Chapter 4 - Forecasting Solution Manual.doc

March 17, 2018 | Author: Erica Uy | Category: Forecasting, Analysis, Statistical Analysis, Statistical Inference, Statistical Theory
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CHAPTER 4 F O R E C A S T I N G

C H A P T E R

Forecasting

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

27

DISCUSSION QUESTIONS 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   obtain, qualitative   models   may   be appropriate. 2. Approaches   are qualitative   and   quantitative. Qualitative   is   relatively subjective;   quantitative   uses numeric models. 3. Short­range   (under   3 months),   medium­range   (3 months   to   3 years), and long­range (over 3 years). 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 forecasting model

 

the

(g) Make the forecast (h) Implement evaluate the results

 

and

5. Any   three   of:   sales planning,   production planning 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 smoothing is a   weighted   moving   average

4

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 preclude 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   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 constant, , is large (close to 1.0), more weight is given to recent   data;   when    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   indices   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.

18. Independent   variable (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   medium­range 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: 

15. Adaptive   forecasting refers   to   computer monitoring   of   tracking signals and self­adjustment if a   signal   passes   its   present limit. 16. Tracking   signals  alert the user of a forecasting 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   would   mean that as X increases, Y tends to   fall.   The   variables   move together,   but   move   in opposite directions.

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.





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

28

CHAPTER 4 F O R E C A S T I N G

alphas   based   on   what results   they   yield, rather   than   what alphas make the most sense.

2. What   happens   to   the graph   when   alpha   equals one? The forecast follows the same pattern as the demand

4. At what level of alpha is the   mean   absolute   deviation (MAD) minimized? alpha = .16

29

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.

4.2 (a) No, the data appear to have no consistent pattern  (see part d for graph).



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 naïve 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.

Year

1

2

3

4

5

6

7

Demand

7

9

5

9.0 7.0 6.4

13.0  7.7  7.8

 8.0  9.0 11.0

12.0 10.0  9.6

(b) 3-year moving (c) 3-year weighted (except   for   the   first forecast) but is offset by one period.   This   is   a   naïve forecast. 3. Generalize what happens to   a   forecast   as   alpha increases. As   alpha   increases   the forecast is more sensitive to changes in demand. *Active Models 4.1, 4.2, 4.3, and   4.4   appear   on   our   Web site, www.pearsonhighered.com/h eizer.

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

8

9

10

11 Forecas t

13.0  9.0 11.0  7.0 11.0 11.0 11.3 11.0 10.9 12.2 10.5 10.6

END-OF-CHAPTER PROBLEMS 4.1   (a)  

374 + 368 + 381  374.33 pints 3

(b)

Weight Week Pint ed of s Moving Use Averag d e August 360Error Forecasti Week of Pints Forecast 31ng  .20 Foreca Septe Error 389 st 381 mber 7 0 0 August 31 360 360 360 .1 =  September 7 389 360 29 5.8 365.8 38.1 September 14 410 365.8 44.2 8.84 374.64 Septe 4101.272 375.91 368 September 21 381 374.64 6.36 mber 2  .3 14 = September 28 368 375.912 –7.912 –1.5824 374.32 96 110. 4 October 5 374 374.3296–.3296 –.06592 374.26 Septe 381 36 374 mber  .6 2. With   beta   set   to   zero, 21 = find   the   best   alpha   and 224. observe the MAD. Now find 4 the   best   beta.   Observe   the Septe 368 MAD. Does the addition of a mber 28 trend improve the forecast? Octobe 374 alpha   =   .11,   MAD   = r5 2.59; beta above .6 changes Forec the   MAD   (by   a   little)   to ast 2.54. 372.9

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?

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

9.0 8.4

(c) The forecast is 374.26.

30

CHAPTER 4 F O R E C A S T I N G

(d) The   three­year moving   average appears   to   give better results.

(c) The   banking industry has a great deal   of   seasonality in   its   processing requirements 4.5   (a)   

4.5 (c) Weighted   2   year M.A. with .6 weight for most recent year. Year

Mileage

1

(b)

4 5

The   coming   January   = December = 23

Forecast

3,000 4,000 3,400 3,800 3,700

3,700 + 3,800 2 =3,750 miles 2 3

(b) [i] Naïve

[ii] 3­month moving   (20 + 21 + 23)/3 = 21.33

3,600 3,640 3,640

[iii] 6­month weighted       [(0.1   17) + (.1  18)

Forecast   for   year   6   is   3,740 miles.

4.3

Year

1

2

3

Demand Naïve Exp. Smoothing

7

9.0 7.0 6.4

5.0 9.0 7.4

6

4

5

6

7

8

9

10

11

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 6.5  7.5  9.7  9.0 10.2 11.3 10.4 10.6

Year

Mileage

1 2 3 4 5

3,000 4,000 3,400 3,800 3,700

MAD 

    +

7.0 9.2

Two-Year � 420 �    Moving Average MAD  140 � � 3 � �

300  100. 3

4.5 (d) 3,500 3,700 Year 3,600 1 2 3 4 5

Mileage Forecast 3,000 4,000 3,400 3,800 3,700

Forecast Error

3,000 3,000 3,500 3,450 3,625 Total

(0.1  20) + (0.2  20)     + (0.2  21) + (0.3  23)]/ 1.0

1,000

The forecast is 3,663 miles.

4.6 Y Sales January February March April May June July August September October November December

Naïve  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.

Sum Average

TEACHING   NOTE: Notice   how   well exponential   smoothing forecasts the naïve.

= 20.6

X Period

20 21 15 14 13 16 17 18 20 20 21 23    218   18.2

(a)

4.4    (a)  FJuly  FJune + 0.2(Forecasting error)                        42 + 0.2(40 �42)  41.6         (b)  FAugust  FJuly + 0.2(Forecasting error)                           41.6 + 0.2(45 - 41.6)  42.3

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

[iv] Exponential smoothing   with alpha = 0.3 FOct  18 + 0.3(20 - 18)  18.6

FNov  18.6 + 0.3(20 - 18.6)  19.0

FDec  19.02 + 0.3(21 - 19.02)  1

FJan  19.6 + 0.3(23 - 19.6)  20.6 10 11 12 78

[v] Trend  � x  78, x  6.5,   � y = 218, y  18. � xy - nx y � x 2 - nx 2 1474 - (12)(6.5)(18.2) 54.4 b   0.38 650 - 12(6.5)2 143 a  y - bx a  18.2 - 0.38(6.5)  15.73 b

Forecast   =   15.73  + .38(1 3)   = 20.67, where next January

CHAPTER 4 F O R E C A S T I N G 4.9

31

(d) Table for Problem 4.9(d):  = .1

Month

Table for Problem 4.9 (a, b, c) Price per Forecast |Error| Chip

January February March April May June July August Septembe r October November December

$1.80 $1.80 1.67 1.80 Price per 1.70 1.79 Month Chip 1.85 1.78 January $1.80 1.90 1.79 February  1.67 1.87 1.80 March1.80  1.70 1.80 April 1.83  1.85 1.80 May  1.90 1.70 1.81 June  1.87 July  1.80 1.65 1.80 August  1.83 1.70 1.78 September  1.70 1.75 1.77 October  1.65 Totals November  1.70 December  1.75 MAD (total/12)

 = .3

 = .5

Forecast |Error| Forecast

$.00 $1.80 Two-Month .13  1.80 .09Moving 1.76 Average .07  1.74 .11  1.77 .07  1.81 .00 1.735  1.83 .03 1.685  1.82 1.775 .11  1.82 1.875 1.885 .15  1.79 1.835 .08  1.75 1.815 .02 1.765  1.73 $.8 1.675 6 1.675 $.072

Forecast

$.00 Three.13 Month Moving .06 Average .11 .13 .06 .03 1.723 .01 1.740 .12 1.817 1.873 .14 1.857 .05 1.833 .02 1.777 $.8 1.727 6 1.683 $.072 Totals

|Error| |Error|

$1.80 $.00 Two-Month  1.80 .13  1.74 Moving .04  1.72 Average .13  1.78 .12  1.84 .03  1.86 .035.06  1.83 .165.00 .125  1.83 .13 .005 .085  1.76 .11 .005  1.71 .01 .115  1.70 .115.05 $.81 .025 .075 . $.067 5

ThreeMonth Moving Average

 = .5 is preferable, using MAD, to  = .1 or  = .3. One could 4.10

Year Demand (a) 3-year moving (b) 3-year weighted

is   the 13th month.

