opm 625
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Solved Problems Chapter 3: Forecasting
Problem 1. The historical demand of a spare part is given below:
|
July |
1,000 |
|
August |
1,200 |
|
September |
900 |
|
October |
1,100 |
a. Use a three-period moving average to make a forecast for November.
Solution
(1,200 + 900 + 1,100)/3 = 3,200/3 = 1,067
b. Make a forecast for November using a weighted moving average with weights of 0.4, 0.3, 0.2, and 0.1
Solution
1,100(0.4) + 900(0.3) + 1,200(0.2) + 1,00(0.1)
= 440 + 270 + 240 + 100
= 1,050
c. Make a forecast for November using single exponential smoothing with α = 0.2 if the forecast for July was 1,100.
Solution
FAug = FJuly + 0.2(AJuly - FJuly)
= 1,100 + 0.2(1,000 – 1,100)
= 1,100 – 20
= 1,080
FSep = FAug + 0.2(AAug – FAug)
= 1,080 + 0.2(1,200 – 1,080)
= 1,080 + 24
= 1,104
FOct = FSep + 0.2(ASep – FSep)
= 1,104 + 0.2(900 – 1,104)
= 1,104 + 40.8
= 1,063.2
FNov = FOct + 0.2(AOct – FOct)
= 1,063.2 + 0.2(1,100 – 1,063.2)
= 1,063.2 + 7.36
= 1,070.56
d. Calculate MAD, RSFE (running sum of forecast errors), and tracking signal for the forecasts for August, September, and October made by the exponential smoothing method.
Solution
|
Month |
Forecast |
Actual Demand |
Actual Deviation |
RSFE |
Absolute Deviation |
|
August |
1,080 |
1,200 |
+120 |
+120 |
120 |
|
September |
1,104 |
900 |
-204 |
– 84 |
204 |
|
October |
1,063.2 |
1,100 |
+36.8 |
– 47.2 |
36.8 |
|
|
|
|
Total absolute deviation |
360.8 |
Mean absolute deviation (MAD) = 360.8/3 = 120.27
Tracking signal
Problem #2 starts on the next page
Problem 2A. Calculate the comparative weights for the preceding 10 periods if α = 0.2 is used in exponential smoothing to make a forecast for period t. Plot these weights on a graph.
Solution
|
Period |
Weight |
|
|
|
|
t – 1 |
α = 0.2000 |
|
t – 2 |
α(1-α) = 0.1600 |
|
t – 3 |
α(1-α)2 = 0.1280 |
|
t – 4 |
α(1-α)3 = 0.1024 |
|
t – 5 |
α(1-α)4 = 0.0819 |
|
t – 6 |
α(1-α)5 = 0.0655 |
|
t – 7 |
α(1-α)6 = 0.0524 |
|
t – 8 |
α(1-α)7 = 0.0419 |
|
t – 9 |
α(1-α)8 = 0.0336 |
|
t – 10 |
α(1-α)9 = 0.0268 |
Figure 2 on the next page gives the plot of these weights.
Figure 2
Problem 2B. Repeat the calculations for the last six periods using α = 0.4, and plot the weights on the same graph.
Solution
|
Period |
Weight |
|
|
|
|
t – 1 |
α = 0.4000 |
|
t – 2 |
α(1-α) = 0.2400 |
|
t – 3 |
α(1-α)2 = 0.1400 |
|
t – 4 |
α(1-α)3 = 0.0864 |
|
t – 5 |
α(1-α)4 = 0.0518 |
|
t – 6 |
α(1-α)5 = 0.0311 |
Problem 2C. What will be the weights for each of the past four periods in a four-year moving average approach? Plot these weights on the same graph.
Solution
The weight of each year (t – 1), (t – 2), (t – 3), (t – 4) will be ¼ or 0.25.
Problem 3. The historical demand for a product is given below:
January 1,850
February 2,000
March 2,200
April 2,300
May 2,500
Use lease squares regression analysis to make a forecast for June. What will be the range of this forecast using two standard errors of estimates?
