Om Heizer Om10 Ism 04

Chapter FORECASTING Discussion Questions 1.? Qualitative clay sculptures incorporate internal factors into the promiseing model. Qualitative models argon useful when subjective factors be important. When quantitative entropy ar hard to obtain, qualitative models may be appropriate. 2.? Approaches argon qualitative and quantitative. Qualitative is relatively subjective quantitative uses numerical models. 3.? Short- snip (under 3 months), medium-range (3 months to 3 age), and long-range (over 3 years). 4.? The step that should be used to cook a apprehending system argon (a)?Determine the purpose and use of the enter (b)? Select the item or quantities that argon to be reckoned (c)? Determine the time horizon of the promise (d)? Select the type of forecasting model to be used (e)? Gather the necessary data (f)? Validate the forecasting model (g)? Make the forecast (h)? Implement and tax the results 5.? Any three of sales planning, production planning and budgeting, cash budgeting, analyzing various direct plans. 6.? There is no mechanism for growth in these models they be built exclusively from historical charter determines. Such manners will always lag sheers. .? exponential smoothing is a chargeed moving average out where wholly previous values are weighted with a set of weights that dec distribution channel exponenti eachy. 8.? frenzied, MSE, and MAPE are common measures of forecast accuracy. To remember the more(prenominal) faultless forecasting model, forecast with each tool for several period of times where the demand egress is known, and calculate MSE, MAPE, or hallucinating for each. The sm completelyer illusion exhibits the better forecast. 9.? The Delphi technique involves (a)? Assembling a root of experts in much(prenominal) a manner as to preclude direct communication between recognisable members of the group (b)?Assembling the responses of each expert to the questions or problems of interest (c)? meaning upmari zing 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 completelittle additional convergence is likely if the process is continued. 0.? 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 mould the quantity being forecast. 11.? A time series is a sequence of evenly spaced data points with the four components of effort, seasonality, cyclical, and random variation. 12.? When the smoothing constant, (, is large (close to 1. 0), more weight is given to recent d ata 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, etcetera , of the application. 14.? exponential smoothing weighs all previous values with a set of weights that decline exponentially. It stop place a full weight on the closely recent period (with an important of 1. 0). This, in effect, is the naive come along, which places all its emphasis on last periods actual demand. 15.? Adaptive forecasting refers to computer monitoring of tracking signals and self- modification 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 signifi shagt 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. 18.? mugwump variable (x) is said to explain variations in the dependent variable (y). 