Chapter 03 - Forecasting
CHAPTER 3
FORECASTING
Solutions
1.
a. Plotting each data set reveals that blueberry muffin orders are stable, varying around an
average. Therefore, the naïve forecast is the last value, 33. The demand for cinnamon buns
has a trend. The last change was from 31 to 33 (33 – 31 = 2). Using the last value and
adding the last trend change, the forecast is 33 + 2 = 35. Demand for cupcakes has an
apparent seasonal variation with peaks every five days. Day 1 = 45, Day 6 = 48 and Day
11 = 47. Since the peaks occur every five days, the next peak would be at Day 16.
b. The use of sales data instead of demand implies that sales adequately reflect demand. We
are assuming that no stock-outs occurred because demand equals sales if there are no
shortages.
2.
a.
Sales
20
0
F
M
A
t
1
Y
19
tY
19
2
18
36
3
15
45
4
20
80
5
6
18
22
90
132
7
20
140
M
Month
J
J
A
S
b. 1)
From Table 3–1 with n = 7, t = 28, t2 = 140
b
n  tY   t  Y 7(542 )  28(132 )

 .50
7(140 )  28(28)
n  t 2  ( t ) 2
a
 Y  b  t 132  .50 (28)

 16 .86
n
7
28 132 542
For Sept., t = 8, and Yt = 16.86 + .50(8) = 20.86 (000)
3-1
Chapter 03 - Forecasting
2)
MA 5 
15  20  18  22  20
 19
5
Solutions (continued)
Month
April
Forecast = F(old) + .20[Actual – F(old) ]
18.8
= 19
+ .20[ 18 – 19
]
May
18.04
= 18.8
+ .20[ 15
– 18.8 ]
June
18.43
= 18.04
+ .20[ 20
– 18.04 ]
July
18.34
= 18.43
+ .20[ 18
– 18.43 ]
August
19.07
= 18.34
+ .20[ 22
– 18.34 ]
= 19.07
+ .20[ 20
– 19.07 ]
4)
September 19.26
20
5)
.6 (20) + .3(22) + .1(18) = 20.4
3)
c. Probably 5 month moving average because the data appear to vary around an average of
about 19 [18.86].
d. Sales are reflective of demand (i.e., no stockouts or backorders occurred).
3.
a. 88 + .1(89.6 – 88) = 88.16
b. 88.16 + .1(92 – 88.16) = 88.54
4.
a. 22
b.
22  18  21  22
 20 .75
4
c. F3 = 20 + .30(22 – 20) = 20.6
F4 = 20.6 + .30(18 – 20.6) = 19.82
F5 = 19.82 + .30(21 – 19.82) = 20.17
F6 = 20.17 + .30(22 – 20.17) = 20.72
5.
a. Annual sales are increasing by 15,000 bottles per year.
b. t = 16, Yt = 80 + 15(16) = 320, which is 320,000 bottles.
6.
Yt  500 
200
t  500  20 t
10
3-2
Chapter 03 - Forecasting
Solutions (continued)
7.
a.
t
1
Y
220
t*Y
220
t2
1
2
245
490
4
3
280
840
9
4
275
1,100
16
5
300
1,500
25
6
310
1,860
36
7
350
2,450
49
8
360
2,880
64
9
400
3,600
81
10
11
380
420
3,800
4,620
100
121
12
450
5,400
144
13
460
5,980
169
14
15
475
500
6,650
7,500
196
225
16
510
8,160
256
17
525
8,925
289
18
171
541
7001
t
i
t
2
i
b
b
a
9,738 324
75,713 2109
 171
 Y  7001
 2109
 t Y  75, 713
i
i i
(n)( tiYi )  ( ti )( Yi )
(n)( ti2 )  ( ti ) 2
(18)(75, 713)  (171)(7001) 165, 663

 19
(18)(2109)  (171) 2
8721
 y  b t or
y  bt
n
 7001 
 171 
a
  19 
  388.944  19(9.5)  208.444
 18 
 18 
Yˆ  208.444  19t
3-3
Chapter 03 - Forecasting
b.
F = 208.444 + (19)(20) = 588.444
F = 208.444 + (19)(21) = 607.444
The forecasted demand for week 20 and 21 is 588.444 and 607.444 respectively.
Solutions (continued)
c.
800  541
 13 .63 weeks
19
It would take approximately 14 more weeks. Since we have just completed week 18, the
loading volume is expected to reach 800 by week 32 (18 + 14).
3-4
Chapter 03 - Forecasting
8.
t
1
Y
200
t*Y
200
t2
1
2
3
214
211
428
633
4
9
4
228
912
16
5
235
1,175
25
6
222
1,332
36
7
248
1,736
49
8
250
2,000
64
9
253
2,277
81
10
267
2,670
100
11
281
3,091
121
12
275
3,300
144
13
280
3,640
169
14
288
4,032
196
15
310
4,650
225
a.
Y  3762  t Y  32076 t
t
120
b
(n)( tiYi )  ( ti )( Yi )
1
b
i
2
1
i i
 1240
(n)( ti2 )  ( ti ) 2
(15)(32076)  (120)(3762) 29700

