53.3 52.3
= .3 produces the smaller absolute error and smaller tracking signal values.
= .1 = .3
12. b. Day Dt Ft et MAD TS Ft et MAD TS
8 39 32.0 7.0 0.7 10.0 32.0 7.0 2.1 3.3
9 24 32.7 8.7 1.5 -1.1 34.1 10.1 4.5 -0.7
10 26 31.8 5.8 1.9 -3.9 31.1 5.1 4.7 -1.7
11 36 31.2 4.8 2.2 -1.3 29.5 6.5 5.2 -0.3
12 43 31.7 11.3 3.1 2.7 31.5 11.5 7.1 1.4
13 46 32.9 13.1 4.1 5.2 34.9 11.1 8.3 2.5
14 29 34.2 5.2 4.2 3.9 38.3 9.3 8.6 1.4
55.9 60.6
Now = .1 produces the smaller absolute error, and tracking signal results are
acceptable for both values of .
c. This example illustrates that variability in the data may cause some forecasting methods
13. a. From the following output for = .2, = .3, and = .4, we see that the smallest
absolute deviation we have at = .2 and = .3, so we look at the bias for these two
values of and we conclude that = .2 gives a model with less bias than = .3.
Thus our final choice is = .2.
c. This problem illustrates that it is not always straightforward to find the best value for
. By splitting the data into two sets and comparing the forecast errors in both cases, we
14. a. The model that uses F1 = 170 provides the smallest absolute error, based on the
analysis of the historical demand for the first seven days only. The starting forecast value
of 170 is then used to answer part b as shown below.