1

2

3

4

6

4

forecast becomes 56.3, or 56 patients.

4

5

6

5.0 10.0 8.0 4.7  5.0 6.3 4.5  5.0 7.3

4.8    (a)   

7

8

9

10

11

Forecast

7.0 7.7 7.8

9.0 8.3 8.0

12.0  8.0  8.3

14.0  9.3 10.0

15.0 11.7 12.3

13.7 14.0

(96 + 88 + 90)  91.3 3

results   appear   in the figure below.

  

(c) Only trend provides an   equation that   can extend beyond   one month 4.7 Present = Period (week) 6. ��1 � �1 � �1 � �1 � � a) So:  F7  ��3 �A6 + � 4 �A5 + �4 �A4 + �6 �A3 � 1.0 � � � � � � � �� �

where

�1 � �1 � �1 � �1 �  � �(52) + � �(63) + � �(48) + � �(70) = 56.76 patients, 3 4 4 � � � � � � �6 � or 57 patients

                                                                         1 1 1 1 1.0 = � weights   ,  ,  ,  3 4 4 6

b) If the weights are 20, 15, 15,   and 10,  there  will be no change in the forecast because   these   are   the same  relative  weights   as in   part   (a),   i.e.,   20/60, 15/60, 15/60, and 10/60. c) If the weights are 0.4, 0.3, 0.2,   and   0.1,   then   the

MAD = 13.5/5 = 2.7

(d) MSE   =   66.75/5   = 13.35 (e) MAPE   =   14.94%/5 = 2.99% 4.9 (a, b) The computations for both the two­ and   three­month averages   appear in   the   table;   the

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

(c) MAD   (two­month moving   average)   =   .750/10 = .075 MAD   (three­month moving average) = .793/9 = . 088 Therefore,   the   two­month moving   average   seems   to have performed better.

.127 .160 .053 .073 .027 .133 .127 .027 .067 .793

32

CHAPTER 4 F O R E C A S T I N G

4.11  (a) Year

             (b) |Error| = |Actual – Forecast|

1 2 3 4 5 6 7 8 9 10 11 Forecast Year 1 2 3 4 5 6 7 8 9 10 11 MAD Demand 4 6.0 4.0 5.0 10.0 8.0 7.0 9.0 12.0 14.0 15.0 Exp. smoothing 5.1 4.6 2.4 Exp. Smoothing 5 4.7 5.1 14.8 1.3  4.8 1.1 6.4 0.2 6.9 5.2 6.9 1.6  7.5 0.1  8.9 2.1 10.4 4.5 11.8 These calculations were completed in Excel. Calculations are slightly different in Excel OM and POM for Windows due to rounding differences.

4.12 

2 3 Actual 4 Demand 5

t

Day

1

Monday

88

2

Tuesday

72

3

Wednesday

68

4

Thursday

48

5

Friday

6

(c) The   forecasts   are about the same.

F2 = 88 + .25(88 – 88) = 88 + 0 = 88 F3 = 88 + .25(72 – 88) = 88 – 4 = 84 F4 = 84 + .25(68 – 84) = 84 – 4 = 80 F5 = 80 + .25(48 – 80) = 80 – 8 = 72 4.13 (a) Exponential smoothing,  = 0.6: Year 1 2 3 4 5 6

Demand 45 50 52 56 58 ?

(45 + 50 + 52)/3 = 4 (50 + 52 + 56)/3 = 52.7 (52 + 56 + 58)/3 = 55.3

?

 = 12.3 MAD = 6.2 (c) Trend projection:

Ft = Ft–1 + (At–1 – Ft–1) Let    =   .25.   Let   Monday forecast demand = 88

50 52 56 58

Year

Demand

1 2 3 4 5 6

45 50 52 56 58 ?

Trend Projection 42.6 42.6 42.6 42.6 42.6 42.6

+ + + + + +

3.2 3.2 3.2 3.2 3.2 3.2

     

1 2 3 4 5 6

= = = = = =

45.8 49.0 52.2 55.4 58.6 61.8

 = 3.2 MAD = 0.64 Y  a + bX b

�XY �nXY �X 2 �nX 2

a  Y �bX

Exponential Smoothing X Y 41 1 41.0 + 0.6(45–41) = 45 43.4 2 5047.4 43.4 + 0.6(50–43.4) = 3 5250.2 47.4 + 0.6(52–47.4) = 4 5653.7 50.2 + 0.6(56–50.2) = 5 5856.3 53.7 + 0.6(58–53.7) =

XY

X2

45 100 156 224 290

1 4 9 16 25

Then:  X   =  15,  Y   =  261, 2 XY = 815, X  = 55,  X = 3, Y = 52.2 Therefore: Exponential   smoothing,    = 815 �5 �3 �52.2 0.9: b  3.2 55 �5 �3 �3 Exponential Year Demand Smoothing a  52.20 �3.20 �3  42.6 Y6  42.6 + 3.2 �6  61.8 1 45 41  = 25.3 MAD = 5.06

2 3 4 5 6

50 52 56 58 ?

41.0 + 0.9(45–41) =(d) 44.6     Comparing   the 44.6 + 0.9(50–44.6 ) = 49.5 results   of   the   forecasting 49.5 + 0.9(52–49.5) = 51.8 methodologies   for   parts   (a), 51.8 + 0.9(56–51.8) = 55.6 (b), and (c). 55.6 + 0.9(58–55.6) = 57.8 Forecast Methodology

 = 18.5 MAD = 3.7 (b) 3­year   average: Year 1

Demand

Exponential smoothing,  = 0.6 Exponential smoothing,  = 0.9 moving 3-year moving average Trend projection Three-Year Based   on   a   mean   absolute Moving Average

45

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deviation  criterion,   the  trend

MAD 5.06 3.7 6.2 0.64

CHAPTER 4 F O R E C A S T I N G

projection   is   to   be   preferred over   the   exponential smoothing   with    =   0.6, exponential   smoothing   with  = 0.9, or the 3­year moving average   forecast methodologies. 4.14 Method 1: MAD:   (0.20   + 0.05 + 0.05 + 0.20)/4 = .125  better        MSE : (0.04 + 0.0025 + 0.0025 + 0.04)/4 = . 021 Method 2: MAD:   (0.1   + 0.20   +   0.10  +   0.11)  /   4  =   . 1275        MSE : (0.01 + 0.04 + 0.01 + 0.0121) / 4 = .018   better

33

4.15 Year

Sales

2005 2006 2007 2008

450 495 518 563

2009

584

Forecast Three-Year Moving Average

(450 + 495 + 518)/3 = 487.7 (495 + 518 + 563)/3 = 525.3 (518 + 563 + 584)/3 = 555.0

2010

Forecast Expon Year

Sales

2005 2006 2007 2008

450 495 518 563

2009

584

2010

Smoothing  = 410.0 410 + 446 + 490.1 515.2 515.2 558.2 558.2 581.4

0.9(450 – 410) 0.9(495 – 446) + 0.9(518 – + 0.9(563 – + 0.9(584 –

                                     4.16 Year

Time Period X

2005 2006 2007 2008 2009

1 2 3 4 5

Sales

2610

X  3,  Y  522

b

�X 2 - nX 2

MAD 0.3  74.6 MAD 0.6  51.8 MAD 0.9  38.1

Y  a + bX �XY - nXY

(Refer   to   Solved   Problem 4.1) For    =   0.3,   absolute deviations for 2005–2009 are 40.0,   73.0,   74.1,   96.9,   88.8, respectively. So the MAD = 372.8/5 = 74.6.

