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C H A P T E R
Forecasting
DISCUSSION Q UESTIONS
1. Qualitative models incorporate subjective factors into the
forecasting model. Qualitative models are useful when subjective
factors are important. When quantitative data are difficult to ob-
tain, qualitative models may be appropriate.
2. Approaches are qualitative and quantitative. Qualitative is
relatively subjective; quantitative uses numeric models.
3. Short-range (under 3 months), medium-range (3 to 18 months),
and long-range (over 18 months).
4. The steps that should be used to develop a forecasting
system are:
(a) Determine the purpose and use of the forecast
(b) Select the item or quantities that are to be forecasted
(c) Determine the time horizon of the forecast
(d) Select the type of forecasting model to be used
(e) Gather the necessary data
(f) Validate the forecasting model
(g) Make the forecast
(h) Implement the results
(i) Evaluate the results
5. Any three of: sales planning, production planning and budget-
ing, cash budgeting, analyzing various operating plans.
6. There is no mechanism for growth in these models; they are
built exclusively from historical demand values. Such methods
will always lag trends.
7. Exponential smoothing is a weighted moving average where
all previous values are weighted with a set of weights that decline
exponentially.
8. MAD, MSE, and MAPE are common measures of forecast
accuracy. To find the more accurate forecasting model, forecast
with each tool for several periods where the demand outcome is
known, and calculate MSE, MAPE, or MAD for each. The
smaller error indicates the better forecast.
9. The Delphi technique involves:
(a) Assembling a group of experts in such a manner as to pre-
clude direct communication between identifiable members
of the group
(b) Assembling the responses of each expert to the questions
or problems of interest
(c) Summarizing these responses
(d) Providing each expert with the summary of all responses
(e) Asking each expert to study the summary of the responses
and respond again to the questions or problems of interest.
(f) Repeating steps (b) through (e) several times as necessary
to obtain convergence in responses. If convergence has
not been obtained by the end of the fourth cycle, the re-
sponses at that time should probably be accepted and the
process terminated—little additional convergence is
likely if the process is continued.
10. A time series model predicts on the basis of the assumption
that the future is a function of the past, whereas a causal model
incorporates into the model the variables of factors that might
influence the quantity being forecast.
11. A time series is a sequence of evenly spaced data points with the
four components of trend, seasonality, cyclical, and random variation.
12. When the smoothing constant, D, is large (close to 1.0),
more weight is given to recent data; when D is low (close to 0.0),
more weight is given to past data.
13. Seasonal patterns are of fixed duration and repeat regularly.
Cycles vary in length and regularity. Seasonal indexes allow
“generic” forecasts to be made specific to the month, week, etc.,
of the application.
14. Exponential smoothing weighs all previous values with a set
of weights that decline exponentially. It can place a full weight on
the most recent period (with an alpha of 1.0). This, in effect, is the
naïve approach , which places all its emphasis on last period’s
actual demand.
15. Adaptive forecasting refers to computer monitoring of track-
ing signals and self-adjustment if a signal passes its present limit.
16. Tracking signals alert the user of a forecasting tool to peri-
ods in which the forecast was in significant error.
17. The correlation coefficient measures the degree to which the
independent and dependent variables move together. A negative
value would mean that as X increases, Y tends to fall. The vari-
ables move together, but move in opposite directions.
18. Independent variable ( x ) is said to cause variations in the
dependent variable ( y ).
19. Nearly every industry has seasonality. The seasonality must
be filtered out for good medium-range planning (of production
and inventory) and performance evaluation.
20. There are many examples. Demand for raw materials and
component parts such as steel or tires is a function of demand for
goods such as automobiles.
21. Obviously, as we go farther into the future, it becomes more
difficult to make forecasts, and we must diminish our reliance on
the forecasts.
ETHICAL DILEMMA
This exercise, derived from an actual situation, deals as much with
ethics as with forecasting. Here are a few points to consider:
No one likes a system they don’t understand, and most
college presidents would feel uncomfortable with this one.
It does offer the advantage of depoliticizing the funds al-
location if used wisely and fairly. But to do so means all
parties must have input to the process (such as smoothing
constants) and all data need to be open to everyone.
The smoothing constants could be selected by an agreed-
upon criteria (such as lowest MAD ) or could be based on
input from experts on the board as well as the college.
Abuse of the system is tied to assigning alphas based on
what results they yield, rather than what alphas make the
most sense.
Regression is open to abuse as well. Models can use many
years of data yielding one result, or few years yielding a
totally different forecast. Selection of associative variables
can have a major impact on results as well.
ACTIVE MODEL EXERCISES
ACTIVE MODEL 4.1: Moving Averages
1. What does the graph look like when n = 1
The forecast graph mirrors the data graph but one period
later.
2. What happens to the GRAPH as the number of periods in the
moving average increases?
The forecast graph becomes shorter and smoother.
3. What value for n minimizes the MAD for this data?
n = 1 (a naive forecast)
ACTIVE MODEL 4.2: Exponential Smoothing
1. What happens to the graph when alpha equals zero?
The graph is a straight line. The forecast is the same in
each period.
2. What happens to the graph when alpha equals one?
The forecast follows the same pattern as the demand (ex-
cept for the first forecast) but is offset by one period. This is a
naive forecast.
3. Generalize what happens to a forecast as alpha increases.
As alpha increases the forecast is more sensitive to
changes in demand.
4. At what level of alpha is the mean absolute deviation (MAD)
minimized?
Alpha =.
ACTIVE MODEL 4.3: Exponential Smoothing with
Trend Adjustment
1. Scroll through different values for alpha and beta. Which
smoothing constant appears to have the greater affect on the graph?