19.? Nearly every industry has seasonality. The seasonality essential be filtered out for safe medium-range planning (of production and inventory) and performance evaluation. 20.? There are many examples.Demand for birthday suit 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 limit forecasts, and we essential diminish our conviction on the forecasts. Ethical Dilemma This exercise, derived from an actual situation, deals as much with ethics as with forecasting. Here are a few points to consider No one likes a system they dont understand, and around 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 rescue input to the process (such as smoothing constants) and all data conduct to be open to everyone. The smoothing constants could be selected by an agreed-upon criteria (such as lowest MAD) or could be based on input from experts on the board as well as the college. Abuse of the system is tied to assigning alphas based on what results they yield, rather than what alphas make the most sense. retroflection 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 impingement on results as well. Active Model Exercises* active MODEL 4. 1 Moving Averages 1.? What does the graphical recordical record side like when n = 1? The forecast graph mirrors the data graph but one period later. 2.? What happens to the graph as the twist of periods in the moving average increases? The forecast graph becomes shorter and smo different. 3.? What value for n minimizes the MAD for this data? n = 1 (a naive forecast) ACTIVE MODEL 4. 2 Exponential Smoothing 1.? What happens to the graph when alpha equals zero? The graph is a straight line.The forecast is the equivalent in each period. 2.? What happens to the graph when alpha equals one? The forecast follows the same pattern as the demand (except for the first forecast) but is graduation by one period. This is a naive forecast. 3.? Generalize what happens to a forecast as alpha increases. As alpha increases the forecast is more sensitive to changes in demand. *Active Models 4. 1, 4. 2, 4. 3, and 4. 4 appear on our entanglement site, www. pearsonhighered. com/heizer. 4.? At what level of alpha is the mean positive deviation (MAD) minimized? alpha = . 16 ACTIVE MODEL 4. 3 Exponential Smoothing with inclination Adjustment .? Scroll through different values for alpha and beta. Which smoothing constant appears to have the keene r effect on the graph? alpha 2.? With beta set to zero, find the better(p) alpha and observe the MAD. Now find the best beta. Observe the MAD. Does the addition of a trend improve the forecast? alpha = . 11, MAD = 2. 59 beta above . 6 changes the MAD (by a little) to 2. 54. ACTIVE MODEL 4. 4 apparent motion Projections 1.? What is the annual trend in the data? 10. 54 2.? mathematical function the scroll criterions 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. End-of-Chapter Problems pic (b) dull calendar week of Pints Used Moving Average distinguished 31 360 family line 7 389 381 ( . 1 = ? 38. 1 September 14 410 368 ( . 3 = cx. 4 September 21 381 374 ( . 6 = 224. 4 September 28 368 372. October 5 374 depend 372. 