24
CHAPTER 4 F O R E C A S T I N G
C H A P T E R
Forecasting
DISCUSSION Q UESTIONS 1. Qualitative incorporate subjective subjective factors factors into the Qualitative models incorporate forecasting model. Qualitative models are useful when subjective factors factors are important. important. When quantitative quantitative data are difficult to obtain, qualitative models may be appropriate. 2. Approaches Approaches are qualitati qualitative ve and quantitati quantitative. ve. Qualitati Qualitative ve is relatively subjective; quantitative uses numeric models. 3. Short-ra Short-range nge (under 3 months), months), medium-r medium-range ange (3 months months to 3 years), and long-range (over 3 years). 4.
The steps that should should be used used to develo develop p a forecas forecastin ting g system are: (a) Determine Determine the the purpose and and use of the forecast forecast (b) Select the item or quantities quantities that are to be forecasted forecasted (c) Determine Determine the the time horizon horizon of the forecast forecast (d) Select Select the type of forecast forecasting ing model to to be used (e) Gather Gather the necessary necessary data data (f) Validate Validate the foreca forecasting sting model (g) Make Make the fore forecas castt (h) Implement Implement and evaluat evaluatee the results results
5. Any three three of: sales sales planni planning, ng, product production ion planni planning ng and budgeting, 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 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 several periods where the demand demand outcome outcome is known, known, and calcul calculate ate MSE, MAPE, or MAD for each. each. The smaller error indicates the better forecast. 9.
The Delphi technique technique involves: (a) (a) Asse Assemb mbli ling ng a group group of expe expert rtss in such a manne mannerr as to preclude preclude direct direct commun communica icatio tion n betwee between n identi identifia fiable ble members of the group (b) Assembling Assembling the responses responses of each each expert to the questions questions or problems of interest (c) Summarizin Summarizing g these responses responses (d) Providing Providing each expert expert with the summary summary of all responses responses
(e) Asking each expert to study the summary of the responses and respond again to the questions or problems of interest. (f) Repeating Repeating steps steps (b) through (e) several several times times as necessary necessary to obtain convergence in responses. If convergence has not been obtained by the end of the fourth cycle, the responses 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 an associative 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 smoothing constant, constant, α, is large (close to 1.0), more weight is given to recent data; when α is low (close to 0.0), more weight is given to past data. 13. Seasonal patterns are of fixed duration and repeat regularly. Cycles Cycles vary in length length and regularity. regularity. Seasonal Seasonal indexes allow “generic” forecasts to be made specific to the month, week, etc., of the application. 14. Exponential Exponential smoothing 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 naï ve ve approach , which places all its emphasis on last period’s actual demand. 15. Adapti Adaptive ve foreca forecasti sting ng refers refers to comput computer er monito monitorin ring g of tracking signals and self-adjustment if a signal passes its present limit. 16. Tracking signals alert alert the user user of a forecas forecastin ting g tool tool to periods 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 value would would mean that that as X increa increases ses,, Y tends tends to fall. The variables move together, but move in opposite directions. 18. Independent Independent variable ( x x) is said to explain 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.
25
CHAPTER 4 F O R E C A S T I N G
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.
*Active Models 4.1, 4.3, and 4.4 appear on the CD-ROM only. Active Model 4.2 also appears in the text.
ETHICA D IE!!A 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 system they don’t understand, understand, and most college presidents would feel uncomfortable with this one. It does offer the advantage of depoliticizing the funds allocation 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 agreedupon criteria criteria (such as lowest 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.
Acti"e !o#e$ E%ercises& ACTIVE MODEL 4.1: Moving Averages What does the graph look like when n = 1? The forecast graph mirrors the data graph but one period later. 1.
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 = n = 1 (a naive forecast)
ACTIVE MODEL 4.2: Exponential Smoothing What happens to the graph when alpha equals zero? The graph is a straight line. The forecast is the same in each period. 1.
What happens to the graph when alpha equals one? The forecast forecast follows follows the same patter pattern n as the demand demand (except for the first forecast) but is offset by one period. This is a naive forecast. 2.
Generalize what happens to a forecast as alpha increases. As alpha alpha increa increases ses the foreca forecast st is more more sensit sensitive ive to changes in demand. 3.
4.2
'ear 'ear
(b) (c)
Demand 3-year moving 3-year eig!"ed
(a) No, the data data appear appear to have no consistent consistent pattern pattern (see part d for graph).
(
2
)
4
5
*
+
7
9
5
9.0 7.0 #.$
13.0 7.7 7.8
8.0 9.0 11.0
12.0 10.0 9.#
,
-
(.
(( Forecas t
13.0 9.0 11.0 7.0 11.0 11.0 11.3 11.0 10.9 12.2 10.5 10.#
9.0 8.$
CHAPTER 4 F O R E C A S T I N G
4. At what level of alpha is the mean absolute deviation (MAD) minimized? alpha = .16
2*
(d) The three-year moving average appears to give better results.
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 effect 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 addition of a trend improve the for ecast? 4.3 'ear Demand /ave E(. 'moo"!ing
(
2
)
7
9.0 7.0 #.$
5.0 9.0 7.$
#
4
5
*
+
,
-
(.
((
Forecast
9.0 13.0 8.0 12.0 13.0 9.0 11.0 7.0 5.0 9.0 13.0 8.0 12.0 13.0 9.0 11.0 #.5 7.5 9.7 9.0 10.2 11.3 10.$ 10.#
7.0 9.2
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? 10.54
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 intercept of 57.81 with a slope of 9.44 yields a MAD of 7.17.
RO0E!S END/OF/CHAPTER P 4.1
(a)
374 + 368 + 381 3
TEACHING NOTE: Notice smoothing forecasts the naive.
= 374.33 pints
4.4 (a) FJuly
(b) 1ee o3 A%g%&" 31 'e("em)er 'e("em)er 'e("em)er 'e("em)er +,"o)er 5
Naive tracks the ups and downs best but lags the data by one period. Exponential smoothing is probably better because it smoothes the data and does not have as much variation.
7 1$ 21 28
Pints Use# 3#0 389 $10 381 3#8 37$
1eigte# !o"ing A"erage
381 × .1 * 38.1 3#8 × .3 * 110.$ 37$ × .# * 22$.$ 372.9
(c) The forecast is 374.26.
well
exponential
= F June + 0.2(Forecasting error) = 42 + 0.2(40 – 42) = 41.6
(b) FAugust
= F July + 0.2(Forecasting error) = 41.6 + 0.2(45 − 41.6) = 42.3
(c) The banking industry has a great deal of seasonality in its processing requirements 1ee o3
ore,a&" 372.9
how
A%g%&" 31 'e("em)er 'e("em)er 'e("em)er 'e("em)er +,"o)er 5
4.5 (a)
3#0 389 $10 381 3#8
Forecasti Error ng 72. Error 0 0 3#0 3#0 29 5.8 3#5.8 $$.2 8.8$ 37$.#$ #.3# 1.272 375.912 7.912 1.582$
37$
37$.329#.329#
Pints Forecas t 7 1$ 21 28
3,700 + 3,800 2
Foreca st
3#0 3#5.8 37$.#$ 375.912 37$.329 # .0#592 37$.2#3 #
*3,750 miles
(b) 'ear
!i$eage
To/'ear !o"ing A"erage
Error
6Error6
2+
1 2 3 $ 5
CHAPTER 4 F O R E C A S T I N G
3000 $000 3$00 3800 3700
4.5 3500 3700 3#00 To"a&
MAD =
300 3
100 100 100 10 0
100 100 100 300
(c) Weighted 2 year M.A. with .6 weight for most recent year.
'ear
!i$eage
Forecast
Error
6Error6
3000 $000 3$00 3800 3700
3#00 3#$0 3#$0
200 1#0 #0
200 1#0 #0 $2 0
1 2 3 $ 5
= 100. Forecast for year 6 is 3,740 miles.
MAD = 140 = 4.5
420
÷
3
(d)
'ear
!i$eage Forecast
1 2 3 $ 5
3000 $000 3$00 3800 3700
3000 3000 3500 3$50 3#25 To"a
Forecast Error 0 1000 100 350 75 132 5
Error Ne 8 75. Forecast 0 500 50 175 38
3000 3500 3$50 3#25 3##3
The forecast is 3,663 miles.
4.6 Y Sa$es X Perio#
4an%ary e)r%ary ar,! A(ri ay 4%ne 4%y A%g%&" 'e("em)er +,"o)er /ovem)er De,em)er '%m Average
(a)
20 21 15 1$ 13 1# 17 18 20 20 21 23 6 218 6 18.2
1 2 3 $ 5 # 7 8 9 10 11 12 78 #.5
X 2 1 $ 9 1# 25 3# $9 #$ 81 100 121 1$$
XY 20 $2 $5 5# #5 9# 119 1$$ 180 200 231 27#
#50
1$7$
CHAPTER 4 F O R E C A S T I N G
(b) [i] Naive
The coming January = December = 23
[ii] 3-month moving
(b)
(20 + 21 + 23)/3 = 21.33
(88 + 90)
2,
= 89
2
(c)
[iii] 6-month weighted [(0.1 × 17) + (.1 × 18) Table for Problem 4.9 (a, b, c)
Forecast To/!ont
!ont 4an%ary e)r%ary ar,! A(ri ay 4%ne 4%y A%g%&" 'e("em)er +,"o)er /ovem)er De,em)er
Price :er Ci: <1.80 1.#7 1.70 1.85 1.90 1.87 1.80 1.83 1.70 1.#5 1.70 1.75
+ (0.1 × 20) + (0.2 × 20) + (0.2 × 21) + (0.3 × 23)]/1.0 = 20.6
!o"ing A"erage
1.735 1.#85 1.775 1.875 1.885 1.835 1.815 1.7#5 1.#75 1.#75
Te9:erat;r 2 #a< e !7A7 93 9$ 93
93.5
Tree/ !ont !o"ing A"erage
1.723 1.7$0 1.817 1.873 1.857 1.833 1.777 1.727 1.#83 To"a& 6 =Error Error6 >2 0.5
95
93.5
9#
9$.0
88
95.5
∑ x = 78, x = 6.5, ∑ y = 218, y = 18.17 1474 − (12)(6.5)(18.2) 54.4 = = 0.38 b= 650 − 12(6.5)2 143 a = 18.2 − 0.38(6.5) = 15.73
90
92.0
4.7
Present = Period (week) 6.
