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Solved Problems

Solved Problem 5-1 Demand for patient surgery at Washington General Hospital has increased steadily in the past few years, as seen in the following table:

YEAR	OUTPATIENT SURGERIES PERFORMED
1	45
2	50
3	52
4	56
5	58
6	—

The director of medical services predicted 6 years ago that demand in year 1 would be 42 surgeries. Using exponential smoothing with a weight of $α = 0.20,$ $α = 0.20,$ develop forecasts for years 2 through 6. What is the MAD?

Solution

YEAR	ACTUAL	FORECAST (SMOOTHED)	ERROR	\|ERROR\|
1	45	42	$+$ $+$ 3	3
2	50	$42.6 = 42 + 0.2 (45 - 42)$ $42.6 = 42 + 0.2 (45 - 42)$	$+$ $+$ 7.4	7.4
3	52	$44.1 = 42.6 + 0.2 (50 - 42.6)$ $44.1 = 42.6 + 0.2 (50 - 42.6)$	$+$ $+$ 7.9	7.9
4	56	$45.7 = 44.1 + 0.2 (52 - 44.1)$ $45.7 = 44.1 + 0.2 (52 - 44.1)$	$+$ $+$ 10.3	10.3
5	58	$47.7 = 45.7 + 0.2 (56 - 45.7)$ $47.7 = 45.7 + 0.2 (56 - 45.7)$	$+$ $+$ 10.3	10.3
6	—	$49.8 = 47.7 + 0.2 (58 - 47.7)$ $49.8 = 47.7 + 0.2 (58 - 47.7)$	—	—
		$MAD = \frac{\sum \| Errors \|}{n} = \frac{38.9}{5} = 7.78$ $MAD = \frac{\sum \| Errors \|}{n} = \frac{38.9}{5} = 7.78$		38.9

Solved Problem 5-2 Quarterly demand for the Jaguar XJ8 at a New York auto dealership is forecast with the equation

$\hat{Y} = 10 + 3 X$ $\hat{Y} = 10 + 3 X$

where

$\begin{array}{l} X = time period (quarter): & quarter 1 of last year & = & 0 \\ quarter 2 of last year & = & 1 \\ quarter 3 of last year & = & 2 \\ quarter 4 of last year & = & 3 \\ quarter 1 of last year & = & 4, and so on \end{array}$ $\begin{array}{l} X = time period (quarter): & quarter 1 of last year & = & 0 \\ quarter 2 of last year & = & 1 \\ quarter 3 of last year & = & 2 \\ quarter 4 of last year & = & 3 \\ quarter 1 of last year & = & 4, and so on \end{array}$

and

$\hat{Y} = predicted quarterly demand$ $\hat{Y} = predicted quarterly demand$

The demand for luxury sedans is seasonal, and the indices for quarters 1, 2, 3, and 4 are 0.80, 1.00, 1.30, and 0.90, respectively. Using the trend equation, forecast the demand for each quarter of next year. Then adjust each forecast for seasonal (quarterly) variations.

Solution

Quarter 2 of this year is coded $X = 5;$ $X = 5;$ quarter 3 of this year, $X = 6;$ $X = 6;$ and quarter 4 of this year, $X = 7.$ $X = 7.$ Hence, quarter 1 of next year is coded $X = 8;$ $X = 8;$ quarter 2, $X = 9;$ $X = 9;$ and so on.

$\begin{array}{l} \hat{Y} (next year quarter 1) & = & 10 + (3) (8) = 34 Adjusted forecast = (0.80) (34) = 27.2 \\ \hat{Y} (next year quarter 2) & = & 10 + (3) (9) = 37 Adjusted forecast = (1.00) (37) = 37 \\ \hat{Y} (next year quarter 3) & = & 10 + (3) (10) = 40 Adjusted forecast = (1.30) (40) = 52 \\ \hat{Y} (next year quarter 4) & = & 10 + (3) (11) = 43 Adjusted forecast = (0.90) (43) = 38.7 \end{array}$ $\begin{array}{l} \hat{Y} (next year quarter 1) & = & 10 + (3) (8) = 34 Adjusted forecast = (0.80) (34) = 27.2 \\ \hat{Y} (next year quarter 2) & = & 10 + (3) (9) = 37 Adjusted forecast = (1.00) (37) = 37 \\ \hat{Y} (next year quarter 3) & = & 10 + (3) (10) = 40 Adjusted forecast = (1.30) (40) = 52 \\ \hat{Y} (next year quarter 4) & = & 10 + (3) (11) = 43 Adjusted forecast = (0.90) (43) = 38.7 \end{array}$

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Solved Problems