Alpha
2. With beta set to zero, find the best alpha and observe the
MAD. Now find the best beta. Observe the MAD. Does the addi-
tion of a trend improve the forecast?
Alpha = .11, MAD = 2.59; Beta above .6 changes the MAD
(by a little) to 2.54.
ACTIVE MODEL 4.4: Trend Projections
1. What is the annual trend in the data?
2. Use the scrollbars for the slope and intercept to determine the
values that minimize the MAD. Are these the same values that
regression yields?
No they are NOT the same values. For example, an inter-
cept of 57.81 with a slope of 9.44 yields a MAD of 7.17.
END- OF-CHAPTER PROBLEMS
(a) 374.33 pints
(b)
Week of Pints Used
Weighted Moving Average August 31 360 September 7 389 381 u .1 = 38. September 14 410 368 u .3 = 110. September 21 381 374 u .6 = 224. September 28 368 372. October 5 374 Forecast 372.
(c)
The forecast is 374.26.
Week of Pints Forecast
Forecasting Error
Error u .20 Forecast August 31 360 360 0 0 360 September 7 389 360 29 5.8 365. September 14 410 365.8 44.2 8.84 374. September 21 381 374.64 6.36 1.272 375. September 28 368 375.912 –7.912 –1.5824 374. October 5 374 374.3296 –.3296 –.06592 374.
4.2 (a) No, the data appear to have no consistent pattern.
Year 1 2 3 4 5 6 7 8 9 10 11 Forecast Demand 7 9 5 9.0 13.0 8.0 12.0 13.0 9.0 11.0 7.
(b) 3-year moving 7.0 7.7 9.0 10.0 11.0 11.0 11.3 11.0 9.
(c) 3-year weighted 6.4 7.8 11.0 9.6 10.9 12.2 10.5 10.6 8.
(b) x Naive The coming January = December = 23
x 3-month moving (20 + 21 + 23)/3 = 21.
x 6-month weighted (0.1 u 17) + (.1 u 18)
+ (0.1 u 20) + (0.2 u 20)
+ (0.2 u 21) + (0.3 u 23) = 20.
x Exponential smoothing with alpha = 0.
Oct Nov Dec Jan
F
F
F
F
x Trend ¦ x 78, x 6.5, ¦ y = 218, y 18.
2
b
a
Forecast = 15.73 + .38(13) = 20.67, where next January
is the 13th month.
(c) Only trend provides an equation that can extend beyond
one month
4.7 Using MAD for this problem,
(3) Marketing (5) Operations (1) (2) Marketing VP’s Error Operations VP’s Error Year Sales VP’s Forecast [(2)–(3)] VP’s Forecast [(2)–(5)] 1 167,325 170,000 2,675 160,000 7, 2 175,362 170,000 5,362 165,000 10, 3 172,536 180,000 7,464 170,000 2, 4 156,732 180,000 23,268 175,000 18, 5 176,325 165,000 11,325 165,000 11, Totals 50,094 49,
MAD (marketing VP) = 50,094/5 = 10,018.8. MAD (operations VP) = 49,816/5 = 9,963.2. Therefore, based on past data, the VP of operations has been presenting better forecasts.
(a) 91.
(b) 89
(c)
Temperature 2 day M.A. |Error| (Error)^2 Absolute % Error 93 — — — — 94 — — — — 93 93.5 0.5 0.25 100(.5/93) = 0.54% 95 93.5 1.5 2.25 100(1.5/95) = 1.58% 96 94.0 2.0 4.00 100(2/96) = 2.08% 88 95.5 7.5 56.25 100(7.5/88) = 8.52% 90 92.0 2.0 4.00 100(2/90) = 2.22% 13.5 66.75 14.94% MAD = 13.5/5 = 2.
(d) MSE = 66.75/5 = 13.
(e) MAPE = 14.94%/5 = 2.99%
4.9 (a, b) The computations for both the two- and three-month
averages appear in the table; the results appear in the
figure below.
(c) MAD (two-month moving average) = .750/10 =.
MAD (three-month moving average) = .793/9 =.
Therefore, the two-month moving average seems to have per-
formed better.
Table for Problem 4.9 (a, b, c)
Forecast (^) |Error| Two-Month Three-Month Two-Month Three-Month Price per Moving Moving Moving Moving Month Chip Average Average Average Average January $1. February 1. March 1.70 1.735. April 1.85 1.685 1.723 .165. May 1.90 1.775 1.740 .125. June 1.87 1.875 1.817 .005. July 1.80 1.885 1.873 .085. August 1.83 1.835 1.857 .005. September 1.70 1.815 1.833 .115. October 1.65 1.765 1.777 .115. November 1.70 1.675 1.727 .025. December 1.75 1.675 1.683 .075. Totals .750.
(c) The forecasts are about the same.
4.9 (d) Table for Problem 4.9(d).
D = .1 D = .3 D =.
Month Price per Chip Forecast |Error| Forecast |Error| Forecast |Error| January $1.80 $1.80 $.00 $1.80 $.00 $1.80 $. February 1.67 1.80 .13 1.80 .13 1.80. March 1.70 1.79 .09 1.76 .06 1.74. April 1.85 1.78 .07 1.74 .11 1.72. May 1.90 1.79 .11 1.77 .13 1.78. June 1.87 1.80 .07 1.81 .06 1.84. July 1.80 1.80 .00 1.83 .03 1.86. August 1.83 1.80 .03 1.82 .01 1.83. September 1.70 1.81 .11 1.82 .12 1.83. October 1.65 1.80 .15 1.79 .14 1.76. November 1.70 1.78 .08 1.75 .05 1.71. December 1.75 1.77 .02 1.73 .02 1.70. Totals $.86 $.86 $. MAD (total/12) $.072 $.072 $.
