Question

use the following gasoline sales time series. If needed, round your answers to two-decimal digits. Week Sales (1,000s of gallons) 0.1, and a -0.2. (a) Show the exponential smoothing forecasts using a Exponential Smoothing Week a = 0.1 -0.2 (b) Applying the MSE measure of forecast accuracy, would you prefer a smoothing constant of a 0.1 or 0.2 for the gasoline sales time series? An - Select your answer- smoothing constant provides the more accurate forecast, with an overall MSE of (c) Are the results the same if you apply MAE as the measure of accuracy? An - Select your answer- smoothing constant provides the more accurate forecast, with an overall MAE of (d) What are the results if MAPE is used? An Select your answer smoothing constant provides the more accurate forecast, with an overall MAPE of

Answer 1

SOLUTION:

For exponential smoothing forecast, we use the formula

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

where $$\alpha$$ is the smoothing forecast and $$F_{t}$$ is the forecast value at t and $$A_{t}$$ is the actual value at t.

1. $$\alpha$$ =0.1

assume $$F_{1}$$ = $$A_{1}$$ , if no information is given for  $$F_{1}$$ , then find $$F_{2}$$ using the formula

$$F_{2}$$ = 0.1*$$A_{1}$$ + (1-0.1)* $$F_{1}$$ = 0.1*16+0.9*16=16

Similarly, apply the same formula for every week.

$$F_{3}$$ =0.1*22+0.9*16=16.6

$$F_{4}$$ =0.1*17+0.9*16.6=16.6

$$F_{5}$$ =0.1*23+0.9*16.6=17.3

$$F_{6}$$ =0.1*15+0.9*17.3=17

$$F_{7}$$ =0.1*17+0.9*17=17

$\\ $F_{8}$$ =0.1*21+0.9*17=17.4

$\\ $F_{9}$$ =0.1*19+0.9*17.4=17.6

$\\ $F_{10}$$ =0.1*20+0.9*17.6=17.8

$\\ $F_{11}$$ =0.1*18+0.9*17.8=17.9

$\\ $F_{12}$$ =0.1*15+0.9*17.9=17.6

$\\ $F_{13}$$=0.1*23+0.9*17.6=18.1

similarly for $$\alpha$$ =0.2, $\\ $F_{13}$$=0.2*23+0.8*17.9=18.9

week	sales ($$A_{t}$$)	$$F_{t}$$ ($$\alpha$$ =0.1)	$$F_{t}$$ ($$\alpha$$ =0.2)
1	16	16	16
2	22	16	16
3	17	16.6	17.2
4	23	16.6	17.2
5	15	17.3	18.3
6	17	17	17.7
7	21	17	17.5
8	19	17.4	18.2
9	20	17.6	18.4
10	18	17.8	18.7
11	15	17.9	18.6
12	23	17.6	17.9

(b) Now to find the MSE, formula is $MSE= \frac{\sum_{t=2}^{N}{(A_{t}-F_{t})}^2}{N}$

Therefore for ($$\alpha$$ =0.1) MSE= 13.03

and for ($$\alpha$$ =0.2) MSE= 12.4

Since the value of MSE is less for ($$\alpha$$ =0.2) , therefore it is chosen for the gasoline sales time series.