Chapter 5, Problem 5E

A firm experienced the demand shown in the following table.

*Unkown future value to be forecast

1. Fill in the table by preparing forecasts based on a five-year moving average, a three-year moving average, and exponential smoothing (with a $w=0.9$ and a $w=0.3$). Note The exponential smoothing forecasts may be begun by assuming ${\hat{Y}}_{}$
2. t + 1 = Y t .
3. Using the forecasts from 2005 through 2009, compare the accuracy of each of the forecasting methods based on the RMSE criterion.
4. Which forecast would you have used for 2010? Why?

To determine

To complete: The missing values in the table.

Explanation of Solution

  

To fill 5^th year column. Use below formula.

  $\begin{array}{l}\text{ Moving average for 2005 = }\frac{\text{Actual price of year 2000 + 2001 +2002 +2003+2004}}{\text{N}}\\ \text{N = Number}\text{of}\text{years}\end{array}$

Similarly, for 3year column.

  $\begin{array}{l}\text{ Moving average for 2003 = }\frac{\text{Actual price of year 2000 + 2001 +2002}}{\text{N}}\\ \text{N = Number}\text{of}\text{years}\end{array}$

For exponential smoothing:

Let the value forecasted value for 2000 is zero and assume the forecast value of 2001 is 800.

  $\begin{array}{l}\text{Forcasted value = last year forcasted value + w}\times \text{ (last year actual - last year forecast)}\\ \text{For example, }\\ \text{Forcast of 2002 = 925 + 0}\text{.9(925-800) = 912}\text{.5}\end{array}$

To determine

To compare: The accuracy of forecast methods.

Explanation of Solution

Root mean square error is used to find the accuracy.

  $RMSE=\sqrt{\frac{{\displaystyle \sum (Y_t}-\widehat{Y_t}{)}^2}{N}}$

Year	Actual value (Yt)	Forecast ( $\widehat{Y}$ t)	$\left(Y_t-\widehat{Y_t}\right)$	${\left(Y_t-\widehat{Y_t}\right)}^2$
2005	1160	960	200	40000
2006	1200	1032	168	28224
2007	1150	1087	63	3969
2008	1270	1137	133	17689
2009	1290	1186	104	10816
				${\displaystyle \sum {\left(Y_t-\widehat{Y_t}\right)}^2=100698}$

Now calculate the RMSE.

  $\begin{array}{c}RMSE=\sqrt{\frac{{\displaystyle \sum (Y_t}-\widehat{Y_t}{)}^2}{N}}\\ =\sqrt{\frac{100698}{5}}\\ =141.91\end{array}$

Now calculate the RMSE for 3 year moving average.

Year	Actual value (Yt)	Forecast ( $\widehat{Y}$ t)	$\left(Y_t-\widehat{Y_t}\right)$	${\left(Y_t-\widehat{Y_t}\right)}^2$
2005	1160	1025	135	18225
2006	1200	1112	88	7744
2007	1150	1170	-20	400
2008	1270	1170	100	10000
2009	1290	1207	83	6889
				${\displaystyle \sum {\left(Y_t-\widehat{Y_t}\right)}^2=43258}$

Calculate RMSE.

  $\begin{array}{c}RMSE=\sqrt{\frac{{\displaystyle \sum (Y_t}-\widehat{Y_t}{)}^2}{N}}\\ =\sqrt{\frac{43258}{5}}\\ =93.01\end{array}$

Now calculate the RMSE when w = 0.9

Year	Actual value (Yt)	Forecast ( $\widehat{Y}$ t)	$\left(Y_t-\widehat{Y_t}\right)$	${\left(Y_t-\widehat{Y_t}\right)}^2$
2005	1160	1136.26	23.74	563.5876
2006	1200	1157.63	42.37	1795.217
2007	1150	1195.76	-45.76	2093.978
2008	1270	1154.58	115.42	13321.78
2009	1290	1258.46	31.54	994.7716
				${\displaystyle \sum {\left(Y_t-\widehat{Y_t}\right)}^2=18769.33}$

  $\begin{array}{c}RMSE=\sqrt{\frac{{\displaystyle \sum (Y_t}-\widehat{Y_t}{)}^2}{N}}\\ =\sqrt{\frac{18769.33}{5}}\\ =61.27\end{array}$

Now calculate the RMSE when w = 0.3

Year	Actual value (Yt)	Forecast ( $\widehat{Y}$ t)	$\left(Y_t-\widehat{Y_t}\right)$	${\left(Y_t-\widehat{Y_t}\right)}^2$
2005	1160	979.82	180.18	32464.83
2006	1200	1033.87	166.13	27599.18
2007	1150	1083.71	66.29	4394.364
2008	1270	1103.71	166.4	27688.96
2009	1290	1153.52	136.48	18626.79
				${\displaystyle \sum {\left(Y_t-\widehat{Y_t}\right)}^2=110774.12}$

  $\begin{array}{c}RMSE=\sqrt{\frac{{\displaystyle \sum (Y_t}-\widehat{Y_t}{)}^2}{N}}\\ =\sqrt{\frac{110774.12}{5}}\\ =148.84\end{array}$

Conclusion: The lowest value of RMSE reflects the greater accuracy. Therefore, when w = 0.9 then RMSE shows most accurate model.

To determine

To Find: The method used to forecast.

Explanation of Solution

Since the greater accuracy come for the exponential smoothening with w = 0.9. Thus, same will be used to forecast the demand of 2010.