D = .5 is preferable, using MAD, to D = .1 or D = .3. One could
also justify excluding the January error and then dividing by
n = 11 to compute the MAD. These numbers would be $.
(for D = .1), $.078 (for D = .3), and $.074 (for D = .5)
4.10 Year 1 2 3 4 5 6 7 8 9 10 11 Forecast
Demand 4 6 4 5.0 10.0 8.0 7.0 9.0 12.0 14.0 15.
(a) 3-year moving 4.7 5.0 6.3 7.7 8.3 8.0 9.3 11.7 13.
(b) 3-year weighted 4.5 5.0 7.3 7.8 8.0 8.3 10.0 12.3 14.
4.11 Year 1 2 3 4 5 6 7 8 9 10 11 Forecast
Demand 4 6.0 4.0 5.0 10.0 8.0 7.0 9.0 12.0 14.0 15. Exp. Smoothing 5 4.7 5.1 4.8 4.8 6.4 6.9 6.9 7.5 8.9 10.4 11.
Year Time Period X Sales Y X^2 XY
2001 1 450 1 450 2002 2 495 4 990 2003 3 518 9 1554 2004 4 563 16 2252 2005 5 584 25 2920
6 = 2610 6 = 55 6 = 8166
X = 3, Y = 522
u
2 2
Y a bX
XY nXY
b
X nX
a Y bX
y x
y
Year Sales Forecast Trend Absolute Deviation
2001 450 454.8 4. 2002 495 488.4 6. 2003 518 522.0 4. 2004 563 555.6 7. 2005 584 589.2 5. 2006 622.
MAD = 5.
Forecast Exponential Absolute Year Sales Smoothing D = 0.6 Deviation
2001 450 410.0 40. 2002 495 410 + 0.6(450 – 410) = 434.0 61. 2003 518 434 + 0.6(495 – 434) = 470.6 47. 2004 563 470.6 + 0.6(518 – 470.6) = 499.0 64. 2005 584 499 + 0.6(563 – 499) = 537.4 46. 2006 537.4 + 0.6(584 – 537.4) = 565.
MAD = 51.
Forecast Exponential Absolute Year Sales Smoothing D = 0.9 Deviation 2001 450 410.0 40. 2002 495 410 + 0.9(450 – 410) = 446.0 49. 2003 518 446 + 0.9(495 – 446) = 490.1 27. 2004 563 490.1 + 0.9(518 – 490.1) = 515.2 47. 2005 584 515.2 + 0.9(563 – 515.2) = 558.2 25. 2006 558.2 + 0.9(584 – 558.2) = 581.
MAD = 38.
(Refer to Solved Problem 4.1)
For D = 0.3, absolute deviations for 2001–2005 are: 40.0, 73.0,
74.1, 96.9, 88.8, respectively. So the MAD = 372.8/5 = 74.6.
MAD
MAD
MAD
D
D
D
Because it gives the lowest MAD, the smoothing constant of
D = 0.9 gives the most accurate forecast.
3 year moving average
trend
MAD
MAD
MAD
D
One would use the trend (regression) forecast because it has the
lowest MAD.
4.19 Trend adjusted exponential smoothing: D = 0.1, E = 0.
Unadjusted Adjusted Month Income Forecast Trend Forecast |Error| Error 2 February 70.0 65.0 0.0 65 5.0 25. March 68.5 65.5 0.1 65.6 2.9 8. April 64.8 65.9 0.16 66.05 1.2 1. May 71.7 65.92 0.13 66.06 5.6 31. June 71.3 66.62 0.25 66.87 4.4 19. July 72.8 67.31 0.33 67.64 5.2 26. August 68.16 68.60 24.3 113.
MAD = 24.3/6 = 4.05, MSE = 113.2/6 = 18.87: note all numbers
are rounded
Note: To use POM for Windows to solve this problem, a period 0,
which contains the initial forecast and initial trend, must be added.
Unadjusted Adjusted Month Demand (y) Forecast Trend Forecast Error |Error| Error 2
February 70.0 65.0 0 65.0 5.00 5.0 25. March 68.5 65.5 0.4 65.9 2.60 2.6 6. April 64.8 66.16 0.61 66.77 –1.97 1.97 3. May 71.7 66.57 0.45 67.02 4.68 4.68 21. June 71.3 67.49 0.82 68.31 2.99 2.99 8. July 72.8 68.61 1.06 69.68 3.12 3.12 9. Totals 419.1 16.42 20.36 76. Average 69.85 2.74 3.39 12. August Forecast 71.30 (Bias) (MAD) (MSE)
Based upon the MSE criterion, the exponential smoothing with D = 0.1, E = 0.8 is to be preferred
over the exponential smoothing with D = 0.1, E = 0.2. Its MSE of 12.70 is lower. Its MAD of 3.39 is
also lower than that in Problem 4.19.
4.20 Trend adjusted exponential smoothing: D = 0.1, E = 0.
D D
5 4 1 4 4 0.2^19 0.8^ 20.
4.21 F A F T
5 5 4 1 4 0.4^ 19.91^ 17.
T E F F E T
FIT 5 F 5 T 5 19.91 2.23 22.
6 5 1 5 5 0.2^24 0.8^ 22.
F D A D F T
6 6 5 1 5 0.4 22.51^ 19.91^ 0.6 2.
T E F F E T
FIT 6 F 6 T 6 22.51 2.38 24.
7 6 6 6
7 7 6 6
7 7 7
F A F T
T F F T
FIT F T
D D
E E
8 7 (1^ )(^7 7 )^ (0.2)(31)
F D A D F T
8 8 7 1 7 0.4 27.14^ 24.