 7.07
(15)(1240)  (120) 2
4200
  Yi
a  
 n
   ti
  b 
  n



 3772 
 120 
a
  (7) 
  250.8  (7)(8)  194.23
 15 
 15 
Y  194.23  7.07t
3-5
Chapter 03 - Forecasting
Solutions (continued)
Forecasted demand for periods 16 through 19 are:
Y16 = 194.23 + (7)(16) = 307.37
Y17 = 194.23 + (7)(17) = 314.44
Y18 = 194.23 + (7)(18) = 321.51
Y19 = 194.23 + (7)(19) = 328.59
b. Initial Trend =
228  200
 9.33
3
St + Tt = TAFt
228 + 9.33 = 237.33
TAFt + .3(A – TAFt) = St
237.33 + .3(235 – 237.33) = 236.63
Tt–1 + .2 (TAFt – TAFt–1 – Tt–1) = Tt
9.33
232
236.63 + 9.33 = 245.96
245.96 + .3(232 – 245.96) = 241.77
9.33 + .2(245.96 – 237.33 – 9.33)
= 9.19
7
248
241.77 + 9.19 = 250.96
250.96 + .3(248 – 250.96) = 250.07
9.19 + .2(250.96 – 245.96 – 9.19)
= 8.352
8
250
250.07 + 8.352 = 258.42
258.42 + .3(250 – 258.42) = 255.89
8.352 + .2(258.42 – 250.96 – 8.352) = 8.174
9
253
255.89 + 8.174 = 264.06
264.06 + .3(253 – 264.06) = 260.74
8.174 + .2(264.06 – 258.42 – 8.174) = 7.667
10
267
260.74 + 7.667 = 268.41
268.41 + .3(267 – 268.41) = 267.99
7.667 + .2(268.41 – 264.06 – 7.667) = 7.004
11
281
267.99 + 7.004 = 274.99
274.99 + .3(281 – 274.99) = 276.79
7.004 + .2(274.99 – 268.41 – 7.004) = 6.92
12
275
276.79 + 6.92 = 283.71
283.71 + .3(275 – 283.71) = 281.10
6.92 + .2(283.71 – 274.99 – 6.92)
= 7.28
13
280
281.10 + 7.28 = 288.38
288.38 + .3(280 – 288.38) = 285.87
7.28 + .2(288.38 – 283.71 – 7.28)
= 6.758
14
288
285.87 + 6.758 = 292.63
292.63 + .3(288 – 292.63) = 291.24
6.758 + .2(292.63 – 288.38 – 6.758) = 6.256
15
310
291.24 + 6.256 = 297.50
297.50 + .3(310 – 297.50) = 301.25
6.256 + .2(297.5 – 292.63 – 6.256) = 5.98
Period
5
Actual
235
6
16
9.
301.25 + 5.98 = 307.23
The initial estimate of trend is based on the net change of 30 for the three periods from 1 to 4,
for an average of +10 units. Use  = .5 and  = .4.
Initial trend = (240 – 210)/3 = 10
3-6
Chapter 03 - Forecasting
t Period At Actual
1
210
Model
2
224
Development
3
229
4
240
TAFt
Model Test
Next
TAFt + (At – TAFt)
= St
Tt–1
+ ( TAFt–1 – TAFt–1 – Tt–1)
= Tt
10
+ .4(0)
= 10
+ .4(262.5 – 250 – 10)
= 11.00
5
255
250*
250
+ .5(255 – 250)
= 252.5
6
265
262.5
262.5
+ .5(265 – 262.5)
= 263.75 10
7
272
274.75 274.75
+ .5(272 – 274.75) = 272.37 11.00
+ .4(274.75 – 262.5 – 11.00) = 11.50
8
285
284.87 284.87
+ .5(285 – 284.87) = 284.94 11.50
+ .4(284.87 – 274.75 – 11.50) = 10.95
9
294
295.89 295.89
+ .5(294 – 295.89) = 294.95 10.95
+ .4(295.89 – 284.87 – 10.95) = 10.98
10
305.92
Forecast
* Estimated by the manager
Solutions (continued)
3-7
Chapter 03 - Forecasting
10.
Yt = 70 + 5t
t = 0 (June of last year)
t = 1 (July of last year)
t = 7 (January of this year)
t = 8 (February of this year)
t = 9 (March of this year)
t = 19 (January of next year)
t = 20 (February of next year)
t = 21 (March of next year)
YJan = 70 + (5)(19) = 165
YFeb. = 70 + (5)(20) = 170
YMar. = 70+ (5)(21) = 175
Forecast = (Trend) * (Seasonal Relative)
Month
January
Trend * Seasonal Relative
165 * 1.10
Forecast (Trend * Seasonal Rel)
181.5
February
170 * 1.02
173.4
175 * .95
166.25
March
11.
Quarter
Value of t
Trend component, Ft
Quarter relative
Forecast
I
8
116
1.1
127.6
II
9
143.5
1.0
143.5
3-8
III
10
175
0.6
105
IV
11
210.5
1.3
273.65
I
12
250
1.1
275
Chapter 03 - Forecasting
12.
a.
Week Day
1
2
3
4
5
6
Sales
Fri
149
Sat
250
Sun
166
Fri
Sat
Moving
Total
Centered
Moving Av. Sales/MA5
188.3
1.33 Sat
565
190
0.87
154
570
191.7
0.80
255
575
190.3
1.34 Sat
Sun
162
571
189.7
0.85
Fri
152
569
191.3
0.79
Sat
260
574
194.3
1.33 Sat
Sun
171
583
193.7
0.88
Fri
150
581
196.3
0.76
Sat
268
589
197
1.36 Sat
Sun
173
591
200
0.87
Fri
159
600
201.7
0.79
Sat
273
605
202.7
1.35 Sat
Sun
176
608
204
0.86
Fri
163
612
205
0.80
Sat
276
615
207.3
1.33 Sat
Sun
183
622
x : Fri = 0.79; Sat = 1.34; Sun = 0.87
b. Fri = 163+(163-159)=167, Sat = 276+(276-273)=279, Sun = 183+(183-176)=190
13.
14.
Wednesday
= .15 x 4 = 0.60
Thursday
= .20 x 4 = 0.80
Friday
= .35 x 4 = 1.40
Saturday
= .30 x 4 = 1.20
a. There appears to be a long-term upward increasing trend in the data. The forecast will
underestimate when data values increase.
3-9
Chapter 03 - Forecasting
Solutions (continued)
b.
Trend Analysis for Passengers
Linear Trend Model
Yt = 396.974 + 4.59340*t
480
 
470
 Actual
Fits
Actual
Fits
 
460
Passengers

450
440

430

420
410

400



 





MAPE:
MAD:
MSD:
0
10
20
Time
Solutions (continued)
t
1
Y
405
t*Y
405
t2
1
2
3
410
420
820
1260
4
9
4
415
1660
16
5
412
2060
25
6
420
2520
36
7
424
2968
49
8
433
3464
64
9
438
3942
81
10
440
4400
100
11
446
4906
121
12
451
5412
144
13
455
5915
169
14
464
6496
196
15
466
6990
225
16
474
7584
256
17
476
8092
289
18
482
8676
324
3-10
0.6765
2.9086
14.6487
Chapter 03 - Forecasting
t
1
b
b
 171
Y  7931 t Y  77570 t
i
2
1
i i
(n)( tiYi )  ( ti )( Yi )
(n)( ti2 )  ( ti ) 2
(18 )(77570 )  (171)(7931 ) 40059

