CHAPTER 8: Solutions to Selected Exercises
8.1
a.
 3   y3  1     2
= .1 (317) + .9 (362.82)
= 358.2380
b.
One-period-ahead forecast error for period 4 is
c.
 4   y 4  1     3
y 4  yˆ 4 3  y 4   3  297 - 358.238  - 61.2380 .
= .1 (297) + .9 (358.238)
= 352.1142
8.2
8.3
d.
One-period ahead forecast error for period 5 is
a.
All values in spreadsheet should agree with the values in Figure 8.1
b.
When   .4 , SSE = 35,688
a.
The point forecast for the cod catch in time period 28 is
y5  yˆ 5 4  y5   4  399 - 352.1142  46.8858
yˆ 28 24  yˆ 24 4 24   24  354.5438
The 95% prediction interval is

24
 z.025 s 1  3 2

= 354.5438 + 1.96 (34.95)
= 354.5438 + 68.6207
= [285.9231, 423.1645]
b.
1 3.034
2
The point forecast for the cod catch in time period 29 is
yˆ 29 24  yˆ 245 24   24  354.5438
The 95% prediction interval is
 24  z.025 s 1  4 2
= [354.5438 + 1.96 (34.95) 1  4(.034) 2 ]
= [354.5438 + 68.6602]
= [285.8836, 423.2040]
8.4
a.
b.
 26   y 26  1     25
= .034(375) + .966(355.5453)
= 356.2068
A point forecast made in time period 26 for any future monthly cod catch is
yˆ 26 26   26  356.2068
58
95% prediction interval for March  y 27  y 261  is

26
 z .025 s 
= [356.2068 + 1.96(34.95)]
= [356.2068 + 68.5020]
= [287.7048, 424.7088]
95% prediction interval for April  y 28  y 26 2  is

26
 z.025 s 1   2

2
 356.2068  1.9634.95 1  .034 


 356.2068  68.5416
 287.6652,424.7484
95% prediction interval for May  y 29  y 263  is