T E F F E T
D D
8 8 8
9 8 8 8
FIT F T
F A F T
9 9 8 1 8 0.4^ 29.28^ 27.
T E F F E T
FIT 9 F 9 T 9 29.28 2.32 31.
4.23 Students must determine the naive forecast for the four
months. The naive forecast for March is the February actual of 83,
etc.
(a) Actual Forecast |Error| |% Error|
March 101 120 19 100 (19/101) = 18.81% April 96 114 18 100 (18/96) = 18.75% May 89 110 21 100 (21/89) = 23.60% June 108 108 0 100 (0/108) = 0% 58 61.16%
E.S. MAD (for manager) 14.
MAPE (for manager) 15.29%
(b) Actual Naive |Error| |% Error| March 101 83 18 100 (18/101) = 17.82% April 96 101 5 100 (5/96) = 5.21% May 89 96 7 100 (7/89) = 7.87% June 108 89 19 100 (19/108) = 17.59% 49 48.49%
MAD (for naive) 12.
MAPE (for naive) 12.12%.
(c) MAD for the manager’s technique is 14.5, while MAD for the
naive forecast is only 12.25. MAPEs are 15.29% and 12.12%,
respectively. So the naive method is better.
4.24 (a) Graph of Demand
The observations obviously do not form a straight line, but do
tend to cluster about a straight line over the range shown.
Note: To use POM for Win-
dows to solve this problem, a
period 0, which contains the
initial forecast and initial
trend, must be added.
4.29 2007 is 25 years beyond 1982. Therefore, the quarter num-
bers are 101 through 104.
(2) (3) (4) Adjusted (1) Quarter Forecast Seasonal Forecast Quarter Number (77 + .43 Q ) Factor [(3) u (4)]
Winter 101 120.43 .8 96. Spring 102 120.86 1.1 132. Summer 103 121.29 1.4 169. Fall 104 121.72 .7 85.
4.30 Given Y = 36 + 4.3 X
(a) Y = 36 + 4.3(70) = 337
(b) Y = 36 + 4.3(80) = 380
(c) Y = 36 + 4.3(90) = 423
4.31 (a)
Sales Year Season ( y ) ( x ) ( xy ) x^2
1 SS 26,825 1 26,825 1 FW 5,722 2 11,444 4 2 SS 28,630 3 85,890 9 FW 7,633 4 30,532 16 3 SS 30,255 5 151,275 25 FW 8,745 6 52,470 36 Totals 107,810 21 358,436 91
y = 17,968.33 x = 3.
2
b
a
y x
(b) The problem with this line is that it shows a decreasing
trend when sales have been rising each year.
(c) Two separate forecast lines should be generated—one
for Spring/Summer and one for Fall/Winter—or the
analysis can be performed as a multiple regression.
4.32 (a)
2 2 2
920 4(15)^20
x
y
xy nx y
b
x nx
a y bx
Y x
(b) If the forecast is for 20 guests, the bar sales forecast
is 50 + 18(20) = $410. Each guest accounts for an
additional $18 in bar sales.
4.33 (a) See the table below.
2
55 5(3)^55
b
a
y x
For next year ( x = 6), the number of transistors (in millions) is
forecasted as y = 126 + 18(6) = 126 + 108 = 234.
(b) MSE = 160/5 = 32
(c) MAPE = 13.23%/4 = 2.65%
4.34 Y = 7.5 + 3.5 X 1 + 4.5 X 2 + 2.5 X 3
(a) 28
(b) 43
(c) 58
4.35 (a) Y ˆ = 13,473 + 37.65(1860) = 83,
(b) The predicted selling price is $83,502, but this is the
average price for a house of this size. There are other
Table for Problem 4. Year Transistors ( x ) ( y ) xy x^2 126 + 18 x Error Error 2 |% Error| 1 140 140 1 144 –4 16 100 (4/140) = 2.86% 2 160 320 4 162 –2 4 100 (2/160) = 1.25% 3 190 570 9 180 10 100 100 (10/190) = 5.26% 4 200 800 16 198 2 4 100 (2/200) = 1.00% 5 210 1,050 25 216 –6 36 100 (6/210) = 2.86% Totals 15 900 2,800 55 160 13.23%
x = 3 y =
x y xy x^2 16 330 5,280 256 12 270 3,240 144 18 380 6,840 324 14 300 4,200 196 60 1,280 19,560 920
factors besides square footage that will impact the sell-
ing price of a house. If such a house sold for $95,000,
then these other factors could be contributing to the
additional value.
(c) Some other quantitative variables would be age of the
house, number of bedrooms, size of the lot, and size of
the garage, etc.
(d) Coefficient of determination = (0.63)^2 = 0.397. This
means that only about 39.7% of the variability in the
sales price of a house is explained by this regression
model that only includes square footage as the explana-
tory variable.
4.36 (a) Given: Y = 90 + 48.5 X 1 + 0.4 X 2 where:
1 2
expected travel cost
number of days on the road
distance traveled, in miles
0.68 (coefficient of correlation)
Y
X
X
r
If:
Number of days on the road o X 1 = 5 and distance traveled
o X 2 = 300
then:
Y = 90 + 48.5 u 5 + 0.4 u 300 = 90 + 242.5 + 120 = 452.
Therefore, the expected cost of the trip is $452.50.
(b) The reimbursement request is much higher than pre-
dicted by the model. This request should probably be
questioned by the accountant.