 4.593
8721
(18 )(2109 )  (171) 2
  Yi 
 t 
  (b)  i 
a
 n 
 n 




 7931 
 171 
a
  (4.5933 )
  396 .974
 18 
 18 
Y  396 .974  4.593 t
Forecasted demand for the next three weeks are:
Y19 = 396.974 + (4.593)(19) = 484.25
Y20 = 396.974 + (4.593)(20) = 488.84
Y21 = 396.974 + (4.593)(21) = 493.44
3-11
 2109
Chapter 03 - Forecasting
Solutions (continued)
15.
Day
(Data)
No. Served
Moving
Total
(Relative estimates)
Centered Average Data Centered Average
1=1
2=2
3=3
4=4
5=5
80
75
78
95
130
90.57
90.86
6=6
7=7
136
40
634
91.14
91.43
95/90.57 = 1.0489
130/90.86 = 1.4308
136/91.14 =
1.49922
40/91.43 = 0.4375
8=1
9=2
10 = 3
11 = 4
12 = 5
13 = 6
14 = 7
82
77
80
94
131
137
42
636
638
640
639
640
641
643
91.29
91.43
91.57
91.86
92.14
92.29
92.71
82/91.29 = .8983
77/91.43 = .8422
80/91.57 = .8736
94/91.86 = 1.0233
125/91.14 = 1.4217
135/92.29 = 1.4845
42/92.71 = .4530
15 = 1
16 = 2
17 = 3
18 = 4
19 = 5
20 = 6
21 = 7
84
78
83
96
135
140
44
645
646
649
651
655
658
660
93.00
93.57
94.00
94.29
94.71
95.29
96.00
84/93.00 = .9032
77/93.57 = .8336
83/94.00 = .8830
96/94.29 = 1.0182
135/94.71 = 1.4253
140/95.29 = 1.4693
37/96.00 = .4583
22 = 1
23 = 2
24 = 3
25 = 4
26 = 5
27 = 6
28 = 7
87
82
88
99
144
144
48
663
667
672
675
684
688
692
96.43
97.71
98.29
98.86
87/96.43 = .9022
82/97.71 = .8392
98/98.29 = .8953
103/98.86 = 1.0014
Group and average the relative estimates:
1’s
2’s
3’s
4’s
5’s
6’s
7’s
1.0489
1.4301
1.4922
.4375
.8983
.8422
.8736
1.0233
1.4217
1.4845
.4530
.9032
.8336
.8830
1.0182
1.4253
1.4693
.4583
.9022
.8392
.8953
1.0014
2.7037
x : .9012
2.5150
.8383
2.6520
.8840
4.0919
1.0230
4.2779
1.4260
4.4459
1.4820
1.3488
.4496
3-12
Chapter 03 - Forecasting
16.
a. The trend may be non-linear (although most students will view it as linear). Trendadjusted smoothing would have a slight edge over a linear trend line.
b. Yes. This would cause concern because you wouldn’t know the actual demand for the
pain reliever.
Sales
60
50
40
30
2
4
6
8
Day
10
12
14
50
40
c.
9
10
30
TAF
51.7
53.7
11
53.9370
12
54.7730
13
56.0920
14
56.7360
15
57.5976
16
58.4344
MSE=
6.088
3-13
Chapter 03 - Forecasting
Day Demand TAFt
TAFt + .3(At – TAFt) = St
Tt–1 + .3(TAFt – TAFt–1 – Tt–1)
= Tt
8
49
50
50 + .3(49 – 50) = 49.7
2 + .3(50 – 50 – 0 )
=2
9
52
51.7
51.7 + .3(52 – 51.7) = 51.79
2 + .3(51.7 – 50 – 2 )
= 1.91
10
48
53.7
53.7 + .3(48 – 53.7) = 51.99
1.91 + .3(53.7 – 51.7 – 1.91)
11
52
53.927 53.927 + .3(52 – 53.9) = 53.349 1.937 + .3(53.927 – 53.7 – 1.937)
12
ei2
ei
–1
1
.3
.09
= 1.937
–5.7
32.49
= 1.424
–1.9270
3.713329
55
54.773 54.773 + .3(55 – 54.6) = 54.841 1.424 + .3(54.773 – 53.927 – 1.424) = 1.251
.2270
.051529
13
54
56.092 56.093 + .3(54 – 56.093) =
55.465
1.251 + .3(56.093 – 54.773 – 1.251) = 1.271
–2.0920
4.376464
14
56
56.736 56.736 + .3(56 – 56.736) =
56.515
1.271 + .3(56.736 – 56.093 – 1.271)
=
1.0826
–.7360
.541696
15
57
57.5976 57.597 + .3(57 – 57.597) =
57.418
1.0826 + .3(57.597 – 56.735 – 1.0826) =
1.0164
–.5976
.357126
16
-
MSE 
58.4344
42.62014
 6.088592
7
3-14
Chapter 03 - Forecasting
Solutions (continued)
17.
Month
Jan .........
Units
Sold
640
Index
0.80
Month
Jul. ..........
Units
Sold
765
Index
0.90
Feb.........
648
0.80
Aug. ........
805
1.15
Mar. .......
630
0.70
Sept. ........
840
1.20
Apr. .......
761
0.94
Oct. .........
828
1.20
May .......
735
0.89
Nov. ........
840
1.25
Jun. ........
850
1.00
Dec. ........
800
1.25
Month
Jan .........
Units
Sold
640
Index
0.80
Deseasonalized
800
Month
Jul............
Units
Sold
765
Index
0.90
Deseasonalized
850
Feb.........
648
0.80
810
Aug. ........
805
1.15
700
Mar. .......
630
0.70
900
Sept. ........
840
1.20
700
Apr. .......
761
0.94
809.6
Oct. .........
828
1.20
690
May .......
735
0.89
825.8
Nov. ........
840
1.25
672
Jun. ........
850
1.00
850
Dec. .........
800
1.25
640
Solution
d. Part b indicated a downward trend in sales. Fit a monthly trend line to the
deseasonalized values using t = 0 in December of the previous year. Then predict
trend values for the first three months of next year (t = 13, 14, 15). Finally, multiply
each month’s trend value by the appropriate monthly seasonal index.
e. Advertising and sales promotions.
3-15
Chapter 03 - Forecasting
Solutions (continued)
18.
Deseasonalize the values, where:
Deseasonalized sales = (Actual sales) / (Seasonal relative)
Deseasonalized sales for quarter 1 is (88) / (1.1) = 80
Deseasonalized sales for quarter 2 is (99) / (.99) = 100
Deseasonalized sales for quarter 3 is (108) / (.90) = 120
Deseasonalized sales for quarter 4 is (141.4) / (1.01) = 140
There is a trend of +20 from previous quarter, hence the trend forecast would be 160 units.
Multiplying the trend forecast by the seasonal relative for quarter 1 yields a forecast for the
next quarter: T*S = (140 + 20) x (1.1) = 176.
19.
1
20.
2
Quarter
3
4
0.83
0.98
0.84
1.34
0.82
0.99
0.82
1.38
0.83
0.99
0.83
1.34
0.82
1.00
0.84
1.33
3.30
3.96
3.33
5.39
x 0.83 0.99
0.83
1.35
Units
t sold Naive
e | e | e2
11 147

