Question

In a forecasting mode, at the end of period 13, the forecasted value for period 14 is 75. Actual value in the periods 14 to 16 are constant at 100. If the assumed simple exponential smoothing parameter is 0.5, then the MSE at the end of period 16 is

• A. 820.31
• B. 43.75
• C. 273.44
• D. 14.58

Answer 1

The correct option is C 273.44

$F_t=F_{t-1}+\alpha (D_{t-1}-F_{t-1})$

$F_{14}=75;D_{14}=100;\alpha =0.5$

$F_{15}=75+0.5(100-75)=87.5$

$F_{16}=87.5+0.5(100-87.5)=93.75$

Now,

${\text{(error)}}^2=(D_t-F_t{)}^2$

$(\text{error}{)}_{14}^2=(25{)}^2=625$

$(\text{error}{)}_{15}^2=(12.5{)}^2=156.25$

$(\text{error}{)}_{16}^2=(6.25{)}^2=39.06$

$MSE=\frac{\sum (error{)}^2}{n}=\frac{625+156.25+39.06}{3}=273.44$