So: F7
= 1÷ 3
A6
+ 1 A + 1 A + 1 ÷ ÷ ÷ 4 5 4 4 6
1 1 = ÷ (52) + ÷ 3 4 where 1.0 = ∑ weights 4.8
(a)
1 1 1 1 , , , 3 4 4 6
(96 + 88 + 90) 3
1 1 ÷ (48) + ÷ 4 6
(63) +
= 91.3
A3
1.0
(70) = 56.75 patients
1.5
2.25 * 1001.595: 1.58; 2.0 $.00 10029#: * 2.08; 7.5 * 5#.25 1007.588: 8.52; 2.0 $.00 100290: * 2.22; 13. ##.7 1$.9$ 5 5 ;
MAD = 13.5/5 = 2.7
(d) MSE = 66.75/5 = 13.35
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
* 0.5$;
= 18 + 0.3(20 − 18) = 18.6 F Nov = 18.6 + 0.3(20 − 18.6) = 19.02 F Dec = 19.02 + 0.3(21 − 19.02) = 19.6 F Jan = 19.6 + 0.3(23 − 19.6) = 20.62 ≈ 21 [v] Trend
.035 .1#5 .125 .005 .085 .005 .115 .115 .025 .075 . A?so$;te @ Error
0.25 100.593:
[iv] Exponential smoothing with alpha = 0.3 F Oct
6Error6 To/!ont Tree/ !ont !o"ing !o"ing A"erage A"erage
(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.
.127 .1#0 .053 .073 .027 .133 .127 .027 .0#7 .793
2-
CHAPTER 4 F O R E C A S T I N G 4.9
(d) Table for Problem 4.9(d): 8 7( !ont 4an%ary e)r%ary ar,! A(ri ay 4%ne 4%y A%g%&" 'e("em)e r +,"o)er /ovem)er De,em)er
8 7)
Price :er Ci:
Forecast
6Error6
Forecast
6Error6
Forecast
6Error6
<1.80 1.#7 1.70 1.85 1.90 1.87 1.80 1.83 1.70
<1.80 1.80 1.79 1.78 1.79 1.80 1.80 1.80 1.81
<.00 .13 .09 .07 .11 .07 .00 .03 .11
<1.80 1.80 1.7# 1.7$ 1.77 1.81 1.83 1.82 1.82
<.00 .13 .0# .11 .13 .0# .03 .01 .12
<1.80 1.80 1.7$ 1.72 1.78 1.8$ 1.8# 1.83 1.83
<.00 .13 .0$ .13 .12 .03 .0# .00 .13
1.#5 1.70 1.75 To"a&
1.80 1.78 1.77
1.7# 1.71 1.70
.11 .01 .05 <.81
.15 .08 .02 <.8 # <.072
AD "o"a12:
4.10
8 75
'ear Demand
(
2
)
$
#
$
(a) 3-year moving (b) 3-year
4
5
1.79 1.75 1.73
*
5.0 10.0 8.0 $.7 5.0 #.3 $.5 5.0 7.3
.1$ .05 .02 <.8 # <.072
<.0#7 5
+
,
-
(.
((
Forecast
7.0 7.7 7.8
9.0 8.3 8.0
12.0 8.0 8.3
1$.0 9.3 10.0
15.0 11.7 12.3
13.7 1$.0
eig!"ed
(c) MAD (two-month moving average) = .750/10 = .075 MAD (three-month moving average) = .793/9 = .088 Therefore, the two-month moving average seems to have performed better.
(c) The forecasts are about the same. 4.11 (a) 'ear Demand E(. 'moo"!ing
(
2
)
$ 5
#.0 $.7
$.0 5.1
4
5
*
5.0 10.0 8.0 $.8 $.8 #.$
+
,
-
(.
((
Forecast
7.0 #.9
9.0 #.9
12.0 7.5
1$.0 8.9
15.0 10.$
11.8
CHAPTER 4 F O R E C A S T I N G
).
(b) |Error| = |Actual – Forecast| 'ear E(. &moo"!ing
( 1
2 1.3
) 1.1
4 0.2
5 5.2
* 1.#
+ 0.1
, 2.1
$.5
(. 5.1
(( $.#
!AD 2.$
These calculations were completed in Excel. Calc ulations are slightly different in Excel OM and POM for Windows due to rounding differences.
Σ = 12.3 MAD = 6.2
4.12 Act;a$ De9an#
Forecast De9an#
t
Da<
1
onday
88
88
2
T%e&day
72
88
3
=edne&day
#8
8$
$
T!%r&day
$8
80
5
riday
(c)
Trend projection:
'ear
72
← An&er
F t = F t– 1 + α( At– 1 – F t– 1)
De9an#
1 2 3 $ 5 #
Tren# Proection
$5 50 52 5# 58 ?
$2.# $2.# $2.# $2.# $2.# $2.#
> > > > > >
3.2 3.2 3.2 3.2 3.2 3.2
Let α = .25. Let Monday forecast demand = 88
× 1 * $5.8 × 2 * $9.0 × 3 * 52.2 × $ * 55.$ × 5 * 58.# × # * #1.8
= a + bX ∑ XY – nXY b= ∑ X 2 – nX 2 a = Y – bX
Y
F 3 = 88 + .25(72 – 88) = 88 – 4 = 84 F 4 = 84 + .25(68 – 84) = 84 – 4 = 80 F 5 = 80 + .25(48 – 80) = 80 – 8 = 72 4.13 (a) Exponential smoothing, α = 0.6: De9an# $5 50 52 5# 58 ?
E%:onentia$ A?so$;te S9ooting 8 .7* De"iation $1 $.0 $1.0 > 0.#$5$1: * $3.$ #.# $3.$ > 0.#50$3.$: * $7.$ $.# $7.$ > 0.#52$7.$: * 50.2 5.8 50.2 > 0.#5#50.2: * 53.7 $.3 53.7 > 0.#5853.7: * 5#.3
Σ = 25.3 MAD = 5.06 Exponential smoothing, α = 0.9: 'ear
De9an#
E%:onentia$ S9ooting 8 .7-
A?so$;te De"iation
$5 50 52 5# 58 ?
$1 $1.0 > 0.9$5$1: * $$.# $$.# > 0.950$$.# : * $9.5 $9.5 > 0.952$9.5: * 51.8 51.8 > 0.95#51.8: * 55.# 55.# > 0.95855.#: * 57.8
$.0 5.$ 2.5 $.2 2.$
1 2 3 $ 5 #
Σ = 18.5 MAD = 3.7 (b) 'ear
3-year moving average: De9an#
1 2 3 $ 5
$5 50 52 5# 58
#
?
