CHAPTER 4
DISCUSSION QUESTIONS
3.
A time series model uses only historical values of the quantity of interest to predict future values of
that quantity. The associative model, on the other hand, attempts to identify underlying causes or
factors that control the variation of the quantity of interest, predict future values of these factors, and
use these predictions in a model to predict future values of the specific quantity of interest.
4.
Qualitative models incorporate subjective factors into the forecasting model. Qualitative models are
useful when subjective factors are important. When quantitative data are difficult to obtain,
qualitative models may be appropriate.
5.
The term least squares refers to the holding of the sum of the square of the difference between the
observed values and the regression line to a minimum.
6.
The disadvantages of moving average forecasting models are that the averages always stay within
past ranges, that they require extensive record keeping of past data, and that they cannot be used to
develop a forecast several periods into the future.
7.
When the smoothing constant, , is large (close to 1.0), more weight is given to recent data; when 
is low (close to 0.0), more weight is given to past data.
15.
Measures of forecast accuracy:
(a)
MAD (mean absolute deviation). This is a sum of the absolute values of individual errors
divided by the number of periods of data.
(b)
MSE (mean squared error). This is the average of the squared differences between the forecast
and observed values.
16.
Independent variable (x) is said to cause variations in the dependent variable (y).
17.
Coefficient of determination is the percent of variation in the dependent variable (y) that is explained
by a regression analysis.
18.
Tracking signals alert the user of a forecasting tool to periods in which the forecast was in significant
error.
END-OF-CHAPTER PROBLEMS
4.2
(a)
No, the data appear to have no consistent pattern.
(b)
(c)
Year
Demand
3-year moving
3-year weighted
Chapter 4: Forecasting
1
7
2
9
3
5
4
9.0
7.0
6.4
5
6
7
8
9
10
11 Forecast
13.0 8.0 12.0 13.0 9.0 11.0 7.0
7.7 9.0 10.0 11.0 11.0 11.3 11.0
9.0
7.8 11.0 9.6 10.9 12.2 10.5 10.6
8.4
1
(d)
The three-year moving average appears to give better results.
14
12
10
8
6
Demand
3-year moving
3-year weighted
4
2
0
1
4.3
2
Year
Demand
Naïve
Exp. Smoothing
3
1
7
6
2
9.0
7.0
6.5
4
5
3
5.0
9.0
7.8
6
4
9.0
5.0
6.4
7
8
9
10
11 Forecast
5
6
7
8
9
10
11 Forecast
13.0 8.0 12.0 13.0 9.0 11.0 7.0
9.0 13.0 8.0 12.0 13.0 9.0 11.0
7.0
7.7 10.3 9.2 10.6 11.8 10.4 10.7
8.8
14
12
10
8
6
Demand

Naive
Exp. Smoothing
4
2
0
1
2
3
4
5
6
7
8
9
10
11 Forecast
Naïve tracks the ups and downs best, but lags the data by one period. Thus, it gives quite large errors.
Exponential smoothing is much better because it smoothens the data and does not have as much
variation.
4.4
4.5
2
(a)
(b)
(c)
FJuly  FJune  0.2Forecasting error  42  0.240  42  416
.
FAugust  FJuly  0.2Forecasting error  416
.  0.2(45  416)
.  42.3
Because the banking industry has a great deal of seasonality in its processing requirements
Period
1
2
3
4
5
6
7
Demand
7
9
5
9
13
8
Forecast
Exponentially Smoothed Forecast
5
5  0.2  (7  5)  5.4
5.4  0.2  9  5.4  612
.
612
.  0.2  5  612
.   590
.
590
.  0.2  9  590
.   6.52
6.52  0.2  13  6.52  782
.
782
.  0.2  8  782
.   786
.
Instructor’s Solutions Manual t/a Operations Management
4.6
January
February
March
April
May
June
July
August
September
October
November
December
Sum
Average
(a)
Y Sales
20
21
15
14
13
16
17
18
20
20
21
23
218
18.2
X Period
1
2
3
4
5
6
7
8
9
10
11
12
78
6.5
X2
1
4
9
16
25
36
49
64
81
100
121
144
650
XY
20
42
45
56
65
96
119
144
180
200
231
276
1474
25
20
Trend line
15
10
5
0
Jan Feb Mar Apr May Jun
(b)
Jul Aug Sep Oct Nov Dec
Naïve
The coming January = December = 23
20  21  23 / 3  2133
3-month moving
.
01
6-month weighted
.  17  .01  18  01
.  20  0.2  20  0.2  21  0.3  23  20.6
Exponential smoothing with alpha = 0.3
FOct  18  0.320  18  18.6
FNov  18.6  0.320  18.6  19.02
FDec  19.02  0.321  19.02   19.6
FJan  19.6  0.323  19.6  20.62  21
Trend
.
 x  78 , x  6.5 ,  y  218 , y  1817
1474  12 6.518.2  54.4