8166 - (5)(3)(522) 336    33.6 55 - (5)(9) 10

a  Y - bX  522 - (33.6)(3)  421.2 y  421.2 + 33.6 x y  421.2 + 33.6 �6  622.8 Year

Sales

Forecast Trend

2005 2006 2007 2008 2009 2010

450 495 518 563 584

454.8 488.4 522.0 555.6 589.2 622.8

Because   it   gives   the   lowest MAD,   the   smoothing constant   of   =   0.9   gives   the   most accurate forecast. 4.18 We   need   to   find   the smoothing   constant  .   We know   in  general   that  Ft  = Ft–1 + (At–1 – Ft–1); t = 2, 3, 4. Choose either t = 3 or t = 4 (t = 2 won’t let us find  because F2 = 50 = 50 + (50 – 50) holds for any ). Let’s pick t = 3. Then F3 = 48 = 50 + (42 – 50) or                    48 = 50 + 42  – 50 or                    –2 = –8

4.17 Year

Sales

2005 2006 2007 2008

450 495 518 563

2009 2010

584

Forecast Exponential Smoothing 410.0 410 + 434 + 470.6 499.0 499 + 537.4 565.6

0.6(450 – 410) = 434.0 0.6(495 – 434) = 470.6 + 0.6(518 – 470.6) = 0.6(563 – 499) = 537.4 + 0.6(584 – 537.4) =

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So,                 .25 =  Now we can find F5 : F5 =  50 + (46 – 50)       F5 = 50 + 46 – 50 = 50 – 4 For           = .25, F5 = 50 –  4(.25) = 49 The forecast for time period 5 = 49 units.

34

CHAPTER 4 F O R E C A S T I N G

4.19 Trend adjusted  exponential smoothing:  =  0.1,  = 0.2

4.20 Trend

Unadjust ed Month Income Forecast Trend Februar y March April May June July August

70.0

65.0

0.0

68.5 64.8 71.7 71.3 72.8

65.5 65.9 65.92 66.62 67.31 68.16

0.1 0.16 0.13 0.25 0.33

MAD  = 24.3/6 = 4.05, MSE = 113.2/6 = 18.87. Note that 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.

Month February March April May June July Totals Average August Forecast

Demand (y) 70.0 68.5 64.8 71.7 71.3 72.8 419.1

 

adjusted

Unadjusted Forecast

Trend

65.0 65.5 66.16 66.57 67.49 68.61

Adjusted Forecast

Error

|Error|

65.0 65.9 66.77 67.02 68.31 69.68

5.00 2.60 –1.97 4.68 2.99 3.12

5.0 2.6 1.97 4.68 2.99 3.12 20.3 6 3.39 (MAD)

0 0.4 0.61 0.45 0.82 1.06

16.42

  69.85

2.74 (Bias)

71.30

Error2

25.00 6.76 3.87 21.89 8.91 9.76 76.19

12.70 (MSE)

Based   upon   the   MSE   criterion,   the   exponential   smoothing   with    =   0.1,    =   0.8   is   to   be   preferred over the exponential smoothing with    = 0.1,    = 0.2.  Its MSE of 12.70 is lower.  Its MAD of 3.39 is also lower than that in Problem 4.19. FIT8  F8 + T8  27.14 + 2.45  29.59 exponential   smoothing:    = 0.1,  = 0.8 F  A + ( 1 � ) ( F + T )  ( 0.2 ) ( 28) 9

8

4.21  F5   A4 + ( 1 � ) ( F4 + T4 )  ( 0.2 ) ( 19) + ( 0.8) ( 20.14 )

8

8

              + ( 0.8) ( 29

 3.8 + 16.11  19.91 T9   ( F9 �F8 ) + ( 1 �) T8  ( 0.4 ) ( 29.28 T5   ( F5 �F4 ) + ( 1 � ) T4  ( 0.4 ) ( 19.91 �17.82 )               + ( 0.6 ) ( 2.4            + ( 0.6 ) ( 2.32 )  0.4 ( 2.09 ) FIT9  F9 + T9  29.28 + 2.32  31.60            + 1.39  0.84 + 1.39  2.23          

FIT5  F5 + T5  19.91 + 2.23  22.14 F6  A5 + ( 1 � ) ( F5 + T5 )  ( 0.2) ( 24) + ( 0.8) ( 22.14 )         

 4.8 + 17.71  22.51

T6   ( F6 �F5 ) + ( 1 � ) T5  0.4 ( 22.51 �19.91) + 0.6 ( 2.23)              

    0.4 ( 2.6 ) + 1.34

 1.04 + 1.34  2.38

FIT6  F6 + T6  22.51 + 2.38  24.89 4.22   F7  A6 + (1 �)( F6 + T6 )  (0.2)(21) + (0.8)(24.89)  4.2 + 19.91  24.11           T7  ( F7 �F6 ) + (1 �)T6  (0.4)(24.11 �22.51) + (0.6)(2.38)  2.07          FIT7  F7 + T7  24.11 + 2.07  26.18 F8  A7 + (1 � )( F7 + T7 )  (0.2)(31) + (0.8)(26.18)  27.14 T8   ( F8 �F7 ) + ( 1 � ) T7  0.4 ( 27.14 �24.11)

               + 0.6 ( 2.07)  2.45

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CHAPTER 4 F O R E C A S T I N G

Assume

4.23 Students   must determine   the   naïve   forecast for   the   four   months.   The naïve   forecast   for   March   is the February actual of 83, etc. (a)

Appearances X

Demand Y

3 4 7 6 8 5 9

3 6 7 5 10 7 ?

Actual Forecas |Error| t March

101

120

April

 96

114

May

 89

110

 Averages

y = 55

108

101

 83

18

April May

 96  89

101  96

   

June

108

 89

19

The   observations   obviously do not form a straight line but do   tend   to   cluster   about   a straight   line   over   the   range shown. (b) Least­squares regression: Y  a + bX b

�XY - nXY �X 2 - nX 2

a  Y - bX

30 �4(2.5)

2



650 30

S Year1 Year2 Average Seaso Deman DemanYear1-Year2 D n d d Demand Fall Winter Spring Summe r

200 350 150 300

250 300 165 285

225.0 325.0 157.5 292.5

Sum of Ave Yr1 to Yr2  Demand 4 Average Yr1 to Yr2  Demand Seasonal index =  Average Seasonal Demand New Annual Demand Yr3  �Seasonal index 4 1200  �Seasonal index 4

Average seasonal demand 

49  12.25 4 48.49% MAPE (for na� ve)   12.12%. 4 (c) If there are nine  Naïve outperforms  performances   by   Stone management. Temple  Pilots,   the  estimated (c) MAD   for   the   manager’s sales are: technique   is   14.5,   while Y9  .676 + 1.03 �9  .676 + 9.27  9.93 drums MAD   for   the   naïve

4.24 (a) Graph of demand

650 �4(2.5)(55)

Average Yr1 to Yr2 � Yr1  Demand + Yr2  Demand � � � 2 Demand for season � �

MAD (for na� ve) 

forecast   is   only   12.25. MAPEs   are   15.29%   and 12.12%,   respectively.   So the naïve method is better.



