Question

Lenovo uses the​ zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as​ follows:                                                                                                              month price per chip month price per chip january ​$1.901.90 july ​$1.801.80 february ​$1.611.61 august ​$1.821.82 march ​$1.601.60 september ​$1.601.60 april ​$1.851.85 october ​$1.571.57 may ​$1.901.90 november ​$1.621.62 june ​$1.951.95 december ​$1.751.75 this exercise contains only part

d. with alphaα ​= 0.1 and the initial forecast for october of ​$1.831.83​, using exponential​ smoothing, the forecast for periods 11 and 12 is ​(round your responses to two decimal​ places): month oct nov dec forecast ​$1.831.83 1.801.80 1.791.79 with alphaα ​= 0.3 and the initial forecast for october of ​$1.761.76​, using exponential​ smoothing, the forecast for periods 11 and 12 is ​(round your responses to two decimal​ places): month oct nov dec forecast ​$1.761.76 1.701.70 1.681.68 with alphaα ​= 0.5 and the initial forecast for october of ​$1.721.72​, using exponential​ smoothing, the forecast for periods 11 and 12 is ​(round your responses to two decimal​ places): month oct nov dec forecast ​$1.721.72 1.651.65 1.631.63 based on the months of​ october, november, and​ december, the mean absolute deviation using exponential smoothing where alphaα ​= 0.1 and the initial forecast for octoberequals=​$1.831.83 is ​$ . 160.160 ​(round your response to three decimal​ places). based on the months of​ october, november, and​ december, the mean absolute deviation using exponential smoothing where alphaα ​= 0.3 and the initial forecast for octoberequals=​$1.761.76 is ​$ 0.1130.113 ​(round your response to three decimal​ places). based on the months of​ october, november, and​ december, the mean absolute deviation using exponential smoothing where alphaα ​= 0.5 and the initial forecast for octoberequals=​$1.721.72 is ​$ nothing ​(round your response to three decimal​ places).

Answer 1

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months


`\begin{tabular} c Month&Price per Chip&Month&Price per Chip\\[1ex] January&\$1.90&July&\$1.80\\ February&\$1.61&August&\$1.83\\ March&\$1.60&September&\$1.60\\ April&\$1.85&October&\$1.57\\ May&\$1.90&November&\$1.62\\ June&\$1.95&December&\$1.75 \end{tabular}`

The forcast for a period
`F_(t+1)` is given by the formular


`F_(t+1)=\alpha A_t+(1-\alpha)F_t`

where
`A_t` is the actual value for the preceding period and
`F_t` is the forcast for the preceding period.

Part 1A:
Given α ​= 0.1 and the initial forecast for october of ​$1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:


`F_(11)=\alpha A_(10)+(1-\alpha)F_(10) \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804`

Therefore, the foreast for period 11 is $1.80


Part 1B:

Given α ​= 0.1 and the forecast for november of ​$1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:


`F_(12)=\alpha A_(11)+(1-\alpha)F_(11) \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ =0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782`

Therefore, the foreast for period 12 is $1.78



Part 2A:

Given α ​= 0.3 and the initial forecast for october of ​$1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:


`F_(11)=\alpha A_(10)+(1-\alpha)F_(10) \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ =0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703`

Therefore, the foreast for period 11 is $1.70


Part 2B:

Given α ​= 0.3 and the forecast for November of ​$1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:


`F_(12)=\alpha A_(11)+(1-\alpha)F_(11) \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ =0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676`

Therefore, the foreast for period 12 is $1.68




Part 3A:

Given α ​= 0.5 and the initial forecast for october of ​$1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:


`F_(11)=\alpha A_(10)+(1-\alpha)F_(10) \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ =0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645`

Therefore, the forecast for period 11 is $1.65


Part 3B:

Given α ​= 0.5 and the forecast for November of ​$1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:


`F_(12)=\alpha A_(11)+(1-\alpha)F_(11) \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ =0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635`

Therefore, the forecast for period 12 is $1.64



Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:


`(|1.57-1.83|+|1.62-1.80|+|1.75-1.78|)/(3) = (|-0.26|+|-0.18|+|-0.03|)/(3) \\ \\ = (0.26+0.18+0.03)/(3) = (0.47)/(3) \approx0.16`

Therefore, the mean absolute deviation using exponential smoothing where α ​= 0.1 of October, November and December is given by: 0.157



Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:


`(|1.57-1.76|+|1.62-1.70|+|1.75-1.68|)/(3) = (|-0.17|+|-0.08|+|-0.07|)/(3) \\ \\ = (0.17+0.08+0.07)/(3) = (0.32)/(3) \approx0.107`

Therefore, the mean absolute deviation using exponential smoothing where α ​= 0.3 of October, November and December is given by: 0.107




Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:


`(|1.57-1.72|+|1.62-1.65|+|1.75-1.64|)/(3) = (|-0.15|+|-0.03|+|0.11|)/(3) \\ \\ = (0.15+0.03+0.11)/(3) = (29)/(3) \approx0.097`

Therefore, the mean absolute deviation using exponential smoothing where α ​= 0.5 of October, November and December is given by: 0.097