
would then replace $2.875.
Here is the result of a linear regression using weighted average admission fees as the predicting (independent) variable.
Weights are obtained each year by taking 35% of adult fees, plus
50% of children’s fees, plus 15% of group fees. The weighted fees
each year (1989–1998) are $0.975, $0.975, $0.975, $0.975,
$1.275, $1.775, $1.775, $2.275, $2.20, $2.875.
Gate admissions ⫽ 31,451 ⫹ 39,614 ⫻ (average fee in
given year)
r ⫽ 0.847
MAD ⫽ 13,212
MSE ⫽ 254,434,912
If we assume that admission fees are not raised in 1999 and
2000, expected gate admissions 145,341 in each year and
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revenues $417,856. Comparing the earlier time-series model to
this second regression, we note that the r is higher and MAD and
MSE are lower in the time-series approach.
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.
Kwik Lube
1. The relationship between Kwik Lube sales (y), average industry sales (x), and year (t with t 1 corresponding to 1972) is
shown in the table below. The x and y values are in thousands of
dollars. One could try a multiple regression analysis but the correlation of y with just x is 0.998, leading one to use the simple linear
regression equation: y 2.99x 1.42.
t
x
y
1
2
3
4
5
6
7
8
22
25
24
26
33
35
39
44
68
75
75
78
99
104
120
133
The year 1971 was excluded since the Kwik Lube revenues were
not for an entire year. 1979 (t 8) was the last year of Kwik Lube
operation without the competition from Speedy Lube. The forecasted sales for 1980 would be estimated using the average industry sales of $47,000 for x:
y 2.99(47) 1.42 141.95
and the forecasted sales for 1981 would use the industry sales of
$52,000:
y 2.99(52) 1.42 156.90
The estimated lost sales is the difference between the forecasted
and actual sales: (141,950 $156,900) ($111,000 $111,000) $76,850.
A 95% prediction interval for 1980 is 141.95 5.20 and for
1981 is 156.90 5.80. Thus, despite the danger of extrapolation,
the results of a regression outside the range of the data, one can be
reasonably certain that the lost sales were at least $65,850.
2. Without the questionnaire study, the best estimate of lost sales
would be from the regression of y on t:
y 9.38t 51.8
with a somewhat lower correlation. The estimated lost sales would
be $59,820, about $20,000 less than the estimate based on average
industry sales. Even recovering as little as 10 percent of this difference would pay for the study.
3. The lawsuit filed by Dick Johnson should discuss two basic
areas which will build a sound case for damages being awarded in
his favor.
The first factor involves the concept behind setting up a franchise. Franchises are designed so that independent owners can
start a business with a well-known name (and consequently, with
an already-captured market). This, coupled with proven strategies
and expertise given to a franchise purchaser by the franchise
seller, reduces the usually high probability of a new business
going under in its infancy stage. The franchise fee is the cost paid
for the reduced risk of a new enterprise.
Naturally, the franchising firm will protect itself against competition in a franchise contract. A franchise holder who violates such
clauses has, in essence, gained free proven strategies and has capitalized on them. Thus, the franchising firm has been damaged by the
fact that a competitor has gained information without paying for it.
This is the case with Kwik Lube. A franchise owner, T. A.
Williams, has benefited from Johnson’s expertise more than is
justified by the monetary gains earned from franchise fees. This is
not simply an economic issue, however, for such a situation was
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thought of before by Johnson. He had sought to protect himself
with a noncompetition clause in his franchised contract. Thus,
Williams is legally in the wrong for his breach of contract.
What this first area of discussion in the lawsuit does is to determine that there, in fact, has been damage done to the plaintiff, Johnson. The second area to be discussed in the lawsuit should deal with
how those damages can be mitigated by the defendant, Williams.
FORECASTING MODELS
65
Usually in lawsuits, there is a problem with measuring the
damage done. Johnson, however, can measure his loss by forecasting sales and then comparing actual sales to predicted sales.
In summary, the lawsuit should discuss how damage was incurred to plaintiff, Johnson, and how said damage should and/or
could be mitigated. A well-presented lawsuit or petition to the court
should result in a favorable judgment for the owner of Kwik Lube.