Question: Consider the stylized example in Figure 7.13. Repeat the forecasts using the additive model with trend, using the same smoothing constants. FIGURE 7.13 Stylized Model
Consider the stylized example in Figure 7.13. Repeat the forecasts using the additive model with trend, using the same smoothing constants.
FIGURE 7.13 Stylized Model with Trend and Cyclic Fluctuations
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a. What is the value of MSE corresponding to the multiplicative model in Figure 7.13?
b. What is the value of MSE corresponding to the additive model with the same smoothing constants?
1 Stylized Example for the Holt-Winters Method Initial value of S() for t- 1 to 4 Alpha Gamma 3 0.6 0.2 Beta 0.5 Initial value T(1) Period Observed s(t) T(t) Ct) 1.10 Forecast Error 100.0 0.0 110 Initial value C(1) 0.0 100.0 2 120 1.20 8. 90 100.0 0.0 0.90 80 100.0 0.0 0.80 110.0 102.0 10 1.15 121 1.0 -11.0 11 106.4 144 2.7 1.29 123.6 -20.4 12 98.2 94.5 149 C= TH(t/S(t)+(1-y)(C{t-p)) 117 113.3 4.8 0.98 F(t+1D=(S()+T()Ct+1-p) 7.5 13 112 122.5 7.0 0.87 14 165 132.2 8.4 1.21 150 9.2 15 10 192 142.2 1.33 181. 16 11 153 152.3 9.7 0.99 148.3 -4.7 T(0=B(S(t-S(t-1)+(1-BIT(t-1) 17 12 144 162.7 10.0 18 13 209 172.8 10.0 1.21 208.9 -0.1 19 14 240 182.4 2.6 S() = ax(t)/C[t-p}+(1-a](S(t-1)+Tt-1) 20 15 189 191.8 2.2 21 16 176 201.2 9.5 0.88 176.9 0.9 22 17 253 210.4 9.4 1.21 254.9 1.9 23 290.1 18 288 219.4 9.2 1.32 2.1 24 226.1 19 225 228.4 9.1 0.99 1.1 25 20 208 237.5 9.1 0.88 208.1 0.1 26 297.1 0.1 21 297 246.5 9.1 1.21 27 22 336 255.5 9.0 1.32 336.2 0.2 28 23 261 264.6 9.0 0.99 261.0 0.0 29 273.7 24 240 9.1 0.88 239.7 -0.3 30 25 341 282.8 9.1 1.21 340.7 -0.3 31 7.10 7.11 7.12 7.7/7.8 7.9 7.13
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The forecasted value obtained from exponential smoothing with trend and cyclical factor method is of less error comparing to other methods This method is of two types they are additive method and mult... View full answer
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