12 148
147
1
1
1
13 151
148
3
3
9
14 145
151
–6
6 36
15 155
145
10 10 100
16 152
155
–3
3
9
17 155
152
3
3
9
18 157
155
2
2
4
19 160
157
3
3
9
20 165
160
5
5 25
18 36 202
MAD :
36
 4.0
9
MAD :
MSE :
202
 25 .25
8
MSE :
Trend
146
e | e | e2
1 1 1
148
150
0
1
0
1
0
1
152
–7
154
1
156
–4
4 16
158
–3
3
9
160
–3
3
9
162
–2
2
4
164
1
1
1
7 49
1
1
–16 23 91
23
 2.3
10
91
 10 .11
9
Linear trend provides forecasts with less average error and less average squared error.
3-16
Chapter 03 - Forecasting
Solutions (continued)
21.
Period
Demand
F1
e
e
F2
e
e
2
e2
4
66
2
2
e2
4
1
68
66
2
2
75
68
7
7
49
68
7
7
49
3
70
72
–2
2
4
70
0
0
0
4
74
71
3
3
9
72
2
2
4
5
69
72
–3
3
9
74
–5
5
25
6
72
70
+2
2
4
76
–4
4
16
7
80
71
9
9
81
78
2
2
4
8
78
74
4
4
16
80
–2
2
4
32
176
24
106
a. MAD F1: 32/8 = 4.0
MAD F2: 24/8 = 3.0
F2 appears to be more accurate.
b. MSE F1: 176/7 = 25.14
MSE F2: 106/7 = 15.14
F2 appears to be more accurate.
c. Either one might already be in use, familiar to users, and have past values for comparison.
If control charts are used, MSE would be natural; if tracking signals are used, MAD would
be more natural.
d. MAPE F1 = .4269/8 = .0534
MAPE F2 = .3284/8 = .0411
22.
Since .0411 < .0534, F2 appears to be more accurate.
a.
Forecast #1
A
(A–F)
Month Sales F Forecast Error
Error2
|e|
1
770
771
–1
1
1
Forecast #2
Forecast
769
Error
1
Error2
1
|e|
1
2
3
789
794
785
790
4
4
16
16
4
4
787
792
2
2
4
4
2
2
4
780
784
–4
16
4
798
–18
324
18
5
768
770
–2
4
2
774
–6
36
6
6
772
768
4
16
4
770
2
4
2
7
760
761
–1
1
1
759
1
1
1
8
775
771
4
16
4
775
0
0
0
9
786
784
2
4
2
788
–2
4
2
10
790
788
2
4
2
788
2
4
2
12
94
28
–16
382
36
3-17
Chapter 03 - Forecasting
Solutions (continued)
b.
Period
Absolute Percentage Error (F1)
Absolute Percentage Error (F2)
1
1 / 770 = .00130
1 / 770 = .0013
2
4 / 789 = .00507
2 / 789 = .00253
3
4 / 794 = .00504
2 / 794 = .00252
4
4 / 780 = .00513
18 / 780 = .02308
5
2 / 768 = .00260
6 / 768 = .00781
6
4 / 772 = .00518
2 / 772 = .00259
7
8
1 / 760 = .00132
4 / 775 = .00516
1 / 760 = .00132
0 / 775 = 0
9
2 / 786 = .00254
2 / 786 = .00254
10
2 / 790 = .00253
2 / 790 = .00253
Total
.03587
.04622
MAPE F1 = .03587 / 10 = .003587
MAPE F2 = .04622 /10 = .004622
Since .003587 < .004622, choose forecasting method 1.
Forecast #1:
(A – F) 2
n–1
94/9 = 10.44
Forecast #2
382/9 = 42.44
MSE =
| e |
N
28/10 = 2.8
MAD =
36/10 = 3.6
3-18
Chapter 03 - Forecasting
Solutions (continued)
c.
Naïve
Forecast
(A – F)
Error
|e|
Error2
789
770
19
19
361
3
794
789
5
5
25
4
780
794
–14
14
196
5
768
780
–12
12
144
6
772
768
4
4
16
7
760
772
–12
12
144
8
775
760
15
15
225
9
786
775
11
11
121
10
790
786
4
4
16
20
96
1,248
MAD =
96
9
Month
1
Sales
770
2
11
790
MSE =
1,248
8
Control limits: 0  2
= 156
= 10.67
20
Tracking
=
Signal
10.67
=1.87
156 = 0  25 [in control]
It appears that naïve forecast is in control because all of the tracking signal values are
well within  4 and all of the error terms are well within 2 sigma control limits. However,
MAD and MSE values are much higher than MAD and MSE values for the other two
forecasting methods. Therefore, naïve forecasting method does not appear to be
performing as well as the other two forecasting methods.
23.
a.
150
100
Insurance
needed ($000)
50
0
0
000
10
20
30
Current Age of Head of Household (years)
b. y = 150 – .1(30) = 147. Thus, a 30-year old will need $147,000 of life insurance.
3-19
40
Chapter 03 - Forecasting
Solutions (continued)
24.
a. Let x1 = weight in lb.
x2 = distance in miles
y
y = $.10 x1 + $.15 x2 + $10
= delivery charge
b. y = $.10(40) + $.15(26) + $10 = $17.90
25.
a.
X = Price
6.00
Y = Sales
200
X*Y
1200.00
Y2
40,000
X2
36.0000
6.50
6.75
190
188
1235.00
1269.00
36,100
35,344
42.2500
45.5625
7.00
180
1260.00
32,400
49.0000
7.25
170
1232.50
28,900
52.5625
7.50
162
1215.00
26,244
56.2500
8.00
160
1280.00
25,600
64.0000
8.25
155
1278.00
24,025
68.0625
8.50
156
1326.00
24,336
72.2500
8.75
148
1295.00
21,904
76.5625
9.00
140
1260.00
19,600
81.0000
9.25
133
1230.25
17,689
85.5625
X
b
b
1
 92.75
Y  1982  X Y 15081 Y
i
i i
i
2
 332142
X
(n)( X iYi )  ( X i )( Yi )
(n)( X i2 )  ( X i ) 2
(12)(15081)  (92.75)(1982) 2852.5

