22-Mec-B4 Integrated Manufacturing Systems · May 2017

Question 6 of 6: Four Forecasts from One Five-Month Demand Series

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2017 — 16-Mec-B4 Integrated Manufacturing Systems. Three hours, open book, any non-communicating calculator permitted. Six questions are printed and any five constitute a complete paper; all questions are of equal value, so each is treated below as a 20-mark question. Every question is solved here, because the set is a study resource rather than a sitting.

Reference texts. M. P. Groover, Automation, Production Systems, and Computer-Integrated Manufacturing, 5th ed. (process planning, CAPP, group technology, discrete control and programmable logic controllers); R. B. Chase and F. R. Jacobs, Operations and Supply Chain Management, 16th ed. (forecasting); S. Nahmias and T. L. Olsen, Production and Operations Analysis, 7th ed. (inventory models and lot sizing); C. E. Ebeling, An Introduction to Reliability and Maintainability Engineering, 3rd ed. (series and parallel reliability); S. Kalpakjian and S. R. Schmid, Manufacturing Engineering and Technology, 8th ed. (machinability data and cutting conditions).

Question 6: Four Forecasts from One Five-Month Demand Series (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 observed demand for one product, with the smoothing parameters supplied by the question.

Given data
Period $t$	Month	Demand $D_t$ (units)
1	January	80
2	February	100
3	March	60
4	April	80
5	May	90
Smoothing constant		$\alpha = 0.20$, with $F_1 = 70$
Weights, most recent first		0.30, 0.25, 0.20, 0.15, 0.10

Find. The June forecast by four-month moving average and the July forecast once June is known; the exponentially smoothed forecasts for February through June; the regression forecasts for June, July and August; and the weighted moving average forecast for June.

Figure 6.1 — The demand history and the four June forecasts. Three of the four methods land within half a unit of 82, which is the series mean; only exponential smoothing differs materially, because it is still recovering from a low starting value.

Approach. Apply each of the four models in turn to the same series, then compare them, since the comparison is the point of the question: a series with no trend and no seasonality is where the simple models are hardest to separate.

Part (a) — four-month moving average for June. The moving average forecast is the arithmetic mean of the most recent $n$ observations, $$F_{t+1}=\frac{1}{n}\sum_{i=t-n+1}^{t}D_i ,$$ so with $n = 4$ the June forecast uses February through May: $$F_{\text{Jun}}=\frac{100+60+80+90}{4}=\frac{330}{4}=\boxed{82.5\ \text{units}}$$
Roll the window forward for July. Once June is observed at 100 units, January drops out of the window and June enters it, so the July forecast averages March through June: $$F_{\text{Jul}}=\frac{60+80+90+100}{4}=\frac{330}{4}=82.5\ \text{units}.$$ The two forecasts are identical, which is not a coincidence worth hiding: the value leaving the window, February's 100, happens to equal the value entering it, June's 100, so the mean cannot move.
Part (b) — set up the exponential smoothing recursion. Single exponential smoothing updates the previous forecast by a fraction of its own error, $$F_{t+1}=F_t+\alpha\left(D_t-F_t\right)=\alpha D_t+\left(1-\alpha\right)F_t ,$$ which with $\alpha = 0.20$ weights the newest observation at one fifth and retains four fifths of the old forecast. The recursion is started from the given January forecast of 70 units.
Run the recursion month by month. Substituting each observation in turn: $$F_{\text{Feb}} = 70 + 0.20(80-70) = 72.00, \qquad F_{\text{Mar}} = 72.00 + 0.20(100-72.00) = 77.60,$$ $$F_{\text{Apr}} = 77.60 + 0.20(60-77.60) = 74.08, \qquad F_{\text{May}} = 74.08 + 0.20(80-74.08) = 75.264,$$ and one more step carries it to the month asked for: $$\boxed{\;F_{\text{Jun}} = 75.264 + 0.20(90-75.264) = 78.21\ \text{units}\;}$$ The forecast is still climbing towards the series mean because the initial value of 70 was 12 units below it and, at $\alpha = 0.20$, roughly a fifth of the remaining error is removed each month.
Part (c) — fit the least-squares line. Numbering the months $t = 1$ to $5$, the regression line $F_t = a + bt$ has slope and intercept $$b=\frac{\sum\left(t-\bar t\right)\left(D_t-\bar D\right)}{\sum\left(t-\bar t\right)^{2}}, \qquad a=\bar D - b\,\bar t ,$$ with $\bar t = 3$ and $\bar D = (80+100+60+80+90)/5 = 82$ units.
Evaluate the slope, and check the cancellation. Forming the deviation products term by term gives $$\sum\left(t-\bar t\right)\left(D_t-\bar D\right) =(-2)(-2)+(-1)(18)+(0)(-22)+(1)(-2)+(2)(8) =4-18+0-2+16=0,$$ against a denominator of $\sum(t-\bar t)^{2}=4+1+0+1+4=10$. The numerator is exactly zero, so $b = 0$. This looks like an arithmetic slip and is not: the two early deviations and the two late deviations cancel term for term, and the sum is displayed above so that a marker can see the cancellation is genuine.
State the regression forecasts. With $b = 0$ the intercept is $a = \bar D = 82$ and the fitted line is flat, so every future period takes the same value: $$\boxed{\;F_{\text{Jun}}=F_{\text{Jul}}=F_{\text{Aug}}=82.0\ \text{units}\;}$$ The correct interpretation is that the data contain no detectable trend over these five months; a regression forecast of a flat line is the model telling you that, not the model failing.
Part (d) — apply the weighted moving average. The weights are listed heaviest first and must be assigned to the months most recent first, which is the convention that makes the method a smoothing model at all; they sum to $0.30+0.25+0.20+0.15+0.10=1.00$, as they must. Then $$F_{\text{Jun}}=\sum_i w_i D_i =0.30(90)+0.25(80)+0.20(60)+0.15(100)+0.10(80)$$ $$=27.0+20.0+12.0+15.0+8.0=\boxed{82.0\ \text{units}}$$
Compare the four forecasts. Three of the four methods return 82.0 to 82.5 units, and the fourth, exponential smoothing, returns 78.21 only because its starting value was set 12 units low and it has had five periods to recover. Measured against the actual June demand of 100 units the errors are 17.5, 21.79, 18.0 and 18.0 units respectively, so on this single period the four-month moving average is marginally best and exponential smoothing worst. No conclusion about the methods should be drawn from one period; the honest reading is that a level series with this much random variation cannot be forecast closely by any of them, and the choice between them should rest on the tracking signal computed over many periods.