Solution
Let x = 1, 2, 3, 4, 5 represent the months from January to May, and let y represent the corresponding demand.
|
x |
Y |
X2 |
Y2 |
xy |
|
1 |
1,850 |
1 |
3,422,500 |
1,850 |
|
2 |
2,000 |
4 |
4,000,000 |
4,000 |
|
3 |
2,200 |
9 |
4,840,000 |
6,600 |
|
4 |
2,300 |
16 |
5,290,000 |
9,200 |
|
5 |
2,550 |
25 |
6,502,500 |
12,750 |
|
15 |
10,900 |
55 |
24,055,000 |
34,400 |
a and b in equation, Y = a + bX, are determined as follows:
= 2,180 – 170(3)
= 2,180 – 510
= 1,670
Therefore, Y = 1,670 + 170x
The forecast for June is calculated at x = 6
Y = 1,670 + 170 (6)
= 1,670 + 1,020
= 2,690
The standard error of estimate, Syx, is calculated by the formula
The range of forecast using two standard errors of estimate will be:
Problem 4. The Midwest Fashion Company manufactures men’s suits for which demand is highly seasonal. The company has been experiencing difficulty in making accurate forecasts. The following table gives the number of suits sold by quarters in 2004, 2005, and 2006. Make forecasts for the four quarters of 2007.
|
Demand |
|||
|
Quarter |
2004 |
2005 |
2006 |
|
I |
540 |
590 |
649 |
|
II |
70 |
72 |
80 |
|
III |
110 |
126 |
130 |
|
IV |
443 |
460 |
510 |
Solution
The first step is to determine the trend line
|
(1) Quarter and Year |
(2) Actual Demand |
(3) Periodic Average for 2004, 2005, and 2006 |
(4) Seasonal Factor: (3) ÷ Average of (3) (315 in this problem) |
(5) Deseasonalized Demand, Y, (2)/(4) |
(6) Period, X |
(7) XY |
X2 |
|
I, 04 |
540 |
(540 + 590 + 649)/3 = 593 |
1.883 |
287 |
1 |
287 |
1 |
|
II, 04 |
70 |
( 70 + 72 + 80)03 = 74 |
0.235 |
298 |
2 |
596 |
4 |
|
III, 04 |
110 |
(110 + 126 + 130)/3 = 122 |
0.387 |
284 |
3 |
852 |
9 |
|
IV, 04 |
443 |
(443 + 460 + 510)/3 = 471 |
1.495 |
296 |
4 |
1,184 |
16 |
|
I, 05 |
590 |
|
1.883 |
313 |
5 |
1,565 |
25 |
|
II, 05 |
72 |
|
0.235 |
306 |
6 |
1,836 |
36 |
|
III, 05 |
126 |
|
0.387 |
326 |
7 |
2,282 |
49 |
|
IV, 05 |
460 |
|
1.495 |
308 |
8 |
2,464 |
64 |
|
I, 06 |
649 |
|
1.883 |
345 |
9 |
3,105 |
81 |
|
II, 06 |
80 |
|
0.235 |
340 |
10 |
3,400 |
100 |
|
III, 06 |
130 |
|
0.387 |
336 |
11 |
3,696 |
121 |
|
IV, 06 |
510 |
|
1.495 |
341 |
12 |
4,092 |
144 |
|
Total |
3,780 |
1,260 |
|
3,780 |
78 |
25,359 |
650 |
|
Average |
315 |
315 |
|
315 |
6.5 |
|
|
Least Squares Estimate
= 315 – 5.52(6.5)
= 315 – 35.88 = 279.12
Thus the trend equation will be: Y = 279.12 + 5.52X
Forecasts
|
Quarter and Year |
Trend (from equation Y = 279.12 + 5.52X) |
Seasonal Factor |
Forecast = Trend x Seasonal Factor |
|
I, 2007 |
350.88 (for X = 13) |
1.883 |
661 |
|
II, 2007 |
356.40 (for X = 14) |
0.235 |
79 |
|
III, 2007 |
361,92 (for X = 15) |
0.387 |
140 |
|
IV, 2007 |
367.44 (for X = 16) |
1.495 |
549 |
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