(c) A number of other variables should be included, such as:
1. the type of travel (air or car)
2. conference fees, if any
3. costs of entertaining customers
4. other transportation costs—cab, limousine, special
tolls, or parking
In addition, the correlation coefficient of 0.68 is not exceptionally
high. It indicates that the model explains approximately 46% of
the overall variation in trip cost. This correlation coefficient
would suggest that the model is not a particularly good one.
4.37 (a), (b)
Period Demand Forecast Error Running sum |error|
1 20 20 0.00 0.00 0. 2 21 21.5 –0.50 –0.50 0. 3 28 21.25 6.75 6.25 6. 4 37 24.63 12.38 18.63 12. 5 25 30.81 –5.81 12.82 5. 6 29 27.91 1.09 13.91 1. 7 36 28.45 7.55 21.46 7. 8 22 32.23 –10.23 11.23 10. 9 25 27.11 –2.11 9.12 2. 10 28 26.06 1.94 11.06 1. MAD 4. RSFE = 11.06; MAD = 4.84 Tracking = 11.06/4.84 2.
4.38 (a) least squares equation: Y = –0.158 + 0.1308 X
(b) Y = – 0.158 + 0.1308(22) = 2.
(c) coefficient of correlation = r = 0.
coefficient of determination = r^2 = 0.
Year X Patients Y X^2 Y^2 XY 1 36 1 1296 36 2 33 4 1089 66 3 40 9 1600 120 4 41 16 1681 164 5 40 25 1600 200 6 55 36 3025 330 7 60 49 3600 420 8 54 64 2916 432 9 58 81 3364 522 10 61 100 3721 610 55 478 385 23892 2900
Given: Y = a + bX where:
2 2
XY nXY
b
X nX
a Y bX
and 6 X = 55, 6 Y = 478, 6 XY = 2900, 6 X^2 = 385, 6 Y^2 = 23892,
X = 5.5, Y = 47.8, Then:
2
385 10 5.5^385 302.5^ 82.
b
a
u u
u
u
and Y = 29.76 + 3.28 X. For:
X Y
X Y
u
u
Therefore:
Year 11 o 65.8 patients
Year 12 o 69.1 patients
The model “seems” to fit the data pretty well. One should, how-
ever, be more precise in judging the adequacy of the model.
Two possible approaches are computation of (a) the correlation
coefficient, or (b) the mean absolute deviation. The correlation
coefficient:
2 2 2 2
2 2
2
n XY X Y
r
n X X n Y Y
r
u u
ª u º ª u º
u
The coefficient of determination of 0.853 is quite respectable—
indicating our original judgment of a “good” fit was appropriate.
(f) The correlation coefficient and the coefficient of deter-
mination are given by:
2 2 2 2
2 2
2
4128 124.67 64.25^ 11.
and 0.
n XY X Y
r
n X X n Y Y
r
u u
ª u º ª u º
u u
4.42 (a) This problem gives students a chance to tackle a realis-
tic problem in business, i.e., not enough data to make a
good forecast. As can be seen in the accompanying fig-
ure, the data contains both seasonal and trend factors.
Averaging methods are not appropriate with trend, seasonal,
or other patterns in the data. Moving averages smooth out season-
ality. Exponential smoothing can forecast January next year,
but not further. Because seasonality is strong, a naïve model that
students create on their own might be best.
One model might be: F t +1 = A t –
That is forecast next period = actual one year earlier to account for
seasonality. But this ignores the trend.
One very good approach would be to calculate the increase
from each month last year to each month this year, sum all 12
increases, and divide by 12. The forecast for next year would
equal the value for the same month this year plus the average
increase over the 12 months of last year.
Using this model, the January forecast for next year becomes:
25
F
where 148 = total increases from last year to this year.
The forecasts for each of the months of next year then become:
Jan 29 July 56 Feb 26 Aug 53 Mar 32 Sep 45 Apr 35 Oct 35 May 42 Nov 38 Jun 50 Dec 29
Both history and forecast for the next year are shown in the ac-
companying figure:
4.43 (a) and (b) See the following table.
Actual Smoothed Smoothed Week Value Value Forecast Value Forecast t A ( t ) Ft ( D = 0.2) Error Ft ( D = 0.6) Error 1 50 +50.0 +0.0 +50.0 +0. 2 35 +50.0 –15.0 +50.0 –15. 3 25 +47.0 –22.0 +41.0 –16. 4 40 +42.6 –2.6 +31.4 +8. 5 45 +42.1 –2.9 +36.6 +8. 6 35 +42.7 –7.7 +41.6 –6. 7 20 +41.1 –21.1 +37.6 –17. 8 30 +36.9 –6.9 +27.1 +2. 9 35 +35.5 –0.5 +28.8 +6. 10 20 +35.4 –15.4 +32.5 –12. 11 15 +32.3 –17.3 +25.0 –10. 12 40 +28.9 +11.1 +19.0 +21. 13 55 +31.1 +23.9 +31.6 +23. 14 35 +35.9 –0.9 +45.6 –10. 15 25 +36.7 –10.7 +39.3 –14. 16 55 +33.6 +21.4 +30.7 +24. 17 55 +37.8 +17.2 +45.3 +9. 18 40 +41.3 –1.3 +51.1 –11. 19 35 +41.0 –6.0 +44.4 –9. 20 60 +39.8 +20.2 +38.8 +21. 21 75 +43.9 +31.1 +51.5 +23. 22 50 +50.1 –0.1 +65.6 –15. 23 40 +50.1 –10.1 +56.2 –16. 24 65 +48.1 +16.9 +46.5 +18. 25 +51.4 +57.
MAD = 11.8 MAD = 13.
(c) Students should note how stable the smoothed values
are for D = 0.2. When compared to actual week 25 calls
of 85, the smoothing constant, D = 0.6, appears to do a
slightly better job. On the basis of the standard error
of the estimate and the MAD, the 0.2 constant is
better. However, other smoothing constants need to be
examined.
