Home Work; Chapter 8; Forecasting Supply Chain Requirements Book reference: Ballou, Ronald H. (2004). “Business Logistics/ Supply Chain Management: Planning, Organizing, and Controlling the Supply Chain.” (5th Edition). Original reference of this document: http://wweb.uta.edu/insyopma/prater/ballou08_im.pdf

1. 73
CHAPTER 8
FORECASTING SUPPLY CHAIN REQUIREMENTS
4
(a) The answer to this question is aided by using the FORECAST module in
LOGWARE. A sample calculation is shown as carried out by FORECAST. The
results are then summarized from FORECAST output. An example calculation for an
α = 0.1 is shown. Other α values would be used, ranging from 0.01 to 1.0.
We first calculate a starting forecast by averaging the first four weekly
requirements. That is,
[2,056 + 2,349 + 1,895 + 1,514]/4 = 1,953.50
Now, we back cast this value and start the forecast at time 0. Thus, the forecasts and
the associated errors would be:
The standard error of the forecast is:
S
N
F = = =
Total squared error 1 677 71376
6
52879
, , .
.
Note: FORECAST does not use N-1 in the denominator.
Repeating this type of analysis, the following table can be developed. The results
from FORECAST are shown.
Forecast Error
Squared
error
F1 = 1953.50
F2 = .1(2056) + .9(1953.50) = 1963.75
F3 = .1(2349) + .9(1963.75) = 2002.28
F4 = .1(1895) + .9(2002.20) = 1991.48
F5 = .1(1514) + .9(1991.48) = 1943.73
F6 = .1(1194) + .9(1943.73) = 1868.76 -749.73 562,095.07
F7 = .1(2268) + .9(1868.76) = 1908.00 399.24 159,392.58
F8 = .1(2653) + .9(1908.00) = 1982.50 745.00 555,025.00
F9 = .1(2039) + .9(1982.50) = 1988.15 56.50 3,192.25
F10 = .1(2399) + .9(1988.15) = 2029.24 410.85 168,797.72
F11 = .1(2508) + .9(2029.24) = 2077.12 478.76 229,211.14
Total squared error 1,677,713.76

2. 74
α SF
.01 528.72
.05 528.42
.1 528.46
.2 528.89
.5 535.55
1.0 566.07
The α that minimizes SF is 0.05.
(b) The forecast errors are computed in part a.
(c) If we assume that the errors are normally distributed around the forecast, we can then
construct a 95 percent confidence band on the forecast. That is, if Y is the actual
volume in period 11, then the range of the forecast (F11 = 2,017.81 for α = 0.05) will
be:
Y = F11 + z× SF
= 2,017.81 + 1.96×528.42
Then,
982.11 ≤ Y ≤ 3,053.51
All values are in thousands.
5
(a) & (b) The solution to this problem was aided by the use of the exponential smoothing
module in FORECAST. Using the first four week's data to initialize the level/trend
version of the exponential smoothing model and setting α and β equal to 0.2, the
forecast for next week is F11 = 2,024.47, with a standard error of the forecast of SF =
171.28.
(c) Assuming that the forecast errors are normally distributed around F11, a 95 percent
statistical confidence band can be constructed. The confidence band is:
Y = F11 + z× SF
= 2,024.47 + 1.96×171.28
where z = 1.96 for 2.5 percent of the area under the two tails of a normal distribution.
The range of the actual weekly volume is expected to be:
1,688.76 ≤ Y ≤ 2,360.18

