Chapter 3 Workload Assessment (Forecasting)
Review Questions
1.
What is suboptimization and why is it a concern of P/OM?
Answer: Suboptimization is doing less good than the best. It is a worse result than would
have been possible if the planner had known how to optimize. Often suboptimals occur
because the decision-maker is not aware of a better solution. However, the resources
including information may not be available. If suboptimization is planned, then it is hoped
that the result will be good enough and better than no optimization at all. The issue is often
related to achieving a local optimum such as the top of a nearby hill. In fact, the global
optimum may be Mt. Everest which is far away and unassailable to novice climbers. The
systems approach is a needed protection against serious suboptimization.
2.
What is meant by the statement ‘‘P/OM is a scarce resource’’?
Answer: “P/OM is a scarce resource” means that time and talent are limited. Good P/OM
professionals are often assigned more tasks than they can handle. For example, when a
company goes from one shift to two shifts without increasing the operations management
team, the resource has to be spread further. Any scarce resource deserves to be used to best
advantage. For operations management, this means including P/OM in formulation of
strategy.
3.
What is a time series? How would it be used to make predictions about emerging
technological developments?
Answer: A time series is the data of any variable classified by time, or in which the values of
the variable are functions of the time periods. One use of time series is to forecast a variable
in some future time period on the basis of that variable’s values in some past time periods. If
the variable were concerning the rate or extent of technological development, then the time
series might indicate future amounts of technological development. External causes are not
part of a time series. Thus, such an historical series would not be able to forecast the nature
or the success of any technological development. Historical forecasting is more suitable to
mature products than newly introduced products. If emerging technological developments
behave much like new product introductions, historical forecasting will be of little use.
1
4.
Is there a basis for predicting periods of prosperity and of economic slowdowns on the
basis of cycles?
Answer: Cycles are one form of time series often used by business for other events related to
that time series (e.g., sales). Some periods of prosperity and slowdowns, such as that which is
associated with tourism activity, may follow seasonal cycles, and therefore it is predictable
from cycle histories. For other types of events, any basis for prediction would be found in the
extent to which the cycles are repeated in the data history. In the larger economic sense, we
have ample evidence that recessions are seemingly always a surprise.
5. What are life cycle stages? Why does the concept of life cycle stages unite P/OM with
other members of the organization in using the systems approach?
Answer: Life cycle stages are four stages in new product development that appear in a
regular way over time, and that all products and services go through. They provide a
classification for understanding the kinds of trends that can be expected in demand. Life
cycle stages have different effects on various parts of the organization, and thus have systems
implications. Marketing uses life cycle stages to select appropriate pricing, advertising, and
promotion activities; operations uses life cycle stages to adjust the production system’s
capabilities. Finance must adjust to the way in which revenues and profits vary with the life
cycle stage.
6. Detail the life cycle stages of new products and explain them for both goods and services.
Answer: The four life cycle stages are (1) introduction to the market, which is preceded by
the idea of the product, and by feasibility studies and design; (2) growth of volume and share;
(3) maturation, which ultimately results in market saturation, (i.e., the new product stops
growing; market shares are stable; organizations wish to prolong this profitable stage); and
(4) decline or decay, leading to restaging or withdrawal, when updates or new products
replace retiring ones.
7. What is extrapolation and how is it used?
Answer: Extrapolation is the process of moving from observed data to unknown values or
future points. It is the main function of forecasting. To base a sales forecast for 2016 on
2
historical data from 2013, 2014, and 2015 would be an extrapolation from the historical to
the unknown. If the relationship between two variables is known over a certain range of
values (for example, the effects of temperature and pressure on the strength of a material),
then it is an extrapolation to extend the relationship beyond the current experience (to
determine the material’s strength if subjected to temperatures that have not been tested).
8. What is correlation? What is autocorrelation? Distinguish between them.
Answer: Correlation is the generally imperfect relation between two variables, measuring
the extent to which the variables move together (one increases as the other increases), move
opposite (one decreases as the other increases), or have no association (as one increases,
there is no suggestion as to how the other will behave). Autocorrelation is the extent to
which values in a data set are correlated with one another; in this case there is only one
variable, not two, but the time series covers two different periods, e.g., 1, 2 and 3 with 9, 10,
and 11.