To evaluate the trend adjusted exponential smoothing model,
actual week 25 calls are compared to the forecasted value. The
model appears to be producing a forecast approximately mid-
range between that given by simple exponential smoothing using
D = 0.2 and D = 0.6. Trend adjustment does not appear to give
any significant improvement.
4.45 We begin by reordering the numbers in the table to account
for the fact that enrollment lags birth by 5 years. Notice that the
table in the problem contains some extraneous information.
Enrollment Births 5 Years Later Year ( x ) ( y ) xy x^2
1 131 148 19,388 17, 2 192 188 36,098 36, 3 158 155 24,490 24, 4 93 110 10,230 8, 5 107 124 13,268 11, Totals 681 725 103,472 99,
x = 136.2 y = 145
2
b
a
We now can use this equation for the next 2 years.
Births Projected 5 Years Enrollment Year Earlier (43.3948 + .746 x ) 11 130 140. 12 128 138.
4.46 X Y X^2 Y^2 XY
Column totals 6885 36.96 3857893 110.26 20299.
Given: Y = a + bX where:
2 2
XY nXY
b
X nX
a Y bX
Week Actual Value Smoothed Value Trend Estimate Forecast Forecast t At Ft ( D = 0.3) T (^) t ( E = 0.2) FIT t Error
1 50.000 50.000 0.000 50.000 0. 2 35.000 50.000 0.000 50.000 –15. 3 25.000 45.500 –0.900 44.600 –19. 4 40.000 38.720 –2.076 36.644 3. 5 45.000 37.651 –1.875 35.776 9. 6 35.000 38.543 –1.321 37.222 –2. 7 20.000 36.555 –1.455 35.101 –15. 8 30.000 30.571 –2.361 28.210 1. 9 35.000 28.747 –2.253 26.494 8. 10 20.000 29.046 –1.743 27.303 –7. 11 15.000 25.112 –2.181 22.931 –7. 12 40.000 20.552 –2.657 17.895 22. 13 55.000 24.526 –1.331 23.196 31. 14 35.000 32.737 0.578 33.315 1. 15 25.000 33.820 0.679 34.499 –9. 16 55.000 31.649 0.109 31.758 23. 17 55.000 38.731 1.503 40.234 14. 18 40.000 44.664 2.389 47.053 –7. 19 35.000 44.937 1.966 46.903 –11. 20 60.000 43.332 1.252 44.584 15. 21 75.000 49.209 2.177 51.386 23. 22 50.000 58.470 3.594 62.064 –12. 23 40.000 58.445 2.870 61.315 –21. 24 65.000 54.920 1.591 56.511 8. 25 59.058 2.100 61.
Note: To use POM for Windows
to solve this problem, a period 0,
which contains the initial fore
cast and initial trend, must be
added.
4.49 (a) Continued
Method o Exponential Smoothing 0.6 = D Year Deposits ( Y ) Forecast |Error| Error
(MSE)
Method o Least Squares–Simple Regression on GSP Coefficients: GPS Deposits Year ( X ) ( Y ) Forecast |Error| Error
- 30 18.10 12.6350 5.46498 29.
- 31 24.10 15.9140 8.19 67.
- 32 25.60 20.8256 4.774 22.
- 33 30.30 23.69 6.60976 43.
- 34 36.00 27.6561 8.34390 69.
- 35 31.10 32.6624 1.56244 2.
- 36 31.70 31.72 0.024975 0.
- 37 38.50 31.71 6.79 46.
- 38 47.90 35.784 12.116 146.
- 39 49.10 43.0536 6.046 36.
- 40 55.80 46.6814 9.11856 83.
- 41 70.10 52.1526 17.9474 322.
- 42 70.90 62.9210 7.97897 63.
- 43 79.10 67.7084 11.3916 129.
- 44 94.00 74.5434 19.4566 378.
- TOTALS 787.30 150.3 1513.
- AVERAGE 17.8932 3.416 34.
- Next period forecast = 86.2173 Standard error = 6. (MAD) (MSE)
- Year Period ( X ) Deposits ( Y ) Forecast Error Method o Linear Regression (Trend Analysis)
- 1 1 0.25 –17.330 309.
- 2 2 0.24 –15.692 253.
- 3 3 0.24 –14.054 204.
- 4 4 0.26 –12.415 160.
- 5 5 0.25 –10.777 121.
- 6 6 0.30 –9.1387 89.
- 7 7 0.31 –7.50 61.
- 8 8 0.32 –5.8621 38.
- 9 9 0.24 –4.2238 19.
- 10 10 0.26 –2.5855 8.
- 11 11 0.25 –0.947 1.
- 12 12 0.33 0.691098 0.
- 13 13 0.50 2.329 3.
- 14 14 0.95 3.96769 9.
- 15 15 1.70 5.60598 15.
- 16 16 2.30 7.24427 24.
- 17 17 2.80 8.88257 36.
- 18 18 2.80 10.52 59.
- 19 19 2.70 12.1592 89.
- 20 20 3.90 13.7974 97.
- 21 21 4.90 15.4357 111.
- 22 22 5.30 17.0740 138.
- 23 23 6.20 18.7123 156.
- 24 24 4.10 20.35 264.
- 25 25 4.50 21.99 305.
- 26 26 6.10 23.6272 307.
- 27 27 7.70 25.2655 308.
- 28 28 10.10 26.9038 282.
- 29 29 15.20 28.5421 178.
- 30 30 18.10 30.18 145.
- 31 31 24.10 31.8187 59.
- 32 32 25.60 33.46 61.
