Question: With = 0 . 3 and the initial forecast for January of $ 1 . 9 0 , using exponential smoothing, the forecast is: (

With =0.3 and the initial forecast for January of $1.90, using exponential smoothing, the forecast is: (Round your responses to two
decimal places.)
With =0.5 and the initial forecast for January of $1.90, using exponential smoothing, the forecast is: (Round your responses to two
decimal places.)
Based on all of the months in the forecast, the mean absolute deviation using exponential smoothing where =0.1 and the initial
forecast for January of $1.90 is $,.(Round your response to three decimal places.)
Based on all of the months in the forecast, the mean absolute deviation using exponential smoothing where =0.3 and the initial
forecast for January of $1.90 is $,.(Round your response to three decimal places.)
Based on all of the months in the forecast, the mean absolute deviation using exponential smoothing where =0.5 and the initial
forecast for January of $1.90 is $,.(Round your response to three decimal places.)
Based on the mean absolute deviation, the better forecast is achieved using = Dell uses the CR5 chip in some of its laptop computers. The prices for the chip during the past 12 months were as follows.
\table[[Month,Price Per Chip,Month,Price Per Chip],[January,$1.90,July,$1.90],[February,$1.61,August,$1.85],[March,$1.60,September,$1.70],[April,$1.85,October,$1.60],[May,$1.88,November,$1.50],[June,$1.95,December,$1.75]]
This exercise contains only part d.
With =0.1 and the initial forecast for January of $1.90, using exponential smoothing, the forecast is: (Round your responses to two decimal places.)
\table[[Month,Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec],[Forecast,$1.90,,,,,,,,,,,]]
 With =0.3 and the initial forecast for January of $1.90, using

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