OM Project
Question: Forecasting problem 3.49, page 106 (Chapter 3) of Text Book Heizer,Render & Rajashekhar
(a) Using exponential smoothing, with α = .6, then trend analysis, and finally linear regression discuss which forecasting model fits best for Salinas’s strategic plan. Justify the selection of one model over another.
Answer:
We have done forcasting using exponential smoothing and linear regression methods.
Below are the forcast values:
Method
Exponential smoothing
MAD
3.5
Linear Regression
10.6
Year
1
1
2
3
4
5
Forecast value
86.22
54.72
56.36
58
59.63
61.27
Since the MAD for liner regression is large compared to MAD of Exponential smoothing.
Therefore Exponential smoothing is better method in this case.
Refer Sheets exp smoothing (a) and regression (a) for details of calculation
(b) Carefully examine the data. Can you make a case for excluding a portion of the information?
Why? Would that change your choice of model?
Answer:
We have removed the data for the following periods year 1983, 1988, 1991 & 2001.
We have removed this data as outliers because this data correspondes to ression period as given in the problem statement.
Below are the new forcast values:
Method
Exponential smoothing
MAD
3.84
Linear Regression
11.74
Year
1
1
2
3
4
5
Forecast value
86.19
55.67
57.5
59.34
61.18
63.02
Choice of model does not changes because the MAD value again for Exponential smoothing remains lower than MAD of Li
Refer Sheets exp smoothing (b) and regression (b) for details of calculation
Rajashekhar
on discuss which
en in the problem statement.
remains lower than MAD of Linear regression.
Year
1964
1965
1966
1967
1968
Deposits Actual
0.25
0.24
0.24
0.26
0.25
GSP
0.4
0.4
0.5
0.7
0.9
-
0.25
-0.01
0.01
0.24
0.00
0.00
0.24
0.02
0.02
0.25
0.00
0.00
Deposite Forcast ( alpha = 0.6)
Forcast Error
Absolute
(a) Using exponential smoothing, with α = .6, then trend analysis, and finally linear regression discuss which forecasting model fits best for Salinas’s strategic plan. Justify the selection of one model over another.
Answer:
We have done forcasting using exponential smoothing and linear regression methods.
Below are the forcast values:
Method
Exponential smoothing
MAD
3.5
Linear Regression
10.6
Year
1
1
2
3
4
5
Forecast value
86.22
54.72
56.36
58
59.63
61.27
Since the MAD for liner regression is large compared to MAD of Exponential smoothing.
Therefore Exponential smoothing is better method in this case.
Refer Sheets exp smoothing (a) and regression (a) for details of calculation
(b) Carefully examine the data. Can you make a case for excluding a portion of the information?
Why? Would that change your choice of model?
Answer:
We have removed the data for the following periods year 1983, 1988, 1991 & 2001.
We have removed this data as outliers because this data correspondes to ression period as given in the problem statement.
Below are the new forcast values:
Method
Exponential smoothing
MAD
3.84
Linear Regression
11.74
Year
1
1
2
3
4
5
Forecast value
86.19
55.67
57.5
59.34
61.18
63.02
Choice of model does not changes because the MAD value again for Exponential smoothing remains lower than MAD of Li
Refer Sheets exp smoothing (b) and regression (b) for details of calculation
Rajashekhar
on discuss which
en in the problem statement.
remains lower than MAD of Linear regression.
Year
1964
1965
1966
1967
1968
Deposits Actual
0.25
0.24
0.24
0.26
0.25
GSP
0.4
0.4
0.5
0.7
0.9
-
0.25
-0.01
0.01
0.24
0.00
0.00
0.24
0.02
0.02
0.25
0.00
0.00
Deposite Forcast ( alpha = 0.6)
Forcast Error
Absolute