26
 z.025s 1  2 2

2
 356.2068  1.9634.95 1  2.034 


 356.2068  68.5811
 287.6257,424.7879
8.5
a.
 2  y2  1    1  b1 
 .2245  .8203.0051   .2933
 211.1694
b2    2   1   1   b1
 .1211.1694  203.0051  .9 .2933
 .5525 (Slight round - off error. Excel carries values with
more decimal places)
b.
c.
yˆ 3 2   2  b2  21.1694  .5525  211.7219
y3  yˆ 3 2  185  211.7219  26.7219
 3  y3  1    2  b2 
 .2(185)  .8[211.1694  .5525]
 206.3775
b3    3   2   1   b2
 .1[206.3775 - 211.1694]  .9(.5525)
 .0181
(Round - off error)
59
d.
yˆ 4 3   3  b3  206.3775  .0181  206.3956
y4  yˆ 4 3  169  206.3956  37.3956
8.6
8.8
(Round - off error)
a.
Assuming  0  202.6246 and b0  .3682 (exactly 4 decimal place accuracy) all
values in the spreadsheet should be same as values in Figure 8.6.
b.
SSE  46,387 when   .1 and   .1
e.
Values in spreadsheet should be same as values in Part a.
a.
Point forecast for sales in week 56
yˆ 56 52   52  4b52  315.9460  4(4.5040)  333.962
95% prediction interval for sales in week 56
2
2
2
1+  2 1      2 1  2    2 1  3 
 1  (.247) 2 (1  .095) 2  (.247) 2 (1  2.095) 2  (.247) 2 1  3.095
 1.2603
2
[ yˆ 56 52  z .025 s 1   2 1      2 1  2    2 1  3  ]
2
2
2
 [333.962  1.96(27.89) 1.2603 ]
 [333.962  61.37]
 [272.59, 395.33]
b.
Point forecast for sales in week 57
yˆ 57 52   52  5b52  315.9460  5(4.5040)  338.466
95% prediction interval for sales in week 57
1+  2 1      2 1  2    2 1  3    2 (1  4 ) 2
2
 1.2603   2 1  4 
2
2
2
(From first part of Part a)
 1.2603  (.247) 2 1  4.095
 1.3765
2
[ yˆ 57 52  z .025 s 1   2 1      2 1  2    2 1  3    1  4  ]
2
2
 [338.466  1.96(27.89) 1.3765 ]
 [338.466  64.13]
 [274.34, 402.60]
8.9
a.
Revised point forecast for sales in week 56 when y53  300
yˆ 56 53   53  3b53  322.8089  3(4.7281)  336.9932
60
2
2
Revised 95% prediction interval for sales in week 56 when y53  300
1+  2 1      2 (1  2 ) 2
2
 1  .247  1  .095  .247  1  2.095
 1.1595
2
2
2
2
[ yˆ 56 53  z .025 s 1   2 1      2 1  2  ]
2
2
 [336.9932  1.96(27.89) 1.1595 ]
 [336.9932  58.86]
 [278.13, 395.85]
Revised point forecast for sales in week 57 when y53  300
yˆ57 53   53  4b53  322.8089  4(4.7281)  341.7213
Revised 95% prediction interval for sales in week 57 when y53  300
1+  2 1      2 (1  2 ) 2   2 1  3 
2
2
 1.1595   2 1  3 
2
 1.1595  (.247) 2 (1  3(.095)) 2
= 1.2602
(Should be same as 8.8a; round-off)
[ yˆ 57 53  z .025 s 1   2 1      2 1  2    2 (1  3 ) 2 ]
2
2
 [341.7213  1.96(27.89) 1.2602 ]
 [341.7213  61.37]
 [280.35, 403.09]
b.
Revised level and growth rate when y54  320
 54  y54  1   [ 53  b53 ]
 .247(320)  .753[322.8089  4.7281]
 325.6754
b54   [ 54   53 ]  (1   )b53
 (.095)[325.6754 - 322.8089]  .905(4.7281)
 4.5512
c.
Revised point forecast for sales in week 55 when y54  320
yˆ55 (54)   54  b54  325.6754  4.5512  330.2266
Revised 95% prediction interval for sales in week 55 when y54  320
61
[ yˆ 55 (54)  z[.025] s ]
 [330.2266  (1.96)( 27.89)]
 [330.2266  54.66]
 [275.57, 384.89]
Revised point forecast for sales in week 56 when y54  320