Tree/'ear !o"ing A"erage
0.8 1.0 0.2 0.# 0.#
Σ = 3.2 MAD = 0.64
F 2 = 88 + .25(88 – 88) = 88 + 0 = 88
'ear 1 2 3 $ 5 #
A?so$;te De"iation
A?so$;te De"iation
X
Y
XY
X2
1 2 3 $ 5
$5 50 52 5# 58
$5 100 15# 22$ 290
1 $ 9 1# 25
2
Then: Σ X = 15, ΣY = 261, Σ XY = 815, Σ X = 55, X = 3, Y = 52.2 Therefore: 815 – 5 × 3 × 52.2 = 3.2 b= 55 – 5 × 3 × 3 a = 52.20 – 3.20 × 3 = 42.6 Y 6
= 42.6 + 3.2 × 6 = 61.8
(d) Comparing the results of the forecasting methodologies for parts (a), (b), and (c). Forecast !eto#o$og< E(onen"ia &moo"!ing α * 0.# E(onen"ia &moo"!ing α * 0.9 3-year moving average Trend (ro@e,"ion
!AD 5.0# 3.7 #.2 0.#$
Based on a mean absolute deviation criterion, the trend projection is to be preferred over the exponential smoothing with α = 0.6, exponential smoothing with α = 0.9, or the 3-year moving average forecast methodologies. 4.14
$5 > 50 > 52:3 * $9 50 > 52 > 5#:3 * 52.7 52 > 5# > 58:3 * 55.3
7 5.3
Method 1:
MAD: 0.20 + 0.05 + 0.05 + 0.20 = 0.5 ← better MSE : 0.04 + 0.0025 + 0.0025 + 0.04 = 0.085
Method 2:
MAD: 0.1 + 0.20 + 0.10 + 0.11 = 0.51 MSE : 0.01 + 0.04 + 0.01 + 0.0121 = 0.072 ← better
)(
CHAPTER 4 F O R E C A S T I N G
4.15 'ear
Sa$es
2003 200$ 2005 200#
$50 $95 518 5#3
2007
58$
2008
Forecast T ree/'ear !o"ing A"erage
A?so$;te De"iatio n
$50 > $95 > 518:3 * $87.7 $95 > 518 > 5#3:3 * 525.3 518 > 5#3 > 58$:3 * 555.0
75.3 58.7
Σ
= 134 MAD = 67 4.16 'ear
Ti9e Perio# X
Sa$es Y
X 2
1 2 3 $ 5
$50 $95 518 5#3 58$
1 $ 9 1# 25
2003 200$ 2005 200# 2007
Σ
*
XY
$50 990 155$ 2252 2920
Σ * 55
Σ * 81##
2#10
X
= 3, Y = 522 = a + bX ∑ XY − nXY 8166 − (5)(3)(522) = = b= 55 − (5)(9) ∑ X 2 − nX 2 a = Y − bX = 522 − (33.6)(3) = 421.2 y = 421.2 + 33.6 x y = 421.2 + 33.6 × 6 = 622.8 Y
'ear 2003 200$ 2005 200# 2007 2008
336 10
= 33.6
Sa$es
Forecast Tren#
A?so$;te De"iation
$50 $95 518 5#3 58$
$5$.8 $88.$ 522.0 555.# 589.2 #22.8
$.8 #.# $.0 7.$ 5.2
Σ = 28 MAD = 5.6 4.17 'ear
Sa$es
2003 200$ 2005 200#
$50 $95 518 5#3
2007 2008
58$
Forecast E%:onentia$ S9ooting 8 .7* $10.0 $10 > $3$ > $70.# $99.0 $99 > 537.$ 5#5.#
A?so$;te De"iation
0.#$50 $10: * $3$.0 0.#$95 $3$: * $70.# > 0.#518 $70.#: * 0.#5#3 $99: * 537.$ > 0.#58$ 537.$: *
$0.0 #1.0 $7.$ #$.0 $#.#
Σ = 259 MAD = 51.8
CHAPTER 4 F O R E C A S T I N G
Forecast E %:onentia$ 'ear
Sa$es
2003 200$ 2005 200#
$50 $95 518 5#3
2007
58$
2008
S9ooting $10.0 $10 > $$# > $90.1 515.2 515.2 558.2 558.2 581.$
A?s o$;t e De"iatio n
8 .7-
0.9$50 $10: * $$#.0 0.9$95 $$#: * $90.1 > 0.9518 $90.1: *
$0.0 $9.0 27.9 $7.8
> 0.95#3 515.2: *
25.8
> 0.958$ 558.2: *
Σ = 190.5 MAD = 38.1 (Refer to Solved Problem 4.1) For α = 0.3, absolute deviations for 2003–2007 are 40.0, 73.0, 74.1, 96.9, 88.8, respectively. So the MAD = 372.8/5 = 74.6. MADα= 0.3 MADα= 0.6
= 74.6 = 51.8 α= 0.9 MAD = 38.1 Because it gives the lowest MAD, the smoothing constant of α = 0.9 gives the most accurate forecast. 4.18 We need to find the smoothing constant α. We know in general that F t = F t– 1 + α( At– 1 – F t– 1); t = 2, 3, 4. Choose either t = 3 or t = 4 (t = 2 won’t let us find α because F 2 = 50 = 50 + α(50 – 50) holds for any α). Let’s pick t = 3. Then F 3 = 48 = 50 + α(42 – 50)
or
48 = 50 + 42α – 50α
or
–2 = –8α
So,
.25 = α
Now we can find F 5 : F 5 = 50 + α(46 – 50) F 5 = 50 + 46 α – 50α = 50 – 4 α
α = .25, F 5 = 50 – 4(.25) = 49
For
The forecast for time period 5 = 49 units. 4.19
Trend adjusted exponential smoothing:
α = 0.1, β = 0.2
Una#;st A#;ste# e# !ont Inco9e Forecast Tren# Forecast 6 Error 2 Error6 e)r%ar y ar,! A(ri ay 4%ne 4%y A%g%&"
70.0
#5.0
0.0
#8.5 #$.8 71.7 71.3 72.8
#5.5 #5.9 #5.92 ##.#2 #7.31 #8.1#
0.1 0.1# 0.13 0.25 0.33
#5 #5.# ##.05 ##.0# ##.87 #7.#$ #8.#0
5.0
25.0
2.9 1.2 5.# $.$ 5.2 2$.3
8.$ 1.# 31.9 19.7 2#.# 113. 2
MAD = 24.3/6 = 4.05, MSE = 113.2/6 = 18.87. Note that all numbers are rounded.
)2
Note: To use POM for Windows to solve this problem, a period 0, which contains the initial forecast and initial trend, must be added.
))
4.20
CHAPTER 4 F O R E C A S T I N G
Trend adjusted exponential smoothing:
4.21 F 5 = !ont
α = 0.1, β = 0.8
Una#;ste# α A4 + ( 1 – α) ( F4 + T4 ) = ( 0.2 ) ( 19) + ( 0.8) ( 20.14) De9an# =<> Forecast Tren# = 3.8 + 16.11 = 19.91
e)r%ary 70.0 #5.0 0 ar,! T5 = β ( F5 – F4 ) + ( 1#8.5 #5.5 – 17.82) 0.$ – β) T4 = ( 0.4 ) ( 19.91 A(ri #$.8 ##.1# 0.#1 + ( 0.6##.57 ) ( 2.32) = 0.40.$5 ( 2.09) ay 71.7 4%ne 71.3 + 1.39#7.$9 = 0.84 + 1.390.82 = 2.23 4%y 72.8 #8.#1 1.0# + = FIT5 = F5 + T5 = 19.91 To"a& $19.12.23 22.14 Average #9.85 F6 = αA5 + ( 1 – α) ( F5 + T5 ) A%g%&" ore,a&"
= ( 0.2 ) ( 24) + ( 0.8) ( 22.14) = 4.8 + 17.71 = 22.51
4.23 Students must determine the naive forecast for the four A#;ste# months. The naive forecast for March is the February actual of 83, Forecast Error 6Error6 Error2
etc.
#5.0 #5.9 ##.77 #7.02 #8.31 #9.#8
5.00 2.#0 1.97 $.#8 2.99 3.12
5.0 2.# 1.97 $.#8 2.99 3.12 20.3 # 3.39 AD:
1#.$2 71.30
2.7$ ia&:
25.00 #.7# 3.87 21.89 8.91 9.7# 7#.19 12.70 'E:
Based upon the MSE criterion, the exponential smoothing with α = 0.1, β = 0.8 is to be preferred over the α = 0.1, β =) +0.2. MSE β ( F – F ) smoothing + ( 1 – β) T =with 0.4 ( 22.51 – 19.91 0.6Its T =exponential ( 2.23 ) of 12.70 is lower. Its MAD of 3.39 is 6
6
5
5
= 0.4 ( 2.6) + 1.34 = 1.04 + 1.34 = 2.38 FIT6 = F6 + T6 = 22.51 + 2.38 = 24.89
(a)
= αA6 + (1 – α)(F6 + T6 ) = (0.2)(21) + (0.8)(24.89) = 4.2 + 19.91 = 24.11 – 22.51) T7 = β( F7 – F6 ) + (1 – β)T6 = (0.4)(24.11 + (0.6)(2.38) = 2.07 FIT7 = F7 + T7 = 24.11 + 2.07 = 26.18 F8
= β ( F8 – F7 ) + ( 1 – β) T7 = 0.4 ( 27.14 – 24.11) + 0.6 ( 2.07) = 2.45 FIT8 = F8 + T8 = 27.14 + 2.45 = 29.59
T9
= β ( F9 – F8 ) + ( 1 – β) T8 = ( 0.4 ) ( 29.28 – 27.14) + ( 0.6) ( 2.45) = 2.32
FIT9
120
19
A(ri
9#
11$
18
ay
89
110
21
4%ne
108
108
0
MAD (for manager) =
100 19101: * 18.81; 100 189#: * 18.75; 100 2189: * 23.#0; 100 0108: * 66 0; #1.1# ;
58
MAPE (for manager) =
T8
= αA8 + ( 1 – α) ( F8 + T8 ) = ( 0.2 ) ( 28) + ( 0.8) ( 29.59) = 29.28
101
58
= αA7 + (1 – α )(F7 + T7 ) = (0.2)(31) + (0.8)(26.18) = 27.14
F9
Act;a$ Forecas 6Error6 6@ Error6 t ar,!
4.22 F 7
Note: To use POM for Windows to solve this problem, a period 0, which contains the initial forecast and initial trend, must be
= 14.5 4 61.16% 4
= 15.29%
= F9 + T9 = 29.28 + 2.32 = 31.60
Act;a$ Nai"e 6Error6 6@ Error6
(b)
ar,!
101
83
18
A(ri ay
9# 89
101 9#
5 7
4%ne
108
89
19 $ 9
100
18101: * 17.82; 100 59#: 6 * 5.21; 100 789: 6 * 7.87; 100 19108: * 17.59; $8.$9 ;
CHAPTER 4 F O R E C A S T I N G
MAD (for naive) =
(d) R = .82 is the correlation coefficient, and R 2 = .68 means 68% of the variation in sales can be explained by TV appearances.
