Bill and Joe leave their house at the same time heading in opposite directions. If Bill is driving three times faster than Joe a
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Given the table below of the prices for the Lenovo zx-81 chip during the last 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}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7Cc%7Cc%7C%7D%0AMonth%26Price%20per%20Chip%26Month%26Price%20per%20Chip%5C%5C%5B1ex%5D%0AJanuary%26%5C%241.90%26July%26%5C%241.80%5C%5C%0AFebruary%26%5C%241.61%26August%26%5C%241.83%5C%5C%0AMarch%26%5C%241.60%26September%26%5C%241.60%5C%5C%0AApril%26%5C%241.85%26October%26%5C%241.57%5C%5C%0AMay%26%5C%241.90%26November%26%5C%241.62%5C%5C%0AJune%26%5C%241.95%26December%26%5C%241.75%0A%5Cend%7Btabular%7D)
The forcast for a period
is given by the formular
where
is the actual value for the preceding period and 
is the forcast for the preceding period.
Part 1A:
Given <span>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:
Therefore, the foreast for period 11 is $1.80
Part 1B:
</span>Given <span>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:
Therefore, the foreast for period 12 is $1.78</span>
Part 2A:
Given <span>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:
Therefore, the foreast for period 11 is $1.70
</span>
<span><span>Part 2B:
</span>Given <span>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:
Therefore, the foreast for period 12 is $1.68
</span></span>
<span>Part 3A:
Given <span>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:
Therefore, the forecast for period 11 is $1.65
</span>
<span><span>Part 3B:
</span>Given <span>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:
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 </span></span></span><span>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:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157
</span><span><span>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 </span><span>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:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107
</span></span>
<span><span>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 </span><span>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:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>
The forcast for a period
where
Part 1A:
Given <span>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:
Therefore, the foreast for period 11 is $1.80
Part 1B:
</span>Given <span>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:
Therefore, the foreast for period 12 is $1.78</span>
Part 2A:
Given <span>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:
Therefore, the foreast for period 11 is $1.70
</span>
<span><span>Part 2B:
</span>Given <span>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:
Therefore, the foreast for period 12 is $1.68
</span></span>
<span>Part 3A:
Given <span>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:
Therefore, the forecast for period 11 is $1.65
</span>
<span><span>Part 3B:
</span>Given <span>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:
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 </span></span></span><span>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:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157
</span><span><span>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 </span><span>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:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107
</span></span>
<span><span>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 </span><span>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:
Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>
Given:
Anita's fixed salary = Rs. W
Bonus = Rs. Z for every bedcover sold by her.
To find:
Her total earnings if she sold 46 bedcovers.
Solution:
We have,
Bonus if she sold 1 bedcover = Rs. Z
Bonus if she sold 46 bedcover = Rs. Z × 46
We know that,
Total earnings = Fixed salary + Bonus
=
Therefore, the correct option is (b).