yˆ56 (54)   54  2b54  325.6754  2(4.5512)  334.7778
Revised 95% prediction interval for sales in week 56 when y54  320
2
yˆ56 54  z.025s 1   2 1   
 [344.7778  1.96(27.89) 1  (.247) 2 (1  .095) 2 ]
 [334.7778  56.63]
 [278.15, 391.41]
Revised point forecast for sales in week 57 when y54  320
yˆ57 (54)   54  3b54  325.6754  3(4.5512)  339.329
Revised 95% prediction interval for sales in week 57 when y54  320
2
yˆ57 54  z.025s 1   2 1      2 (1  2 )2
 [339.329  1.96(27.89) 1  (.247) 2 (1  .095) 2  (.247) 2 (1  2(0 .095)) 2 ]
 [339.329  58.86]
 [280.47, 398.19]
8.10
8.11
a.
Using Excel,  0  204.8030 and b0  6.9406
c.
  .034,   0, SSE  24,750, s  33.54
a.
Using the error correction form of the smoothing equations,
 2   1  b1  [ y2  ( 1  b1  sn24 )]
  1  b1   [ y 2  yˆ 2 (1)]
 22.3079  1.0286  .2[1.1105]
 23.5586
b2  b1   [ y 2  ( 1  b1  sn 24 )]
 b1   [ y 2  yˆ 2 (1)]
 1.0286  (.2)(.1)[1 .1105]
 1.0508
62
sn 2  sn 2 4  (1   ) [ y 2  ( 1  b1  sn 2 4 )]
 sn  2  (1   ) [ y 2  yˆ 2 (1)]
 6.5529  (.8)(.1)[1 .1105]
 6.6417
(Round - off error because spreadshee t carried
more decimal places)
b.
yˆ 3 (2)   2  b2  sn34
  2  b2  sn 1
 23.5586  1.0508  18.5721
 43.1815
One-period-ahead forecast error in period 3 is
y3  yˆ 3 2  43  43.1815  0.1815
c.
 3   2  b2   [ y3  ( 2  b2  sn34 )]
  2  b2   [ y3  yˆ 3 (2)]
 23.5586  1.0508  .2[-0.1815]
 24.5731
b3  b3   [ y 3  ( 2  b2  sn3 4 )]
 b 3   [ y 3  yˆ 3 (2)]
 1.0508  (.2)(.1)[- 0.1815]
 1.0472
sn3  sn3 4  (1   ) [ y 3  ( 2  b2  sn3 4 )]
 sn 1  (1   ) [ y 3  yˆ 3 (2)]
 18.5721  (.8)(.1)[0 .1815]
 18.5576
d.
(Round - off error)
yˆ 4 (3)   3  b3  sn44
  3  b3  sn0
 24.5731  1.0472  (10.9088)
 14.7115
One-period-ahead forecast error in period 4 is
y4  yˆ 4 3  16  14.7115  1.2885
8.12
a.
Assuming 4-decimal place accuracy for initial values, all values in your Excel
spreadsheet should agree with those in Figure 8.9
b.
All values in your optimization result should agree with the values in Figure 8.10.
63
8.14
a.
yˆ 20 (16)   16  4b16  sn204
  16  4b16  sn16
 36.3426  4(.9809)  (-10.9088)
 29.3574
95% prediction interval for bike sales in period 20.
c  c 4  1   2 1      2 1  2    2 1  3 
2
2
2
 1  (.561) 2 (1  0) 2  (.561) 2 (1  20) 2  (.561) 2 (1  30) 2
 1  3(.561) 2
 1.9442
[ yˆ 20 (16)  z[.025] s c4 ]
 [29.3574  1.96(1.2025) 1.9442 ]
 [29.3574  3.2863]
[26.0711, 32.6437]
b.
yˆ 21 (16)   16  5b16  sn214
  16  5b16  sn17
  16  5b16  sn13
( sn13 is last estimate for seasonal factor in quarter 1)
 36.3426  5(.9809) - 14.2162
 27.0309
95% prediction interval for bike sales in period 21
c  c5  1   2 1      2 1  2    2 1  3   [ (1  4 )  (1   ) ] 2
2
2
 c 4  [ (1  4 )  (1   ) ]2
 1.9442  [.561(1  4(0)  .439(0)] 2
 2.2589
[ yˆ 21 (16)  z[.025] s c5 ]
 [27.0309  1.96(1.2025) 2.2589 ]
 [27.0309  3.5423]
 [23.4886, 30.5732]
8.15
a.
S 2  y2 / yˆ 2  116 / 100.1912  1.1578
S 6  y 6 / yˆ 6  123 / 110.0735  1.1174
S10  y10 / yˆ10  131 / 119.9559  1.0921
S14  y14 / yˆ14  140 / 129.8382  1.0783
64
2
S 2  S 6  S10  S14
4
1.1578  1.1174  1.0921  1.0783