9 (c) projecting wrongdoing Week of Pints porten d misunderstanding ( . 20 promise August 31 360 360 0 0 360 September 7 389 360 29 5. 8 365. 8 September 14 410 365. 8 44. 2 8. 84 374. 64 September 21 381 374. 64 6. 36 1. 272 375. 12 September 28 368 375. 912 7. 912 1. 5824 374. 3296 October 5 374 374. 3296 . 3296 . 06592 374. 2636 The forecast is 374. 26. (d)? The three-year moving average appears to give better results. pic pic credulous 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. TEACHING NOTE Notice how well exponential smoothing forecasts the naive. pic (c)? The banking industry has a great deal of seasonality in its processing requirements pic b) Two- course of instruction course of instruction Mileage Moving Average Error Error 1 3,000 2 4,000 3 3, cd 3, vitamin D century light speed 4 3,800 3,700 100 100 5 3,700 3,600 100 100 Totals 100 ccc pic 4. 5? (c)? Weighte d 2 year M. A. ith . 6 weight for most recent year. family Mileage Forecast Error Error 1 3,000 2 4,000 3 3,400 3,600 two hundred 200 4 3,800 3,640 160 160 5 3,700 3,640 60 60 420 Forecast for year 6 is 3,740 miles. pic 4. 5? (d) Forecast Error ( New Year Mileage Forecast Error ( = . 50 Forecast 1 3,000 3,000 ?0 0 3,000 2 4,000 3,000 1,000 500 3,500 3 3,400 3,500 100 50 3,450 4 3,800 3,450 350 one hundred seventy-five 3,625 5 3,700 3,625 75 ? 38 3,663 Total 1,325 The forecast is 3,663 miles. 4. 6 Y Sales X intent X2 XY January 20 1 1 20 February 21 2 4 42 March 15 3 9 45 April 14 4 16 56 May 13 5 25 65 June 16 6 36 96 July 17 7 49 119 August 18 8 64 144 September 20 9 81 180 October 20 10 100 200 November 21 11 121 231 celestial latitude 23 12 144 276 Sum 18 78 650 1,474 Average ? 18. 2 6. 5 (a) pic (b)? i? NaiveThe coming January = December = 23 ii? 3-month moving (20 + 21 + 23)/3 = 21. 33 iii? 6-month weighted (0. 1 ( 1 7) + (. 1 ( 18) + (0. 1 ( 20) + (0. 2 ( 20) + (0. 2 ( 21) + (0. 3 ( 23)/1. 0 = 20. 6 iv? Exponential smoothing with alpha = 0. 3 pic v? Trend? pic pic Forecast = 15. 73? +?. 38(13) = 20. 67, where next January is the 13th month. (c)? single trend provides an equation that can extend beyond one month 4. 7? Present = achievement (week) 6. a) So where pic )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, thus the forecast becomes 56. 3, or 56 patients. pic pic Temperature 2 day M. A. Error(Error)2 strong % Error 93 94 93 93. 5 0. 5 ? 0. 25 100(. 5/93) = 0. 54% 95 93. 5 1. 5 ? 2. 25 100(1. 5/95) = 1. 58% 96 94. 0 2. 0 ? 4. 00 100(2/96) = 2. 08% 88 95. 5 7. 56. 25 100(7. 5/88) = 8. 52% 90 92. 0 2. 0 ? 4. 00 100(2/90) = 2. 22% 13. 5 66. 75 14. 94% MAD = 13. 5/5 = 2. 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 three-month averages appear in the table the results appear in the figure below. pic (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. pic (c)? The forecasts are about the same. pic 4. 12? t sidereal day Actual Forecast Demand Demand 1 Monday 88 88 2 Tuesday 72 88 3 Wednesday 68 84 4 Thursday 48 80 5 Friday 72 ( reaction Ft = Ft1 + ((At1 Ft1) allow ( = . 25. Let Monday forecast demand = 88 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 Exponential Absolute Year Demand Smoothing ( = 0. aberration 1 45 41 4. 