A

b

March

2

4.26

Actual Naïve |Error|

(b)

2

The regression line is y = 5 + 20x. The forecast for May (x = 5) is y = 5 + 20(5) = 105.

X  =   33,  Y  =   38,  XY  = 227, X2 = 199,  X = 5.5,  Y = 6.33. Therefore:

108

�xy - n x y

�x �nx 100   20 5 a  y �bx  55 �(20)(2.5) 5

227 �6 �5.5 �6.333  1.0286 199 �6  �5.5 �5.5 a  6.333 �1.0286  �5.5  .6762 58 Y  .676 + 1.03 x  (rounded) MAD (for management)   14.5 4 The   following   figure   shows 61.16% MAPE (for management)  both the data and the resulting  15.29% 4 equation: June

b

35

�10 drums

(d)  R   =   .82   is   the correlation   coefficient,   and R2  =   .68  means  68% of  the variation   in   sales   can   be explained   by   TV appearances.

20,000  1, 250 16 6,000 Average over spring:   1,500 4 1,500 Spring index:   1.2 1, 250 Average over all seasons: 

�5,600 � Answer:  � (1.2)  1,680 sailboats � 4 � � 4.27

4.25 

Month January February March April  Totals

Winter Number of Accidents (y) 30 40 60 90 220

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2006 2007 2008 2009

1,400 1,200 1,000 900 4,500

Spring

Sum

1,500 1,400 1,600 1,500 6,000

1, 2, 2, 1, 7,

4.28

Avera

36

CHAPTER 4 F O R E C A S T I N G

Quarte 2007 2008 r Winter Spring Summer Fall

73 104 168 74

65 82 124 52

2009

89 146 205 98

4.29 2011   is   25   years beyond  1986.   Therefore,   the 2011 quarter numbers are 101 through 104.

Then  y  =  a  +  bx, where  y  = number sold, x = price, and

(1) Quarter

(2) Quarter Number

(3) Forecast (77 + . 43Q

Winter Spring Summer Fall

101 102 103 104

120.43 120.86 121.29 121.72

4.30 Given Y = 36 + 4.3X (a)            Y  =   36   + 4.3(70) = 337 (b)           Y  =   36   + 4.3(80) = 380 (c)            Y  =   36   + 4.3(90) = 423 4.31

Let x1,  x2 , K ,  x6  be the prices and y1,  y2 , K , y6 be the number sold. 4.33 (a) See the table below. 6 �xi X  i 12,880   Average price = 3.2583 - 5(3)(180) 2,880 - 2,700                (1) b6  55 - 45 55 - 5(3)2 6 yi �180 18 Y  i1  Average number sold = 550 (2) 610 6 a  180 - 3(18)  180 - 54  126                 (3) �xi yi  9,783 i 1 y  126 + 18 x 6

2 �xi  = 67.1925

i1

6

b

�xi yi - nx y

i 1 6

2 2 �x i - nx



(9, 783) - 6(3.25833)(550) 67.1925 - 6 (3.25833)2

i 1

-969.489   -277.6 3.49222 a  y - bx  550 - [(-277.6)(3.25)]  1, 454.6

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

                (4)

x

4.32 (a)

y

xy

x2

16 330 5,280 12 270 3,240 18 380 6,840 14 300 CHAPTER 4 F O R E C A S4,200 TING 60 1,280 19,560

So at x = 2.80, y = 1,454.6 – 277.6($2.80) =  677.32. Now round  to  the  nearest  integer: Answer: 677 lattes.

the   additional value. (c) Some   other quantitative variables   would 60 be   age   of   the x  15 4 house,   number   of 1,280 bedrooms,   size   of y  320 the lot, and size of 4 �xy - nx y 19,560 - 4(15)(320) the garage, etc. 360 b    18   of 20 �x 2 - nx 2 920 - 4(15)2 (d) Coefficient determination   = a  y - bx  320 - 18(15)  50 (0.63)2  =   0.397. Y  50 + 18 x This   means   that only   about   39.7% (b) If   the   forecast   is of   the   variability for   20   guests,   the in   the   sales   price bar   sales   forecast of   a   house   is is   50   +   18(20)   = explained   by   this $410.   Each   guest regression   model accounts   for   an that   only   includes additional   $18   in square   footage   as bar sales. the   explanatory variable. (b) MSE = 160/5 = 32 4.36 (a)   Given:  Y  =   90   + (c) MAPE = 13.23%/5 48.5X1 + 0.4X2 where:  = 2.65%

4.34 Y  =   7.5   +   3.5X1  + 4.5X2 + 2.5X3

Y  expected travel cost X1  number of days on the road

(a) 28 (b) 43 (c) 58 4.35 (a)  Y�  =   13,473   + 37.65(1860) = 83,502

X2  distance traveled, in miles

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. (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   entertaining customers

of

4. other transportation costs—cab, limousine, special   tolls,   or parking

In   addition,   the   correlation r  0.68 (coefficient of correlation)coefficient   of   0.68   is   not exceptionally   high.   It If: indicates   that   the   model Number of days on the explains   approximately   46% road    X1  =   5   and of the overall variation in trip distance   cost.   This   correlation traveled  X2 = 300 coefficient   would   suggest then: that   the   model   is   not   a

(b) The   predicted selling   price   is $83,502,   but   this is   the   average price for a Table for Problem 4.33 house   of Year Transistors this   size. (x) (y) There   are  1 140 other 

xy

x2

126 + 18x

Error

 140

 1

144

–4

 4

162

–2

 9

180

10

16

198

 2

25

216

–6

factors  2 160  320 besides square  3 190  570 footage that   will  4 200  800 impact  5 210 1,050 the selling 90 2,80 price of a Totals 1 5 0 0 house.   If y = 180 x = 3 such   a  Y = 90 + 48.5  5 + house   sold   for 0.4  300 = 90 + 242.5 + 120 = $95,000,   then 452.5 these other factors could   be contributing   to

5 5

particularly good one. 4.37  (a, b) Perio Demand Forecas

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256 144 324 196 37 920

d 1 2 3 4 5 6 7 8 9 10

t 20 21 28 37 25 29 36 22 25 28

20 20 20.5 24.25 30.63 27.81 28.41 32.20 27.11 26.05

Cumulative error = 14.05; MAD = 5 Tracking = 14.05/5  2.82

4.38 (a) least   squares equation:  Y  =   –0.158   + 0.1308X (b) Y  =  – 0.158   + 0.1308(22)   =   2.719 million (c) coefficient   of correlation = r = 0.966 coefficient   of determination =  r2 = 0.934 4.39 Ye Pat X 2 ar ien X ts Y  1  36   1

Y2 X Y

     1,29 36 6  2  33   4      1,08 66 9  3  40   9     1,60 12 0 0  4  41  16     1,68 16 Error 2 |% Error| 1 4  16 100 (4/140)  =  5  40  25     2.86% 1,60 20   4 100 (2/160)  = 0 0  6  55  36 1.25%     100 1003,02 (10/190) 33 = 5.26% 5 0   4 100 (2/200) =  7  60  49     1.00% 3,60 42  36 100 (6/210)  = 0 0  8  54  64 2.86%     16 2,91 43 13.23 0 % 6 2  9  58  81     3,36 52 4 2 10  61 100     3,72 61 1 0

