Question

Problem 08-06 Algo (Moving Averages and Exponential Smoothing)

Consider the following time series data:

Month	1	2	3	4	5	6	7
Value	23	13	21	13	19	21	17

(a) Choose the correct time series plot Month (iv) Month Select your answer What type of pattern exists in the data? Select your answer-

Answer 1

q(a) The correct time series plot is (i) (YOU CAN SEE THE INTERSECTION OF BOTH THE VALUES AND THE FIRST GRAPH IS APPROPRIATE ACCORDING TO THE VALUES).

(b)

Months	Value	3-month moving average Forecast	Absolute Value of Forecast Error	Squared Forecast Error
1	23	-	-	-
2	13
3	21
4	13	19	| 13-19| = 6	36
5	19	15.67	19-15.67 = 3.33	11.0889
6	21	17.67	21-17.67| 3.33	11.0889
7	17	17.67	17-17.671-0.67	0.4489

Forecast for 8th month =   

1921 17

MSE =

sumo f squaredforecasterror 11.0889+11.0889 +0.448936 14.66 6

(c) Calculate forecast using this formula S_t=αy_t−1+(1−α)S_t−1

0.2

Months	Value (Yt)	0.2 Forecast (St)	Absolute Value of Forecast Error	Squared Forecast Error
1	23	-	-	-
2	13	23	13-23 10	100
3	21	21	0	0
4	13	21	13-21-8	64
5	19	19.4	$\left | 19 -19.4 \right | = 0.4$	0.16
6	21	19.32	$\left | 21 -19.32 \right | = 1.68$	2.8224
7	17	17.67	17-17.671-0.67	0.4489

Forecast for month 8 = 0.2(17)+(1-0.2)(17.67) = 17.536

MSE = $\frac{sum of squared forecast error}{6} = \frac{100+0+64+0.16+2.8224+0.4489}{6}$

= 27.90

(d) The three month moving average MSE is better as it is less than the exponential smoothing forecast.

(e) Let MSE = 0.1 then,

Months	Value (Yt)	Forecast (St)	Absolute Value of Forecast Error	Squared Forecast Error
1	23	-	-	-
2	13	23	13-23 10	100
3	21	22	$\left | 21-22 \right | = 1$	1
4	13	21.9	$\left | 13-21.9 \right | = 8.9$	79.21
5	19	21.01	$\left | 19 -21.01 \right | = 2.01$	4.0401
6	21	20.809	21 -20.809-0.191	0.036481
7	17	20.8281	17-17.671-0.67	0.4489

MSE = 30.12

MSE with alpha = 0.2 is the smallest MSE.