0 2 50 41. 0 + 0. 6(4541) = 43. 4 6. 6 3 52 43. 4 + 0. 6(5043. 4) = 47. 4 4. 6 4 56 47. 4 + 0. 6(5247. 4) = 50. 2 5. 8 5 58 50. 2 + 0. 6(5650. 2) = 53. 7 4. 3 6 ? 53. 7 + 0. 6(5853. 7) = 56. 3 ( = 25. 3 MAD = 5. 06 Exponential smoothing, ( = 0. 9 Exponential Absolute Year Demand Smoothing ( = 0. exit 1 45 41 4. 0 2 50 41. 0 + 0. 9(4541) = 44. 6 5. 4 3 52 44. 6 + 0. 9(5044. 6 ) = 49. 5 2. 5 4 56 49. 5 + 0. 9(5249. 5) = 51. 8 4. 2 5 58 51. 8 + 0. 9(5651. 8) = 55. 6 2. 4 6 ? 55. 6 + 0. 9(5855. 6) = 57. 8 ( = 18. 5 MAD = 3. 7 (b)? 3-year moving average Three-Year Absolute Year Demand Moving Average deviance 1 45 2 50 3 52 4 56 (45 + 50 + 52)/3 = 49 7 5 58 (50 + 52 + 56)/3 = 52. 7 5. 3 6 ? (52 + 56 + 58)/3 = 55. 3 ( = 12. 3 MAD = 6. 2 (c)? Trend protrusion Absolute Year Demand Trend Projection difference of opinion 1 45 42. 6 + 3. 2 ( 1 = 45. 8 0. 8 2 50 42. 6 + 3. 2 ( 2 = 49. 0 1. 0 3 52 42. 6 + 3. 2 ( 3 = 52. 2 0. 2 4 56 42. 6 + 3. 2 ( 4 = 55. 4 0. 5 58 42. 6 + 3. 2 ( 5 = 58. 6 0. 6 6 ? 42. 6 + 3. 2 ( 6 = 61. 8 ( = 3. 2 MAD = 0. 64 pic X Y XY X2 1 45 45 1 2 50 100 4 3 52 156 9 4 56 224 16 5 58 290 25 Then (X = 15, (Y = 261, (XY = 815, (X2 = 55, pic= 3, pic= 52. 2 Therefore pic (d)? Comparing the results of the forecasting methodologies for parts (a), (b), and (c). Forecast Methodology MAD Exponential smoothing, ( = 0. 5. 06 Exponential smoothing, ( = 0. 9 3. 7 3-year moving average 6. 2 Trend projection 0. 64 Based on a mean absolute deviation criterion, the trend projection is to be preferred over the exponential smoothing with ( = 0. 6, exponential smoothing with ( = 0. 9, or the 3-year moving average forecast methodologies. 4. 14 Method 1MAD (0. 20 + 0. 05 + 0. 05 + 0. 20)/4 = . 125 ( better MSE (0. 04 + 0. 0025 + 0. 0025 + 0. 04)/4 = . 021 Method 2MAD (0. 1 + 0. 20 + 0. 10 + 0. 11) / 4 = . 1275 MSE (0. 01 + 0. 04 + 0. 01 + 0. 0121) / 4 = . 018 ( better 4. 15 Forecast Three-Year Absolute Year Sales Moving Average Deviation 2005 450 2006 495 2007 518 20 08 563 (450 + 495 + 518)/3 = 487. 7 75. 3 2009 584 (495 + 518 + 563)/3 = 525. 3 58. 7 2010 (518 + 563 + 584)/3 = 555. 0 ( = 134 MAD = 67 4. 16 Year Time Period X Sales Y X2 XY 2005 1 450 1 450 2006 2 495 4 990 2007 3 518 9 1554 2008 4 563 16 2252 2009 5 584 25 2920 ( = 2610 ( = 55 ( = 8166 pic pic Year Sales Forecast Trend Absolute Deviation 2005 450 454. 8 4. 8 2006 495 488. 4 6. 2007 518 522. 0 4. 0 2008 563 555. 6 7. 4 2009 584 589. 2 5. 2 2010 622. 8 ( = 28 MAD = 5. 6 4. 17 Forecast Exponential Absolute Year Sales Smoothing ( = 0. 6 Deviation 2005 450 410. 0 40. 2006 495 410 + 0. 6(450 410) = 434. 0 61. 0 2007 518 434 + 0. 6(495 434) = 470. 6 47. 4 2008 563 470. 6 + 0. 6(518 470. 6) = 499. 0 64. 0 2009 584 499 + 0. 6(563 499) = 537. 4 46. 6 2010 537. 4 + 0. 6(584 537. 4) = 565. 6 ( = 259 MAD = 51. 8 Forecast Exponential Absolute Year Sales Smoothing ( = 0. Deviation 2005 450 410. 0 40. 0 2006 495 410 + 0 . 9(450 410) = 446. 0 49. 0 2007 518 446 + 0. 9(495 446) = 490. 1 27. 9 2008 563 490. 1 + 0. 9(518 490. 1) = 515. 2 47. 8 2009 584 515. 2 + 0. 9(563 515. 2) = 558. 2 25. 8 2010 558. 2 + 0. 9(584 558. 2) = 581. 