Problem 15-05 (Algorithmic)
Consider the following time series data.
Week | 1 | 2 | 3 | 4 | 5 | 6 |
Value | 16 | 13 | 18 | 11 | 15 | 14 |
Week | Time Series Value |
Forecast |
---|---|---|
1 | 16 | |
2 | 13 | |
3 | 18 | |
4 | 11 | |
5 | 15 | |
6 | 14 |
Week | Time Series Value |
Forecast |
---|---|---|
1 | 16 | |
2 | 13 | |
3 | 18 | |
4 | 11 | |
5 | 15 | |
6 | 14 |
a).
Formula | (An-1 + An-2 + An-3)/3 | A-F | (A-F)^2 | |
Week (n) | Time Series Value (A) | Forecast (F) | Error (E) | Squared Forecast Error |
1 | 16 | |||
2 | 13 | |||
3 | 18 | |||
4 | 11 | 15.67 | (4.67) | 21.78 |
5 | 15 | 14.00 | 1.00 | 1.00 |
6 | 14 | 14.67 | (0.67) | 0.44 |
7 | 13.33 |
Forecast for week 7 = 13.33
MSE = Sum of squared forecast error/number of errors = (21.78 + 1.00 + 0.44)/3 = 7.74
b). For forecast at n = 2, F2 is assumed to be the same as the actual value for n = 1, A1. For the remaining forecasts, using alpha = 0.2, the forecast Fn is calculated as Fn-1 + alpha*Errorn-1
Formula |
F2 = A1; Fn = Fn-1 + (a*En-1) |
E = A-F | (A-F)^2 | |
Week (n) | Time Series Value (A) | Forecast (F) | Forecast Error (E) | Sqaured Forecast Error |
1 | 16 | |||
2 | 13 | 16.00 | (3.00) | 9.00 |
3 | 18 | 15.40 | 2.60 | 6.76 |
4 | 11 | 15.92 | (4.92) | 24.21 |
5 | 15 | 14.94 | 0.06 | 0.00 |
6 | 14 | 14.95 | (0.95) | 0.90 |
7 | 14.76 |
Forecast for week 7 = 14.76
MSE = Sum of squared forecast error/number of errors = (9.00 + 6.76 + 21.21 + 0.00 + 0.90)/5 = 8.17
c). Using the table in part (b) and excel solver, the value of alpha at which MSE is minimized is 0.142 and MSE = 8.108
alpha | 0.142 | |||
Formula |
F2 = A1; Fn = Fn-1 + (a*En-1) |
E = A-F | (A-F)^2 | |
Week (n) | Time Series Value (A) | Forecast (F) | Forecast Error (E) | Sqaured Forecast Error |
1 | 16 | |||
2 | 13 | 16.00 | (3.00) | 9.00 |
3 | 18 | 15.57 | 2.43 | 5.88 |
4 | 11 | 15.92 | (4.92) | 24.19 |
5 | 15 | 15.22 | (0.22) | 0.05 |
6 | 14 | 15.19 | (1.19) | 1.42 |
7 | 15.02 | |||
MSE | 8.108 |
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