 19.51
(12)(729.06)  (92.75) 2
146.1875
  Yi
a  
 n

  Xi
  (b) 

 n



 1982 
 92.75 
a
  (19.56) 
  165.1667  (19.56)(7.7292)  315.98
 12 
 12 
Y  316.35  19.51X i
3-20
2
i
 729.06
Chapter 03 - Forecasting
Solutions (continued)
Actual data are represented by circles.
Predicted values are represented by pluses.
200
+
190
+
+
180
+
sales
+
170
+
160
+
+
150
+
+
140
+
+
130
6
7
8
9
price
r
r
n( xy)  ( x)( y )
n(  x )  (  x ) 2 n(  y 2 )  (  y ) 2
2
(12)(15081)  (92.75)(1982)
(12)(729.06)  (92.75) 2 (12)(332142)  (1982) 2
 .9849
b. r = –.9848. Implies a high, negative relationship between price and demand.
r2 = (–.9848) 2 = .97. It appears that approximately 97% of the variation in sales can be
accounted for by the price of our product. This indicates that price is a good predictor of
sales.
26.
a.
100


y
 








50
0
10
20
30
3-21
x
40
50
Chapter 03 - Forecasting
Solutions (continued)
b.
c.
x
y
x2
xy
y2
15.00
74.00
1110.0
225.0
5476.0
25.00
80.00
2000.0
625.0
6400.0
40.00
84.00
3360.0
1600.0
7056.0
32.00
81.00
2592.0
1024.0
6561.0
51.00
96.00
4896.0
2601.0
9216.0
47.00
95.00
4465.0
2209.0
9025.0
30.00
83.00
2490.0
900.0
6889.0
18.00
78.00
1404.0
324.0
6084.0
14.00
70.00
980.0
196.0
4900.0
15.00
72.00
1080.0
225.0
5184.0
22.00
85.00
1870.0
484.0
7225.0
24.00
88.00
2112.0
576.0
7744.0
33.00
90.00
2970.0
1089.0
8100.0
366.00
1076.00
31329.0
12078.0
89860.0
b
n  xy   x  y (13)(31329 )  (366 )(1076 )

 .584
n  x 2  ( x ) 2
(13)(12078 )  (366 ) 2
a
 y  b  x 1076  (.584 )(366 )

 66 .33
n
13
r
r
n( xy)  ( x)( y )
n(  x 2 )  (  x ) 2 n(  y 2 )  (  y ) 2
(13)(31329 )  (366 )(1076 )
(13)(12078 )  (366 ) 2 (13)(89860 )  (1076 ) 2
 .869
r 2  (.869 ) 2  .755
Approximately 75% of the variation in the dependent variable is explained by the
independent variable.
d. y = 66.33 + .584 (41) = 90.268.
3-22
Chapter 03 - Forecasting
Solutions (continued)
27.
a. Fertilizer
(X)
1.6

Mower
(Y)
10
(X2)
2.56
(Y2)
100
(X)*(Y)
16.0
1.3
1.8
8
11
1.69
3.24
64
121
10.4
19.8
2.0
12
4.00
144
24.0
2.2
12
4.84
144
26.4
1.6
9
2.56
81
14.4
1.5
8
2.25
64
12.0
1.3
7
1.69
49
9.1
1.7
10
2.89
100
17.0
1.2
6
1.44
36
7.2
1.9
11
3.61
121
20.9
1.4
8
1.96
64
11.2
1.7
10
2.89
100
17.0
1.6
9
2.56
81
14.4
22.8
131
38.18
1269
219.8
Answers to parts a and b
r
r
r
n( X iYi )  ( X i )( Yi )
n( X i2 )  ( X i ) 2 n( Yi 2 )  ( Yi ) 2
(14 )(219 .8)  (22 .8)(131)
(14 )(38 .18)  (22 .18) 2 (14 )(1269 )  (131) 2
90 .4
14 .68 605

90 .4
 .95924
94 .241
Since the value of r is close to 1, there is a strong positive relationship between these
variables.
3-23
Chapter 03 - Forecasting
Solutions (continued)
b
b
(n)( X iYi )  ( X i )( Yi )
(n)( X i2 )  ( X i ) 2
(14 )(219 .8)  (22 .8)(131) 90 .4