4 ( = 190. 5 MAD = 38. 1 (Refer to Solved Problem 4. 1)For ( = 0. 3, absolute deviations for 20052009 are 40. 0, 73. 0, 74. 1, 96. 9, 88. 8, respectively. So the MAD = 372. 8/5 = 74. 6. pic Because it gives the lowest MAD, the smoothing constant of ( = 0. 9 gives the most accurate forecast. 4. 18? We remove to find the smoothing constant (. We know in general that Ft = Ft1 + ((At1 Ft1) t = 2, 3, 4. Choose either t = 3 or t = 4 (t = 2 wont let us find ( because F2 = 50 = 50 + ((50 50) holds for any (). Lets pick t = 3. Then F3 = 48 = 50 + ((42 50) or 48 = 50 + 42( 50( or 2 = 8( 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. 4. 19? Trend set exp onential smoothing ( = 0. 1, ( = 0. 2 Unadjusted Adjusted Month Income Forecast Trend Forecast ErrorError2 February 70. 0 65. 0 0. 0 65 ? 5. 0 ? 25. 0 March 68. 5 65. 5 0. 1 65. 6 ? 2. 9 ? 8. 4 April 64. 8 65. 9 0. 16 66. 05 ? 1. 2 ? 1. 6 May 71. 7 65. 92 0. 13 66. 06 ? 5. 6 ? 31. 9 June 71. 66. 62 0. 25 66. 87 ? 4. 4 ? 19. 7 July 72. 8 67. 31 0. 33 67. 64 ? 5. 2 ? 26. 6 August 68. 16 68. 60 24. 3 113. 2 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 net this problem, a period 0, which contains the initial forecast and initial trend, must be added. 4. 20? Trend adjusted exponential smoothing ( = 0. 1, ( = 0. 8 pic pic pic pic pic pic pic pic pic pic pic pic 4. 23? Students must determine the naive forecast for the four months.The naive forecast for March is the February actual of 83, etc. (a) Actual Forecast Error % Error March ci great hundred 19 100 (19/101) = 18. 81% April ? 96 114 18 100 (18/96) ? = 18. 75% May ? 89 cx 21 100 (21/89) ? = 23. 60% June 108 108 ? 0 100 (0/108) ? = 0% 58 61. 16% pic (b) Actual Naive Error % Error March 101 ? 83 18 100 (18/101) = 17. 82% April ? 96 101 ? 100 (5/96) ? = 5. 21% May ? 89 ? 96 ? 7 100 (7/89) ? =? 7. 87% June 108 ? 89 19 100 (19/108) = 17. 59% 49 48. 49% pic Naive outperforms management. (c)? MAD for the managers technique is 14. 5, spell 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 pic Assume Appearances X Demand Y X2 Y2 XY 3 3 9 9 9 4 6 16 36 24 7 7 49 49 49 6 5 36 25 30 8 10 64 100 80 5 7 25 49 35 9 ? (X = 33, (Y = 38, (XY = 227, (X2 = 199, pic= 5. 5, pic= 6. 33. Therefore pic The following figure sho ws both the data and the resulting equation pic (c) If there are nine performances by Stone Temple Pilots, the prognosticated sales are (d) R = . 82 is the correlation coefficient, and R2 = . 68 means 68% of the variation in sales can be explained by TV appearances. 4. 25? Number of Accidents Month (y) x xy x2 January 30 1 30 1 February 40 2 80 4 March 60 3 180 9 April 90 4 360 16 ? Totals 220 pic The regression line is y = 5 + 20x. The forecast for May (x = 5) is y = 5 + 20(5) = one hundred five. 4. 26 Season Year1 Year2 Average Average Seasonal Year3 Demand Demand Year1(Year2 Season major power Demand Demand Demand Fall 200 250 225. 0 250 0. 90 270 Winter 350 300 325. 250 1. 30 390 Spring 150 165 157. 5 250 0. 63 189 Summer 300 285 292. 5 250 1. 17 351 4. 27 Winter Spring Summer Fall 2006 1,400 1,500 1,000 600 2007 1,200 1,400 2,100 750 2008 1,000 1,600 2,000 650 2009 900 1,500 1,900 500 4,500 6,000 7,000 2,500 4. 