3. 75
6
(a) The data may be restated as shown below.
Sales, S t S×t t2
Trend
value,a
St
Seasonal
indexb
27,000 1 27,000 1 41,087 0.66
70,000 2 140,000 4 41,192 1.70
41,000 3 123,000 9 41,298 0.99
13,000 4 52,000 16 41,403 0.32
30,000 5 150,000 25 41,508 0.72
73,000 6 438,000 36 41,613 1.75
48,000 7 336,000 49 41,719 1.15
15,000 8 120,000 64 41,824 0.36
34,000 9 306,000 81 41,929 0.81
82,000 10 820,000 100 42,035 1.95
51,000 11 561,000 121 42,140 1.21
16,000 12 192,000 144 42,245 0.38
500,000 78 3,265,000 650
a
Computed from the linear trend line. For example, for period 1,
S1 = 40,981.6 + 105.3×1 = 41,087.
b
The ratio of the actual sales S to the trend line value St.
For example, for period 1, the seasonal index is 27,000/41,087 = 0.66.
Given the values from the above table and that t = 78/12 = 6.5, N = 12, and S =
500,000/12 = 41,666, the coefficients in the regression trend line would be:
b
S t N S t
t N t
=
× − × ×
− ×
=
− × ×
− ×
=
2 2 2
3 265 000 12 41666 65
650 12 65
1053
, , , .
.
.
and
a S b t= − × = − × =41 666 1053 65 40 9816, . . , .
Therefore, the trend value St for any period t would be:
St = 40,981.6 + 105.3×t
(b) The seasonal factors are determined by the ratio of the actual sales in a period to the
trend value for that period. For example, the seasonal factor for period 12 (4th
quarter of last year) would be 16,000/42,245 = 0.38. This and the seasonal factors for
all past quarters are shown in the previous table.

4. 76
(c) The forecasts using the seasonal factors from the last 4 quarters are as follows:
t St
Seasonal
factors Forecast
13 42,351 0.81 34,304
14 42,456 1.95 82,789
15 42,561 1.21 51,499
16 42,666 0.38 16,213
7
An exponential smoothing model is used to generate a forecast for period 13 (January of
next year). The sales for January through April are used to initialize the model, and an α
= 0.2 is used as the smoothing constant. The FORECAST module is used to generate the
forecast. The results are summarized as follows:
Region 1 Region 2 Region 3 Combined
Forecast, F13 219.73 407.04 303.30 938.26
Forecast error, SE 26.89 25.50 17.54 61.41
Note that the sum of the forecasts by region nearly equals the forecast of the combined
usage. However, whether a by-region forecast is better than an overall forecast that is
disaggregated by region depends on the forecast error. The standard error of the forecast
is the best indicator. A comparison of a bottoms-up forecast developed from regional
forecasts to that of a forecast from combined data can be based on the law of variances.
That is, if the usage rates within the regions are independent of each other, the estimate of
the total error can be built from the individual regions and compared to that of the
combined usage data. The total forecast error (variance) from the individual regions SC
2
might be estimated as the weighted average of the variances as follows:
S
F
F
S
F
F
S
F
F
SC
C
E
C
E
C
E
2 1 2 2 2 3 2
1 2 3
= + +
where
Fi = forecasts of each region
FC = forecast based on combined data
SEi
2
= variance of the forecast in each region
ST
2
= total variance of the forecast based on regional data
Therefore,
ST
2 2 2 2219 73
93007
2689
40704
93007
2550
30330
930 07
1754
0236 72307 0 438 650 25 0326 30765
55574
= + +
= × + × + ×
=
.
.
.
.
.
.
.
.
.
. . . . . .
.

5. 77
Then,
ST = =55574 2357. .
Since ST < SC, it appears that a bottom-up, or regional, forecast will have a lower error
than a top-down forecast.
9
(a) See the plot in Figure 8-1. It shows that there is a seasonal component with a very
slight trend to the data as well as some random, or unexplained, variation.
FIGURE 8-1 Plot of time series data for Problem 9
0
50
100
150
200
250
300
Jan
Apr
Jly
Oct
Jan
Apr
Jly
Oct
Jan
Apr
Jly
Oct
Jan
Apr
Jly
Oct
Jan
Apr
Jly
Oct
Averagemonthlyunitprices
Time, months
(b) A time series model typically will involve only two components: trend and
seasonality. Using 2 years of data should be sufficient to establish an accurate trend
line and the seasonal indices. We can develop the following table for computing a
regression line and seasonal indices.