 6.158
14 .68
(14 )(38 .18)  (22 .18) 2
  Yi 

X 
  (b)  i 
a
 n 
 n 




 131 
 22 .8 
a
  (6.158 )
  9.3571  (6.158 )(1.6286 )  .672
 14 
 14 
Y  .672  6.158 X i
c. Y = – .672 + (6.158)(2) = 11.64
Therefore, we expect 12 mowers to be sold in the first week of August.
28.
t
Period
1
A
Demand
129
F
Predicted
124
E
Error
5
|e|
5
Cum.
Error
5
2
3
194
156
200
150
–6
6
6
6
–1
5
4
91
94
–3
3
2
5
85
80
5
5
7
6
7
132
126
140
128
–8
–2
8
2
8
126
124
2
9
95
100
10
11
149
98
12
MADt
Tracking
Signal
5*
1.40***
–1
–3
5.9**
4.73
–.17
–.63
2
–1
3.911
–.26
–5
5
–6
4.238
–1.42
150
94
–1
4
1
4
–7
–3
3.267
3.487
–2.14
–.86
85
80
5
5
2
3.941
.51
13
137
140
–3
3
–1
3.659
–.27
14
134
128
6
6
5
4.361
1.14
* Initial MAD:
| e |
5
=
25
=5
5
** Updated MAD’s [6 through 14]: MADt = MADt–1 +  (| e |t – MAD t–1)
*** Tracking Signal = Cumulative Error/MADt = 7/5 = 1.40
E.g., MAD6
= MAD5 + .3| e |t – MAD5
= 5.0 + .3(8 – 5.0) = 5.9
3-24
Chapter 03 - Forecasting
Solutions (continued)
Since all tracking signal values are within the limits, the forecast is in control.
3
2
Value of 1
Tracking
0
Signal
–1
–2
–3
Upper Limit
Lower Limit
5
6
7
8
9 10 11 12 13 14
Period
29. Refer to data in problem 22
a.
Tracking signal =
Errors
MAD
#1:
12/2.8 = 4.29
[both slightly beyond limits of  4]
#2: –16/3.6 = –4.44
Since 4.29 > 4, the forecasting method 1 is biased. Using F1 we are underestimating
demand.
Since –4.44 < –4, the forecasting method 2 is also biased. Using F2 we are
underestimating demand.
b.
Control limits are 0  2
#1
02
limits, the forecast is in control.
MSE
10.44  0  6.46.
42.44  0  13.03.
#2
02
these limits, the forecast is not in control.
3-25
Since all errors are within these
Since the error for month 4 exceeds
Chapter 03 - Forecasting
Solutions (continued)
30.
a.
t
Month
1
Cum. MAD t = MAD t–1 T.S. =
| e | + .1[ | e | t – MAD t–1]
8
e
Error
–8
Cum.
Error
–8
| e |t
8
2
–2
–10
2
10
3
4
–6
4
14
4
5
7
9
1
10
7
9
21
30
6
5
15
5
35
7
0
15
0
35
8
9
–3
–9
12
3
3
9
38
47
10
–4
–1
4
51
11
1
0
1
52
12
13
6
8
6
14
14
4
15
16
Cum. Error
MAD t
4.727*
0/4.727 =
0
6
8
4.857
5.171
6/4.857 =
14/5.171 =
1.235
2.707
18
4
5.054
18/5.054 =
3.562
1
–2
19
17
1
2
4.649
4.384
19/4.649 =
17/4.384 =
4.087**
3.878
17
–4
13
4
4.346
13/4.346 =
.334
18
–8
5
8
4.711
5/4.711 =
1.061
19
20
–5
–1
0
–1
5
1
4.740
4.366
0/4.740 =
–14.366 =
0
–.229
* The initial MAD is Cum. | e |  11: 52/11 = 4.727.
** The tracking signal for month 15 is slightly beyond +4, so at that point, the
forecast is suspect.
3-26
Chapter 03 - Forecasting
Solutions (continued)
b.
Month
1
Error
–8
Error2
64
2
–2
4
3
4
16
4
7
49
5
9
81
6
5
25
7
0
0
8
–3
9
9
–9
81
10
–4
16
–1
345
 e2
345
02
02
 0  12 .38
n 1
9
All errors are within these limits.
c. 12
UCL
0
–12
LCL
12
14
16
Month
18
20
The errors may be cyclical, suggesting that there may be a cyclical component in demand
that is being overlooked in the forecast.
31.
a. yt = 35.8 + 4.018t
b.
c.
Year
10
11
12
13
14
Forecast
75.98
80.00
84.02
88.04
92.05
MSE = 15.15/8 = 1.8938, so s =
1.8938 = 1.376. Control limits are 0  2(1.376) = 0  2.75.
Year
10
Sales
77.2
Forecast
75.98
Error
1.22
11
82.1
80.00
2.10
12
87.8
84.02
3.78
13
90.6
88.04
2.56
14
98.9
92.05
6.55
The forecast is not in control; two of the last 5 points are outside of the limits. In an actual
situation, the error in year 12 would have triggered an examination of forecast
performance.
3-27
Chapter 03 - Forecasting
Solutions (continued)
32.
a.
e12
e22
e1
e2
1
1
1
1
1
4
1
2
–1
9
1
3
1
–3
+1
9
1
3
1
–1
+4
1
16
1
4
52
+3
–3
9
9
3
3
47
1
0
1
0
1
0
48
48
1
+1
1
1
1
1
51
52
52
–1
–1
1
1
1
1
54
55
53
–1
+1
1
1
1
1
–2
+4
34
35
16
15
Period
1
Actual
37
Forecast 1
36
Forecast 2
36
error 1
+1
error 2
+1
2
39
38
37
+1
+2
3
37
40
38
–3
4
39
42
38
5
45
46
41
6
49
46
7
47
46
8
49
9
10

MSE 1 
34
 3.78
9
MSE 2 
35
 3.89
9
MAD1 
16
 1.6
10
MAD 2 
15
 1.5
10
Both forecasting methods have MADs that are approximately equal (MAD1 = 1.6, MAD2
= 1.5), and MSEs that are also approximately equal (MSE1 = 3.78, MSE2 = 3.89).
b. The errors for alternative 1 cycle (+1, +1, –3, –3, –1, +3, +1, –1, –1), although all are
within 2s control limits. The errors for alternative 2 (+1, +2, –1, +1, +4, –3, 0, +1, –1, +1)
don’t appear to cycle, but the error of +4 is just beyond the 2s control limits. While the
alternative 1 has a small negative bias (slight overestimation), alternative 2 has a small
positive bias (slight underestimation).
(MFE1 = –0.2, MFE2 = +.4, where MFE is the Mean Forecast Error)
3-28
Chapter 03 - Forecasting
Solutions (continued)
33.
A
F
(sales) (Forecast)
15
15
Cumulative
Error
Error Error
0
0
0
e2
0
MAD
0
.5
TS
0
21
20
1
1
1
1
1
23
30
25
30
–2
0
–1
–1
2
0
3
3
4
0
1
32
35
–3
–4
3
6
9
1.2
–3.333
38
40
–2
–6
2
8
4
1.333
–4.50
42
45
–3
–9
3
11
9
1.571
–5.729
47
50
–3
–12
3
14
9
1.75
–6.85
MSE 
 e2
36

 5.14
8
8 1
s  MSE
34.
A–F
(Error)
0
.75
2
–1
–1.333
No, the forecast is not performing adequately.
As you can see from the tracking signal values,
we overestimate the demand consistently.
(Tracking signal values continue to get larger.)
s  5.14  2.267
A
T = 10 + 5t F = T * S
Cumulative
(sales)
Trend
Forecast Error
Error
(Error)2 Error Error MAD
14
15
13.5
.5
.5
.25
.5
.5
20
20
19
1
24
25
26.25
–2.25
31
30
33
–2
31
35
31.5
–.5
–3.25
.25
37
43
40
45
38
47.25
–1
4.25
–4.25
0
48
50
55
–7
–7
52
55
49.5
2.5
1.5
1
1
1.5
.75
–.75
5.063
2.25
3.75
1.25
4
2
5.75
1.438 –1.912
.5
6.25
1.25
1
18.063
1
4.25
7.25
11.50
1.208 –3.518
1.643 0
49
7
18.50
2.313 –3.026
2.5
21.0
2.333 –1.929
–2.75
–4.5
MSE 
(error ) 2 84 .876