28 Av erage Average Quarterly Seasonal Quarter 2007 2008 2009 Demand Demand Index Winter 73 65 89 75. 67 106. 67 0. 709 Spring 104 82 146 110. 67 106. 67 1. 037 Summer 168 124 205 165. 67 106. 67 1. 553 Fall 74 52 98 74. 67 106. 67 0. 700 4. 29? 2011 is 25 years beyond 1986. Therefore, the 2011 quarter numbers are 101 through 104. (5) (2) (3) (4) Adjusted (1) Quarter Forecast Seasonal Forecast Quarter Number (77 + . 3Q) portion (3) ( (4) Winter 101 great hundred. 43 . 8 96. 344 Spring 102 120. 86 1. 1 132. 946 Summer 103 121. 29 1. 4 169. 806 Fall 104 121. 72 . 7 85. 204 4. 30? give 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 4. 33? (a)? See the table below. For next year (x = 6), the number of transistors (in millions) is forecasted as y = 126 + 18(6) = 126 + 108 = 234. Then y = a + bx, where y = number sold, x = equipment casualty, and 4. 32? a) x y xy x2 16 330 5,280 256 12 270 3,240 144 18 380 6,840 324 14 300 4,200 196 60 1,280 19,560 920 So at x = 2. 80, y = 1,454. 6 277. 6($2. 80) = 677. 32. Now round to the nearest integer Answer 677 lattes. pic (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. Table for Problem 4. 33 Year Transistors (x) (y) xy x2 126 + 18x Error Error2 % Error ? 1 140 ? 140 ? 1 144 4 ? 16 100 (4/140)? = 2. 86% ? 2 160 ? 320 ? 4 162 2 4 100 (2/160)? = 1. 25% ? 3 190 ? 570 ? 9 180 10 100 100 (10/190) = 5. 26% ? 4 200 ? 800 16 198 ? 2 4 100 (2/200) = 1. 00% ? 210 1,050 25 216 6 ? 36 100 (6/210)? = 2. 86% Totals 15 900 2,800 (b)? MSE = 160/5 = 32 (c)? MAPE = 13. 23%/5 = 2. 65% 4. 34? Y = 7. 5 + 3. 5X1 + 4. 5X2 + 2. 5X3 (a)? 28 (b)? 43 (c)? 58 4. 35? (a)? pic = 13,473 + 37. 65(1860) = 83,502 (b)? The predicted change price is $83,502, but this is the average price for a house of this size. There a re other factors besides square footage that will impact the selling price of a house. If such a house sold for $95,000, accordingly these other factors could be bring 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)? precondition Y = 90 + 48. 5X1 + 0. 4X2 where pic If Number of days on the road ( X1 = 5 and distance operateed ( X2 = 300 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. (c)?A number of other variables should be included, such as 1.? the type of travel (air or car) 2.? conference fees, if any 3.? costs of entertaining customers 4.? other transportation costscab, limousine, special tolls, or parking In addition, the correlation coefficient of 0. 68 is not exceptionally high. It indicates that the model explains approximately 46% of the general variation in trip cost. This correlation coefficient would suggest that the model is not a particularly good one. 4. 37? (a, b) Period Demand Forecast Error Running sum error 1 20 20 0. 00 0. 00 0. 00 2 21 20 1. 00 1. 0 1. 00 3 28 20. 5 7. 50 8. 50 7. 50 4 37 24. 25 12. 75 21. 25 12. 75 5 25 30. 63 5. 63 15. 63 5. 63 6 29 27. 81 1. 19 16. 82 1. 19 7 36 28. 41 7. 59 24. 41 7. 59 8 22 32. 20 10. 20 14. 21 10. 20 9 25 27. 11 2. 10 12. 10 2. 10 10 28 26. 05 1. 95 14. 05 1. 95 MADpic5. 00 Cumulative error = 14. 05 MAD = 5? Tracking = 14. 05/5 ( 2. 82 4. 