6. 78
We also have N = 24, t = 300/24 = 12.5, and P = 5575/24 = 232.29.
Now,
b
P t N P t
t N t
=
× − × ×
− ×
=
−
−
= −
∑
∑ 2 2 2
69 678 24 23229 125
4 900 24 125
0 008
, ( . )( . )
, ( . )
.
and
a P b tt= − × = − − =232 29 0 008 125 23239. ( . )( . ) .
Therefore, the trend equation is:
Prices,
Pt
Time,
t P×t t2
Trend,a
Tt
Seasonal
indexb
St
211 1 211 1 232.4 0.91
210 2 420 4 232.4 0.90
214 3 642 9 232.4 0.92
208 4 832 16 232.4 0.90
276 5 1380 25 232.2 1.19
269 6 1614 36 232.3 1.16
265 7 1855 49 232.3 1.14
253 8 2024 64 232.3 1.09
244 9 2196 81 232.3 1.05
202 10 2020 100 232.3 0.87
221 11 2431 121 232.3 0.95
210 12 2520 144 232.2 0.90
215 13 2795 169 232.3 0.93 0.92
225 14 3150 196 232.2 0.97 0.93
230 15 3450 225 232.3 0.99 0.96
214 16 3424 256 232.3 0.92 0.91
276 17 4692 289 232.2 1.19 1.19
261 18 4698 324 232.2 1.12 1.14
250 19 4750 361 232.2 1.08 1.11
248 20 4960 400 232.2 1.07 1.08
229 21 4809 441 232.2 0.99 1.02
221 22 4862 484 232.2 0.95 0.91
209 23 4807 529 232.2 0.90 0.92
214 24 5136 576 232.2 0.92 0.91
5,575 300 69,678 4,900
a
Computed from the trend regression line. For example, the period 1 trend is T1 =
232.39 - 0.008×1 = 232.4.
b
The seasonal index is the ratio of the actual price to the trend for the same period.
For example, the period 1 seasonal index is 211/232 = 0.91.

7. 79
T tt = − ×232 29 0 008. .
Note that the trend is negative for the last two years of data, even though the 5-year
trend would appear to be positive.
Now, computing the trend value Tt for each value of t gives the results as shown
in the previous table. The seasonal index is a result of dividing Pt by Tt for each
period t. The indices are averaged for corresponding periods that are one year apart.
Forecasting into the 5th year shows the potential error in the method. That is, for
January of the 5th
year, the forecast is Ft = Tt×St-12, or F25 = [232.39 − 0.008×25][0.92]
= 213.6. Repeating for each method, we have:
t
Actual
price
Forecast
price
Forecast
error
Squared
error
Revised
seasonala
25 210 213.6 - 3.6 13.0 0.91
26 223 215.6 7.1 50.4
27 204 222.9 -18.9 357.2
28 244 211.3 32.7 1069.3
29 274 276.3 - 2.3 5.3
30 246 264.6 -18.6 345.9
31 237 257.7 -20.7 428.5
32 267 250.7 16.3 265.7
33 212 236.8 -24.8 615.0
34 211 211.2 - 0.2 0.0
35 188 213.5 -25.5 51.0
36 188 211.2 -23.2 538.2
Total squared error 3,739.5
a
The seasonal index for period 25 is .90. The average of the seasonal index for period 25 − 12 = 13,
and this period is (0.92 + 0.90)/2 = 0.91.
The standard error of the forecast is SF = − =3 7395 12 2 1934, . / ( ) . . Now, the forecast
for period 37 would be:
F37 23239 0 008 37 091 21121= − × =( . . )( . ) .
(c) Using the exponential smoothing module in the FORECAST software, the forecast
for the coming period is F = 201.26, with SF = 17.27. The smoothing constants given
in the problem are the "best" that FORECAST could find.
(d) Each model should be combined according to its ability to forecast accurately. We
can give each a weight in proportion to its forecast error, or standard error of the
forecast (SF). Hence, the following table can be developed:

8. 80
Model type
(1)
Forecast error
(2) = (1)/36.61
Proportion of
total error
(3)=1/(2)
Inverse of error
proportion
(4)=(3)/4.013
Model weights
Regression 19.34 0.528 1.894 0.472
Exp. smooth. 17.27 0.472 2.119 0.528
Total 36.61 1.000 4.013 1.000
Therefore, each of the model results is weighted according to the model weights.
The weighted forecast for the upcoming January would be:
Model type
(1)
Forecast
(2)
Model weight
(3)=(1)×(2)
Weighted
proportion
Regression 211.21 0.472 99.69
Exp. smooth. 201.26 0.528 106.27
Weighted forecast 205.96
In a similar fashion, we can weight the forecast error variances to come up with a
weighted forecast error standard deviation SFw. That is,
SFw = × + × =0 472 1934 0528 1727 18 282 2
. . . . .
A 95 percent confidence band using the combined results might be constructed as:
Y = 205.96 ± z×18.28
where z is 1.96 for 95 percent of the area under the normal distribution.
Y = 205.96 ± 1.96×18.28
Hence, we can be 95 percent sure that the actual price Y will be within the following
range:
170.13 ≤ Y ≤ 241.79
10
The plot of the sales data is shown in Figure 8-2. The plot reveals a high degree of
seasonality with a noticeable downward trend. A level-trend-seasonal model seems
reasonable.
(b) Using the search capability within the FORECAST software, a Level-Trend-Seasonal
form of the exponential smoothing model was found to give the lowest forecast error.
A 14-period initialization and 6 periods to compute error statistics were used. The
respective smoothing constants were α = 0.01, β = 0.08, and γ = 0.60. This produced

9. 81
a forecast for the upcoming period (January 2004) of F = 6,327.60 and a standard
error of the forecast of SF = 1,120.81.
FIGURE 8-2 Plot of Time Series Data for Hudson Paper Company
(c) Assuming that the forecast errors are normally distributed around the forecast, a 95
percent confidence band on the forecast is given by:
Y = F + z×SF
Y = 6,327.60 ± 1.96×1,120.81
where z = 1.96 for 95 percent of the area under the normal distribution curve.
Therefore, we can be 95 percent sure that the actual sales Y should fall within the
following limits:
4,130.8 ≤ Y ≤ 8,524.4
11
(a) For A569, the BIAS = −165,698 and the RMSE = 126,567 when using the 3-month
moving average. However, if a level only exponential smoothing model with an α =
0.10, the BIAS drops to –9,556 and the RMSE is 118,689. The model fits the data
better and there is a slight improvement in the forecasting accuracy.
For A366, the BIAS = 18,231 and the RMSE = 144,973 when using the 3-month
moving average. A level-trend-seasonal model offers the best fit, but it is suspect
since the data show a high degree of random variability rather than seasonality.
Overall, a simple level-only model is probably better in practice. The model has an α
= 0.08, a BIAS = −3,227, and a RMSE = 136,256. This is an improvement over the
3-month moving average.
0
50 00
1 00 0 0
1 50 0 0
2 00 0 0
2 50 0 0
3 00 0 0
Jan
Apr
Jly
Oct
Jan
Apr
Jly
Oct
Jan
Apr
Jly
Oct
Jan
Apr
Jly
Oct
Jan
Apr
Jly
Oct
Aggregatesalesin000s
T im e , m o nth s

10. 82
(b) Using the level-only models, the forecast for October for A569 = 193,230 and for
A366 = 603,671.
(c) The 3-sigma (99.7 percent) confidence band on the forecasts would be:
For A569, Y = 193,230 ± 3(118,689), or 0 ≤ Y ≤ 549,297.
For A366, Y = 603,671 ± 3(136,256), or 194,903 ≤ Y ≤ 1,012,439.
The actual October usage falls within the 3-sigma confidence bands for each of
these products. The difference of the actual from the forecast for each product is
attributable to the substantial variability in the data, which is characteristic of
purchasing in the steel processing industry.