9. What is an historical forecast? When is it used?
Answer: Historical forecasting is based on the assumption that a pattern observed by
studying the past behavior of a variable will be repeated in the future. This method works if
a stable pattern exists, such as with seasonal events like rainfall, tourism, or seasonal buying.
10. Why are long life cycle stages considered desirable? Does this apply to start-up and
growth as well as the mature stage of new products?
Answer: The longer the maturity cycle, the longer the organization has to earn profits to
support new product development, and the longer the organization has to recover costs of
assembly lines and other investments made in a product. Introductory and declining stages
are not profitable; hence, lengthening these stages is not advantageous. Recent experience
suggests shorter life cycles overall, resulting in shorter maturity stages and more frequent
product replacement.
11. When are multiple forecasts desirable?
Answer: Usually, stronger forecasts can be obtained if both data and experience are
combined. For example, regression may be combined with use of the Delphi technique.
Multiple forecasts will reveal that some techniques work better than others. There are formal
3
methods for measuring the accuracy of forecasts. Failures should be reworked or discarded;
successes should be repeated. Finally, it may be possible to average the results of different
forecasts. This is especially useful when the most accurate single technique is ambiguous or
unstable—combine first one method—and then another. In other words, the goal of multiple
forecasts is to get better forecast results, and that can be based on the track record.
4
Problems
1: The larger the number of periods in the simple moving average forecasting method, the greater
the method's responsiveness to changes in demand. (True or False)?
Answer: False. The forecast will be smoother with a larger number of periods.
2: The coefficient of correlation can never be negative. (True or False)?
Answer: False. The coefficient of correlation goes from -1 to +1.
3: Most forecasting techniques assume that there is some underlying stability in the system.
(True or False)?
Answer: True. If the system is not stable, the forecast will not be dependable.
4: The percent of variation in the dependent variable that is explained by the regression equation
is measured by which one of the following? (a) mean absolute deviation, (b) slope, (c)
coefficient of determination, (d) correlation coefficient, or (e) intercept.
Answer: (c) coefficient of determination. Read the relevant section.
5: Time series data may exhibit which of the following behaviors? (a) trend, (b) random
variations, (c) seasonality, (d) all of these.
Answer: (d).The time series may consist of all components listed above. Read the relevant
section.
Note: See Excel File SMCh03 for solutions to problems 6 to 10.
6: Weekly sales of bread at a local Foods Market are given in the table below. Forecast sales
for week 7 using a three-week moving average.
Week
Sales
Moving
Average
Answer
1
402
2
385
3
420
4
382
5
410
6
432
7
385
8
411
402.33 395.67 404.00 408.00 409.00
408.00
9
409.33
Forecasts for other months are also given in the table above.
5
7: Weekly sales of ten-grain bread at the local Whole Foods Market are given in the table
below. Forecast demand for week 6 using a four-week moving average.
Week
Sales
Moving
Average
Answer
1
415
2
389
3
420
4
382
5
410
6
432
7
380
8
410
401.50 400.25 411.00 401.00
400.25
9
408.00
Forecasts for other months are also given in the table above.
8: What is the forecast for May based on a weighted moving average applied to the following
past demand data and using the weights: .7, .2, .1 (largest weight is for most recent data)?
Weights
Month
Demand
WMA
Forecast
Answer
0.1
Nov.
37
0.2
Dec.
36
43.7
0.7
Jan.
40
Feb.
42
Mar.
47
April
43
May
38.9
41
45.3
43.7
Forecasts for other months are also given in the table above.
9: What is the forecast for April based on a weighted moving average applied to the following
past demand data and using the weights: .6, .3, .1 (largest weight is for most recent data)?
Weights
Month
Demand
WMA
Forecast
Answer
0.1
Nov.
37
42.9
0.3
Dec.
40
0.6
Jan.
36
Feb.
47
Mar.
42
April
43
May
37.3
43
42.9
43.1
Forecasts for other months are also given in the table above.