Final results — Question 6
Method	Forecast (units)	Error against June = 100
(a) Four-month moving average, June	82.5	17.5
(a) Four-month moving average, July (June = 100)	82.5	—
(b) Exponential smoothing, February	72.00	—
(b) Exponential smoothing, March	77.60	—
(b) Exponential smoothing, April	74.08	—
(b) Exponential smoothing, May	75.264	—
(b) Exponential smoothing, June	78.21	21.79
(c) Regression line	F = 82.0 + 0.0 t	—
(c) Regression, June / July / August	82.0 / 82.0 / 82.0	18.0
(d) Weighted moving average, June	82.0	18.0

Topic. Time-series forecasting for a level (non-trending) series: simple and weighted moving averages, single exponential smoothing, and linear regression.

Key relations. Moving average F_t+1 = (1/n) times the sum of the last n demands. Weighted moving average F_t+1 = sum of w_iD_i with the weights summing to one. Single exponential smoothing F_t+1 = αD_t + (1 − α)F_t, whose average age of data is (1 − α)/α periods, matching a moving average of n = (2 − α)/α periods. Least squares b = sum of (t − tbar)(D − Dbar) divided by sum of (t − tbar)², a = Dbar − b·tbar.

Why this works. All four models estimate the same thing — the current level of a series with no trend — and differ only in how they weight past observations. The simple moving average weights the last four equally and the rest not at all; the weighted moving average tilts those weights towards the present; exponential smoothing applies geometrically declining weights over the whole history, so it never quite forgets its starting value; and regression weights every observation but constrains the fit to a straight line. Because this series is level, the three that estimate a mean all converge on the mean, and the differences between them are noise. At α = 0.20 exponential smoothing is roughly equivalent to a nine-period moving average, which is why it responds so slowly here.

Common pitfall. Assigning the weighted-average weights in chronological order, so that 0.30 lands on January instead of May; the answer then comes out at 80.5 and the method has been inverted. A second pitfall is declaring an arithmetic error when the regression slope comes out at exactly zero — show the deviation sum term by term and the cancellation is evidently real. A third is starting the exponential recursion at February with the January demand rather than the given January forecast of 70, which shifts every subsequent value.

Source. Chase and Jacobs, Operations and Supply Chain Management, 16th ed., ch. 18 (Forecasting), sections on moving averages, exponential smoothing and linear regression analysis; Nahmias and Olsen, Production and Operations Analysis, 7th ed., ch. 2.

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