11. 83
WORLD OIL
Teaching Note
Strategy
The purpose of this case study is to allow students to develop an appropriate forecasting
model for some time series data. Discussion may begin with the nature of this
productone with which most students should be very familiar. Based on the many
available forecasting approaches, students should be encouraged to select several for
consideration. In this note, both exponential smoothing and time series decomposition
are evaluated. Both are appropriate here because (1) they can project from historical time
series data, (2) they can handle seasonality, which appears to be present in the data, (3)
there is enough data to construct and test the models, and (4) the forecast is for a short
period into the future.
Assistance with the computational aspects of this problem is available with the use of
the FORECAST module in the LOGWARE software.
Answers to Questions
(1) Develop a forecasting procedure for this service station. Why did you select your
method?
Both exponential smoothing and time series decomposition forecasting methods are
tested using the FORECAST module in LOGWARE. For exponential smoothing, an
initialization period of one seasonal cycle (52 weeks) plus two weeks are used for a total
of 54 weeks, a minimum requirement in FORECAST. The last 30 weeks of data is used
for computing the error statistics. This number of periods is arbitrary, but seems
reasonably large so as to give stable statistical values. We wish to minimize the forecast
error over time, and FORECAST computes both MAD and RMSE statistics that can be
used to make comparisons among model types. Testing the various exponential
smoothing model types and the time series gives the following statistics.
Model type
Smoothing
constants
α β γ MAD BIAS RMSE
Forecast
week 6 of this
year
Level only.... .4 − − 37.82 -5.27 67.61 817.35
Level-trend... .2 .5 − 45.85 7.13 67.80 860.26
Level-seasonal .3 − 1.0 38.97 11.30 45.71 648.75
Level-trend-
seasonal...... .01 .2 .4 30.27 -6.05 44.17 770.74⇐
TS decomp..... − − − 59.46 37.18 71.85 731.33
The MAD and RMSE statistics show how well the forecast has been able to track
historical fuel usage rates. They are an indication of the accuracy of the forecasting
process in the future on the average. We favor forecasting methods that can minimize
these statistics. In this case, the Level-Trend-Seasonal version of the exponential
smoothing model seems to do this best. Both MAD and RMSE are the lowest for this
model type among the alternatives.

12. 84
Further evidence of the performance of a forecasting method is obtained from a plot
of the forecast against the actual usage rates. This is shown in Figure 1. Note that the
Level-Trend-Seasonal model tracks the usage rates quite well, especially in the more
recent weeks. The modeling process has likely stabilized in the last 30 weeks of the data
and is now tracking quite well.
FIGURE 1 Fit of Level-Trend-Seasonal Exponential Smoothing Model to Fuel
Usage Data on Mondays of the Week
(2) How should the periods of promotions, holidays, or other periods where usage rates
deviate from normal patterns, be handled in the forecast?
If the deviations occur at the same time within the seasonal cycle and with the same
relative intensity, no special procedures are required. The adaptive characteristic of the
exponential smoothing process will automatically incorporate these deviations into the
forecast. However, when the deviations are not regular, as promotions may be timed
irregularly, they may best be handled as outliers in the time series and eliminated from
the time series. The model may be fit without the outliers, and then the effect of them
treated as modifications to the forecast. These modifications can be handled manually.

13. 85
(3) Forecast next Monday's fuel usage and indicate the probable accuracy of the
forecast.
From the Level-Trend-Seasonal exponential smoothing model developed in question 1,
where the smoothing constants are α = 0.01, β = 0.2, and γ = 0.4, the forecast for Monday
of week 6 would be 771 gallons. However, this forecast only represents the average fuel
usage.
Determining the accuracy of the forecast requires that the forecast track the mean of
the actual usage, i.e., a bias of 0, and that the forecast errors be normally distributed.
While the BIAS (sum of the forecast errors over the last 30 weeks) is not exactly 0, and
will not likely ever be so, it is low (-6.05), such that we will assume good tracking by the
forecast model. A histogram of the forecast errors can reveal whether they follow the
familiar bell-shaped pattern. Such a histogram is given below. We can conclude that
while the errors are not precisely normally distributed, we cannot reject the idea that they
did not come from a normally distributed population. A goodness-of-fit test could be
used to check this assumption. Although this test is not performed here, it is quite
forgiving, such that the normal distribution of errors assumption is not likely to be
rejected where the data show a reasonably normal distribution pattern. The distribution
here qualifies.
We can now proceed with developing a 95 percent confidence band around the
forecast. The forecast of the actual fuel usage rate Y will be:
F z Y F zF F− ≤ ≤ +( $ ) ( $ )σ σ
where $σ F is the standard error of the forecast. F is the forecast, and z is the number of
standard deviations for 95 percent of the area under a normal distribution. FORECAST
computes the root mean squared error (RMSE) as:
RMSE
A F
N
t t
t
N
=
( )
=
−∑ 2
1