6
10: What is the forecast for May based on a weighted moving average applied to the following
past demand data and using the weights: 0.5, 0.3, 0.2 (largest weight is for most recent data)?
Weights
Month
Demand
WMA
Forecast
Answer
0.2
Jan
137
0.3
Feb
140
141.3
0.5
Mar
136
Apr
145
May
138
Jun
144
July
137.4
141.3
139.7
142.4
Forecasts for other months are also given in the table above.
11: Given an actual demand of 110, a previous forecast value of 120, and an  = .3, the
exponential smoothing forecast for the next period would be ______.
Answer: 117
Forecast (t) = Forecast (t-1) + α*{Actual Demand (t-1) – Forecast (t-1)}, 0 ≤ α ≤ 1.
Forecast = 120 + 0.3*(110-120) = 120 – 3 = 117.
12: Given an actual demand of 103, a previous forecast value of 99, and an  = .4, the
exponential smoothing forecast for the next period would be _______.
Answer: 100.6
Forecast (t) = Forecast (t-1) + α*{Actual Demand (t-1) – Forecast (t-1)}, 0 ≤ α ≤ 1.
Forecast = 99 + 0.4*(103-99) = 99 + 1.6 = 100.6.
13: For a given product demand, the time series trend equation is 25 + 3.2 X. What is the
demand forecast for period 10?
Answer: 57.
Forecast = 25 + 3.2 x 10 = 57
7
14: The demands for an item for the 12 months of year 2014 are given below.
Period (T) 1
Month
Jan
Demand
151
2
Feb
152
3
4
5
6
Mar Apr May Jun
132 161 182 192
7
8
9
Jul Aug Sep
174 159 183
10
11
12
Oct Nov Dec
169 175 181
The trend projection equation is Y = 148.97 + 2.864T where T is the number of the month. Make
a forecast for June of 2015.
Answer: 200.52
The value of T to be used in the trend equation is 18. June is the 6th month of 2015.
Y = 148.97 + 2.864*18 = 200.52
15: A forecasting method has produced the following data over the past five months. What is the
mean absolute deviation (MAD)?
Actual
10
8
10
6
9
Forecast
11
10
8
6
8
Error
-1
-2
2
0
1
|Error|
1
2
2
0
1
Answer: 1.2
Sum of absolute errors = 1 + 2 + 2 + 0 + 1 = 6
Number of periods = 5
MAD = 6/5 = 1.2
8
16: For the data given below, is the forecasting system overestimating or underestimating the
demand? (a): overestimating (b): underestimating
Actual
10
8
10
6
9
Forecast
11
10
8
6
8
Error
-1
-2
2
0
1
|Error|
1
2
2
0
1
Answer: Neither underestimate nor overestimate.
The demand is being underestimated since the value of MAD is positive (1.2).
We will have to calculate the value of tracking signal to answer this questions as shown below.
Sum of absolute errors = 1 + 2 + 2 + 0 + 1 = 6
Number of periods = 5
MAD = 6/5 = 1.2
Running Sum of Forecast Errors (RSFE) = -1 + (-2) + 2 + 0 + 1 = 0
RSFE is the sum of all errors.
Tracking Signal = RSFE/MAD = 0/1.2 = 0.
It may be noted that the demand is being overestimated if tracking signal is negative; and is
being underestimated if the tracking signal is positive.
17: A forecasting method has produced the following data over the past six months. What is the
mean absolute deviation (MAD) at the end of six months?
Month
1
2
3
4
5
6
Actual
10
8
10
6
9
11
Forecast
9
10
8
6
8
12
Error
1
-2
2
0
1
|Error|
1
2
2
0
1
Answer: 1.17
Error for month 6 will be -1 (11-12). Absolute error for month 6 is 1.
Sum of absolute errors = 1 + 2 + 2 + 0 + 1 + 1= 7
Number of periods = 6
MAD = 7/6 = 1.167
9
18: If the coefficient of correlation is – 0.8, what is the percent of variation in the dependent
variable that is explained by the regression equation?
Answer: 64%
Coefficient of determination = (- 0.8)2 = .64 or 64%.