 0.38
650  126.52
143
a  18.2  0.386.5  15.73
b
Forecast  15.73  .3813  20.67
(c)
Trend: Only trend provides an equation that can extend beyond one month
Chapter 4: Forecasting
3
4.7
Actual
95
108
123
130
Forecast
100
110
120
130
|Error|
5
2
3
0
10
Error2
25
4
9
0
38
MAD  10 4  2.5 , MSE  38 4  9.5
4.8
(a)
(b)
(c)
96  88  90
 91 .3
3
88  90
 89
2
Temperature
93
94
93
95
96
88
90
2 day M.A.
—
—
93.5
93.5
94.0
95.5
92.0
|Error|
—
—
0.5
1.5
2.0
7.5
2.0
13.5
MAD  13.5 5  2.7
4.9
(a)
3-month moving average:
Month
January
February
March
April
May
June
July
August
September
October
November
December
January
February
Sales
11
14
16
10
15
17
11
14
17
12
14
16
11
Three-Month
Moving Average
(11  14  16 ) / 3  13 .67
14  16  10
3  1333
.
(16  10  15) / 3  13 .67
10  15  17
3  14.00
(15  17  11) / 3  14.33
(17  11  14 ) / 3  14 .00
(11  14  17 ) / 3  14 .00
(14  17  12 ) / 3  14 .33
(17  12  14 ) / 3  14 .33
(12  14  16 ) / 3  14 .00
14  16  11
3  13.67
Absolute Deviation
3.67
1.67
3.33
3.00
0.33
3.00
2.00
0.33
1.67
3.00
  22.00
MAD  2.20
4
Instructor’s Solutions Manual t/a Operations Management
(b)
3-month weighted moving average
Month
Sales
January
February
March
April
May
June
July
August
September
October
November
December
January
February
Three- Month Moving
Average Weights = 1, 2, 3
11
14
16
10
15
17
11
14
17
12
14
16
11
Absolute Deviation
1  11  2  14  3  16
6  14.50
6  12.67
1  16  2  10  3  15 6  13.50
1  10  2  15  3  17 6  1517
.
1  15  2  17  3  11 6  13.67
1  17  2  11  3  14 6  13.50
1  11  2  14  3  17 6  15.00
1  14  2  17  3  12 6  14.00
1  17  2  12  3  14 6  1383
.
1  12  2  14  3  16 6  14.67
1  14  2  16  3  11 6  1317
.
4.50
2.33
3.50
4.17
0.33
3.50
3.00
0.00
2.17
3.67
1  14  2  16  3  10
  27.17
MAD  2.72
(c)
(d)
4.10
(a)
(b)
Based on a Mean Absolute Deviation criterion, the 3-month moving average with MAD  2.2
is to be preferred over the 3-month weighted moving average with MAD  2.72 .
Other factors that might be included in a more complex model are interest rates and cycle or
seasonal factors.
Year
Demand
3-year moving
3-year weighted
1
4
2
6
3
4
4
5
6
7
8
9
10
11 Forecast
5.0 10.0 8.0 7.0 9.0 12.0 14.0 15.0
4.67 5.0 6.33 7.67 8.33 8.0 9.33 11.67 13.67
4.5
5.0 7.25 7.75 8.0 8.25 10.0 12.25 14.0
16
14
12
10
8
6
Demand
3-year moving
3-year weighted
4
2
0
1
(c)
2
3
4
5
6
7
8
9
10
11 Forecast
The forecasts are about the same, but the weighted moving average is a little better. For the 3year moving average forecast the MAD = 2.54. For the 3-year weighted moving average
forecast the MAD = 2.31. See problem 12.
4.11 Year
Demand
Exp. Smoothing
Chapter 4: Forecasting
1
4
5
2
3
4
5
6
7
8
9
10
11 Forecast
6.0 4.0 5.0 10.0 8.0 7.0 9.0 12.0 14.0 15.0
4.70 5.09 4.76 4.83 6.38 6.87 6.91 7.54 8.87 10.41 11.79
5
16
14
12
10
8
6
4
Demand
Exp. Smoothing
2
0
1
4.12
2
3
4
5
6
7
8
9
6
1.67
0.75
1.62
7
0.67
0.8
0.13
10
11 Forecast
Error  |Actual – Forecast|
Year
3-year moving
3-year weighted
Exp. smoothing
1
2
3
4
5
0.33 5.0
0.5
5.0
0.237 5.17
8
0.67
1.0
2.09
9
4.0
3.75
4.46
10
4.67
4.0
5.13
11
3.33
2.75
4.59
MAD
2.542
2.313
2.927
The 3-year weighted average (MAD=2.313) is slightly better than the 3-year moving average
(MAD=2.542), and quite a bit better than exponential smoothing. Note that the MAD for
exponential smoothing should be calculated for the same range as the moving averages (periods 411) for a fair comparison. Thus, even though we can calculate a forecast with exponential smoothing
and an error for periods 1-3, that should not be used to calulate the MAD, since MAD is based only
on periods 4-11 with the other two forecasts.
4.16
Year
1996
1997
1998
1999
2000
Time Period X
1
2
3
4
5
Sales Y
450
495
518
563
584
  2610
X2
1
4
9
16
25
  55
XY
450
990
1554
2252
2920
  8166
X  3 , Y  522
Y  a  bX
b
 XY  nXY 8166  53522  336


