Chapter 17.4, Problem 18E

To determine

Develop a forecast for $t=6$ by using the Holt’s linear exponential smoothing method with $\alpha =0.3$ and $\beta =0.5$.

Answer to Problem 18E

The forecast for $t=6$ by using the Holt’s linear exponential smoothing method with $\alpha =0.3$ and $\beta =0.5$ is 21.16.

Explanation of Solution

Calculation:

The given data represents time series data.

The equations for Holt’s linear exponential smoothing method is,

$L_t=\alpha Y_t+\left(1-\alpha \right)\left(L_{t-1}+b_{t-1}\right)$

$b_t=\beta \left(L_t-L_{t-1}\right)+\left(1-\beta \right)b_{t-1}$

$F_{t+1}=L_t+b_{t}k$

Where $L_t$ represents the estimate of the level of the time series in period t, $b_t$ represents estimate of the slope of the time series in period t, $\alpha$ represents the smoothing constant for the level of the time series, $Y_t$ is the value of time series in period t, $\beta$ is the smoothing constant for the slope the time series, $F_{t+1}$ is the forecast for k periods ahead and $k$ represents the number of periods ahead to be forecast.

For week 1:

Consider

$L_1=Y_1$

From given data, it can be observed that $Y_1=6$.

Therefore,

$L_1=6$

Now,

$\begin{array}{c}{b}_1=Y_2-Y_1\\ =11-6\\ =5\end{array}$

Consider k=1 the $F_{t+1}$ is

$F_{t+1}=L_t+b_t$

The equation $F_{t+1}=L_t+b_t$ at $t=1$ is

$\begin{array}{c}{F}_{1+1}=L_1+b_1\\ F_2=6+5\\ =11\end{array}$

For week 2:

The equation $L_t=\alpha Y_t+\left(1-\alpha \right)\left(L_{t-1}+b_{t-1}\right)$ at $t=2$ is,

$\begin{array}{c}{L}_2=\alpha Y_2+\left(1-\alpha \right)\left(L_{2-1}+b_{2-1}\right)\\ =\alpha Y_2+\left(1-\alpha \right)\left(L_1+b_1\right)\end{array}$

Substitute $\alpha =0.3$, $Y_2=11$, $L_1=6$ and $b_1=5$

$\begin{array}{c}{L}_2=\left(0.3\right)\left(11\right)+\left(1-0.3\right)\left(6+5\right)\\ =3.3+7.7\\ =11\end{array}$

The equation $b_t=\beta \left(L_t-L_{t-1}\right)+\left(1-\beta \right)b_{t-1}$ at $t=2$ is,

$\begin{array}{c}{b}_2=\beta \left(L_2-L_{2-1}\right)+\left(1-\beta \right)b_{2-1}\\ =\beta \left(L_2-L_1\right)+\left(1-\beta \right)b_1\end{array}$

Substitute $L_1=6$, $L_2=11$, $\beta =0.5$ and $b_1=5$

$\begin{array}{c}{b}_2=0.5\left(11-6\right)+\left(1-0.5\right)5\\ =0.5\left(5\right)+0.5\left(5\right)\\ =2.5+2.5\\ =5\end{array}$

The equation $F_{t+1}=L_t+b_t$ at $t=2$ is,

$\begin{array}{c}{F}_{2+1}=L_2+b_2\\ =11+5\\ =16\end{array}$

Similarly the remaining estimated levels, estimated trend and forecasts are obtained as follows:

t	Value	Estimated Level	Estimated trend	Forecast
1	6	6	5
2	11	11	5	11
3	9	13.90	3.95	16
4	14	16.70	3.37	17.85
5	15	18.55	2.61	20.07

The forecast for week 6 is,

$\begin{array}{l}{F}_{t+1}=L_t+b_t\\ F_{5+1}=L_5+b_5\\ F_6=L_5+b_5\end{array}$

Substitute $L_5=18.55$ and $b_5=2.61$

$\begin{array}{c}{F}_6=L_5+b_5\\ =18.55+2.61\\ =21.16\end{array}$

Thus, the forecast for $t=6$ by using the Holt’s linear exponential smoothing method with $\alpha =0.3$ and $\beta =0.5$ is 21.16.