 10 .61
n 1
8
MAD 
21
 2.333
9
TS
1
6.25
2
–.6
–2.6
s  MSE  3.257
Since all of the TS (Tracking signal) values fall between 4 and all of the error terms are well
within  3s, the forecasting method appears to be performing adequately.
3-29
Chapter 03 - Forecasting
Case: M & L Manufacturing
1.
The potential benefit of using a formalized approach to forecasting is that it will be easier to
utilize the computer and easier to quantify the information. A less formalized approach is
more likely to utilize personal intuition. For small forecasting problems, intuition may involve
personal bias which may be reflected in the forecast. As the forecasting problem gets larger, it
will be impossible to solely rely on a less formalized approach because a person’s intuition
will be unable to process the large quantity of information.
2.
Product 1
Plotting the data for Product 1 reveals a linear pattern with the expectation of demand in week
7. Demand in week 7 is unusually high and does not fit the linear trend pattern of the
remaining data. Thus, the demand for the 7th week is considered an outlier. There are different
ways of dealing with outliers. A simple and intuitive way is to replace the demand for the
week in question with the average demand from the previous week and the next week in the
time-series. Therefore in this case, the demand of 90 in week 7 will be replaced with 71.5 =
[(67 + 76)/2].
x
1
y
50
x2
1
xy
50
2
54
4
108
3
57
9
171
4
60
16
240
5
64
25
320
6
67
36
402
7
71.5
49
500.5
8
76
64
608
9
79
81
711
10
82
100
820
11
85
121
935
12
87
144
1044
13
92
169
1196
14
96
196
105
1020.5
1015
1344
8449.5
3-30
Chapter 03 - Forecasting
Product 1 (continued)
b
n ( xy )  ( x )( y) (14 )(8449 .5)  (105 (1020 .5)

 3.498
n ( x 2 )  ( x ) 2
(14 )(1015 )  (105 ) 2
a
 y  b( x ) (1020 .5)  (3.498 )(105 )

 46 .658
n
14
The trend line is T = 46.536 + 3.5t
The next four forecasts (t = 15, 16, 17, 18) are:
Period
15
Forecast (T = 46.658 + 3.498t)
T = 46.658 + 3.498 (15) = 99.13
16
T = 46.658 + 3.498 (16) = 102.63
17
T = 46.658 + 3.498 (17) = 106.12
18
T = 46.658 + 3.498 (18) = 109.62
Product 2
Plotting the data for Product 2 yields a more complex pattern: There is a spike once every four weeks;
the values between the spikes are fairly close to each other. In addition, the data appear to be
increasing at the rate of one unit per week. An intuitive approach would be to use the average of the
three nonspike periods plus 1.0 to predict the next three nonspike periods. Doing so for the data up to
period 15 yields a very small average forecast error (MAD = 0.54). Given the fact that we have only
two data points following the last spike, a reasonable forecast might be to use the last three period
average plus 1.0 (i.e., 43.33 to predict orders for period 15, and use the values for periods 13 and 14
plus 1.0 (i.e., 43.5 + 1.0 = 44.5) as a forecast for periods 17 and 18.
The values of the spikes also seem to be increasing. The initial increase was 1.0 and the second
increase was 2.0. A naive forecast here would be 49 + 2 = 51. However, the average increase was 1.5.
Using that would yield a value of 50.5. One might even be tempted to project an increase of 3.0,
although either of the others seem more justifiable. Still, the fact that there is a limited amount of data
makes this forecast more risky.
Hence, the forecasts are:
Period
15
Forecast
43.33
16
50.5 or 51
17
44.5
18
44.5
3-31
Chapter 03 - Forecasting
Case: Highline Financial Services, Ltd.
Aligning data by quarters, we can see (in the tables and in the figures) that demand for service A is
increasing, demand for service B is decreasing, and demand for service C is mixed. Note, though, that
total annual demand for service C has changed only slightly.
A
Year
1
2
Change
Quarter
1
2
3
4
60
45 100 75
72
51 112 85
+12 +6 +12 +10
Forecast
84
57
124
B
95
1
95
85
-10
Quarter
2
3
85
92
75 85
-10 -7
4
65
50
-15
75
65
35
72
C
1
93
102
+9
Quarter
2
3
90 110
75 110
-15
0
4
90
100
+10
121
60
110
110
Freddie should be concerned about service B, because that has declined for every quarter.
Forecasts were made using a simple naïve (additive) approach. An argument could be made for using
a multiplicative approach (i.e., basing the forecast on the percentage change from one year to the next
instead of the actual change).
Service A
Service B
120
100
80
80
Series1
60
Series2
40
20
Demand
Demand
100
60
Year 1
40
Year 2
20
0
1
2
3
0
4
1
Quarter
2
3
Quarter
Service C
120
Demand
100
80
Series1
60
Series2
40
20
0
1
2
3
4
Quarter
3-32
4
Chapter 03 - Forecasting
Enrichment Module: Additional Methods for Evaluating Forecast Accuracy
The major problem in determining which forecast accuracy measure to use is that there is no
universally accepted accuracy measure. In Chapter 3, several different accuracy measures are covered.
In order to develop a better understanding of the forecast accuracy measures, first we must understand
the nature of the forecast errors. There are two types of forecast errors.
The first type of error is called the forecast bias where the direction of the error is the primary
consideration. If the value of the error is negative, then we can conclude that the forecasting method
overestimated sales or demand or whatever is being predicted. If the value of the error is positive, then
we can conclude that the forecasting method underestimated sales or demand because in calculating
the error term, we always subtract the forecasted value from the actual value. In the textbook, three
forecast accuracy measures designed to assess forecast bias are discussed:
1.
Mean Forecast Error (MFE)
2.
Tracking Signal
3.
Control Charts
When we sum the error terms, if there is no bias, positive and negative error terms will cancel each
other out and the MFE will be zero. As was pointed out above, negative MFE is an indication of
overestimation and positive MFE is an indication of underestimation. However, if the positive and
negative error values tend to cancel each other out and the MFE or Tracking Signal value is zero or
near zero, then we can conclude that the forecasting method does not result in bias (underestimation or
overestimation). Even if there is no significant bias, it is possible that the forecasting method results in
too much overall variation from the actual values. We call measurement of this type of variation
overall forecast accuracy.
In measuring forecast bias, we cannot measure the overall accuracy of the forecasting method because
the positive and negative errors will cancel each other out. In measuring the overall accuracy, there is
never an error value with a negative sign. There are different overall accuracy measures. In general, in
determining the overall forecast accuracy, we either square the error terms or take the absolute value
of the error terms. The goal of the overall accuracy is to determine how well the forecasting method
estimates the actual values without evaluating forecast bias. In Chapter 3, we have covered numerous
ways of evaluating the overall accuracy of forecasts including:
1.
Mean Absolute Deviation (MAD)
2.
Mean Squared Error (MSE)
3.
Standard Error of Estimate
In order to be able to assess both the overall accuracy and forecast bias, an analyst should probably
utilize at least one method from each category. In the next section we will discuss two additional
methods for evaluating forecast accuracy.
Relative measures of forecast accuracy:
3-33
Chapter 03 - Forecasting
The utilization of MSE as a criterion in determining the accuracy of forecasts has some drawbacks.
One of the drawbacks is that in many cases it is not appropriate to compare MSE values obtained from
different forecasting models because different methods use different ways of obtaining the forecasted
values. Thus, comparison of methods using a single criterion such as MSE becomes questionable.
Relative measures use percentages in determining the accuracy of forecasts. As a result of the
limitation of MSE mentioned above, analysts may want to use multiple measures to evaluate the
accuracy of forecasting methods. In addition, multiple accuracy measures may be needed to evaluate
both forecast bias and the overall forecast accuracy. Relative measures are generally considered to be
desirable because percentages are easy to interpret.
The relative forecast accuracy measures that we will discuss in the remaining portion of this section
are:
1.
Mean Percentage Error (MPE)
2.
Mean Absolute Percentage Error (MAPE)
MPE measures the forecast bias while MAPE measures overall forecast accuracy. As with any other
forecast bias measure in calculating MPE, negative and positive error terms offset each other.
Therefore, for a given time-series data MPE  MAPE.
Before stating the equations for MPE and MAPE, first we need to define Percentage Error. The
Percentage Error (PE) for a given time-series data measures the percentage points deviation of the
forecasted value from the actual value. The equations for PE in period i, MPE and MAPE are given
below in equations 1, 2 and 3 respectively.
 A  Fi 
(100 )
PEi   i