49
MAPE (for naive) =
= 12.25 4 48.49%
= 12.12%. 4 Naive outperforms management.
4.25
(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. (b) Least-squares regression:
N;9?er o3 Acci#ents =y >
!ont 4an%ary e)r%ary ar,! A(ri To"a& Average&
30 $0 #0 90 220 y * 55
x
1 2 3 $ 10 x * 2.5
= a + bX ∑ XY − nXY b= ∑ X 2 − nX 2 a = Y − bX
Y
Assume A::earances X
De9an# Y
X 2
Y 2
3 $ 7 # 8 5 9
3 # 7 5 10 7 ?
9 1# $9 3# #$ 25
9 3# $9 25 100 $9
XY
9 2$ $9 30 80 35
33, ΣY = 38, Σ XY = 227, Σ X 2 = 199, X = 5.5, Y = 6.33. Therefore:
Σ X =
227 – 6 × 5.5 × 6.333 = 1.0286 199 – 6 × 5.5 × 5.5 a = 6.333 –1.0286 × 5.5 = .6762
b=
Y
= .676 + 1.03 x (rounded)
The following figure shows both the data and the resulting equation:
If there are nine performances by Stone Temple Pilots, the estimated sales are: Y 9
)4
= .676 + 1.03 × 9 = .676 + 9.27 = 9.93 drums ≈ 10 drums
Average over all seasons: Average over spring: Spring index: Answer:
1,500 1,250
20,000
6,000 4
16
= 1,250
= 1,500
= 1.2
5,600 (1.2) = 1,680 sailboats 4 ÷
xy
x 2
30 80 180 3#0 #50
1 $ 9 1# 30
)5
b
= =
a
CHAPTER 4 F O R E C A S T I N G
∑ xy − n x y = ∑ x 2 – nx 2
650 – 4(2.5)(55)
=
2
30 – 4(2.5)
650 – 550 30–25
100 = 20 5 y – bx
= = 55 – (20)(2.5) =5
The regression line is y = 5 + 20 x. The forecast for May ( x = 5) is y = 5 + 20(5) = 105. 4.26 A"erag e Season 'ear( 'ear2 A"erage Season 'ear) Seaso De9an De9an 'ear 'ear a$ De9an ( 2 De9an n # # De9an# # In#e% # a =in"er '(ring '%mme r
200 350 150 300
250 300 1#5 285
225.0 325.0 157.5 292.5
250 250 250 250
0.90 1.30 0.#3 1.17
270 390 189 351
Average Yr1 to Yr2 Yr1 Demand + Yr2 Demand = 2 Demand for season Average seasonal demand = Seasonal index = Yr 3 =
=
Sum of Ave Yr1 toYr2 Demand 4 Average Yr1 to Yr2 Demand Average Seasonal Demand New Annual Demand 4 1200 4
× Seasonal index
× Seasonal index
4.27 200$ 2005 200# 2007
1inter 1$00 1200 1000 900 $500
S:ring
S;99er
1500 1$00 1#00 1500 #000
1000 2100 2000 1900 7000
Fa$$ #00 750 #50 500 2500
4.28
Q;arte 2..5 2..* r =in"er '(ring '%mmer a
73 10$ 1#8 7$
#5 82 12$ 52
2..+
89 1$# 205 98
A"erage A"erag Q;arter$ Season e < a$ De9an De9an# In#e% # 75.#7 110.#7 1#5.#7 7$.#7
10#.#7 10#.#7 10#.#7 10#.#7
0.709 1.037 1.553 0.700
4.32
x
(a)
y
x 2
xy
1# 330 5280 12 270 32$0 18 380 #8$0 1$ 300 CHAPTER 4 F O R E C A S$200 TING #0 1280 195#0
4.29 2009 is 25 years beyond 1984. Therefore, the 2009 quarter numbers are 101 through 104.
=(> Q;arter
=2> Q;arter N;9?er
=)> Forecast =++ B 7 4)Q>
=in"er '(ring '%mmer a
101 102 103 10$
120.$3 120.8# 121.29 121.72
So at x = 1.80, y = 1,454.6 – 277.6($1.80) = 954.90. Now round to the nearest integer: Answer: 955 lattes.
=5> =4> A#;ste# Seasona$ Forecast Factor =)> =4> .8 1.1 1.$ .7
9#.3$$ 132.9$# 1#9.80# 85.20$
(a)
Y = 36 + 4.3(70) = 337
(b)
Y = 36 + 4.3(80) = 380
(c)
Y = 36 + 4.3(90) = 423
4.31
4.34
=
6
Average price = 3.2583
Y = i =1
6
= 320 4 ∑ xy − nx y 19, 560 − 4(15)(320) 2
2
=
2
=
360 20
= 18
= Average number sold4.33 = 550(a)
Y = 7.5 + 3.5 X 1 + 4.5 X 2 + 2.5 X 3
(a) 28 (b) 43
(1)
6
∑ yi
=
= 15 4 1,280
(b) MSE = 160/5 = 32 (c) MAPE = 13.23%/5 = 2.65%
6
X =
y
60
(b) If the forecast is for 20 guests, the bar sales forecast is 50 + 18(20) = $410. Each guest accounts for an additional $18 in bar sales.
be the number sold. i =1
=
∑ x − nx 920 − 4(15) a = y − bx = 320 − 18(15) = 50 Y = 50 + 18 x
Let x1, x2 , B , x6 be the prices and y1, y 2, B , y 6
∑ xi
x
b=
Given Y = 36 + 4.3 X
4.30
25# 1$$ 32$ 19# )* 920
See the (2)table below.
6
2,880 (3) − 5(3)(180) = 2,880 − 2, 700 55 − 45 55 − 5(3)2
∑ xi yi = 9,783
b=
∑ xi2 = 67.1925
180 (4) = 18 10 a = 180 − 3(18) = 180 − 54 = 126
i =1 6
=
i =1
y = 126 + 18 x
Table for Problem 4.33
'ear = x >
Transistors =y >
1
1$0
1$0
2
1#0
x 2
Error 2
(2* B (, x
Error
1
1$$
$
1#
320
$
1#2
2
$
570
9
180
10
100
800
1#
198
2
$
1050
25
21#
#
3#
280 0
5 5
xy
Then y = a + bx, 6
∑ xi yi − nx y
b = i =61
x2i i =1
∑
= a=
− nx
2
=
(9, 783) − 6 (3 .25833 )(550 ) 3 190 67.1925 − 6 (3.25833 )2 $ 200
−969.489
= −277.6 3.49222 y − bx = 1, 454.6 To"a&
5 1 5
x * 3
210 90 0 y * 180
where y = number sold, x = price, and 4.35
1# 0
6@ Error6 100 $1$0: * 2.8#; 100 21#0: * 1.25; 100 10190: * 5.2#; 100 2200: * 1.00; 100 #210: * 2.8#; 13.23 ;
(c) 58 ˆ = 13,473 + 37.65(1860) = 83,502 (a) Y
)+
CHAPTER 4 F O R E C A S T I N G
(b) The predicted selling price is $83,502, but this is the average price for a house of this size. There are other factors besides square footage that will impact the selling 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 explanatory variable. 4.36
(a) Given: Y = 90 + 48.5 X 1 + 0.4 X 2 where: Y = expected travel cost X 1 = number of days on the road
= distance traveled, in miles r = 0.68 (coefficient of correlation)
X 2
4.38
(a) least squares equation: Y = –0.158 + 0.1308 X (b) Y = –0.158 + 0.1308(22) = 2.719 million (c) coefficient of correlation = r = 0.966 coefficient of determination = r 2 = 0.934
4.39 'ear X
Patients Y
X 2
Y 2
XY
3# 33 $0 $1 $0 55 #0 5$ 58 #1 $78
1 $ 9 1# 25 3# $9 #$ 81 100 385
129# 1089 1#00 1#81 1#00 3025 3#00 291# 33#$ 3721 2389 2
3# ## 120 1#$ 200 330 $20 $32 522 #10 2900
1 2 3 $ 5 # 7 8 9 10 55
Given: Y = a + bX where:
∑ XY − nXY ∑ X 2 − nX 2 a = Y − bX b=
If: Number of days on the road traveled → X 2 = 300
→ X 1
= 5 and distance
then: Y = 90 + 48.5 × 5 + 0.4 × 300 = 90 + 242.5 + 120 = 452.5
Therefore, the expected cost of the trip is $452.50. (b) The reimbursement request is much higher than predicted by the model. This request should probably be questioned by the accountant.
and Σ X = 55, ΣY = 478, Σ XY = 2900, Σ X 2 = 385, ΣY 2 = 23892, X = 5.5, Y = 47.8, Then: b=
2900 − 10 × 5.5 × 47.8
3. costs of entertaining customers
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. (a, b)
20 20 20.5 2$.25 30.#3 27.81 28.$1 32.20 27.11 2#.05
R'E * 1$.05 AD * 5
Error R;nning s;9 6error6 0.00 1.00 7.50 12.75 5.#3 1.19 7.59 10.20 2.10 6 1.95
82.5
= 3.28
Therefore: Year 11 → 65.8 patients
4. other transportation costs—cab, limousine, special tolls, or parking
20 21 28 37 25 29 3# 22 25 28
271
= 11: Y = 29.76 + 3.28 × 11 = 65.8 X = 12: Y = 29.76 + 3.28 × 12 = 69.1
2. conference fees, if any
1 2 3 $ 5 # 7 8 9 10
=
X
1. the type of travel (air or car)
Perio De9an# Forecas # t
2900 − 2629
and Y = 29.76 + 3.28 X . For:
(c) A number of other variables should be included, such as:
4.37
=
385 − 302.5 385 − 10 × 5.5 = − × = a 47.8 3.28 5.5 29.76 2
0.00 1.00 7.50 12.75 5.#3 1.19 7.59 10.20 2.10 6 1.95 AD ≈ 5.0 0 Tra,ing * 1$.055 = 2.82
Year 12 → 69.1 patients The model “seems” to fit the data pretty well. One should, however, 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: n ∑ XY
− ∑ X ∑Y n ∑ X 2 − ( ∑ X ) 2 n ∑ Y 2 − ( ∑Y) 2 10 × 2900 − 55 × 478 = 10 × 385 − 552 10 × 23892 − 478 2 29000 − 26290 = 3850 − 3025 238920 − 228484
r =
0.00 1.00 8.50 21.25 15.#3 1#.82 2$.$1 1$.21 12.10 1$.05
= 2
r
2710 825 × 10436
= 0.853
=
2710 2934.3
= 0.924
CHAPTER 4 F O R E C A S T I N G
The coefficient of determination of 0.853 is quite respectable— indicating our original judgment of a “good” f it was appropriate.
