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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
NF
= = =
Total squared error 1 677 71376
6
528 79, , . . 
 
 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 
 
 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 
 
 
 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 41 666 6 5
650 12 6 5
1053, , , .
.
. 
 
 and 
 
 a S b t= − × = − × =41 666 1053 6 5 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. 
 
 
 
 
 
 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 
 S
Ei
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
930 07
26 89 407 04
930 07
2550 30330
930 07
17 54
0 236 72307 0 438 650 25 0 326 307 65
555 74
= + +
= × + × + ×
=
.
.
. .
.
. .
.
.
. . . . . .
.
 
 
 
 77
Then, 
 
 ST = =555 74 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
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150
200
250
300
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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. 
 
 
 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 24 232 29 12 5
4 900 24 12 5
0 008, ( . )( . )
, ( . )
. 
 
 and 
 
 a P b tt= − × = − − =232 29 0 008 12 5 232 39. ( . )( . ) . 
 
 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. 
 
 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 = − =3 739 5 12 2 19 34, . / ( ) . . Now, the forecast 
for period 37 would be: 
 
 F37 232 39 0 008 37 0 91 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: 
 
 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 19 34 0 528 17 27 18 28
2 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 
 
 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. 
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1 0 0 0 0
1 5 0 0 0
2 0 0 0 0
2 5 0 0 0
3 0 0 0 0
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T im e , m o n th s 
 
 
 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. 
 
 
 
 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. 
 
 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. 
 
 
 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 
 
 
 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
46 56
 
 
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 | |
+----+----+----+----+----+----+----+----+----+----+ 
 
 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 6759 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 
 
 
 
 
 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 
 
 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 
 
 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
 periodcurrent for forecast 
 periodcurrent for demand actual
1 periodnext for forecast 1
=
=
+=+
 
 
 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)tFQ ++= 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). 
 
 
 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
C
os
t, 
$
 
 
 
 
 
 
 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
C
as
es
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.