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:

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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).

Mathematics

1 answer:

babunello [11.8K] · Answer 1 · 2026-03-07T23:40:29+03:00

Based on the table below listing the costs for the Lenovo zx-81 chip over the past 12 months

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

The forecast for period $F_{t+1}$ is calculated with the formula

$F_{t+1}=\alpha A_t+(1-\alpha)F_t$

where $A_t$ represents the actual value from the prior period and $F_t$ is the forecast value from the previous period.

Part 1A:
If α = 0.1 and the initial forecast for October is $1.83, with the actual October value being $1.57.

Thus, the forecast for period 11 is calculated as:

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

Consequently, the forecast for period 11 amounts to $1.80

Part 1B:

With α = 0.1 and the forecast for November is $1.80, while the actual November value is $1.62.

Thus, the forecast for period 12 is calculated as:

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

Thus, the forecast for period 12 amounts to $1.78

Part 2A:

Considering α = 0.3 and the initial forecast for October is $1.76, with the actual October value being $1.57.

Thus, the forecast for period 11 is calculated as:

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

Therefore, the forecast for period 11 totals $1.70


Part 2B:

Given α = 0.3 with the forecast for November at $1.70, the actual November value is $1.62.

Thus, the forecast for period 12 is calculated as:

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

Thus, the forecast for period 12 equals $1.68


Part 3A:

Given α = 0.5 and the initial forecast for October stands at $1.72, with the actual value for October being $1.57.

Thus, the forecast for period 11 is:

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

Therefore, the forecast for period 11 is $1.65


Part 3B:

In the case of α = 0.5 and the forecast for November of $1.65, October’s actual value is $1.62.

Thus, the forecast for period 12 is:

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

Consequently, the forecast for period 12 is $1.64

Part 4:

The mean absolute deviation of a forecast equals the sum of absolute values of the difference between actual values and forecasted values, divided by the count of items.

Hence, taking the actual values for October, November, and December as: $1.57, $1.62, $1.75

using α = 0.3 results in forecasted values for October, November, and December: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is:

$\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16$

Therefore, the mean absolute deviation for exponential smoothing where α = 0.1 for October, November, and December is: 0.157

Part 5:

The mean absolute deviation of a forecast is acquired by summing the absolute value of the differences between actual values and forecasted values, then dividing by the number of instances.

So, with actual values for October, November, and December as: $1.57, $1.62, $1.75

using α = 0.3 leads to forecasted values of October, November, and December: $1.76, $1.70, $1.68

Therefore, the mean absolute deviation is:

$
 \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = 
\frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = 
\frac{0.32}{3} \approx0.107$

Thus, the mean absolute deviation for exponential smoothing with α = 0.3 for October, November, and December is: 0.107


Part 6:

The mean absolute deviation of a forecast is computed as the sum of the absolute differences between actual values and forecasted values, divided by the number of observations.

Accordingly, with actual values for October, November, and December as: $1.57, $1.62, $1.75

utilizing α = 0.5 provides forecasted values for October, November, and December: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = 
\frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = 
\frac{29}{3} \approx0.097$

Therefore, the mean absolute deviation in exponential smoothing with α = 0.5 for October, November, and December is: 0.097