 33.6
 X 2  nX 2
55  59
10
a  Y  bX  22  33.6 3  421.2
y  421.2  33.6  x
y  421.2  33.6  6  622.8
6
Instructor’s Solutions Manual t/a Operations Management
Year
1996
1997
1998
1999
2000
2001
Sales
450
495
518
563
584
Forecast Trend
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
MAD  5.6
4.17
Year
1996
1997
1998
1999
2000
2001
Sales
450
495
518
563
584
Forecast Exponential Smoothing   0.6 Absolute Deviation
410.0
40.0
410  0.6(450  410)  434.0
61.0
434  0.6(495  434)  470.6
47.4
470.6  0.6(518  470.6)  499.0
64.0
499  0.6(563  499)  537.4
46.6
537.4  0.6(584  537.4)  565.6
  259
MAD  51.8
Year
1996
1997
1998
1999
2000
2001
Sales
450
495
518
563
584
Forecast Exponential Smoothing   0.9 Absolute Deviation
410.0
40.0
410  0.9(450  410)  446.0
49.0
446  0.9(495  446)  4901
.
27.9
4901
.  0.9(518  4901
. )  5152
.
47.8
5152
.  0.9(563  5152
. )  5582
.
25.8
558.2  0.9(584  558.2)  581.4
  190.5
MAD  38.1
(Refer to Solved Problem 4.1)
MAD  0.3  74.6
MAD  0.6  518
.
MAD  0.9  381
.
Because it gives the lowest MAD, the smoothing constant of   0.9 gives the most accurate
forecast.
4.18
MAD  0.3  74.6
MAD3 year moving average  60.4
MADtrend  5.6
One would use the trend (regression) forecast because it has the lowest MAD.
Chapter 4: Forecasting
7
4.23 (a)
Week
Actual
Miles
1
2
3
4
5
6
7
8
9
10
11
12
(b)
(c)
4.24 (a)
Error
RSFE
 |Error|
Cum.
MAD
0.00
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
0.00
4.00
5.20
10.16
11.19
14.02
15.76
16.37
19.88
20.69
25.04
28.56
0
2
1.73
2.54
2.24
2.34
2.25
2.05
2.21
2.07
2.28
2.38
Forecast
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
Tracking
Signal
2.0
3.0
4.0
4.1
2.7
3.6
3.6
5.0
5.7
3.2
4.6
The MAD  28.56 12  2.38
The RSFE and tracking signals appear to be consistently positive, and at week 10, the tracking
signal exceeds 5 MADs.
Graph of Demand
12
10
8
6
4
2
0
0
1
2
3
4
5
6
Appearances
7
8
9
10
The observations obviously do not form a straight line, but do tend to cluster about a straight line
over the range shown.
(b)
Least Squares Regression:
Y  a  bX
b
 XY  nXY
 X 2  nX 2
a  Y  bX
8
Instructor’s Solutions Manual t/a Operations Management
Assume
Appearances X
3
4
7
6
8
5
9
X2
9
16
49
36
64
25
Demand Y
3
6
7
5
10
8
?
Y2
9
36
49
25
100
64
XY
9
24
49
30
80
40
 X  33,  Y  39 ,  XY  232 ,  X 2  199 , X  5.5 , Y  6.5 . Therefore
232  6  5.5  6.5
1
199  6  5.5  5.5
a  6.5  1.5  5.5  1
Y  1  1X
b
The following figure shows both the data and the resulting equation:
12
10
+
8
+
+
6
+
+
4
+
2
0
0
1
2
3
4
5
6
Appearances
7
8
9
10
If there are nine performances by Green Shades, the estimated sales are:
Y9  1  1  9  1  9  10 drums
4.25
y
7
9
5
11
10
13
55
x
1
2
3
4
5
6
21
x2
1
4
9
16
25
36
91
xy
7
18
15
44
50
78
212
y  917
.
x  3.5
y  5.27  111
. x
Chapter 4: Forecasting
9
Period 7 forecast = 13.07
Period 12 forecast = 18.64, but this is far outside the range of valid data.
4.30 Given
(a)
(b)
(c)
Y  36  4.3X
Y  36  4.3  70  337
Y  36  4.3  80  380
Y  36  4.3  90  423
4.31 (a)
(b)
4000  0.2015,000  7,000
4000  0.2025,000  9,000
4.32 (a)
x
16
12
18
14
60
y
330
270
380
300
1,280
xy
5,280
3,240
6,840
4,200
19,560
x2
256
144
324
196
920
60
 15
4
1,280
y
 320
4
 xy  nx y 19,560  415320  360
b