14. 86
Since RMSE is uncorrected for degrees of freedom lost, we apply a correction factor
(CF) as a multiplier to RMSE to get the unbiased estimate of the standard error of the
forecast ( Fσˆ ):
CF
N
N n
=
-
where n is the number of degrees of freedom lost in the model building process. We
estimate n to be the number of smoothing constants in the model, or three in this case.
Hence,
$
.
. .
.
σ F RMSE CF= ×
=
−
= ×
=
4417
30
30 3
4417 1054
4656
Now, with z@95%
= 1.96 from a normal distribution table, we can be 95 percent confident
that the true 87-octane fuel usage Y on Monday of week 6 will be:
771 − 1.96(46.56) < Y < 771 + 1.96(46.56)
680 < Y < 862 gallons
HISTOGRAM FOR FORECAST ERROR OF LAST 30 WEEKS
Class Width = 20.0000 Number of Classes = 10
0% 50% 100%
MID CLASS +----+----+----+----+----+----+----+----+----+----+
< -80.0000 | |
-70.0000 |****** |
-50.0000 |*** |
-30.0000 |******** |
-10.0000 |******** |
10.0000 |***** |
30.0000 |******** |
50.0000 |****** |
70.0000 |* |
90.0000 |* |
110.0000 | |
>= 120.0000 | |
+----+----+----+----+----+----+----+----+----+----+

15. 87
METRO HOSPITAL
You are the materials manager at Metro Hospital. Approximately one year ago, the
hospital began stocking a new drug (Ziloene) that helps the healing process for wounds
and sutures. It is your responsibility to forecast and order the monthly supply of Ziloene.
The goal is to minimize the combined cost of overstocking and understocking the drug.
Orders are placed and received at the beginning of the month and demand occurs
throughout the month. The following demand and cost data have been compiled.
Costs. If more is ordered than is demanded, a monthly holding cost of $1.00 per case
is incurred. If less is ordered than is demanded, a $2.00 per case lost sales cost is
incurred. The drug has a short shelf life, and any overstocked product at the end of the
month is worthless and no longer available to meet demand.
Demand. The demand for the twelve months of last year was:
Last year's demand
Month 1 2 3 4 5 6 7 8 9 10 11 12
Cases 43 36 24 69 34 75 90 67 59 51 77 50
You believe this demand to be representative of Metro's normal usage pattern.
FIGURE 1 Plot of last year's monthly demand in cases

16. 88
Decision Worksheet
Month
Cases
ordered
Actual
demand
Over @
$1/case
Short @
$2/case Cost, $
1 (13)
2 (14)
3 (15)
4 (16)
5 (17)
6 (18)
7 (19)
8 (20)
9 (21)
10 (22)
11 (23)
12 (24)
Total
Month
Cases
ordered
Actual
demand
Over @
$1/case
Short @
$2/case Cost, $
1 (25)
2 (26)
3 (27)
4 (28)
5 (29)
6 (30)
7 (31)
8 (32)
9 (33)
10 (34)
11 (35)
12 (36)
Total

17. 89
METRO HOSPITAL
Exercise Note
Purpose
Metro Hospital is an in-class exercise designed to illustrate the relationship between good
forecasting and the control of inventory related costs. It shows that accurate forecasting
is a primary factor in minimizing inventory costs. Participants in this exercise use a
variety of methods, often intuition, to forecast demand and to come up with a purchase
quantity. Their performance is measured as over- or understock costs. Using a simple
exponential smoothing forecasting model and an understanding of the standard deviation
of the forecast, an effective purchase plan can be constructed. This process results in
costs that are significantly lower than the majority of the participants are able to achieve
using intuitive methods.
Administration
The descriptive material and the decision worksheet are to be distributed to the class at
the time that the exercise is conducted. To hand out the material ahead of time may take
away much of the drama from the exercise. About one half hour should be scheduled for
running the exercise.
The instructor asks the class to make a decision regarding the size of the order to be
placed in the upcoming period and to record it on the worksheet. The participants are
then informed of the demand for that period from Table 1 after the simulated time of one
month has passed. Given that they now know the actual demand for the period, the
participants are asked to record their costs and then to place an order for the next period.
The pattern is repeated for at least twelve months, a full seasonal cycle. The participants
are asked to sum their costs and to report them to the exercise leader. They are displayed
in a public place, such as a chalkboard, for all to see. Then, the exercise leader
announces his or her cost level that was achieved using a disciplined approach using a
simple forecasting procedure and some basic statistics.
TABLE 1 Actual Demand for Period 13 Through 36
Period 13 14 15 16 17 18 19 20 21 22 23 24
Demand 47 70 55 38 90 24 65 65 23 55 85 66
Period 25 26 27 28 29 30 31 32 33 34 35 36
Demand 53 64 61 63 65 38 80 88 45 70 50 56
Quantitative Analysis
The demand series was generated using a normal distribution with a reasonably high
variance and a very slight upward trend. To illustrate the use of a quantitative approach
to forecasting, an exponential smoothing model was selected, although other methods
such as time series decomposition would also be appropriate. The twelve historical data
points were submitted to the FORECAST module in LOGWARE. A three-month
initialization period and a three-month time period for computing error statistics were
chosen. The smoothing constants for the level, level-trend, and level-trend-seasonal
models were examined. Based on the root mean squared error (RMSE), the best model