),
)-
CHAPTER 4 F O R E C A S T I N G
'ear Patients X Y 1
3#
2
33
3
$0
$
$1
5
$0
#
55
7
#0
8
5$
9
58
10
#1
Tren# Forecast
∑ XY − nXY ∑ X 2 − nX 2 a = Y − bX
A?so$;te De"iati De"iation on
29.8 > 3.28 × 1 * 33.1 29.8 > 3.28 × 2 * 3#.3 29.8 > 3.28 × 3 * 39.# 29.8 > 3.28 × $ * $2.9 29.8 > 3.28 × 5 * $#.2 29.8 > 3.28 × # * $9.$ 29.8 > 3.28 × 7 * 52.7 29.8 > 3.28 × 8 * 5#.1 29.8 > 3.28 × 9 * 59.3 29.8 > 3.28 × 10 * #2.#
2.9
2.9
3.3
3.3
0.$
0.$
b=
and Σ X = 854, ΣY = 478, Σ XY = 42558.6, Σ X 2 = 76129.9, ΣY 2 = 23892, X = 85.4, Y = 47.8. Then: b=
42558.6 − 10 × 85.4 × 47.8 2
76129.9 − 10 × 85.4 1737.4 = = 0.543 3197.3 a = 47.8 − 0.543 × 85.4 = 1.43
1.9
1.9
#.2
#.2
5.#
5.#
and
7.3
7.3
For:
2.1
2.1
1.3
1.3
1.#
1.#
=
42558.6 − 40821.2 76129.9 − 72931.6
Y = 1.43 + 0.543 X
= 131.2: Y = 1.43 + 0.543(131.2) = 72.7 X = 90.6 : Y = 1.43 + 0.543(90.6) = 50.6 X
Σ = 32.6 MAD = 3.26
Therefore: Crime rate = 131.2 → 72.7 patients Crime rate = 90.6 → 50.6 patients Note that rounding differences occur when solving with Excel. 4.41 (a) It appears from the following graph that the points do scatter around a straight line.
The MAD is 3.26—this is approximately 7% of the average number of patients and 10% of the minimum number of patients. We also see absolute deviations, for years 5, 6, and 7 in the range 5.6–7.3. The comparison of the MAD with the average and minimum number of patients and the comparatively large deviations during the middle years indicate that the forecast model is not exceptionally accurate. It is more useful for predicting general trends than the actual number of patients to be seen in a specific year. 4.40
'ear
Cri9e Patient s Rate X Y
X 2
Y 2
1
58.3
3#
3398.9
129#
2
#1.1
33
3733.2
1089
3
73.$
$0
5387.#
1#00
(b) Developing the regression relationship, we have: XY
2098.8 201#.3 293#.0 $
75.7
$1
5730.5
1#81
5
81.1
$0
#577.2
1#00
3103.7 32$$.0 #
89.0
55
7921.0
3025
7
101.1
#0
10221.2
3#00
8
9$.8
5$
8987.0
291#
9
103.3
58
10#70.9
33#$
10
11#.2
#1
13502.$
3721
Co%mn To"a&
85$.0
$78
7#129. 9
$895.0 #0##.0 5119.2 5991.$
Given: Y = a + bX where
7088.2 2389 $2558. 2 #
=S;99e To;rists Ri#ersi: r 9onts> =!i$$ions> =(......s > 'ear = X > =Y > 1 2 3 $ 5 # 7 8 9 10 11 12
7 2 # $ 1$ 15 1# 12 1$ 20 15 7
X 2
Y 2
XY
$9 $ 3# 1# 19# 225 25# 1$$ 19# $00 225 $9
2.25 1.00 1.#9 2.25 #.25 7.29 5.7# $.00 7.29 19.3# 11.5# 2.89
10.5 2.0 7.8 #.0 35.0 $0.5 38.$ 2$.0 37.8 88.0 51.0 11.9
1.5 1.0 1.3 1.5 2.5 2.7 2.$ 2.0 2.7 $.$ 3.$ 1.7
Given: Y = a + bX where:
∑ XY − nXY ∑ X 2 − nX 2 a = Y − bX b=
CHAPTER 4 F O R E C A S T I N G
and Σ X = 132, ΣY = 27.1, Σ XY = 3 52.9, ΣY 2 = 71.59, X = 11, Y = 2.26. Then:
Σ X 2
= 1796,
Then: b=
352.9 − 12 × 11 × 2.26
=
352.9 − 298.3
1796 − 1452 1796 − 12 × 11 a = 2.26 − 0.159 × 11 = 0.511 and
2
=
54.6 344
= 0.159
Y = 0.511 + 0.159 X
(c) Given a tourist population of 10,000,000, the model predicts a ridership of: Y = 0.511 + 0.159 × 10 = 2.101, or 2,101,000 persons.
(d) If there are no tourists at all, the model predicts a ridership of 0.511, or 511,000 persons. One would not place much confidence in this forecast, however, because the number of tourists (zero) is outside the range of data used to d evelop the model. (e) The standard error of the estimate is given by: S yx =
= =
∑ Y 2 − a ∑ Y − b ∑ XY n−2 71.59 − 0.511 × 27.1 − 0.159 ×352.9 12 − 2 71.59 − 13.85 − 56.11 = .163 10
= .404 ( rounded to .407 in POM for Windows software)
4.
4(
CHAPTER 4 F O R E C A S T I N G
(f) The correlation coefficient and the coefficient of determination are given by: n ∑ XY
− ∑ X ∑Y n ∑ X 2 − ( ∑ X ) 2 n ∑ Y 2 − ( ∑Y) 2 12 × 352.9 − 132 × 27.1 = 12 × 1796 − 1322 12 × 71.59 − 27.1 2 4234.8 − 3577.2 = 859.08 − 734.41 21552 − 17424
r =
= and r 2 4.42
657.6 4128 × 124.67
=
657.6 64.25 × 11.166
A(r. ay. 4%n.
35 $2 50
+,". /ov. De,.
Both history and forecast for the next year are shown in the accompanying figure:
= 0.917
= 0.840
(a) This problem gives students a chance to tackle a realistic problem in business, i.e., not enough data to make a good forecast. As can be seen in the accompanying figure, the data contains both seasonal and trend factors.
4.43
(a) and (b) See the following table: Act;a$ S9oote#
Averaging methods are not appropriate with trend, seasonal, or other patterns in the data. Moving averages smooth out seasonality. Exponential smoothing can forecast January next year, but not farther. Because seasonality is strong, a naive model that students create on their own might be best. (b) One model might be: F t+ 1 = At –11 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 y ear. (c) Using this model, the January forecast for next year becomes: 148 F 25 = 17 + 12
= 17 + 12 = 29
where 148 = total monthly increases from last year to this year. The forecasts for each of the months of next year then become: 4an. e). ar.
35 38 29
29 2# 32
4%y. A%g. 'e(.
5# 53 $5
1ee
a$;e
a$;e
t
A=t >
Ft = 8 .72>
1 2 3 $ 5 # 7 8 9 10 11 12 13 1$ 15 1# 17 18 19 20 21 22 23 2$ 25
50 35 25 $0 $5 35 20 30 35 20 15 $0 55 35 25 55 55 $0 35 #0 75 50 $0 #5
>50.0 >50.0 >$7.0 >$2.# >$2.1 >$2.7 >$1.1 >3#.9 >35.5 >35.$ >32.3 >28.9 >31.1 >35.9 >3#.7 >33.# >37.8 >$1.3 >$1.0 >39.8 >$3.9 >50.1 >50.1 >$8.1 >51.$
S9oote # Forecast a$;e Forecas t Error Error Ft = 8 .7*> >0.0 15.0 22.0 2.# 2.9 7.7 21.1 #.9 0.5 15.$ 17.3 >11.1 >23.9 0.9 10.7 >21.$ >17.2 1.3 #.0 >20.2 >31.1 0.1 10.1 >1#.9
AD * 11.8
>50.0 >50.0 >$1.0 >31.$ >3#.# >$1.# >37.# >27.1 >28.8 >32.5 >25.0 >19.0 >31.# >$5.# >39.3 >30.7 >$5.3 >51.1 >$$.$ >38.8 >51.5 >#5.# >5#.2 >$#.5 >57.#
>0.0 15.0 1#.0 >8.# >8.$ #.# 17.# >2.9 >#.2 12.5 10.0 >21.0 >23.$ 10.# 1$.3 >2$.3 >9.7 11.1 9.$ >21.2 >23.5 15.# 1#.2 >18.5
AD * 13.$5
(c) Students should note how stable the smoothed values are for α = 0.2. When compared to actual week 25 calls of 85, the smoothing constant, α = 0.6, appears to do a slightly better job. On the basis of the standard error of the estimate and the MAD, the 0.2 constant is better. However, other smoothing constants need to be examined.
