In: Statistics and Probability
Consider the following time series data:
Month |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
Value |
24 |
13 |
21 |
12 |
20 |
23 |
15 |
25 |
a)
period | demand | forecast | forecast error=demand value-forecast value | absolute forecast error | squared forcast error |
t | Dt | Ft | et=Dt-Ft | | et | | (et)² |
1 | 24 | ||||
2 | 13 | 24 | -11.00 | 11.00 | 121.0 |
3 | 21 | 13 | 8.00 | 8.00 | 64.0 |
4 | 12 | 21 | -9.00 | 9.00 | 81.0 |
5 | 20 | 12 | 8.00 | 8.00 | 64.0 |
6 | 23 | 20 | 3.00 | 3.00 | 9.0 |
7 | 15 | 23 | -8.00 | 8.00 | 64.0 |
8 | 25 | 15 | 10.00 | 10.00 | 100.0 |
forecast for month 9 = 25
MSE= Σ(et)²/n = 71.86
2)
period | demand | forecast | forecast error=demand value-forecast value | absolute forecast error | squared forcast error |
t | Dt | Ft | et=Dt-Ft | | et | | (et)² |
1 | 24 | ||||
2 | 13 | 24 | -11.00 | 11.00 | 121.0 |
3 | 21 | 18.5 | 2.50 | 2.50 | 6.3 |
4 | 12 | 19.33 | -7.33 | 7.33 | 53.8 |
5 | 20 | 17.5 | 2.50 | 2.50 | 6.3 |
6 | 23 | 18 | 5.00 | 5.00 | 25.0 |
7 | 15 | 18.8333333 | -3.83 | 3.83 | 14.7 |
8 | 25 | 18.2857143 | 6.71 | 6.71 | 45.1 |
forecast for month 9 = 18.43
MSE= Σ(et)²/n = 38.86
c)
MSE in part b) is lower than that of part a)
so, average of all the data available as the forecast is better forecast