38? (a)? to the lowest degree 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 Year X Patients Y X2 Y2 XY ? 1 ? 36 1 ? 1,296 36 ? 2 ? 33 ? 1,089 66 ? 3 ? 40 9 ? 1,600 ? 120 ? 4 ? 41 ? 16 ? 1,681 ? 164 ? 5 ? 40 ? 25 ? 1,600 ? 200 ? 6 ? 55 ? 36 ? 3,025 ? 330 ? 7 ? 60 ? 49 ? 3,600 ? 420 ? 8 ? 54 ? 64 ? 2,916 ? 432 ? 9 ? 58 ? 81 ? 3,364 ? 522 10 ? 61 100 ? 3,721 ? 10 55 478 X Y Forecast Deviation Deviation ? 1 36 29. 8 + 3. 28 ( ? 1 = 33. 1 ? 2. 9 2. 9 ? 2 33 29. 8 + 3. 28 ( ? 2 = 36. 3 3. 3 3. 3 ? 3 40 29. 8 + 3. 28 ( ? 3 = 39. 6 ? 0. 4 0. 4 ? 4 41 29. 8 + 3. 28 ( ? 4 = 42. 9 1. 9 1. 9 ? 5 40 29. 8 + 3. 28 ( ? 5 = 46. 2 6. 2 6. 2 ? 6 55 29. 8 + 3. 28 ( ? 6 = 49. 4 ? 5. 6 5. 6 ? 7 60 29. 8 + 3. 28 ( ? 7 = 52. 7 ? 7. 3 7. 3 ? 54 29. 8 + 3. 28 ( ? 8 = 56. 1 2. 1 2. 1 ? 9 58 29. 8 + 3. 28 ( ? 9 = 59. 3 1. 3 1. 3 10 61 29. 8 + 3. 28 ( 10 = 62. 6 1. 6 1. 6 ( = 32. 6 MAD = 3. 26 The MAD is 3. 26this 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. 67. 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 Crime Patients Year Rate X Y X2 Y2 XY ? 1 ? 58. 3 ? 36 ? 3,398. 9 ? 1,296 ? 2,098. 8 ? 2 ? 61. 1 ? 33 ? 3,733. 2 ? 1,089 ? 2,016. 3 ? 3 ? 73. ? 40 ? 5,387. 6 ? 1,600 ? 2,936. 0 ? 4 ? 75. 7 ? 41 ? 5,730. 5 ? 1,681 ? 3,103. 7 ? 5 ? 81. 1 ? 40 ? 6,577. 2 ? 1,600 ? 3,244. 0 ? 6 ? 89. 0 ? 55 ? 7,921. 0 ? 3,025 ? 4,895. 0 ? 7 101. 1 ? 60 10,221. 2 ? 3,600 ? 6,066. 0 ? 8 ? 94. 8 ? 54 ? 8,987. 0 ? 2,916 ? 5,119. 2 ? 9 103. 3 ? 58 10,670. 9 ? 3,364 ? 5,991. 4 10 116 . 2 ? 61 13,502. 4 ? 3,721 ? 7,088. 2 Column 854. 478 Totals months) (Millions) (1,000,000s) Year (X) (Y) X2 Y2 XY ? 1 ? 7 1. 5 ? 49 ? 2. 25 10. 5 ? 2 ? 2 1. 0 4 ? 1. 00 ? 2. 0 ? 3 ? 6 1. 3 ? 36 ? 1. 69 ? 7. 8 ? 4 ? 4 1. 5 ? 16 ? 2. 25 ? 6. 0 ? 5 14 2. 5 196 ? 6. 25 35. 0 ? 6 15 2. 7 225 ? 7. 9 40. 5 ? 7 16 2. 4 256 ? 5. 76 38. 4 ? 8 12 2. 0 144 ? 4. 00 24. 0 ? 9 14 2. 7 196 ? 7. 29 37. 8 10 20 4. 4 400 19. 36 88. 0 11 15 3. 4 225 11. 56 51. 0 12 ? 7 1. 7 ? 49 ? 2. 89 11. 9 Given Y = a + bX where pic and (X = 132, (Y = 27. 1, (XY = 352. 9, (X2 = 1796, (Y2 = 71. 59, pic = 11, pic= 2. 26. Then pic andY = 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 cartel in this forecast, however, because the number of tourists (zero) is out side the range of data used to develop the model. (e)? The standard error of the estimate is given by (f)? The correlation coefficient and the coefficient of determination are given by pic 4. 42? (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. pic 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 Ft+1 = At11 That is forecastnext period = actualone year earlier to account for seasonality. But this ignores the trend. One very good approach would be to calculate the increase from each month last year to each month this year, sum all 12 increases, and divide by 12.The forecast for next year would equal the value for the same month this year plus the average increase over the 12 months of last year. (c) Using this model, the January forecast for next year becomes pic where 148 = total monthly increases from last year to this year. The forecasts for each of the months of next year then become Jan. 29 July. 56 Feb. 26 Aug. 53 Mar. 32 Sep. 45 Apr. 35 Oct. 35 May. 42 Nov. 38 Jun. 50 Dec. 29 Both history and forecast for the next year are shown in the accompanying figure pic 4. 