A dry cleaner uses exponential smoothing to forecast equipment usage at its main
ID: 472822 • Letter: A
Question
A dry cleaner uses exponential smoothing to forecast equipment usage at its main plant. August usage was forecasted to be 50 percent of capacity; actual usage was 53 percent of capacity. A smoothing constant of 0.4 is used.
Prepare a forecast for September. (Round your final answer to 2 decimal places.)
Assuming actual September usage of 56 percent, prepare a forecast for October usage. (Round your answer to 2 decimal places.)
A dry cleaner uses exponential smoothing to forecast equipment usage at its main plant. August usage was forecasted to be 50 percent of capacity; actual usage was 53 percent of capacity. A smoothing constant of 0.4 is used.
Explanation / Answer
Question :
A dry cleaner uses exponential smoothing to forecast equipment usage at its main plant. August usage was forecasted to be 50 percent of capacity; actual usage was 53 percent of capacity. A smoothing constant of 0.4 is used.
St+1=yt+(1)St,0<1,t>0.
a. Prepare a forecast for September. (Round your final answer to 2 decimal places.)
Using exponential smoothing, list all terms for the time series using = 0.4
t = 1
s1 = x0
s1 = 100%
t = 2
st = xt - 1 + (1 - )st - 1
s2 = x2 - 1 + (1 - )s2 - 1
s2 = 0.4(x1) + (1 - 0.4)s1
s2 = 0.4(x1) + (0.6)s1
s2 = 0.4(53%) + (0.6)100%
s2 = 21.2 + 60
s2 = 81.2
t = 3
st = xt - 1 + (1 - )st - 1
s3 = x3 - 1 + (1 - )s3 - 1
s3 = 0.4(x2) + (1 - 0.4)s2
s3 = 0.4(x2) + (0.6)s2
s3 = 0.4() + (0.6)81.2
s3 = 0 + 48.72
s3 = 48.72
Our 3 smoothed values are {100%, 81.2, 48.72}
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b) Assuming actual September usage of 56 percent, prepare a forecast for October usage. (Round your answer to 2 decimal places.)
t = 1
s1 = x0
s1 = 100%
t = 2
st = xt - 1 + (1 - )st - 1
s2 = x2 - 1 + (1 - )s2 - 1
s2 = 0.4(x1) + (1 - 0.4)s1
s2 = 0.4(x1) + (0.6)s1
s2 = 0.4(56%) + (0.6)100%
s2 = 22.4 + 60
s2 = 82.4
t = 3
st = xt - 1 + (1 - )st - 1
s3 = x3 - 1 + (1 - )s3 - 1
s3 = 0.4(x2) + (1 - 0.4)s2
s3 = 0.4(x2) + (0.6)s2
s3 = 0.4() + (0.6)82.4
s3 = 0 + 49.44
s3 = 49.44
Our 3 smoothed values are {100%, 82.4, 49.44}
St+1=yt+(1)St,0<1,t>0.
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