The value of the coefficient of determination explains the variation in the dependent variable due
to variation in the independent variable. It may be noted that the coefficient of determination is
the square of the coefficient of correlation.
19: If you were selecting a forecasting model based on MAD, which of the following MAD
values reflects the most accurate model?
(A) 0.2 (B) 0.8 (C) 2.0 ( D) 4.5 (E) 100.
Answer: A (0.2)
Choose the forecasting model that has the minimum value of MAD.
Note: See Excel File SMCh03 for solutions to problems 20 to 23.
There is a table of data for each problem that is supplied in the text and the spreadsheet matrix
below captures all that data.
20: Compute a weighted 3-month moving average for month 7 by using the following weights: 0.5, 0.3,
and 0.2. The weight is highest for the latest month.
Weights
Month
Actual
Demand
WMA
Forecast
Answer
0.2
1
0.3
2
0.5
3
4
5
6
7
8
9
700
645
660
648
655
760
682
670
756
663.5
651
653.9
706.1
700
691.6
706.1
10
715.4
Forecasts for other months are also given in the table above.
21: The demand forecast for month 7 by using a simple 3-month moving average will be ___.
Month
Actual
Demand
WMA
Forecast
Answer
1
2
3
4
5
6
7
8
9
700
645
660
648
655
760
682
670
756
668.33 651.00 654.33 687.67 699.00
687.67
704.00
Forecasts for other months are also given in the table above.
10
10
702.67
22: Find the forecast for month 9 by using exponential smoothing with a = 0.3.
Month
Actual
Demand
Alpha
Forecast
Answer
1
2
3
4
5
6
7
8
9
700
645
660
648
655
760
682
670
756
0.3
700.00 700.00 684
676
668
664
693
690
684
684
Forecasts for other months are also given in the table above.
10
705
23: Find the value of Mean Absolute Deviation (MAD) at the end of month 8 by using the demand and
forecast data in the table given below.
Month
Actual
Demand
Forecast
Error
Absolute
Error
Cumulative
Error
MAD
Answer
1
2
3
4
5
6
7
8
9
700
645
660
648
655
760
682
670
756
720
-20
675
-30
635
25
587
61
702
-47
718
42
740
-58
753
-83
20
30
25
61
47
42
58
83
20
50
75
136
183
225
283
366
20
45.75
25
25
34
36.6
37.5 40.429 45.75
MAD for other months are also given in the table above.
24: Which of the following statements is true on the basis of the data given in the above table?
a. The demand is being overestimated.
b. The demand is being underestimated.
Answer: b. The demand is being overestimated.
The demand is being overestimated since the value of tracking signal is negative (-2.404).
See the calculations below.
MAD = 45.75 (See Q 23)
RSFE = -110 (Add all error terms in Q 23) = -20 – 30 +25 +61 -47 +42 -58 -83.
Tracking Signal = RSFE/MAD = -110/45.75 = -2.404.
See discussion of tracking signal in Q 16.
11
25: In regression analysis, you determine the intercept value of the model to be 1,000 and the
slope value to be 50. What is the resulting forecast value using this model if the value of the
independent variable X is 12.
Answer: 1,600
Forecast (Y) = 1,000 + 50*12 = 1,600
26: The following trend projection, based on past 36 months (years 2014, 2015, and 2016) data
is used to predict monthly demand:
Y = 250 + 1.5t, where t = 1 is the first month of year 2014 (January 2014)
What is the seasonally adjusted forecast for March 2017? It is given that the seasonal (monthly)
index for March is 0.8.
Answer: 246.8
Forecast (Y) = 250 + 1.5*39 = 308.5.
Seasonally adjusted forecast = 308.5*0.8 = 246.8.
Note: The value of t in the forecasting equation is 39. March 2017 will be the 39th period if
January 2014 is the first period.
27: The following trend projection, based on past 12 quarters (years 2014, 2015, and 2016) data
is used to predict quarterly demand:
Y = 250 - 2.5t, where t = 1 for the first quarter of 2014.
Seasonal (quarterly) indices are Quarter 1=1.5; Quarter 2=0.8; Quarter 3=1.1; and Quarter 4=0.6.
