Question 7 of 7: Forecasting the Demand for a Product

Question 7: Forecasting the Demand for a Product (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Five months of demand history for a single product, together with a smoothing constant, a starting forecast and a set of moving-average weights.

Smoothing constant α = 0.20 with an assumed January forecast of 70 units; weighted-moving-average weights 0.30, 0.25, 0.20, 0.15 and 0.10 applied from the most recent month backwards.

Find. June and July forecasts by four-month moving average; the exponentially smoothed forecasts for February through June; June, July and August forecasts by least-squares regression; and June's forecast by weighted moving average.

Approach. Apply the four standard time-series methods in turn, each to the same history, and compare what they say — the point of the question is as much the comparison as the arithmetic.

1. Part (a) — four-month moving average for June. A four-month moving average forecasts the next period as the mean of the four most recent actuals: $$F_{t+1}=\frac{1}{4}\sum_{i=t-3}^{t}A_i$$ For June the four most recent months are February through May, so $$F_{\mathrm{Jun}}=\frac{100+60+80+90}{4}=\frac{330}{4} \qquad\Rightarrow\qquad \boxed{\ F_{\mathrm{Jun}}=82.5\ \text{units}\ }$$ January is dropped from the window entirely; a moving average has no memory beyond its span.
2. Roll the window forward for July. With June's actual demand known to be 100, the window becomes March through June: $$F_{\mathrm{Jul}}=\frac{60+80+90+100}{4}=\frac{330}{4} \qquad\Rightarrow\qquad \boxed{\ F_{\mathrm{Jul}}=82.5\ \text{units}\ }$$ The forecast is unchanged, which is a useful coincidence to notice rather than an error: February (100) left the window and June (100) entered it, so the sum of the window is identical.
3. Part (b) — set up the exponential smoothing recursion. Single exponential smoothing updates the previous forecast by a fraction α of the previous error: $$F_{t+1}=F_t+\alpha\,(A_t-F_t)=\alpha A_t+(1-\alpha)F_t$$ with $\alpha=0.20$ and the given starting value $F_{\mathrm{Jan}}=70$.
4. Advance the recursion month by month. Each line uses only the previous forecast and the actual just observed: $$F_{\mathrm{Feb}}=70+0.20(80-70)=70+2.00=72.00$$ $$F_{\mathrm{Mar}}=72.00+0.20(100-72.00)=72.00+5.60=77.60$$ $$F_{\mathrm{Apr}}=77.60+0.20(60-77.60)=77.60-3.52=74.08$$ $$F_{\mathrm{May}}=74.08+0.20(80-74.08)=74.08+1.18=75.26$$ $$F_{\mathrm{Jun}}=75.26+0.20(90-75.26)=75.26+2.95=78.21$$ so the smoothed forecasts through June are $$\boxed{\ 72.00,\ 77.60,\ 74.08,\ 75.26,\ 78.21\ \text{units for Feb, Mar, Apr, May, Jun}\ }$$ The series is still climbing towards the data because the assumed January forecast of 70 was well below the true level and a smoothing constant of 0.20 corrects only a fifth of the remaining gap each month.
5. Part (c) — fit a least-squares trend line. Numbering the months $t=1$ (January) to $t=5$ (May) and fitting $F_t=a+bt$, the slope and intercept are $$b=\frac{n\sum tA_t-\left(\sum t\right)\left(\sum A_t\right)}{n\sum t^2-\left(\sum t\right)^2}, \qquad a=\bar{A}-b\,\bar{t}$$ With $n=5$, $\sum t=15$, $\sum A_t=410$, $\sum t^2=55$ and $\sum tA_t=80+200+180+320+450=1{,}230$, $$b=\frac{5(1{,}230)-15(410)}{5(55)-15^2}=\frac{6{,}150-6{,}150}{275-225}=\frac{0}{50}=0$$ $$a=\frac{410}{5}-0\times 3=82$$
6. Project the fitted line forward. Since the slope is exactly zero, the regression line is the horizontal line $F_t=82$, and every future period takes the same value: $$\boxed{\ F_{\mathrm{Jun}}=F_{\mathrm{Jul}}=F_{\mathrm{Aug}}=82\ \text{units}\ }$$ The zero slope is genuine, not a rounding artefact: the deviations of the five demands from their mean of 82 are $-2$, $+18$, $-22$, $-2$ and $+8$, and weighted by their period offsets $-2,-1,0,+1,+2$ they sum to $(-2)(-2)+(18)(-1)+(-22)(0)+(-2)(1)+(8)(2)=4-18+0-2+16=0$. There is no linear trend in this history at all; the apparent rise from March onwards is exactly cancelled by the fall from February.
7. Part (d) — apply the weighted moving average. The five weights are applied to the five months with the largest weight on the most recent data: $$F_{\mathrm{Jun}}=\sum_{i=1}^{5}w_iA_{t+1-i} =0.30(90)+0.25(80)+0.20(60)+0.15(100)+0.10(80)$$ $$=27.0+20.0+12.0+15.0+8.0 \qquad\Rightarrow\qquad \boxed{\ F_{\mathrm{Jun}}=82.0\ \text{units}\ }$$ The weights sum to 1.00, as they must if the forecast is to be unbiased when demand is level.
8. Compare the four answers. The methods give 82.5, 78.21, 82.0 and 82.0 units for June — a spread of only about four units, or five per cent. Three of the four agree closely because the series has no trend and no seasonality, so any method that averages it returns something near the mean of 82. Exponential smoothing is the outlier not because the method is worse but because it is still recovering from the low assumed starting value; given several more months it would converge on the same level. For a series like this one, the weighted moving average or the simple average is the sensible choice, and the practical conclusion is to plan June around 82 units while watching the forecast error, since the month-to-month swings of ±20 units dwarf the differences between the methods.