 18
 x 2  nx 2
920  415 2
20
a  y  bx  320  1815  50
x
Y  50  18x
(b)
If the forecast is for 20 guests, the bar sales forecast is $410 (50+1820). Each guest accounts
for an additional $18 in bar sales.
4.35 Given:  X  15,  Y  20 ,  XY  70 ,  X 2  55 ,  Y 2  130 , X  3 , Y  4
 XY  nXY
(a)
b
 X 2  nX 2
a  Y  bX
70  5  3  4 70  60 10


1
55  5  32
55  45 10
a  4 1 3  4  3  1
Y  1  1X
b
(b)
Correlation coefficient:
r
n  XY   X  Y
n  X 2   X 
2
n  Y 2   Y 
2

5  70  15  20
5  55  152 5  130  202
350  300
50
50


 0.45
.
275  225 650  400
50  250 11180
The correlation coefficient indicates that there is a positive correlation between bank deposits
and consumer price indices in Birmingham, Alabama—i.e., as one variable tends to increase
(or decrease), the other tends to increase (or decrease).
Standard error of the estimate:

(c)
10
Instructor’s Solutions Manual t/a Operations Management
Syx 
 Y 2  a  Y  b  XY
130  1  20  1  70
130  20  70



n2
3
3
40
3
 13.3  3.65
4.36 (a)
Given: Y  90  48.5 X1  0.4 X2 where:
Y  expected travel cost
X1  number of days on the road
X2  distance traveled, in miles
r  0.68 (coefficient of correlation)
If:
Number of days on the road  X1  5 and distance traveled  X2  300
then:
Y  90  48.5  5  0.4  300  90  242.5  120  452.5
(b)
(c)
Therefore, the expected cost of the trip is $452.50.
The reimbursement request is much higher than predicted by the model. This request should
probably be questioned by the accountant.
A number of other variables should be included, such as:
1. the type of travel (air or car)
2. conference fees, if any
3. costs of entertaining customers
4. other transportation costs—cab, limousine, special tolls, or parking
In addition, the correlation coefficient of 0.68 is not exceptionally high. It indicates that the
model explains approximately 50% of the overall variation in trip cost. This correlation
coefficient would suggest that the model is not a particularly good one.
4.37
Column Totals
X
2
1
4
5
3
15
Y
4
1
4
6
5
20
X2
4
1
16
25
9
55
Y2
16
1
16
36
25
94
b
 XY  nXY
 X 2  nX 2
XY
8
1
16
30
15
70
Given: Y  a  bX where:
a  Y  bX
and  X  15,  Y  20 ,  XY  70 ,  X 2  55 ,  Y 2  94 , X  3 , Y  4 . Then:
