Question

you can see in the following​ table, demand for heart transplant surgery at Washington General Hospital has increased steadily in the past few​ years:

Year 1 2 3 4 5

Heart Transplants 48.0 48.0 53.0 57.0 57.0  

Year 1 2 3 4 5

Heart Transplants 44.0

The director of medical services predicted 6 years ago that demand in year 1 would be 44.0 sureries.

A) using exponential smoothing with a of 60.0 anf the given forecast for year 1, the forecasts for 2 years through 6 are? (round your answers to one decimal place)

For the forecast made using expotienal smoothing with a=60.0 and the given forecasts for year 1, MAD= _ surgeries (round your response to one decimal place)

Using expotiental smoothing with a of 90.0 and the given forecast for year 1, the forcasts for years 2 through 6 are (round your responses to one decimal place)

For the forecast made using expotiental smoothing with a=0.90 and given forecast for year 1, MAD= _ surgeries (round your response to one decimal place)

b) forecasts for years 4 through 6 using a 3 year moving average are? (round your response to one decimal place)

For forecast made using a 3 year moving average MAD = _ surgeries (round your answer to one decimal place)

C) forecasts for years 1 through 6 using the trend-projection method are (round your response to one decimal place)

For forecasts made using the trend-projection method, MAD= _ surgeries (round your response to one decimal place)

D) based on the comparison of MAD, the best forecast is archived using the _ method

Answer 1

FORECAST(T + 1) = FORECAST + (ALPHA * (ACTUAL DEMAND - FORECAST))

FORECAST 2 = 44 + (0.6 * (48 - 44) = 46.4

FORECAST 3 = 46.4 + (0.6 * (48 - 46.4) = 47.4

FORECAST 4 = 47.4 + (0.6 * (53 - 47.4) = 50.8

FORECAST 5 = 50.8 + (0.6 * (57 - 50.8) = 54.5

FORECAST 6 = 54.5 + (0.6 * (57 - 54.5) = 56

FORECAST ERROR

PERIOD	ACTUAL DEMAND	FORECAST	DEVIATION(D - F)	ABS DEVIATION
1	48	44	4	4
2	48	46.4	1.6	1.6
3	53	47.4	5.6	5.6
4	57	50.8	6.2	6.2
5	57	54.5	2.5	2.5
SIGMA			19.9	19.9

MAD = SIGMA(ABSOLUTE DEVIATION) / N, WHERE N = 5

MAD = 19.9 / 5 = 4.0

2. USING ALPHA = 0.9

FORECAST(T + 1) = FORECAST + (ALPHA * (ACTUAL DEMAND - FORECAST))

FORECAST 2 = 44 + (0.9 * (48 - 44) = 47.6

FORECAST 3 = 47.6 + (0.9 * (48 - 47.6) = 48

FORECAST 4 = 48 + (0.9 * (53 - 48) = 52.5

FORECAST 5 = 52.5 + (0.9 * (57 - 52.5) = 56.6

FORECAST 6 = 56.6 + (0.9 * (57 - 56.6) = 57

PERIOD	ACTUAL DEMAND	FORECAST	DEVIATION(D - F)	ABS DEVIATION
1	48	44	4	4
2	48	47.6	0.4	0.4
3	53	48	5	5
4	57	52.5	4.5	4.5
5	57	56.6	0.4	0.4
SIGMA			14.3	14.3

MAD = SIGMA(ABSOLUTE DEVIATION) / N, WHERE N = 5

MAD = 14.3 / 5 = 2.9

3. LINEAR REGRESSION

PERIOD (X)	DEMAND (Y)	X	Y	X * Y	X^2
1	48	1	48	48	1
2	48	2	48	96	4
3	53	3	53	159	9
4	57	4	57	228	16
5	57	5	57	285	25
SIGMA		15	263	816	55

INTERCEPT = (SIGMA(Y) * SIGMA(X^2) - SIGMA(X) * SIGMA(XY)) / (N * SIGMA(X^2) - SIGMA(X)^2)

INTERCEPT = (263 * 55) - (15 * 816) / ((5 * 55) - 15^2) = 44.5

SLOPE = ((N * SIGMA(XY)) - (SIGMA(X) * SIGMA(Y))) - (N * SIGMA(X^2) - SIGMA(X)^2)

SLOPE = ((5 * 816) - (15 * 263) / ((5 * 55) - 15^2) = 2.7

LINE EQUATION = A + B(x), WHERE A IS THE INTERCEPT, B IS THE SLOPE, x IS THE PERIOD = 44.5 + (2.7 * X)

FOR THE VALUE OF X = 6 FORECAST = 44.5 + (2.7 * 6) = 60.7

FORECAST ERROR

PERIOD	ACTUAL DEMAND	FORECAST	DEVIATION(D - F)	ABS DEVIATION
1	48	47.2	0.8	0.8
2	48	49.9	-1.9	1.9
3	53	52.6	0.4	0.4
4	57	55.3	1.7	1.7
5	57	58	-1	1
SIGMA			0	5.8

MAD = SIGMA(ABSOLUTE DEVIATION) / N, WHERE N = 5

MAD = 5.8 / 5 = 1.2

4.

SIMPLE MOVING AVERAGE

FORECAST = SIGMA(PREVIOUS N DEMANDS) / N

WHERE N = 2

FORECAST 3 = (48 + 48) / 2 = 48

FORECAST 4 = (48 + 53) / 2 = 50.5

FORECAST 5 = (53 + 57) / 2 = 55

FORECAST 6 = (57 + 57) / 2 = 57

PERIOD	ACTUAL DEMAND	FORECAST	DEVIATION(D - F)	ABS DEVIATION
1	48
2	48
3	53	48	5	5
4	57	50.5	6.5	6.5
5	57	55	2	2
SIGMA			13.5	13.5

MAD = SIGMA(ABSOLUTE DEVIATION) / N, WHERE N = 3

MAD = 13.5 / 3 = 4.5

5. USING LINEAR REGRESSION PROVIDES THE LOWEST MAD VALUE AND IS THEREFORE BETTER.

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