CHAPTER 4 F O R E C A S T I N G
42
4.44 1ee
Act;a$ a$;e
t
At
1 2 3 $ 5 # 7 8 9 10 11 12 13 1$ 15 1# 17 18 19 20 21 22 23 2$ 25
S9oote# a$;e Ft = 8 .7)>
Tren# E sti9ate
50.000 50.000 $5.500 38.720 37.#51 38.5$3 3#.555 30.571 28.7$7 29.0$# 25.112 20.552 2$.52# 32.737 33.820 31.#$9 38.731 $$.##$ $$.937 $3.332 $9.209 58.$70 58.$$5 5$.920 59.058
0.000 0.000 0.900 2.07# 1.875 1.321 1.$55 2.3#1 2.253 1.7$3 2.181 2.#57 1.331 0.578 0.#79 0.109 1.503 2.389 1.9## 1.252 2.177 3.59$ 2.870 1.591 2.100
50.000 35.000 25.000 $0.000 $5.000 35.000 20.000 30.000 35.000 20.000 15.000 $0.000 55.000 35.000 25.000 55.000 55.000 $0.000 35.000 #0.000 75.000 50.000 $0.000 #5.000
T t =
8 .72>
Forecast FITt 50.000 50.000 $$.#00 3#.#$$ 35.77# 37.222 35.101 28.210 2#.$9$ 27.303 22.931 17.895 23.19# 33.315 3$.$99 31.758 $0.23$ $7.053 $#.903 $$.58$ 51.38# #2.0#$ #1.315 5#.511 #1.158
Forecas t Error 0.000 15.000 19.#00 3.35# 9.22$ 2.222 15.101 1.790 8.50# 7.303 7.931 22.105 31.80$ 1.#85 9.$99 23.2$2 1$.7## 7.053 11.903 15.$1# 23.#1$ 12.0#$ 21.315 8.$89
4.46
To evaluate the trend adjusted exponential smoothing model, actual week 25 calls are compared to the forecasted value. The model appears to be producing a forecast approximately midrange between that given by simple exponential smoothing using α = 0.2 and α = 0.6. Trend adjustment does not appear to give any significant improvement.
Note: To use POM for Windows to solve this problem, a period 0, which contains the initial fore cast and initial trend, must be added.
X 2
Y 2
X
Y
$21
2.90
1772$1
8.$1
377
2.93
1$2129
8.58
585
3.00
3$2225
9.00
#90
3.$5
$7#100 11.90
#08
3.##
3#9##$ 13.$0
390
2.88
152100
8.29
$15
2.15
172225
$.#2
$81
2.53
2313#1
#.$0
729
3.22
531$$1 10.37
501
1.99
251001
3.9#
#13
2.75
3757#9
7.5#
709
3.90
502#81 15.21
3##
1.#0
13395#
XY
1220.9 110$.# 1755.0 2380.5
n
Tracking signal =
4.45 !ont
At
ay 4%ne 4%y A%g%&" 'e("em) er +,"o)er /ovem)e r De,em)e r
∑ ( At − F t )
Ft
100 80 110 115 105
100 10$ 99 101 10$
110 125
10$ 105
120
2225.3
t =1
109
MAD
1123.2
6 At Ft 6 0 2$ 11 1$ 1
1
#
#
20
20
11 '%m 87
So: MAD:
87 8
39 10.875
= 10.875 = 3.586
= t Ft > A 0 2$ 11 1$
892.3 121#.9 23$7.$ 997.0 1#85.8
11
27#5.1
'%m 39 Co%mn "o"a&
#885 3#.9#
Given: Y = a + bX where:
2.5#
385789 110.2 3 #
585.# 20299. 5
4)
CHAPTER 4 F O R E C A S T I N G
∑ XY − nXY ∑ X 2 − nX 2 a = Y − bX b=
CHAPTER 4 F O R E C A S T I N G
and Σ X = 6885, ΣY = 36.96, Σ XY = 20299.5, Σ X 2 = 3857893, ΣY 2 = 110.26, X = 529.6, Y = 2.843, Then: b=
20299.5 − 13 × 529.6 × 2.843 2
3857893 − 13 × 529.6 726 = = 0.0034 211703 a = 2.84 − 0.0034 × 529.6 = 1.03
=
20299.5 − 19573.5 3857893 − 3646190
and Y = 1.03 + 0.0034 X As an indication of the u sefulness of this relationship, we can calculate the correlation coefficient: n ∑ XY
− ∑ X ∑Y n ∑ X 2 − ( ∑ X ) 2 n ∑Y 2 − ( ∑Y) 2 13 × 20299.5 − 6885 × 36.96 = 13 × 3857893 − 68852 13 × 110.26 − 36.96 2 263893.5 − 254469.6 = 50152609 − 47403225 1433.4 − 1366.0
r =
= = 2
r
9423.9 2749384× 67.0 9423.9 = 0.692 1658.13 × 8.21
= 0.479
A correlation coefficient of 0.692 is not particularly high. The 2 coefficient of determination, r , indicates that the model explains only 47.9% of the overall variation. Therefore, while the model does provide an estimate of GPA, there is considerable variation in GPA, which is as yet u nexplained. For
= 350: Y = 1.03 + 0.0034 × 350 = 2.22 X = 800: Y = 1.03 + 0.0034 × 800 = 3.75 X
Note: When solving this problem, care must be taken to interpret significant digits. 4.47 (a) There is not a strong linear trend in sales over time. (b, c) Amit wants to forecast by exponential smoothing (setting February’s forecast equal to January’s sales) with alpha = 0.1. Barbara wants to use a 3-period moving average.
4an%ary e)r%ary ar,! A(ri ay
Sa$es
A9it
$00 380 $10 375 $05
$00 398 399.2 39#.8
0ar?a A9it Error 0ar?ara ra Error 39#.#7 388.33 AD 8
20.0 12.0 2$.2 8.22 1#.1 1
21.#7 1#.#7 19.1 7
(d) Note that Amit has more forecast observations, while Barbara’s moving average does not start until month 4. Also note that the MAD for Amit is an average of 4 numbers, while Barbara’s is only 2.
44
Amit’s MAD for exponential smoothing (16.11) is lower than that of Barbara’s moving average (19.17). So his forecast seems to be better.