Chapter 4: Forecasting
11
70  5  4  3 70  60

 1.0
55  5  32
55  45
a  4  1  3  1.0
b
and Y  1.0  1.0 X . The correlation coefficient:
r

n  XY   X  Y
n
X2
  X 
2
n Y2
  Y 
350  300

275  225 470  400
2

5  70  15  20
5  55  152 5  94  20 2
50
50

 0.845
.
50  70 5916
Standard error of the estimate:
Syx 
4.41 (a)
 Y 2  a  Y  b  XY

n2
94  1  20  1  70

52
94  20  70
 1333
.
 115
.
3
It appears from the following graph that the points do scatter around a straight line.
50
40
30
20
10
0
12
0
5
10
15
20
Num. Tourists (in 1,000,000)
25
Instructor’s Solutions Manual t/a Operations Management
(b)
Developing the regression relationship, we have:
Tourists
(Millions) (X)
7
2
6
4
14
15
16
12
14
20
15
7
Year
1
2
3
4
5
6
7
8
9
10
11
12
Ridership
(1,000,000s) (Y)
1.5
1.0
1.3
1.5
2.5
2.7
2.4
2.0
2.7
4.4
3.4
1.7
X2
49
4
36
16
196
225
256
144
196
400
225
49
Y2
2.25
1.00
1.69
2.25
6.25
7.29
5.76
4.00
7.29
19.36
11.56
2.89
XY
10.5
2.0
7.8
6.0
35.0
40.5
38.4
24.0
37.8
88.0
51.0
11.9
Given: Y  a  bX where:
b
 XY  nXY
 X 2  nX 2
a  Y  bX
. ,  XY  352.9 ,  X 2  1796 ,  Y 2  71.59 , X  11, Y  2.26 .
and  X  132 ,  Y  271
Then:
352.9  12  11  2.26 352.9  298.3 54.6


 0.159
1796  12  112
1796  1452
344
a  2.26  0.159  11  0.511
b
and
(c)
Y  0.511  0.159 X
Given a tourist population of 10,000,000, the model predicts a ridership of:
y  0.511  0159
.
 10  2101
.
or 2,101,000 persons.
(d)
(e)
If there are no tourists at all, the model predicts a ridership of 0.511, or 511,000 persons. One
would not place much confidence in this forecast, however, because the number of tourists is
outside the range of data used to develop the model.
The standard error of the estimate is given by:
 Y 2  a  Y  b  XY

n2
Syx 
71.59  13.85  5611
.
 .163  .404  rounded to .407 in POM for Windows software 
10

(f)
71.59  0.511  271
.  0.159  352.9
12  2
The correlation coefficient and the coefficient of determination are given by:
r

n  XY   X  Y
n  X2
  X 
2
n Y2
  Y 
2

4234.8  3577.2

21552  17424 859.08  734.41
12  352.9  132  271
.
12  1796  1322 12  71.59  271
.2
657.6
657.6

 0.917
.
4128  124.67 64.25  11166
and r 2  0.840
Chapter 4: Forecasting
13