Consider the data shown below for sales pattern
of a popular brand of oil over the past 12 weeks.
| Week |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
Sales
(in ' 000 ) Litres |
17 |
21 |
19 |
23 |
18 |
16 |
20 |
18 |
22 |
20 |
15 |
22 |
|
(a) Use 3-peroiod moving average to predict the
forecast. Compute the forecast error.
(b) Use Exponential smoothening to forecast sales. (alpha =
0.2)
(a) Moving average for (weeks 1-3)= (17+21+19)/3
= 19
Moving average for (weeks 2-4)= (21+19+23)/3= 21
Similarly the forecast for remaining weeks can be computed.
| Week |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
Sales
(in '000 ) Litres |
17 |
21 |
19 |
23 |
18 |
16 |
20 |
18 |
22 |
20 |
15 |
22 |
| Moving average Forecast |
|
|
|
19 |
21 |
20 |
19 |
18 |
18 |
20 |
20 |
19 |
| Forecast Error |
|
|
|
23-19=4 |
18-21=-3 |
16-20=-4 |
20-19=1 |
18-18=0 |
22-18=4 |
20-20=0 |
15-20=-5 |
22-19=3 |
| Square of Error |
|
|
|
16 |
9 |
16 |
1 |
0 |
16 |
0 |
25 |
9 |
|
Sum of squared error = 92
Average sum of squared error = 92/9=10.22
(b) Using Exponential smoothening method. Assume F2= 17
F3= 0.2 Y2+ 0.8F2 = 0.2 x (21) + ).8 x (17) = 17.8
Once the actual; time series vale in week 3, Y3 = 19 is known,
we can generate a forecast for week 4 as follows:
F4= 0.2 Y3+ 0.8 F3= 0.2 x (19)+ 0.8 x (17.8) = 18.04
By continuing in this manner we obtain the following table:
| Week |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
| Sales (in '000)
Litres |
17 |
21 |
19 |
23 |
18 |
16 |
20 |
18 |
22 |
20 |
15 |
22 |
| Exponential Smoothening
Forecast |
|
17 |
17.8 |
18.04 |
19.03 |
18.83 |
18.26 |
18.61 |
18.49 |
19.19 |
19.35 |
18.48 |
| Forecast Error |
|
4 |
1.20 |
4.96 |
-1.03 |
-2.83 |
1.74 |
-0.61 |
3.51 |
0.81 |
-4.35 |
3.52 |
| Square of Error |
|
16 |
1.44 |
24.60 |
1.060 |
8.00 |
3.02 |
0.372 |
12.32 |
0.656 |
18.92 |
12.39 |
|
Sum of squared error = 98.778
Average sum of squared error = 98.778/11=8.978 (which is less
compared to Moving average method in (a)).
|