What is the seasonally adjusted forecast for the second quarter of 2017?
Answer: 172
Forecast (Y) = 250 - 2.5*14 = 215.
Seasonally adjusted forecast = 215*0.8 = 172.
Note: The value of t in the forecasting equation is 14. Second quarter of 2017 will be the 14th
period when quarter 1 of 2014 is period 1.
12
Note: See Excel File SMCh03 for solutions to problem 28.
28: The demand data for an item for three years is given in the
Table below. Find the forecast for the Winter quarter of the
fourth year using the seasonal forecasting model. Total demand
in the fourth year is expected to be 2212 units.
Quarter
Fall
Winter
Spring
Summer
Answer
Year 1
110
240
600
450
Year 2
135
355
750
540
Year 3
180
500
830
610
Year 2
135
355
750
540
1780
445
Year 3
180
500
830
610
2120
530
447.4
Solution:
Quarter
Fall
Winter
Spring
Summer
Total
Average
Year 1
110
240
600
450
1400
350
Quarter
Fall
Winter
Spring
Summer
Find Seasonal Index (SI)
Year 1
Year 2
Year 3
0.314
0.303
0.340
0.686
0.798
0.943
1.714
1.685
1.566
1.286
1.213
1.151
Total Demand for next year
Average Demand per quarter
2212
553
Average SI
0.319
0.809
1.655
1.217
Given
Forecast for Next Year
Quarter
Forecast
Fall
176.5
Winter
447.4
Spring
915.3
Summer
672.8
13
Note: See Excel File SMCh03 for solution to problem 29.
29. The Highway Department is considering using a 6-month moving average to forecast crew hours needed to repair
roads month by month for the coming year. To test whether this method is accurate, use last year’s data (shown in the
following table) to predict the last six months’ crew hours to compare with the observed values. Start with the
average of January through June to compare with the actual value of 200 for July and so on through December.
Month
Crew
Hours
Jan.
Feb.
March
April
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
110
120
140
180
250
200
200
220
280
120
100
80
6-Month
Moving
Average
166.7
181.7
198.3
221.7
211.7
186.7
Error
Absolute
Error
33.3
38.3
81.7
-101.7
-111.7
-106.7
Total
MAD
33.3
38.3
81.7
101.7
111.7
106.7
473.3
78.9
Answer: 6-month moving averages are shown in the table above. The error terms are also shown
in the table.
30. Referring to problem 29, prepare an analysis of the errors in using the 6-month moving
average as a predictor.
Answer: The errors are positive for July, August and September and negative for the last three
months. The positive error means that the demand is being underestimated whereas the negative
error indicated an overestimation of demand. The value of mean absolute deviation (MAD) is
78.9. Six months moving average appears to have too much inertia to provide close
approximations. We will try other options to compare with this large MAD.
It is necessary to compare this method (6-month moving average) with other methods to
determine the best method of forecasting for this problem. The next 2 problems do look at other
methods.
14
Note: See Excel File SMCh03 for solutions to problems 31 to 33.
31. Referring to Q 29 and Q 30, use a 3-period moving average to forecast the last six months
and compare directly to the 6-month forecast developed there. Here, the first forecast is the
average of April, May, and June to compare with the actual value of 200 for July and so on
through December.
Month
Crew
Hours
Jan.
Feb.
March
April
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
110
120
140
180
250
200
200
220
280
120
100
80
3-Month
Moving
Average
210.0
216.7
206.7
233.3
206.7
166.7
Error
Absolute
Error
-10.0
3.3
73.3
-113.3
-106.7
-86.7
Total
MAD
10.0
3.3
73.3
113.3
106.7
86.7
393.3
65.6
Answer: The value of MAD for 3-month moving average is 65.6. MAD (3) is an improvement
over MAD (6). MAD (6) is 78.9 as calculated in problem 29. None of the error terms are
satisfactory. This was to be expected, since the data were created to be volatile. Momentum
seemed evident in the first four months and then the system became sporadic and difficult to
predict. Moving average and weighted moving average (WMA) do a better job when there is
stability in the system’s changes from month to month.
