Management Science Chapter 7 Homework Solutions

Chapter 7

Problem Summary

Problem Solutions

7.1 See file Ch7.1.xls

a.

Yes, a stationary model seems appropriate

b.

Coefficients
Standard Error t Stat
P-value
Lower 95%
Upper 95%
Lower 95.0%
Upper 95.0%
Intercept
20.16667
1.373732
14.6802
4.3E-08
17.1058
23.22753
17.1058
23.22753
Period
-0.07692
0.186653
-0.41212
0.688949
-0.49281
0.338967
-0.49281
0.338967

From regression output, t = -.412 and p = .689. A stationary model seems appropriate since the linear term, Period, is not significant.

7.1 c.

Forecast for January -- 19, for upcoming year – 12*19 = 228

7.1 d.

Forecast for January -- 20.4

e. 4 month moving average. MAD is 1.72
7.2 See files Ch7.2a.xls and Ch7.2b.xls

a.

Forecast for January -- 18.86

7.2 b. See file Ch7.2b.xls

Forecast for January -- 20.28

c.  = .6 gives the lower MSE

7.3 See file Ch7.3.xls

a.

b.

Coefficients
Standard Error t Stat
P-value
Lower 95%
Upper 95%
Lower 95.0%
Upper 95.0%
Intercept
406.6014
3.916368
103.8211
4.22E-31
398.4794
414.7235
398.4794
414.7235
Month
10.17522
0.274089
37.12382
2.44E-21
9.606792
10.74364
9.606792
10.74364

From regression output, t = 37.1238, p = 0

7.3 c.

From the above output we see that the forecast is as follows:

25 -- 660, 26 -- 671, 27 -- 681, 28 -- 692, 29 -- 702, 30 -- 712, 31 -- 722, 32 -- 732, 33 -- 742, 34 -- 753, 35 -- 763, 36 -- 773.

7.4 See file Ch7.4a.xls and Ch7.4b.xls

Forecast for upcoming 12 months:

Period
25
26
27
28
29
30
31
32
33
34
35
36
Forecast
665
675
685
695
706
716
726
736
746
756
766
776

7.4b See file Ch7.4b.xls

Forecast for upcoming 12 months:

Period
25
26
27
28
29
30
31
32
33
34
35
36
Forecast
659
665
671
676
682
688
693
699
705
710
716
722

c.  = .5 and  = .7

d.  = .5 and  = .7

7.5 See file Ch7.5.xls

Week 5: Su 318.89, M 371.72, Tu 353.72, W 397.54, Th 404.22, F 362.08, Sa 366.49,

Week 6: Su 326.62, M 380.70, Tu 362.24, W 407.08, Th 413.89, F 370.71, Sa 375.20.

7.6