18. 90
was the level-trend-seasonal (RMSE = 13.59), but the level model with α = 0.19 and
RMSE = 13.89 performed very well and is used here. The model is:
ttt F.A.F 8101901 +=+
where
tF
tA
tF
t
t
t
periodcurrentforforecast
periodcurrentfordemandactual
1periodnextforforecast1
=
=
+=+
LOGWARE gives a forecast value of 58.1 and this is used as the forecast value for
period 13. Applying this simple, level only model to the second year demand as it is
revealed in each period gives the following forecast values:
TABLE 2 Simple Exponential
Smoothing Forecast Values
for the Next Year
Period
Actual
demand Forecast
13 47 58.1
14 70 56.0
15 55 58.7
16 38 58.0
17 90 54.2
18 24 61.0
19 65 54.0
20 65 56.1
21 23 57.8
22 55 51.2
23 85 51.9
24 66 54.6
Now we must determine the order quantity. It can be calculated from
z(RMSE)
t
FQ +
+
=
1
Recall the RMSE was 13.89 for this model. To be precise, we calculate z by trial and
error. The following order quantity and cost computations can be made for a z value of
0.8 (Table 3).

19. 91
TABLE 3 Purchase Order Quantity and Associated Inventory Costs
Period
Actual
demand Forecast
Order
quantity
Units
over
Units
short Cost, $
13 47 58.1 69*
22 22
14 70 56.0 67 3 6
15 55 58.7 70 15 15
16 38 58.0 69 31 31
17 90 54.2 65 25 50
18 24 61.0 72 48 48
19 65 54.0 65 0
20 65 56.1 67 2 2
21 23 57.8 69 46 46
22 55 51.2 62 7 7
23 85 51.9 63 22 44
24 66 54.6 66 0
271
*
Q = 58.1 + 0.8(13.89) = 69.21, or 69
We see from the following graph (Figure 1) that z = 0.8 is optimal.
FIGURE 1 Plot of Total Annual Costs Against the Factor z
265
270
275
280
285
290
295
300
305
310
0 0.2 0.4 0.6 0.8 1 1.2 1.4
Z
Cost,$

20. 92
Figure 2 graphically shows the good purchase pattern of Table 3.
FIGURE 2 Plot of Forecast and Purchase Order Quantity on Product Demand
0
10
20
30
40
50
60
70
80
90
100
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Time period
Cases
Order
quantity
Demand
Forecast
Summary
The exercise leader should discuss that one of the problems with intuitively forecasting
demand is overreacting to randomness in the demand pattern. This has the effect of
causing extreme over and short costs in inventories. A model for short term forecasting
that is integrated into the purchasing and inventory control process can help to avoid
these extremes and give lower costs. Several forecasting models may perform well, such
as exponential smoothing, a simple moving average, a regression model, or a times series
decomposition model. One of the most practical for inventory control purposes is the
exponential smoothing model. The results from a simple, level only model were
illustrated above using the same information that was available to the participants.
Recognizing that it is less costly to order too much than to order too little, the
purchase quantity should exceed the forecast by some margin. The astute participant will
likely approximate the standard deviation of demand from the range of the demand
values, that is, σ = (Max - Min)/6. Then, one or two σ might be used to add a margin of
safety to the forecast and size of the purchase order. This simple approximation
procedure can lead to reasonable results.