4
 1.1114
b.
S 2 
c.
sn2  sn24  S2 (CF )
CF 
L
S 1  S 2   S 3  S 4 
4
.7062  1.1114  1.2937  .8886
 1.0000
Hence, sn  2 =1.1114 (1) = 1.1114

d.
S 3  y3 / yˆ 3  136 / 102.6618  1.3247
S 7  y 7 / yˆ 7  146 / 112.5441  1.2973
S11  y11 / yˆ11  158 / 122.4265  1.2906
S15  y15 / yˆ15  167 / 132.3088  1.2622
S 3 
8.16
a.
S 3  S 7  S11  S15 1.3247  1.2973  1.2906  1.266

 1.2937
4
4
 3   ( y3 / sn34 )  (1   )( 2  b2 )
  ( y 3 / sn 1 )  (1   )( 2  b2 )
 .2(136/1.29 37)  .8(101.7726  2.6203)
 104.5393
b3   ( 3   2 )  (1   )b2
 .1(104.5393 - 101.7726)  .9(2.6203)
 2.6349
sn3   ( y 3 /  3 )  (1   ) sn3 4
  ( y 3 /  3 )  (1   ) sn 1
 .1(136/104. 5393)  .9(1.2937)
 1.2944
b.
yˆ 4 (3)  ( 3  b3 ) sn44
65
 ( 3  b3 ) sn0
 (104.5393  2.6349)(.8886)
 95.2350
One-period-ahead forecast error in period 4
y4  yˆ 4 (3)  96  95.2350  .7650
c.
 4   ( y 4 / sn44 )  (1   )( 3  b3 )
  ( y 4 / sn 0 )  (1   )( 3  b3 )
 .2(96 / .8886)  .8(104.5393  2.6349)
 107.3464
b4    4   3   1   b3
 .1107.3464 - 104.5393  .92.6349
 2.6521
sn 4   ( y 4 /  4 )  (1   ) sn 44
  ( y 4 /  4 )  (1   ) sn0
 .1(96/107.3 464)  .9(.8886)
 .8892
d.
yˆ 5 (4)  ( 4  b4 ) sn54
 ( 4  b4 ) sn1
 (107.3464  2.6521)(.7086)
 77.9449
One-period-ahead forecast error in time period 5 is
y5  yˆ 5 (4)  77.9449 (Round-off error because spreadsheet
carries more decimal places)
8.17
8.19
a.
In order for the spreadsheet to match all values in Figure 8.22, the initial values should
have exactly the same 4 decimal accuracy as given in this exercise. If the initial
values are linked to the cells of a regression output, the initial values will be slightly
different (even though not showing on spreadsheet with 4 decimal places) and all later
values will be somewhat different.
b.
See comments in Part a. Results can be affected by initial values.
a.
yˆ 35 (32)  ( 32  3b32 ) sn354
66
 ( 32  3b32 ) sn31
 [168.1213  3(2.3028)](1.2934)
 226.3834
yˆ 36 (32)  ( 32  4b32 ) sn364
 ( 32  4b32 ) sn32
 [168.1213  4(2.3028)](. 8908)
 157.9678
b.
95% prediction interval for sports drink sales in time period 35
c  c3  a 2 1  2   32  b32   a 2 1     32  2b32    32  3b32 
2
2
2
 .336 1  2.046 168.1213  2.3028
2
2
2
2
2
 (.336) 2 1  .046 168.1213  22.3028  (168.1213  3(2.3028)) 2
 3,910.0819  3,685.2069  30,635.3959
2
2
 38,230.6847
[ yˆ 35 (32)  z[.025] s r c3 ( sn31 )]
 [226.3834  1.96(.0193) 38,230.6847 (1.2934)]
 [226.3834  9.5665]
 [216.8169, 235.9499]
95% prediction interval for sports drink sales in time period 36.
2
2
c  c4   2 1  3   32  b32    2 1  2  ( 32  2b32 ) 2  (1   ) 2 ( 32  3b32 )  ( 32  4b32 ) 2
 (.336) 2 (1  3(.046) 2 )(168.1213  2.3028) 2  (.336) 2 (1  2(.046)) 2 (168.1213  2(2.3028)) 2
 (.336) 2 (1  .046) 2 (168.1213  3(2.3028)) 2  (168.1213  4(2.3028)) 2
 4,246.4410  4,016.4632  3,784.1245  31,446.8156