15
32. Referring to Q 29, Q 30 and Q 31, use a 6-period weighted moving average to forecast the last six months and
compare directly to the un-weighted forecasts for 6-month and 3-month periods. Use the weights 0.1, 0.1, 0.1, 0.2,
.0.2, and 0.3 with the 0.3 weighting the most recent datum point.
Month
Crew
Hours
Jan.
Feb.
March
April
May
June
July
Aug.
Sept.
Oct.
Nov.
Dec.
110
120
140
180
250
200
200
220
280
120
100
80
6-period
Weighted
Moving
Average
Error
Absolute
Weights
Error
0.1
0.1
0.1
0.2
0.2
0.3
183.0
194.0
203.0
231.0
201.0
172.0
17.0
26.0
77.0
-111.0
-101.0
-92.0
Total
MAD
17.0
26.0
77.0
111.0
101.0
92.0
424.0
70.7
Solution:
The weights given are: {0.1, 0.1, 0.1, 0.2, 0.2, 0.3}. Students should check that they sum to one.
The 0.3 applies to the last month in the time series. The 6-month weighted moving average
prediction for July is the weighted average of January through June, i.e., [(0.1)110 + . . . +
(0.3)200]/6 = 183. The error for July is 200 – 183 = +17.
The value of MAD is 70.7. The MAD statistics is large and on a par with the other moving
average statistics. None seem to do a good job, which is to be expected since the data are
volatile and moving averages are better for less volatile data.
16
33. The following table presents data concerning the sales of minivans and minivan tires for 10
years. It is believed that the demand for minivan replacement tires is highly correlated with the
sales figures of minivans for the previous years. Based on these data which is a better predictor of
minivan tire sales - sales of minivans three years prior or four years prior?
Year
(t)
1
2
3
4
5
6
7
8
9
10
Minivan
Sales in
Year t (in
millions)
10
12
11
9
10
12
10
9
8
7
Minivan Tire
Sales in Year t
(in millions)
4
6
7
5
8
7
5
7
8
6
See Excel file SMCh03 for detailed calculations.
Answer: Regression analysis will be used to solve this problem. Sales of minivans is
independent variable and sales of minivan tires is dependent variable.
(a) Regression analysis when forecast of minivan tire sales is based on minivan sales in prior
three-year period.
Sale of minivan tires in period t is based on sale of minivans in period t-3. For example, sale of
tires in year 4 is based on the sales of minivans in year 1.
In the regression equation the sales of minivans in years 1 through 7 will be used as independent
variables and sales of minivan tires for years 4 through 10 will be used as dependent variables.
Using Excel functions, the value of intercept (a) = - 4.00; and the value of slope (b) = 1.00
Regression line: yt = – 4 + xt-3
y11 = – 4 + x8 = – 4 + 9 = 5.
y12 = – 4 + x9 = – 4 + 8 = 4.
y13 = – 4 + x10 = – 4 + 7 = 3.
The value of the coefficient of correlation, r, = 0.891 indicating a strong positive correlation and
a reasonable fit.
17
(b) Regression analysis when forecast of minivan tire sales is based on minivan sales in prior
four-year period.
Sale of minivan tires in period t is based on sale of minivans in period t-4. For example, sale of
tires in year 5 is based on the sale of minivans in year 1.
In the regression equation the sales of minivans in years 1 through 6 will be used as independent
variables and sales of minivan tires for years 5 through 10 will be used as dependent variables.
Using Excel functions, the value of intercept (a) = 11.682; and the value of slope (b) = -0.455.
Regression line: yt = 11.682 – (0.455)xt-4
y11 = 11.682 – (0.455)x7 = 11.68 – (0.455)10 = 7.14
y12 = 11.682 – (0.455)x8 = 11.68 – (0.455)9 = 7.59
y13 = 11.682 – (0.455)x9 = 11.68 – (0.455)8 = 8.05
y14 = 11.682 – (0.455)x10 = 11.68 – (0.455)7 = 8.50
The value of the coefficient of correlation, r, = – 0.471, indicating a poor fit.
It can be concluded that the three year analysis is quite good but extending it to four years it goes
out of alignment.
18