 Ai 
(1)
n
MPE 
 PE
i
i 1
(2)
n
n
MAPE 
 PE
i 1
n
i
(3)
where:
Ai is the actual value from period i, and;
Fi is the forecasted (estimated) value from period i.
Both MPE and MAPE are more intuitive and easier to understand and interpret than most of the other
measures because 4% has far more meaning to the user than MSE of 224 or Tracking Signal value of
3.5.
Problems for the enrichment module
3-34
Chapter 03 - Forecasting
Problem 1
An analyst must decide between two different forecasting techniques for weekly sales of bicycles: a
linear trend equation and the naïve approach. The linear trend equation is: Yˆi  12  2 X i , and it was
developed using data from periods 1 through 10. Based on the data from periods 11 through 20,
calculate the MPE and MAPE. Based on the values of MPE and MAPE, comment on which of the two
methods has the greater overall accuracy. Compare the two methods in terms of the forecast bias.
T
11
Units Sold
25
T
16
Units Sold
39
12
28
17
48
13
14
34
40
18
19
50
47
15
44
20
54
Problem 2
In solving problem 26 in the textbook, we calculated both MAD and MSE values. In this exercise we
are going to use the same data and information and calculate MPE and MAPE values. The revised
problem is stated as follows:
Two different forecasting techniques (F1 and F2) were used to forecast demand for cases of bottled
water. Actual demand and the two sets of forecasts are as follows:
Time Period
1
Actual Demand
68
Forecasted demand 1
(F1)
66
Forecasted demand 2
(F2)
66
2
3
75
70
68
72
68
70
4
74
71
72
5
69
72
74
6
7
72
80
70
71
76
78
8
78
74
80
a.
Compute MPE for both sets of forecasts. Which of the two forecasting methods has a higher
forecast bias? Explain.
b.
Compute the MAPE for the two sets of forecasts. Which of the two forecasting methods
provides higher overall accuracy with this data set?
3-35
Chapter 03 - Forecasting
Solutions to enrichment module problems
Solution to problem 1
Percentage Error Calculations using the Naïve Method
Period
Actual
Forecast
Error
PE i
PE i
11
25
–
–
–
–
12
28
25
3
.1071
.1071
13
34
28
6
.1765
.1765
14
40
34
6
.15
.15
15
44
40
4
.0909
.0909
16
39
44
–5
–.1282
–.1282
17
48
39
9
.1875
.1875
18
50
48
2
.04
.04
19
20
47
54
50
47
–3
7
–.0638
.1296
–.0638
.1296
 PE
i
 PE
 .6895
i
 1.0735
MPE 
.6895
 7.66 %
9
MAPE 
1.0735
 11 .93 %
9
Percentage Error Calculations using the Linear Regression Equation
Period
Actual
Forecast
Error
PE i
PE i
11
12
25
28
34
36
–9
–8
–.36
–.2857
.36
.2857
13
34
38
–4
–.1176
.1176
14
40
40
0
0
0
15
44
42
2
.0455
.0455
16
39
44
–5
–.1282
.1282
17
48
46
2
.0417
.0417
18
50
48
2
.04
.04
19
47
50
–3
–.0638
.0638
20
54
52
2
.037
.037
 PE
i
 PE
 .7911
i
 1.1195
MPE 
 .7911
 7.91 %
10
MAPE 
1.1195
 11 .12 %
10
3-36
Chapter 03 - Forecasting
Since the MAPE values for the two methods are approximately equal, the overall accuracy of the two
methods is about the same. Both methods are predicting approximately 11% away from the actual.
Since the MPE is –7.91% for the linear trend equation, the trend equation is overestimating sales by
7.91%. On the other hand, since the MPE is positive for the naïve forecasting method, it is
overestimating the sales by 7.66%. In this situation, in order to decide between the two methods, we
have to compare the consequences and the cost of overestimation with the consequences and the cost
of underestimation. If the cost of underestimation is less than the cost of overestimation, then the
linear trend equation should be selected. However, if the cost overestimation is less than cost of
underestimation, then the naïve method should be used. The cost of overestimation involves the cost
of carrying excess inventories, while cost of underestimation includes the cost of shortages,
backordering and lost sales.
Solution to problem 2
a.
Forecast Errors and Percentage Errors Using the First Forecasting Method
PE i
PE i
Time Period
ei
1
2
.02941
.02941
2
7
.09333
.09333
3
4
–2
3
–.02857
.04054
.02857
.04054
5
–3
–.04348
.04348
6
7
2
9
.02778
.11250
.02778
.11250
8
4
.05128
.05128
 PE
i
 PE
.28279
 3.535 %
8
 .28279
MPE 
 .42686
MAPE 
i
.42686
 5.336 %
8
3-37
Chapter 03 - Forecasting
b.
Forecast Errors and Percentage Errors Using the Second Forecasting Method
Time Period
ei
PE i
PE i
1
2
.02941
.02941
2
7
.09333
.09333
3
–2
0
0
4
5
3
–3
.02702
–.07246
.02702
.07246
6
2
–.05556
.05556
7
9
.025
8
4
–.02564
 PE
i
 PE
i
 .0211
 .32842
MPE 
.025
.02564
.0211
 0.264 %
8
MAPE 
.32842
 4.1%
8
We recommend the second forecasting method for two reasons:
1. The second forecasting method has less forecast bias (MPE1 > MPE2);
2. The second forecasting method results in higher overall forecast accuracy because
(MAPE1 = 5.34%) > (MAPE2 = 4.1%)
3-38