45
CHAPTER 4 F O R E C A S T I N G
(a)
4.48
(Continued )
Q;arter Contracts Sa$es Y X 1 2 3 $ 5 # 7 8 To"a&
153 172 197 178 185 199 205 22# 151 5 189.375
Y 2
23$09 2958$ 38809 31#8$ 3$225 39#01 $2025 5107# 290$1 3
XY
#$ 100 225 81 1$$ 1#9 1$$ 25# 1183
122$ 1720 2955 1#02 2220 2587 2$#0 3#1# 1838$
b = (18384 – 8 × 189.375 × 11.875)/(290,413 – 8 × 189.375) = 0.1121 a = 11.875 – 0.1121 × 189.375 = –9.3495 Sales (y) = –9.349 + 0.1121 (Contracts)
× 189.375
Average
8 10 15 9 12 13 12 1# 95
X 2
11.875
(b) r = (8
Sxy 2
× 18384 − 1515 × 95)
((8 × 290,413 − 15152 )(8 × 1183 −
= 0.8963 = (1183 − (−9.3495 × 95) − (0.112 × 18384 / 6) = 1.3408
r
= .8034
4.49
(a) !eto#
'ear 1 2 3 $ 5 # 7 8 9 10 11 12 13 1$ 15 1# 17 18 19 20 21 22 23 2$ 25 2# 27 28 29
E%:onentia$ S9ooting .7* 8 De:osits Forecast 6Error6 Error 2 =Y > 0.25 0.2$ 0.2$ 0.2# 0.25 0.30 0.31 0.32 0.2$ 0.2# 0.25 0.33 0.50 0.95 1.70 2.30 2.80 2.80 2.70 3.90 $.90 5.30 #.20 $.10 $.50 #.10 7.70 10.10 15.20
0.25 0.25 0.2$$ 0.2$1 0.252 0.251 0.280 0.298 0.311 0.2#8 0.2#3 0.255 0.300 0.$20 0.738 1.315 1.90# 2.$$2 2.#5# 2.#82 3.$13 $.305 $.90 5.#80 $.732 $.592 5.$97 #.818 8.787
0.00 0.01 0.00$ 0.018 0.002 0.0$8 0.029 0.021 0.071 0.008 0.013 0.07$ 0.199 0.529 0.9#1 0.98$ 0.893 0.357 0.0$3 1.217 1.$8# 0.99$ 1.297 1.580 0.232 1.507 2.202 3.281 #.$12
0.00 0.0001 0.0000 0.0003 0.00 0.0023 0.0008 0.000$ 0.0051 0.0000 0.0002 0.0055 0.0399 0.2808 0.925 0.9#98 0.7990 0.1278 0.0018 1.$81# 2.2108 0.9895 1.#8$5 2.$99 0.05$0 2.2712 $.852$ 10.7#58 $1.1195
CHAPTER 4 F O R E C A S T I N G
4.49
(a) (Continued ) !eto#
'ear
E%:onentia$ S9ooting .7* 8 De:osits Forecast 6Error6 Error 2 =Y >
30 31 32 33 3$ 35 3# 37 38 39 $0 $1 $2 $3 $$ T+TAF' AGERAE
18.10 2$.10 25.#0 30.30 3#.00 31.10 31.70 38.50 $7.90 $9.10 55.80 70.10 70.90 79.10 9$.00 787.30 17.8932
12.#350 15.91$0 20.825# 23.#9 27.#5#1 32.##2$ 31.72 31.71 35.78$ $3.053# $#.#81$ 52.152# #2.9210 #7.708$ 7$.5$3$
/e" (eriod Iore,a&" * 8#.2173
5.$#$98 8.19 $.77$ #.#097# 8.3$390 1.5#2$$ 0.02$975 #.79 12.11# #.0$# 9.1185# 17.9$7$ 7.97897 11.391# 19.$5## 150.3 3.$1# AD: '"andard #.07519
29.8##0 #7.01 22.79$9 $3.#9 #9.#2 2.$$121 0.000#2$ $#.10$2 1$#.798 3#.5# 83.1$81 322.11 #3.## 129.7#8 378.5#1 1513.22 3$.39 'E: error *
!eto# inear Regression =Tren# Ana$ 'ear Perio# = X > De:osits =Y > Forecast Error2 1 2 3 $ 5 # 7 8 9 10 11 12
1 2 3 $ 5 # 7 8 9 10 11 12
0.25 0.2$ 0.2$ 0.2# 0.25 0.30 0.31 0.32 0.2$ 0.2# 0.25 0.33
13 1$ 15 1# 17 18 19 20 21 22 23 2$ 25 2# 27 28 29 30
13 1$ 15 1# 17 18 19 20 21 22 23 2$ 25 2# 27 28 29 30
0.50 0.95 1.70 2.30 2.80 2.80 2.70 3.90 $.90 5.30 #.20 $.10 $.50 #.10 7.70 10.10 15.20 18.10
17.330 15.#92 1$.05$ 12.$15 10.777 9.1387 7.50 5.8#21 $.2238 2.5855 0.9$7 0.#91098 2.329 3.9#7#9 5.#0598 7.2$$27 8.88257 10.52 12.1592 13.797$ 15.$357 17.07$0 18.7123 20.35 21.99 23.#272 25.2#55 2#.9038 28.5$21 30.18
309.0#1 253.823 20$.31 1#0.##2 121.59$ 89.0883 #1.0019 38.2181 19.925$ 8.09#81 1.$3328 0.130392 3.3$##7 9.10#$2 15.25#7 2$.$$58 3#.997# 59.#117 89.$75# 97.959$ 111.0 138.#28 15#.558 2#$.083 305.8#2 307.203 308.5$7 282.3#7 178.011 1$5.93#
31 32 33 3$ 35 3# 37 38 39 $0 $1 $2 $3 $$
31 32 33 3$ 35 3# 37 38 39 $0 $1 $2 $3 $$
T+TAF'
2$.10 25.#0 30.30 3#.00 31.10 31.70 38.50 $7.90 $9.10 55.80 70.10 70.90 79.10 9$.00
990.0 0
AGERAE
22.50
787.3 0 17.893
31.8187 33.$# 35.0953 3#.733# 38.3718 $0.01 $1.#$8$ $3.28#7 $$.9250 $#.5#33 $8.201# $9.8$ 51.$782 53.11#5
4*
59.58 #1.73 22.99$5 0.5381 52.8798 #9.0585 9.912## 21.2823 17.$3 85.31#3 $79.5$ $$3.528 7#2.9#$ 1#71.$# 7559.9 5 171.817 'E:
!eto#
east s;aresSi9:$e Regression on GSP a ? (+7*)* ()75-)* Coecients GPS De:osit J s 'ear = X > =Y > Forecast 6Error6 Error 2 1 2 3 $ 5 # 7 8 9
0.$0 0.$0 0.50 0.70 0.90 1.00 1.$0 1.70 1.30
10 11 12 13 1$ 15 1# 17
1.20 1.10 0.90 1.20 1.20 1.20 1.#0 1.50
18 19 20 21 22 23 2$ 25 2# 27 28 29 30 31 32 33 3$
1.#0 1.70 1.90 1.90 2.30 2.50 2.80 2.90 3.$0 3.80 $.10 $.00 $.00 3.90 3.80 3.80 3.70
0.25 12.198 12.$$82 15$.957 0.2$ 12.198 12.$382 15$.71 0.2$ 10.839 11.0788 122.7$0 0.2# 8.12 8.38 70.22# 0.25 5.$01$ 5.#5137 31.9$ 0.30 $.0$20 $.3$2 18.8530 0.31 1.395$5 1.085$5 1.17820 0.32 5.$735$ 5.1535$ 2#.5# 0.2$ 0.03#08#0.20391$ 0.0$1581 0.2# 1.3233 1.58328 2.50#7# 0.25 2.#82# 2.932#$ 8.#0038 0.33 5.$01$ 5.73137 32.8$8# 0.50 1.3233 1.82328 3.32$3$ 0.95 1.3233 2.27328 5.1#779 1.70 1.3233 3.02328 9.1$020 2.30 $.11$18 1.81$18 3.2912$ 2.80 2.75$81 0.0$518# 0.0020$2 2.80 $.11$18 1.31$18 1.727 2.70 5.$735$ 2.7735$ 7.#9253 3.90 8.19227 $.29227 18.$23# $.90 8.19227 3.29227 10.8390 5.30 13.#297 8.32972 #9.38$3 #.20 1#.3$8$ 10.1$8$ 102.991 $.10 20.$2#5 1#.32#5 2##.55# $.50 21.79 17.29 298.80 #.10 28.5827 22.$827 505.$73 7.70 3$.02 2#.32 #92.752 10.10 38.0983 27.9983 783.90 15.20 3#.7$ 21.5$ $#3.92$ 18.10 3#.7$ 18.#$ 3$7.$1 2$.10 35.3795 11.2795 127.228 25.#0 3$.02 8.$2018 70.899$ 30.30 3$.02 3.72018 13.8397 3#.00 32.## 3.33918 11.15
4+
CHAPTER 4 F O R E C A S T I N G
35 3# 37 38 39 $0 $1 $2 $3 $$ T+TAF' AGERAE
$.10 $.10 $.00 $.50 $.#0 $.50 $.#0 $.#0 $.70 5.00
31.10 31.70 38.50 $7.90 $9.10 55.80 70.10 70.90 79.10 9$.00
38.0983 38.0983 3#.7$ $3.5357 $$.8951 $3.5357 $$.8951 $$.8951 $#.25$$ 50.3325
#.99827 $8.9757 #.39827 $0.9378 1.7# 3.101$# $.3#$28 19.05 $.20$91 17.#813 12.2#$3 150.$12 25.20 #35.288 2#.00 #7#.25# 32.8$5# 1078.83 $3.##75 190#.85 $51.223 901#.$5 10.2551 20$.92 AD: 'E:
2 3 $ 5 To"a
y * y * y * y *
x 37#$0 > 21$# x 3#9$0 > 15#0 x 225#7 > 21$3 x 30$$0 > 31$#
52##0 $78#0 375#7 52$#0 23900 0
5$80# $9$20 39710 55#0# 25053 0
0.90 0.91 0.88 0.93
(where y = attendance and x = time) Revenue in 2008 = (239,000) ($20/ticket) = $4,780,000
2.
Revenue in 2009 = (250,530) ($21/ticket) = $5,261,130 3. In games 2 and 5, the forecast for 2009 exceeds stadium capacity. With this appearing to be a continuing trend, the time
Forecasting S;99ar< Ta?$e E%:onentia$ S9ooting
!eto# ;se#J
inear R egress ion =Tren# Ana$
inear Regression
Y * 18.9#8 > 66666 1.#38 KEAR
AD 'E '"andard error %&ing n 2 in denomina"or Correa"ion ,oeL,ien"
3.$1# 3$.39 #.075
6 10.587 171.817 6 13.$1# 0.8$#
Given that one wishes to develop a five-year forecast, trend analysis is the appropriate choice. Measures of error 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 forecast 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 considerably more accurate. Our argument would be that a change that caused an increase in the rate of growth appears to have taken place at the end of that period. Exclusion 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.
TUDIES CASE S 1
SOUTHWESTERN UNIVERSITY: B
This is the second in 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. Ga9e 1
!o#e$ y * 30713 > 253$ x
Y * 17.#3# > 6 13.593#$ 'P 6 10.255 20$.919 1$.#51
Forecasts 2.., 2.. $8$53
50988
R2
0.92
6 0.813
has come for a new or expanded stadium.
2
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 quantitative model in situations where technological change is occurring.