- 33 33 30.30 35.0953 22.
- 34 34 36.00 36.7336 0.
- 35 35 31.10 38.3718 52.
- 36 36 31.70 40.01 69.
- 37 37 38.50 41.6484 9. - 38 38 47.90 43.2867 21. - 39 39 49.10 44.9250 17. - 40 40 55.80 46.5633 85. - 41 41 70.10 48.2016 479. - 42 42 70.90 49.84 443. - 43 43 79.10 51.4782 762. - 44 44 94.00 53.1165 1671. - TOTALS 990.00 787.30 7559. - AVERAGE 22.50 17.893 171. - –17.636 13. a b - 1 0.40 0.25 –12.198 12.4482 154. - 2 0.40 0.24 –12.198 12.4382 154. - 3 0.50 0.24 –10.839 11.0788 122. - 4 0.70 0.26 –8.12 8.38 70. - 5 0.90 0.25 –5.4014 5.65137 31. - 6 1.00 0.30 –4.0420 4.342 18. - 7 1.40 0.31 1.39545 1.08545 1. - 8 1.70 0.32 5.47354 5.15354 26. - 9 1.30 0.24 0.036086 0.203914 0. - 10 1.20 0.26 –1.3233 1.58328 2. - 11 1.10 0.25 –2.6826 2.93264 8. - 12 0.90 0.33 –5.4014 5.73137 32. - 13 1.20 0.50 –1.3233 1.82328 3. - 14 1.20 0.95 –1.3233 2.27328 5. - 15 1.20 1.70 –1.3233 3.02328 9. - 16 1.60 2.30 4.11418 1.81418 3. - 17 1.50 2.80 2.75481 0.045186 0. - 18 1.60 2.80 4.11418 1.31418 1. - 19 1.70 2.70 5.47354 2.77354 7. - 20 1.90 3.90 8.19227 4.29227 18. - 21 1.90 4.90 8.19227 3.29227 10. - 22 2.30 5.30 13.6297 8.32972 69. - 23 2.50 6.20 16.3484 10.1484 102. - 24 2.80 4.10 20.4265 16.3265 266. - 25 2.90 4.50 21.79 17.29 298. - 26 3.40 6.10 28.5827 22.4827 505. - 27 3.80 7.70 34.02 26.32 692. - 28 4.10 10.10 38.0983 27.9983 783. - 29 4.00 15.20 36.74 21.54 463. - 30 4.00 18.10 36.74 18.64 347. - 31 3.90 24.10 35.3795 11.2795 127. - 32 3.80 25.60 34.02 8.42018 70. - 33 3.80 30.30 34.02 3.72018 13. - 34 3.70 36.00 32.66 3.33918 11. - 35 4.10 31.10 38.0983 6.99827 48. - 36 4.10 31.70 38.0983 6.39827 40. - 37 4.00 38.50 36.74 1.76 3. - 38 4.50 47.90 43.5357 4.36428 19. - 39 4.60 49.10 44.8951 4.20491 17. - 40 4.50 55.80 43.5357 12.2643 150. - 41 4.60 70.10 44.8951 25.20 635. - 42 4.60 70.90 44.8951 26.00 676. - 43 4.70 79.10 46.2544 32.8456 1078. - 44 5.00 94.00 50.3325 43.6675 1906. - TOTALS 451.223 9016. - AVERAGE 10.2551 204.
Given that one wishes to develop a five-year forecast,
trend analysis is the appropriate choice. Measures of er-
ror and goodness-of-fit are really irrelevant. Exponential
smoothing provides a forecast only of deposits for the
next year—and thus does not address the five-year fore-
cast problem. In order to use the regression model based
upon GSP, one must first develop a model to forecast
GSP, and then use the forecast of GSP in the model to
forecast deposits. This requires the development of two
models—one of which (the model for GSP) must be
based solely on time as the independent variable (time is
the only other variable we are given).
(b) One could make a case for exclusion of the older data.
Were we to exclude data from roughly the first 25 years,
the forecasts for the later years would likely be consid-
erably more accurate. Our argument would be that a
change that caused an increase in the rate of growth ap-
pears to have taken place at the end of that period. Ex-
clusion of this data, however, would not change our
choice of forecasting model because we still need to
forecast deposits for a future five-year period.
I NTERNET HOMEWORK PROBLEMS
These problems appears on our companion web site at www.prenhall.
com/heizer
Forecasting Summary Table Exponential Linear Regression Method used: Smoothing (Trend Analysis) Linear Regression
Y = –18.968 + Y = –17.636 + 1.638 u YEAR 13.59364 u GSP MAD 3.416 10.587 10. MSE 34.39 171.817 204. Standard Error using 6.075 13.416 14. n – 2 in denominator Correlation coefficient 0.846 0.
Week 1 2 3 4 5 6 7 8 9 10 Forecast Registration 22 21 25 27 35 29 33 37 41 37 (a) Naïve 22 21 25 27 35 29 33 37 41 37 (b) 2-week moving 21.5 23 26 31 32 31 35 39 39 (c) 4-week moving 23.75 27 29 31 33.5 35 37
4.56 To compute seasonalized or adjusted sales forecast, we just
multiply each seasonalized index by the appropriate trend forecast.