0.00 1.00 7.50 12.75 –5.63 1.19 7.59 –10.20 –2.10   1.95

38

CHAPTER 4 F O R E C A S T I N G

55

478

385

Given: Y = a + bX where: b

 XY - nXY 2

 X - nX

2

a  Y - bX and X = 55, Y = 478, XY =   2900,  X 2   =   385,  Y 2   = 23892, X  5.5, Y  47.8,  Then: 2900 - 10  5.5  47.8

2900 - 2629 271 b    3.28 385 - 302.5 82.5 385 - 10  5.52 a  47.8 - 3.28  5.5  29.76 and Y = 29.76 + 3.28X. For: X  11: Y  29.76 + 3.28 � 11  65.8 X  12: Y  29.76 + 3.28 � 12  69.1

Year Patients X Y  1

36

 2

33

 3

40

 4

41

 5

40

 6

55

 7

60

 8

54

 9

58

10

61

Therefore:

Trend Forecast  8

 94.8

 54

 8,987.0

 9

103.3

 58

10,670.9

10

116.2

 61

13,502.4

Column Totals

854.0

478

76,129. 9

29.8 + 3.28 33.1 29.8 + 3.28 36.3 29.8 + 3.28 39.6 29.8 + 3.28 42.9 29.8 + 3.28 46.2 29.8 + 3.28 49.4 29.8 + 3.28 52.7 29.8 + 3.28 56.1 29.8 + 3.28 59.3 29.8 + 3.28 62.6

Given: Y = a + bX where b

Year 12  69.1 patients

r





n  XY -  X  Y n  X 2 - (  X ) 2  n  Y 2 - (  Y ) 2      10  2900 - 55  478 10  385 - 552  10  23892 - 4782     29000 - 26290

3850 - 3025  238920 - 228484 2710 2710    0.924 825  10436 2934.3

and  X  =   854,  Y  =   478, XY  =   42558.6,  X2  = 76129.9, Y 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



42558.6 76129.

The  MAD  is   3.26—this   is approximately   7%   of   the and      Y = 1.43 + 0.543X average   number   of   patients For: and   10%   of   the   minimum X  131.2 :  Y  1.43 + 0.543(131.2)  72 number   of   patients.   We   also X  90.6 :  Y  1.43 + 0.543(90.6)  50.6 see   absolute   deviations,   for years 5, 6, and 7 in the range  Therefore: 5.6–7.3.   The   comparison   of Crime rate = 131.2 the  MAD  with   the   average  72.7 patients and   minimum   number   of Crime   rate   =   90.6 patients   and   the  50.6 patients comparatively   large Note   that   rounding deviations   during  the  middle differences   occur   when Also note that a crime rate of 131.2 is outside the range of the years indicate that the forecast solving with Excel. data set used to determine the regression equations, so caution is model   is   not   exceptionally 4.41 (a) It appears from the advised. accurate. It is more useful for following   graph predicting general trends than that   the   points   do the actual number of patients scatter   around   a to be seen in a specific year. straight line. 4.40

Year

Crime Patient s Rate X Y

r  0.853

 1

 58.3

 36

The   coefficient   of determination   of   0.853   is quite   respectable—indicating our   original   judgment   of   a “good” fit was appropriate.

 2

 61.1

 33

 3

 73.4

 40

 4

 75.7

 41

 5

 81.1

 40

 6

 89.0

 55

 7

101.1

 60 (Summe

2

 X 2 - nX 2

a  Y - bX

Year 11  65.8 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:

 XY - nXY

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(b) Developing   the regression relationship, we have: Tourists

Ridership

CHAPTER 4 F O R E C A S T I N G

r months) (Millions) (1,000,000s ) Year (X) (Y)  1  2  3  4  5  6  7  8  9 10 11 12

 7  2  6  4 14 15 16 12 14 20 15  7

1.5 1.0 1.3 1.5 2.5 2.7 2.4 2.0 2.7 4.4 3.4 1.7

Syx  

 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) 

(f)  The   correlation coefficient and the coefficient   of determination   are given by:

Given: Y = a + bX where: b

 XY - nXY  X 2 - nX 2

a  Y - bX and  X  =   132,  Y  =   27.1, XY  =   352.9,  X2  =   1796, Y 2   =  71.59,   X   =  11,   Y = 2.26.  Then: 352.9 - 298.3 54.6    0.159 1796 - 1452 344 1796 - 12  112 a  2.26 - 0.159  11  0.511 b

and

352.9 - 12  11  2.26

Y = 0.511 + 0.159X (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 develop   the model. (e) The standard error of   the   estimate   is given by:

39

the   data.   Moving   averages smooth   out   seasonality. Exponential   smoothing   can forecast   January   next   year,   but   not   farther.   Because seasonality is strong, a naïve model that students create on their own might be best. (b) One model might be: Ft+1 = At–11 That   is   forecastnext period = actualone year earlier to account for 

seasonality.   But   this   ignores the trend. n  XY -  X One very good approach Y r   be   to 2  calculate   the  n  X 2 - (  X ) 2  would n Y 2 - ( Y )    increase from each month last  year to each month this year, 12  352.9 - 132  27.1  sum   all   12   increases,   and 12  1796 - 1322  12  71.59 - 27.12    divide by 12. The forecast for   next   year   would   equal   the 4234.8 - 3577.2  value for the same month this - 734.41 year plus the average increase 21552 - 17424  859.08 over   the   12   months   of   last 657.6 657.6   year.  0.917 4128  124.67 64.25  11.166 (c) Using this model, the January forecast for next year and r 2  0.840 becomes: 4.42 (a) This   problem 148 gives   students   a F25  17 +  17 + 12  29 12 chance to tackle a realistic   problem where   148   =   total   monthly in   business,   i.e., increases   from   last   year   to not enough data to this year. make   a   good The forecasts for each of forecast. As can be the months of next year seen   in   the then become: accompanying Jan. 29 figure,   the   data Feb. 26 contains   both Mar. 32 seasonal and trend Apr. 35 factors. May. 42 Jun.

50

Both history and forecast for the next year are shown in the accompanying figure:

Averaging   methods   are not   appropriate   with   trend, seasonal, or other patterns in Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

July. Aug. Sep. Oct. Nov. Dec.

40

CHAPTER 4 F O R E C A S T I N G

standard   error of the estimate and the  MAD,   the   0.2 constant   is better.   However, other   smoothing constants   need   to be examined.

