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Problem 15-05 (Algorithmic) Consider the following time series data. Week 1 2 3 4 5 6...

Problem 15-05 (Algorithmic)

Consider the following time series data.

Week 1 2 3 4 5 6
Value 16 13 18 11 15 14
  1. Develop a three-week moving average for this time series. Compute MSE and a forecast for week 7. Round your answers to two decimal places.
    Week Time Series
    Value
    Forecast
    1 16
    2 13
    3 18
    4 11
    5 15
    6 14

    MSE: _______

    The forecast for week 7: ______
  2. Use  = 0.2 to compute the exponential smoothing values for the time series. Compute MSE and a forecast for week 7. Round your answers to two decimal places.
    Week Time Series
    Value
    Forecast
    1 16
    2 13
    3 18
    4 11
    5 15
    6 14

    MSE: ___

    The forecast for week 7: _____
  3. Use trial and error to find a value of the exponential smoothing coefficient that results in a smaller MSE than what you calculated for  = 0.2. Find a value of  for the smallest MSE. Round your answer to three decimal places.

    = ______

Homework Answers

Answer #1

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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