Seasonal Trend forecast
Y ˆ^ Index u Y ˆ
Hence, for
Quarter I: ˆ 1.25 120,000 150,
Quarter II: ˆ 0.90 140,000 126,
Quarter III: ˆ 0.75 160,000 120,
Quarter IV: ˆ 1.10 180,000 198,
I II
III
IV
Y
Y
Y
Y
u
u
u
u
(a) Seasonal indexes
1.066 (Mon) 0.873 (Tue) 1.25 (Wed) 1.07 (Thu) 0.828 (Fri) 0.913 (Sat)
(b) To calculate for Monday of Week 5 = 201.74 +
0.18(25) u 1.066 = 219.9 rounded to 220
Forecast 220 (Mon) 180 (Tue) 258 (Wed) 221 (Thu) 171 (Fri) 189 (Sat)
4.58 (a) 4000 + 0.20(15,000) = 7,
(b) 4000 + 0.20(25,000) = 9,
4.59 (a) 35 + 20(80) + 50(3.0) = 1,
(b) 35 + 20(70) + 50(2.5) = 1,
4.60 Given: 6 X = 15, 6 Y = 20, 6 XY = 70, 6 X^2 = 55, 6 Y^2 = 130,
X = 3, Y = 4
2 2
2
a XY nXY
b
X nX
a Y bX
b
a
Y X
u u
u
u
(b) Correlation coefficient:
2 2 2 2
2 2
n XY X Y
r
n X X n Y Y
u u
ª u º ª u º
ª¬ º ª¼ ¬ º¼ u
The correlation coefficient indicates that there is a positive
correlation between bank deposits and consumer price indices in
Birmingham, Alabama—i.e., as one variable tends to increase (or
decrease), the other tends to increase (or decrease).
4.60 (c) Standard error of the estimate:
yx
Y a Y b XY
S
n
¦ ¦ ¦ u u
X Y X
2 Y 2 XY 2 4 4 16 8 1 1 1 1 1 4 4 16 16 16 5 6 25 36 30 3 5 9 25 15 Column Totals 15 20 55 94 70
Given: Y = a + bX where:
2 2
XY nXY
b
X nX
a Y bX
and 6 X = 15, 6 Y = 20, 6 XY = 70, 6 X^2 = 55, 6 Y^2 = 94, X = 3,
Y = 4. Then:
2
b
a
u u
u
u
and Y = 1.0 + 1.0 X. The correlation coefficient:
Mon Tue Wed Thu Fri Sat
Week 1 210 178 250 215 160 180 Week 2 215 180 250 213 165 185 Week 3 220 176 260 220 175 190 Week 4 225 178 260 225 176 190 Averages 217.5 178 255 218.3 169 186.3 Overall average = 204
2 2 2 2
2 2
n XY X Y
r
n X X n Y Y
u u
ª u º ª u º ª¬ ^ º ª¼ ¬ º¼
u
Standard error of the estimate:
yx
Y a Y b XY
S
n
¦ ¦ ¦ u u
4.62 Using software, the regression equation is: Games lost =
6.41 + 0.533 u days rain.
CASE STUDIES
SOUTHWESTERN UNIVERSITY: B
This is the second of a series of integrated case studies that run
throughout the text.
1. One way to address the case is with separate forecasting models
for each game. Clearly, the homecoming game (week 2) and the
fourth game (craft festival) are unique attendance situations.
Forecasts Game Model 2006 2007 R 2
1 y = 30,713 + 2,534 x 48,453 50,988 0. 2 y = 37,640 + 2,146 x 52,660 54,806 0. 3 y = 36,940 + 1,560 x 47,860 49,420 0. 4 y = 22,567 + 2,143 x 37,567 39,710 0. 5 y = 30,440 + 3,146 x 52,460 55,606 0. Total 239,000 250,
(where y = attendance and x = time)
2. Revenue in 2006 = (239,000) ($20/ticket) = $4,780,
Revenue in 2007 = (250,530) ($21/ticket) = $5,261,
3. In games 2 and 5, the forecast for 2007 exceeds stadium ca-
pacity. With this appearing to be a continuing trend, the time has
come for a new or expanded stadium.
DIGITAL CELL PHONE, INC.
Objectives:
Selection of an appropriate time series forecasting model based
upon a plot of the data.
The importance of combining a qualitative model with a quan-
titative model in situations where technological change is
occurring.
A plot of the data indicates a linear trend (least squares) model
might be appropriate for forecasting. Using linear trend you obtain
the following:
x y x^2 xy y^2 1 480 1 480 230400 2 436 4 872 190096 3 482 9 1446 232324 4 448 16 1792 200704 5 458 25 2290 209464 6 489 36 2934 239121 7 498 49 3486 248004 8 430 64 3440 184900 9 444 81 3996 197136 10 496 100 4960 246016 11 487 121 5357 237169 12 525 144 6300 275625 13 575 169 7475 330625 14 527 196 7378 277729 15 540 225 8100 291600 16 502 256 8032 252004 17 508 289 8636 258064 18 573 324 10314 328329 19 508 361 9652 258064 20 498 400 9960 248004 21 485 441 10185 235225 22 526 484 11572 276676 23 552 529 12696 304704 24 587 576 14088 344569 25 608 625 15200 369664 26 597 676 15522 356409 27 612 729 16524 374544 28 603 784 16884 363609 29 628 841 18212 394384 30 605 900 18150 366025 31 627 961 19437 393129 32 578 1024 18496 334084 33 585 1089 19305 342225 34 581 1156 19754 337561 35 632 1225 22120 399424 36 656 1296 23616 430336 Totals 666 19,366 16,206 378,661 10,558, Average 18.5 537.9 450.2 10,518.4 293,284.
2 2 2
xy nx y
b
x nx
a y bx
u u
u
u
2 2 2 2
2 2
[ ( ) ][ ( ) ]
[(36) (16,206) (666) ][(36)(10,558,246) (19,366) ]
[(583,416) (443,556)][380,096,856) (375,041,956)]
[139,
n xy x y
r
n x x n y y
u
][5,054,900] 706,978,314,