4.43 (a)   and   (b)   See   the following table: Actual Smoothed Week

Value

Value

t

A(t)

Ft ( = 0.2)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

50 35 25 40 45 35 20 30 35 20 15 40 55 35 25 55 55 40 35 60 75 50 40 65

+50.0 +50.0 +47.0 +42.6 +42.1 +42.7 +41.1 +36.9 +35.5 +35.4 +32.3 +28.9 +31.1 +35.9 +36.7 +33.6 +37.8 +41.3 +41.0 +39.8 +43.9 +50.1 +50.1 +48.1 +51.4

Forecast

MAD

(c) Students   should note   how   stable the   smoothed values are for    = 0.2.   When compared to actual week   25   calls   of 85,   the   smoothing constant,    =   0.6, appears   to   do   a slightly better job. On the basis of the

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CHAPTER 4 F O R E C A S T I N G

4.44 Week t  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Smoothed 87Trend Estimate So: MAD:   10.875 Value 8 Ft ( = 0.3) Tt (  = 0.2) 39

Actual Value At

 

50.000 35.000 25.000 40.000 45.000 35.000 20.000 30.000 35.000 20.000 15.000 40.000 55.000 35.000 25.000 55.000 55.000 40.000 35.000 60.000 75.000 50.000 40.000 65.000

50.00010.875 50.000 45.500 38.720 37.651 38.543 36.555 30.571 28.747 29.046 25.112 20.552 24.526 32.737 33.820 31.649 38.731 44.664 44.937 43.332 49.209 58.470 58.445 54.920 59.058

 3.586

 0.000  0.000 –0.900 –2.076 –1.875 –1.321 –1.455 –2.361 –2.253 –1.743 –2.181 –2.657 –1.331  0.578  0.679  0.109  1.503  2.389  1.966  1.252  2.177  3.594  2.870  1.591  2.100

Forecast FITt 50.000 50.000 44.600 36.644 35.776 37.222 35.101 28.210 26.494 27.303 22.931 17.895 23.196 33.315 34.499 31.758 40.234 47.053 46.903 44.584 51.386 62.064 61.315 56.511 61.158

Forecas t Error

Y

 377  2.93  585  3.00  690  3.45  608  3.66  390  2.88  415  2.15  481  2.53  729  3.22

n

Tracking signal 

X

 421  2.90

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   = 0.2 and    = 0.6. Trend adjustment does not appear to give   any significant improvement. �( At - Ft )

 501  1.99

t 1

Month

At

Ft

May June July August Septemb er October Novembe r Decembe r

100 80 110 115 105

100 104 99 101 104

110 125

104 105

120

109

MAD

 613  2.75  709  3.90  366  1.60 Column totals

 XY - nXY  X 2 - nX 2

a  Y - bX

  0.000 –15.000 –19.600   3.356   9.224  –2.222 –15.101   1.790   8.506  –7.303  –7.931  22.105  31.804   1.685  –9.499  23.242  14.766  –7.053 –11.903  15.416  23.614 –12.064 –21.315   8.489

4.46 (a)

4.45

b

6885 36.96

Given: Y = a + bX where:

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41

42

CHAPTER 4 F O R E C A S T I N G

and  X  = 6885,  Y  = 36.96, XY  =   20299.5,  X 2   = 3857893, Y 2   =   110.26,   X   =   529.6, Y  = 2.843, Then:

(b,   c) Amit   wants   to forecast  by exponential smoothing   (setting February’s   forecast equal   to   January’s 20299.5 - 13  529.6  2.843 20299.5 - 19573.5 sales) with alpha   b  0.1.   Barbara   wants 3857893 - 3646190 3857893 - 13  529.62 to   use   a   3­period 726   0.0034 moving average. 211703 a  2.84 - 0.0034  529.6  1.03

Sales

Amit

400 380 410 375 405

— 400 398 399.2 396.8

and Y = 1.03 + 0.0034X As   an   indication   of   the usefulness   of   this relationship, we can calculate the correlation coefficient: r



 

January February March April May

n  XY -  X  Y 2

 n  X 2 - (  X )  n  Y 2 - ((d) Y )     forecast   observations, while Barbara’s moving 13  20299.5 - 6885  36.96 average   does   not   start 13  3857893 - 68852  13  110.26 - 36.962  until month 4. Also note    that the MAD for Amit 263893.5 - 254469.6 is   an   average   of   4 50152609 - 47403225 1433.4 - 1366.0  numbers,   while Barbara’s is only 2. 9423.9

2749384  67.0 9423.9   0.692 1658.13  8.21 r 2  0.479 A   correlation   coefficient   of 0.692 is not particularly high. The   coefficient   of 2 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 unexplained. For

2 Note that Amit has more

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.

(b) X  350: Y  1.03 + 0.0034 � 350  2.22 (c) X  800: Y  1.03 + 0.0034 � 800  3.75 Note:   When   solving   this problem, care must be taken to interpret significant digits. Also note that X = 800 is outside the range  of  the   data   set  used  to determine   the   regression relationship,   so   caution   is advised. 4.47 (a) There   is  not  a strong   linear   trend   in   sales over time.

4.48 (a)

27 28 Quarter Contracts Sales Y 29 X 1 2 3 4 5 6 7 8 Barba Totals

  153   172   197   178   185   199   205   226 1,51 5 189.375

Average

 8 10 15  9 12 13 12 16 95

 7.70 10.10 15.20

5.497 6.818 8.787

(Continued)

    11.875

396.67 b  = (18384 – 8    189.375   388.33 11.875)/(290,413   –   8   MAD

189.375  189.375) = 0.1121 a  =   11.875   –   0.1121   189.375 = –9.3495 Sales ( y) = –9.349 + 0.1121 (Contracts) (b) r  (8 � 18384 - 1515 � 95)

((8 � 290,413 - 15152 )(8 � 1183 - 952 ))

 0.8963 Sxy  1183 - (-9.3495 � 95) - (0.112 � 18384 / 6)  1.3408 r 2  .8034 4.49 (a)

Year 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

Method  Exponential Smoothing 0.6 = Deposits Forecast (Y )  0.25  0.24  0.24  0.26  0.25  0.30  0.31  0.32  0.24  0.26  0.25  0.33  0.50  0.95  1.70  2.30  2.80  2.80  2.70  3.90  4.90  5.30  6.20  4.10  4.50  6.10

0.25 0.25 0.244 0.241 0.252 0.251 0.280 0.298 0.311 0.268 0.263 0.255 0.300 0.420 0.738 1.315 1.906 2.442 2.656 2.682 3.413 4.305 4.90 5.680 4.732 4.592

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2.20 3.28 6.41

CHAPTER 4 F O R E C A S T I N G

4.49 (a)  (Continued)

Year 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 TOTALS AVERAGE

appropriate choice.   Measures of   error   and goodness­of­fit are really   irrelevant. Exponential smoothing provides a forecast only   of   deposits for the next year— and  thus  does  not

Method  Exponential Smoothing 0.6 = Deposits Forecast (Y )  18.10  24.10  25.60  30.30  36.00  31.10  31.70  38.50  47.90  49.10  55.80  70.10  70.90  79.10  94.00 787.30

12.6350 15.9140 20.8256 23.69 27.6561 32.6624 31.72 31.71 35.784 43.0536 46.6814 52.1526 62.9210 67.7084 74.5434

   17.8932

Forecasting Summary Table

 8  9 10 11 12

0.32 0.24 0.26 0.25 0.33

13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

0.50 0.95 1.70 2.30 2.80 2.80 2.70 3.90 4.90 5.30 6.20 4.10 4.50 6.10 7.70 10.10 15.20 18.10

Exponential Smoothing

Method used:

Linear Regression (Trend Analysis)

Linear Regression

Y = –18.968 +       1.638  YEAR MAD MSE Standard error using  n – 2 in denominator Correlation coefficient

31 31 32 32 Next period forecast = 86.2173 33 33 34 34 35 35 36 (Trend36 Method  Linear Regression Analysis) 37 Year Period (X ) Deposits 37 38 38  1  1 0.25 39 39  2  2 0.24 40 40  3  3 0.24 41 41  4  4 0.26 42 42  5  5 0.25 43 43  6  6 0.30 44 44  7  7 0.31  8  9 10 11 12

43

24.10 25.60 30.30 36.00 31.10 31.70 38.50 47.90 49.10 55.80 70.10 70.90 79.10 94.00