 43,493.8443
[ yˆ 36 (32)  z[.025] s r c4 (sn32 )]
 [157.9678  1.96(.0193) 43,493.8443 (.8908)]
 [157.9678  7.0276]
 [150.9402, 164.9954]
8.20
a.
 33   ( y33 / sn29 )  (1   )( 32  b32 )
 (.336)(124 / .7044)  (.664)(168.1213  2.3028)
 172.3098
67
b33   [ 33   32 ]  1   b32
 .046[172.3098 - 168.1213]  .954(2.3028)
 2.3895
sn3   ( y 33 /  33 )  (1   ) sn 29
 .134(124/172.3098)  .866(.7044)
 .7064
b.
Revised point forecast for sports drink sales in time period 34 when y33  124
yˆ 34 (33)  ( 33  b33 ) sn30
 (172.3098  2.3895)(1.1038)
 192.8331
Revised 95% prediction interval for sports drink sales in time period 34 when
y33  124
c  c1   33  b33   172.3098  2.3895  30,519.8454
2
2
[ yˆ 34 (33)  z[.025] s r c1 ( sn30 )]
 [192.8331  1.96(.0193) 30,519.8454 (1.1038)]
 [ 192.8331  7.2945]
 [185.5386, 200.1276]
Revised point forecast for sports drink sales in time period 35 when y33  124
yˆ 35 (33)  ( 33  2b33 )sn31  [172.3098  2(2.3895)](1.2934)
 229.0467
Revised 95% prediction interval for sports drink sales in time period 35 when
y33  124
c  c 2   2 (1   ) 2 ( 33  b33 ) 2  ( 33  2b33 ) 2
 (.336) 2 (1  .046) 2 (172.3098  2.3895) 2  (172.3098  2(2.3895)) 2
 3,769.8516  31,360.4431
 35,130.2947
[ yˆ 35 (33)  z[.025] s r c 2 ( sn51 )]
 [229.0467  1.96(.0193) 35,130.2947 (1.2934)]
 [229.0467  9.1704]
[219.8763, 238.2171]
68
Revised point forecast for sports drink sales in time period 36 when y33  124
yˆ 36 (33)  ( 33  3b33 )sn32  [172.3098  3(2.3895)](. 8908)
 159.8793
Revised 95% prediction interval for sports drink sales in time period 36 when
y33  124
c  c3   2 (1  2 ) 2 ( 33  b33 ) 2   2 (1   ) 2 ( 33  2b33 ) 2  ( 33  3b33 ) 2
 (.336) 2 (1  2(.046)) 2 (172.3098  2.3895) 2
 (.336) 2 (1  .046) 2 (172.3098  2(2.3895)) 2  (172.3098  3(2.3895)) 2
 4,108.7164  3,873.6833  32,212.4602
 40,194.8599
[ yˆ 36 (33)  z[.025] s r c3 ( sn32 )]
 [159.8793  1.96(.0193) 40,194.8599 (.8908)]
 [159.8793  6.7558]
[153.1235, 166.6351]
8.22
b.
 0  310.9667
b0  13.6459
c.
January
February
March
April
May
June
July
August
September
October
November
December
d.
Assuming links to cells with initial value calculations (more than 4 decimal place
accuracy),
  .700
8.23
a.
sn11  .5137
sn10  .6123
sn 9  .6260
sn8  .7106
sn 7  .5980
sn 7  1.0055
sn 6  1.4850
sn5  1.6810
sn 4  1.9563
sn3  1.2613
sn 2  .9781
sn0  .5722
 0
 0
SSE  5249 Sr  .0390
y 25  L24   25  354.5438  20  374.5438
69
L25  L24  25  354.5438  (.034)( 20)  355.2238
y26  L25   26  355.2238  (15)  340.2238
L26  L25  6  355.2238  (.034)( 15)  354.7138
y27  L26   27  354.7138  (5)  349.7138
b.
y 25  L24   25  354.5438  (30)  324.5438
L25  L24  25  354.5438  (.034)( 30)  353.5238
y26  L25   26  353.5238  4  357.5238
L26  L25  26  353.5238  (.034)( 4)  353.6598
y27  L26   27  353.6598  22  375.6598
70