1. 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
1 2 3 $ 5 # 7 8 9 10 11 12 13 1$ 15 1# 17 18 19 20
y
$80 $3# $82 $$8 $58 $89 $98 $30 $$$ $9# $87 525 575 527 5$0 502 508 573 508 $98
x 2
1 $ 9 1# 25 3# $9 #$ 81 100 121 1$$ 1#9 19# 225 25# 289 32$ 3#1 $00
xy
$80 872 1$$# 1792 2290 293$ 3$8# 3$$0 399# $9#0 5357 #300 7$75 7378 8100 8032 8#3# 1031$ 9#52 99#0
y 2
230$00 19009# 23232$ 20070$ 209$#$ 239121 2$800$ 18$900 19713# 2$#01# 2371#9 275#25 330#25 277729 291#00 25200$ 2580#$ 328329 2580#$ 2$800$
CHAPTER 4 F O R E C A S T I N G
To"a&
21 22 23 2$ 25 2# 27 28 29 30 31 32 33 3$ 35 3#
$85 52# 552 587 #08 597 #12 #03 #28 #05 #27 578 585 581 #32 #5#
$$1 $8$ 529 57# #25 #7# 729 78$ 8$1 900 9#1 102$ 1089 115# 1225 129#
###
193##
1#20#
Average 18.5
' xy − nx y
= =
' x
2
2
=
235225 27##7# 30$70$ 3$$5#9 3#9##$ 35#$09 37$5$$ 3#3#09 39$38$ 3##025 393129 33$08$ 3$2225 3375#1 399$2$ $3033#
378##1 105582$#
$50.2
10518.$
378,661 − 36 × 18.5 × 537.9
737,040 [139,860][5,054,900] 734,040 840,820
2
=
=
29328$.#
20390.0
= 5.25 3885.0 − nx 16,206 − (36 × 18.5 ) a = y − bx = 537.9 − 5.25 × 18.5 = 440.85 n' xy − ' x' y r = [ n'x 2 − ( 'x )2 ][ n' y 2 − (' y )2 ] (36)(378,661) − (666)(19,366) = [(36 ) × (16, 206 ) − (666)2 ][(36)(10, 558, 246 ) − (19, 366)2 ] 13,631,796 − 12,897,756 = [(583,416) − (443,556)][380,096,856) − (375,041,956)]
b
=
537.9
10185 11572 12#9# 1$088 15200 15522 1#52$ 1#88$ 18212 18150 19$37 18$9# 19305 1975$ 22120 23#1#
734,040 706,978,314,000
= .873
r 2
= .76 y = 440.85 + 5.25 (time) r = 0.873 indicating a reasonably good fit
The student should report the linear trend results, but deflate the forecast obtained based upon qualitative information about industry and technology trends. Because there is limited seasonality in the data, the linear trend analysis above provides a good r 2 of .76. However, a more precise forecast can be developed addressing the seasonality issue, which is done below. Methods a and c yield r 2 of .85 and .86, respectively, and methods b and d, which also center the seasonal adjustment, yield r 2 of .93 and .94, respectively. 2. Four approaches to decomposition of The Digital Cell Phone data can address seasonality, as follows: a) Multiplicative seasonal model, Cases = 443.87 + 5.08 (time), r 2 = .85, MAD = 20.89 b) Multiplicative Seasonal Model, with centered moving averages (CMA), which is not covered in our text but can be seen in Render, Stair, and Hanna’s Quantitative Analysis for Management , 9th ed., Prentice Hall Publishing. Cases = 432.28 + 5.73 (time), r 2 = .93, MAD = 12.84 c) Additive seasonal model,
4,
Cases = 444.29 + 5.06 (time), r 2 = .86, MAD = 20.02 d) Additive seasonal model, with centered moving averages. Cases = 431.31 + 5.72 (time), r 2 = .94, MAD = 12.28 The two methods that use the average of all data have very similar results, and the two CMA methods also look quite close. As suggested with this analysis, CMA is typically the better technique.
ASE S TUD' IDEO C FORECASTING AT HARD ROCK CAFE There is a short video (8 minutes) available from Prentice Hall and filmed specifically for this text that supplements this case.
4-
CHAPTER 4 F O R E C A S T I N G
A 2-minute version of the video also appears on the student DVD in the text. 1. Hard Rock uses forecasting for (1) sales (guest counts) at cafes, (2) retail sales, (3) banquet sales, (4) concert sales, (5) evaluating managers, and (6) menu planning. They could also employ these techniques to forecast: retail store sales of individual (SKU) product demands; sales of each entr ée; sales at each work station, etc. 2. The POS system captures all the basic sales data needed to drive individual cafe’s scheduling/ordering. It also is aggregated at corporate HQ. Each entr ée sold is counted as one guest at a Hard Rock Cafe. 3. The weighting system is subjective, but is reasonable. More weight is given to each of the past 2 years than to 3 years ago. This system actually protects managers from large sales variations outside their control. One could also justify a 50%–30%–20% model or some other variation. 4. Other predictors of cafe sales could include season of year (weather); hotel occupancy; spring break from colleges; beef prices; promotional budget; etc. 5.
Y = a + bx
!ont A#"ertising G;est Co;nt X Y 1 2 3 $ 5
1$ 17 25 25 35
21 2$ 27 32 29
#
35
37
7
$5
$3
8
50
$3
9
#0
5$
#0 3# # 3#.#
## 37 # 37.#
10 To"a& Averag e
b=
X 2
Y 2
19# $$1 289 57# #25 729 #25 102$ 8$1 1225 13#9 1225 18$9 2025 18$9 2500 291# 3#00 3#00 $35# 1591015950
XY
29$ $08 #75 800 1015 1295 1935 2150 32$0 39#0 1577 2
15,772 − 10
× 36.6 × 37.6 = 0.7996 ≈ .8 15,910 − 10 × 36.62
At $65,000; y = 8.3 + .8 (65) = 8.3 + 52 = 60.3, or 60,300 guests. For the instructor who asks other q uestions than this one: r 2 = 0.8869 Std. error = 5.062
CHAPTER 4 F O R E C A S T I N G
5.
*This case study appears on our companion Web site, at www.prenhall. com/heizer.
INTERNET C & ASE S TUD' THE NORTH-SOUTH AIRLINE Nortern Air$ine Data 'ear 2001 2002 2003 200$ 2005 200# 2007
Air3ra9e Cost :er Aircra3t
Engine Cost :er Aircra3t
A"erage Age =rs>
51.80 5$.92 #9.70 #8.90 #3.72 8$.73 78.7$
$3.$9 38.58 51.$8 58.72 $5.$7 50.2# 79.#0
#512 8$0$ 11077 11717 13275 15215 18390
So;teast Air$ine Data 'ear 2001 2002 2003 200$ 2005 200# 2007
Air3ra9e Cost :er Aircra3t
Engine Cost :er Aircra3t
A"erage Age =rs>
13.29 25.15 32.18 31.78 25.3$ 32.78 35.5#
18.8# 31.55 $0.$3 22.10 19.#9 32.58 38.07
5107 81$5 73#0 5773 7150 93#$ 8259
Utilizing the software package provided with this text, we can develop the following regression equations for the variables of interest: Northern Airlines—Airframe Maintenance Cost: Cost = 36.10 + 0.0026 × Airframe age Coefficient of determination = 0.7695 Coefficient of correlation = 0.8772 Northern Airlines—Engine Maintenance Cost: Cost = 20.57 + 0.0026 × Airframe age Coefficient of determination = 0.6124 Coefficient of correlation = 0.7825 Southeast Airlines—Airframe Maintenance Cost: Cost = 4.60 + 0.0032 × Airframe age Coefficient of determination = 0.3905 Coefficient of correlation = 0.6249 Southeast Airlines—Engine Maintenance Cost; Cost = –0.67 + 0.0041 × Airframe age Coefficient of determination = 0.4600 Coefficient of correlation = 0.6782
The following graphs portray both the actual data and the regression lines for airframe and engine maintenance costs for both airlines.
Note that the two graphs have been drawn to the same scale to facilitate comparisons between the two airlines. Comparison:
Northern Airlines: There seem to be modest correlations between maintenance costs and airframe age for Northern Airlines. There is certainly reason to conclude, however, that airframe age is not the only important factor. Southeast Airlines : The relationships between maintenance costs and airframe age for Southeast Airlines are much less well defined. It is even more obvious that airframe age is not the only important factor—perhaps not even the most important factor.
Overall, it would seem that:
Northern Airlines has the smallest variance in maintenance costs—indicating that its day-to-day management of maintenance is working pretty well. Maintenance costs seem to be more a function of airline than of airframe age. The airframe and engine maintenance costs for Southeast Airlines are not only lower, but more nearly similar than
5(
CHAPTER 4 F O R E C A S T I N G
those for Northern Airlines. From the graphs, at least, they appear to be rising more sharply with age.
From an overall perspective, it appears that Southeast Airlines may perform more efficiently on sporadic or emergency repairs, and Northern Airlines may place more emphasis on preventive maintenance.
Ms. Young’s report should conclude that:
There is evidence to suggest that maintenance costs could be made to be a function of airframe age by implementing more effective management practices. The difference between maintenance procedures of the two airlines should be investigated.
The data with which she is currently working does not provide conclusive results.
Concluding Comment: The question always arises, with this case, as to whether the data should be merged for the two airlines, resulting in two regressions instead of four. The solution provided is that of the consultant who was hired to analyze the data. The airline’s own internal analysts also conducted regressions, but did merge the data sets. This shows how statisticians can take different views of the same data.