 3.416  34.39  6.075

13 14 15 16 17

1.20 1.20 1.20 1.60 1.50

  10.587  171.817   13.416

Y = –17.636 +       13.59364  GSP   10.255  204.919   14.651

  0.846

   0.813

 0.50  0.95  1.70  2.30  2.80

18 1.60  2.80 19 1.70  2.70 20 1.90  3.90 21 1.90  4.90 22 2.30  5.30 23 2.50  6.20 24 2.80  4.10 25 2.90  4.50 26 3.40  6.10 27 3.80  7.70 TOTALS 990.0 787.30 28 4.10 10.10 0 29 4.00 15.20 AVERAGE 22.50 17.893 30 4.00 18.10 31 3.90 24.10 32 3.80 25.60 33 3.80 30.30 Method  Least squares–Simple Regression on 34 3.70 36.00 35 4.10 31.10 a b 36 4.10 31.70 –17.636 13.5936 37 4.00 38.50 Coefficients GSP Deposit 38 4.50 47.90 : s 39 4.60 49.10 Year (X) (Y ) Forecast 40 4.50 55.80 4.60 70.10  1 0.40  0.25 –12 41 4.60 70.90  2 0.40  0.24 –12 42 4.70 79.10  3 0.50  0.24 –10 43 5.00 94.00  4 0.70  0.26 –8 44  5 0.90  0.25 –5 TOTALS  6 1.00  0.30 –4 AVERAGE  7 1.40  0.31  1  8 1.70  0.32  5 Given   that   one  9 1.30  0.24   wishes  to  develop 0.036086 a   five­year 10 1.20  0.26 –1 forecast,   trend 11 1.10  0.25 –2 analysis   is   the 12 0.90  0.33 –5

–1 –1 –1  4  2  4  5  8  8 13 16 20 21 28 34 38 36 36 35 34 34 32 38 38 36 43 44 43 44 44 46 50

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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

44

CHAPTER 4 F O R E C A S T I N G

4.51

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   five­year period.

Demand

Exponentially Smoothed Forecast

1 2 3 4 5 6 7

 7  9  5  9 13  8 Forecast

5 5 + 0.2 5.4 + 0.2 6.12 + 0.2 5.90 + 0.2 6.52 + 0.2 7.82 + 0.2

4.52 Actual

Forecast

95 108 123 130

ADDITIONAL HOMEWORK PROBLEMS

100 110 120 130

(b) 3­month   weighted moving average

MAD =  10/4 = 2.5,  MSE = 38/4 = 9.5 4.53 (a) 3­month   moving average:

These problems, which appear on   www.myomlab.com, provide   an   additional   13 problems that you may wish to assign.

Three-Month Month

4.50

(a) (b) (c)

Period

Week

1

2

3

4

Registration Naïve 2-week moving 4-week moving

22

21 22

25 21 21.5

27 25 23

January 6 7 February 35 March 29 33 27 April 35 29 26 31 32 23.7 May 27 29 5 5

Sales 8 11 14 3716 3310 31 3115

9

Moving Average 10

Forecast

41 37 37 (11 41+ 14 37 + 16)/3 = 35 39 39 33.5(14 35+ 16 37 + 10)/3 =

June

17

(16 + 10 + 15)/3 =

July

11

(10 + 15 + 17)/3 =

August

14

(15 + 17 + 11)/3 =

September

17

(17 + 11 + 14)/3 =

October

12

(11 + 14 + 17)/3 =

November

14

(14 + 17 + 12)/3 =

December

16

(17 + 12 + 14)/3 =

January

11

(12 + 14 + 16)/3 =

February

0 0            M A D   =   2 . 2 0

(14 + 16 + 11)/3 =

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CHAPTER 4 F O R E C A S T I N G

 9.17 Average4.56 To Absolute   compute Three-Month yMoving seasonalized or adjusted sales x  3.5 Moving Deviation

(c) Based   on   a Sales   Mean Month

Absolute Deviation January 11 criterion,   14the February 3­month   moving March 16 average   10 with April MAD  =   2.2   is May 15  to be   preferred   17 over June July 11 the   3­month August weighted   moving 14 September average   17 with October MAD = 2.72. 12

forecast, we just multiply each seasonalized   index   by   the Period   7   forecast   = appropriate trend forecast. 13.07 (1  11 + 2  14 + 3  16)/6 = 14.50 4.50 Y� Y�  Index Trend forecast Period   12+  forecast   = = 12.67Seasonal (1  14 + 2  16 3  10)/6 2.33 18.64,   but   this   is   far Hence, for (1  16 + 2  10 + 3  15)/6 = 13.50 3.50 outside   the   range   of = 15.17Quarter I: Y� (1  10 + 2  15 +3  17)/6 4.17 I  1.25  120,000  150,000 valid data. (1  15 + 2  17 + 3  11)/6 = 13.67 0.33 � Quarter II: Y3.50 (1  17 + 2  11 + 3  14)/6 = 13.50 II  0.90  140,000  126,000 (1  11 + 2  14 + 3  17)/6 = 15.00 3.00 � Quarter III: Y  0.75  160,000  120,000 (1 (1 (1 (1

   

14 17 12 14

complex   model are   interest   rates and   cycle   or seasonal factors. 4.54 (a) Actua Cumulativ l Wee Miles Foreca Error Error k st 1 2 3 4 5

17 21 19 23 18

17.00 17.00 17.80 18.04 19.03

6

16

18.83

7 8

20 18

18.26 18.61

9 10 11

22 20 15

18.49 19.19 19.35

12

22

18.48

0.00 –4.00 –1.20 –4.96 +1.0 3 +2.8 3 –1.74 +0.6 1 –3.51 –0.81 +4.3 5 –3.52

(b) The  MAD  28.56/12 = 2.38

=

(c) The  cumulative error  and tracking signals  appear  to   be   consistently negative,   and   at week   10,   the tracking   signal exceeds 5 MADs. 4.55

y

x

x2

 7  9  5 11 10 13 55

 1  2  3  4  5  6 21

 1  4  9 16 25 36 91

(b)   Correlation coefficient:

y  5.27 + 1.11x

November 14 (d) Other   factors 16   that December January might be included 11 February in   a   more



+ + + +

2 2 2 2

   

17 12 14 16

+ + + +

3 3 3 3

   

Week 1 Week 2 Week 3 Week 4 Averages

III 12)/6 = 14.00 0.00 � Quarter IV: Y2.17 14)/6 = 13.83 IV  1.10  180,000  198,000 16)/6 = 14.67 3.67 4.57 11)/6 = 13.17 Mon Tue Wed Thu Fri Sat

210 215 220 225 217. 5

       = 27.17

178 250 MAD = 2.72 180 250 176 260 178 260 17 255 8

215 213 220 225 218. 3

160 165 175 176 16 9

(a) Seasonal indices: 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)  1.066 = 219.9   rounded   to 220 – –



45

180 185 190 190 186.3

r



 

n  XY -  X  Y

n  X 2 - (  X ) 2  n  Y 2    5  70 - 15  20

5  55 - 152  5  130 - 2   350 - 300  275 - 225  650 - 400  50  0.45 111.80

Overall average = 204

The   correlation   coefficient indicates   that   there   is   a positive correlation   between   bank deposits and consumer price indices in

Forecast 220 (Mon) 180 (Tue) 258 (Wed) 221 (Thu) 171 (Fri) 189 (Sat)

4.58 (a) 4000   0.20(15,000) = 7,000

+

(b) 4000   + 0.20(25,000) = 9,000 4.59 (a) 35   +   20(80)   + 50(3.0) = 1,785 (b) 35   +   20(70)   + 50(2.5) = 1,560 4.60 Given: X = 15, Y = 2 2 20, XY = 70, X  = 55, Y = 130,  X  = 3,  Y =  4  XY - nXY (a)    b   X 2 - nX 2

Birmingham,   Alabama—i.e., as   one   variable   tends   to increase   (or   decrease),   the other   tends   to   increase   (or decrease).

a  Y - bX   error   of 70 - 5  3  4 70 4.60 (c) Standard - 60 10 b    1 the estimate: 55 - 45 10 55 - 5  32 a  4 - 1 3  4 - 3  1 Y  1 + 1X Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

46

CHAPTER 4 F O R E C A S T I N G

Syx  

 Y 2 - a  Y - b  XY 130 - 1  20 - 1  70  n-2 3 130 - 20 - 70  3

40  13.3  3.65 3

4.61

Column Totals

X

Y

2 1 4 5 3

4 1 4 6 5

15

20

days rain.

CASE STUDIES

 XY - nXY  X 2 - nX 2

55 - 5  32 a  4 - 1  3  1.0







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

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.

70 - 60  1.0 55 - 45

and  Y  =   1.0   +   1.0X.   The correlation coefficient: r



SOUTHWESTERN UNIVERSITY: B

1

and X = 15, Y = 20, XY = 2 2 70, X  = 55, Y  = 94,  X  = 3, Y  = 4. Then: 70 - 5  4  3

Selection of an appropriate time   series   forecasting model based upon a plot of 94 the data. - 1  20 - 1  70



The 5  - 2importance   of combining   a   qualitative 94 - 20 - 70   1.333  1.15 model   with   a   quantitative 3 model   in   situations   where technological   change   is occurring. 4.62 Using software, the  Games lost = 6.41 + 0.533  

a  Y - bX

b

 Y 2 - a  Y - b  XY  n-2

Syx 

regression equation is: 

Given: Y = a + bX where: b

Standard   error   of   the estimate:

n  XY -  X  Y

Game

 n  X 2 - (  X ) 2  n  Y 2 - (  Y ) 2      5  70 - 15  20 5  55 - 152  5  94 - 202     50 50   0.845 50  70 59.16



350 - 300 275 - 225 470 - 400 

 1  2  3  4  5 Total

Model y y y y y

= = = = =

30,713 37,640 36,940 22,567 30,440

+ + + + +

2,534x 2,146x 1,560x 2,143x 3,146x

(where  y  = attendance and  x = time) 2. Revenue   in   2010   = (239,000)   ($50/ticket)   = $11,950,000 Revenue   in   2011   = (250,530)   ($52.50/ticket) = $13,152,825 3. In   games   2   and   5,   the forecast   for   2011   exceeds stadium   capacity.   With   this appearing to be a continuing trend, the time has come for a new or expanded stadium. 2

DIGITAL CELL  PHONE, INC.

Objectives:

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

Totals

y

x2

xy

   48    87  1,44  1,79  2,29  2,93  3,48  3,44  3,99  4,96  5,35  6,30  7,47  7,37  8,10  8,03  8,63 10,31  9,65  9,96 10,18 11,57 12,69 14,08 15,20 15,52 16,52 16,88 18,21 18,15 19,43 18,49 19,30 19,75 22,12 23,61

 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

480 436 482 448 458 489 498 430 444 496 487 525 575 527 540 502 508 573 508 498 485 526 552 587 608 597 612 603 628 605 627 578 585 581 632 656

    1     4     9    16    25    36    49    64    81   100   121   144   169   196   225   256   289   324   361   400   441   484   529   576   625   676   729   784   841   900   961 1,024 1,089 1,156 1,225 1,296

666

19,366

16,206

Average 18.5

537.9

450.2

378,6

10,5

CHAPTER 4 F O R E C A S T I N G

378,661 - 36 �18.5Analysis for Management, �537.9 20390.0   5.25 10th  3885.0   Hall �x - nx 16,206 - (36 �18.52 ) ed.,   Prentice Publishing. a  y - bx  537.9 - 5.25 �18.5  440.85 Cases   =   432.28   +   5.73 n �xy - �x �y (time),  r2  =   .93,   MAD   = r [ n �x 2 - ( �x )2 ][ n �y 2 - ( �y)212.84 ]

b

�xy - nx y  

2

2



(36)(378,661) - c) Additive seasonal model, (666)(19,366) Cases   =   444.29   +   5.06 2 [(36) �(16,206) - (666) ][(36)(10,558,246) - (19,366)2 ] (time),  r2  =   .86,   MAD   = 13,631,796 - 12,897,756 20.02  [(583,416) - (443,556)][380,096,856) - (375,041,956)] d)   Additive   seasonal   model, with   centered   moving 737,040 734,040   averages. [139,860][5,054,900] 706,978,314,000 Cases   =   431.31   +   5.72 734,040 (time),  r2  =   .94,   MAD   =   .873 840,820 12.28 

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 r2 of .76. However, a more precise forecast   can   be   developed addressing   the   seasonality issue,   which   is   done   below. Methods a and c yield  r2  of . 85 and .86, respectively, and methods b and d, which also center   the   seasonal adjustment, yield r2 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),  r2  =   .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

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.

VIDEO CASE STUDY 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.  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 entrée; 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   entrée   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.

2003 2004 2005 2006 2007 2008 2009

4. Other   predictors   of   cafe sales could include season of year   (weather);   hotel occupancy; spring break from colleges;   beef   prices; promotional budget; etc.

47

51.80 54.92 69.70 68.90 63.72 84.73 78.74

Southeast Airline D Year

Airframe Cost per Aircraft

2003 13.29 2004 25.15 5. Y  a + bx 2005 32.18 2006 31.78 Month Advertising Guest Count 2007 25.34 X Y 2008 32.78  1 14 21 2009 35.56  2 17 24  3 25 27 Utilizing   the   software  4 25 32 package   provided   with   this  5 35 29 text,   we   can   develop   the  6

35

37

 7

45

43

 8

50

43

following   regression equations for the variables of interest:

Northern   Airlines— Airframe Maintenance Cost:  Cost = 36.10 + 0.0026  9 60 54  Airframe age  Coefficient   of 10 60 66 determination = 0.7695 Totals 36 37  Coefficient   of 6 6 correlation = 0.8772 Averag 36.6   37.6 e Northern   Airlines—Engine Maintenance Cost:  Cost = 20.57 + 0.0026  Airframe age 15,772 - 10 �36.6 �37.6  Coefficient   of b  0.7996 �.8 15,910 - 10 �36.62 determination = 0.6124   of a  37.6 - 0.7996 �36.6  8.3363 �8.3  Coefficient correlation = 0.7825 Y  8.3363 + 0.7996 X At $65,000; y  8.3 + .8 (65)   8.3 +  52 = 60.3, or 60,300 guests.

For   the   instructor   who   asks other questions than this one:          r2  0.8869 Std. error  5.062

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 THE NORTH­SOUTH   Coefficient   of AIRLINE determination = 0.4600  Coefficient   of Northern Airline Data correlation = 0.6782

ADDITIONAL CASE STUDY*

Year

43.4 38.5 51.4 58.7 45.4 50.2 79.6

Airframe Cost per Aircraft

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

Engine per Airc

18.8 31.5 40.4 22.1 19.6 32.5 38.0

48

CHAPTER 4 F O R E C A S T I N G

The   following   graphs portray   both   the   actual   data and   the   regression   lines   for airframe   and   engine maintenance   costs   for   both airlines.

*This case study appears on our companion Web site, at  www.pearsonhighered.com/heize r, and at www.myomlab.com.

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 day­to­day 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  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

Copyright © 2011 Pearson Education, Inc. publishing as Prentice Hall.

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.

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