3? (a) and (b) See the following table Actual Smoothed Smoothed Week Value Value Forecast Value Forecast t A(t) Ft (( = 0. 2) Error Ft (( = 0. 6)Error 1 50 +50. 0 ? +0. 0 +50. 0 ? +0. 0 2 35 +50. 0 15. 0 +50. 0 15. 0 3 25 +47. 0 22. 0 +41. 0 16. 0 4 40 +42. 6 ? 2. 6 +31. 4 ? +8. 6 5 45 +42. 1 ? 2. 9 +36. 6 ? +8. 6 35 +42. 7 ? 7. 7 +41. 6 ? 6. 6 7 20 +41. 1 21. 1 +37. 6 17. 6 8 30 +36. 9 ? 6. 9 +27. 1 ? +2. 9 9 35 +35. 5 ? 0. 5 +28. 8 ? +6. 2 10 20 +35. 4 15. 4 +32. 5 12. 5 11 15 +32. 3 17. 3 +25. 0 10. 0 12 40 +28. 9 +11. 1 +19. 0 +21. 0 13 55 +31. 1 +23. 9 +31. 6 +23. 4 14 35 +35. 9 ? 0. 9 +45. 6 10. 6 15 25 +36. 7 10. 7 +39. 3 14. 3 16 55 +33. 6 +21. 4 +30. 7 +24. 3 17 55 +37. 8 +17. 2 +45. 3 ? +9. 7 18 40 +41. 3 ? 1. 3 +51. 1 11. 1 19 35 +41. 0 ? 6. 0 +44. 4 ? 9. 4 20 60 +39. 8 +20. 2 +38. 8 +21. 2 21 75 +43. 9 +31. 1 +51. 5 +23. 5 22 50 +50. 1 ? 0. 1 +65. 6 15. 23 40 +50. 1 10. 1 +56. 2 16. 2 24 65 +48. 1 +16. 9 +46. 5 +18. 5 25 +51. 4 +57. 6 MAD = 11. 8 MAD = 13. 45 (c)? Students should mention 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 somewhat 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. 4. 4 Week Actual Value Smoothed Value Trend Estimate Forecast F orecast t At Ft (( = 0. 3) Tt (( = 0. 2) FITt Error ? 1 50. 000 50. 000 ? 0. 000 50. 000 0. 000 ? 2 35. 000 50. 000 ? 0. 000 50. 000 15. 000 ? 3 25. 000 45. 500 0. 900 44. 600 19. 600 ? 4 40. 000 38. 720 2. 076 36. 644 3. 56 ? 5 45. 000 37. 651 1. 875 35. 776 9. 224 ? 6 35. 000 38. 543 1. 321 37. 222 ? 2. 222 ? 7 20. 000 36. 555 1. 455 35. 101 15. 101 ? 8 30. 000 30. 571 2. 361 28. 210 1. 790 ? 9 35. 000 28. 747 2. 253 26. 494 8. 506 10 20. 000 29. 046 1. 743 27. 03 ? 7. 303 11 15. 000 25. 112 2. 181 22. 931 ? 7. 931 12 40. 000 20. 552 2. 657 17. 895 ? 22. 105 13 55. 000 24. 526 1. 331 23. 196 ? 31. 804 14 35. 000 32. 737 ? 0. 578 33. 315 1. 685 15 25. 000 33. 820 ? 0. 679 34. 499 ? 9. 499 16 55. 000 31. 649 ? 0. 109 31. 58 ? 23. 242 17 55. 000 38. 731 ? 1. 503 40. 234 ? 14. 766 18 40. 000 44. 664 ? 2. 389 47. 053 ? 7. 053 19 35. 000 44. 937 ? 1. 966 46. 903 11. 903 20 60. 000 43. 332 ? 1. 252 44. 584 ? 15. 416 21 75. 000 49. 209 ? 2. 177 51. 386 ? 23. 61 4 22 50. 000 58. 470 ? 3. 94 62. 064 12. 064 23 40. 000 58. 445 ? 2. 870 61. 315 21. 315 24 65. 000 54. 920 ? 1. 591 56. 511 8. 489 25 59. 058 ? 2. 100 61. 158 To evaluate the trend adjusted exponential smoothing model, actual week 25 calls are compared to the forecasted value. The model appears to be producing a forecast approximately mid-range between that given by simple exponential smoothing using ( = 0. 2 and ( = 0. 6.Trend adjustment does not appear to give any significant improvement. 4. 45 Month At Ft At Ft (At Ft) May 100 100 0 0 June 80 104 24 24 July 110 99 11 11 August 115 101 14 14 September 105 104 1 1 October 110 104 6 6 November 125 105 20 20 December 120 109 11 11 Sum 87 Sum 39 4. 46 (a) X Y X2 Y2 XY ? 421 ? 2. 90 ? 177241 8. 41 ? 1220. 9 ? 377 ? 2. 93 ? 142129 8. 58 ? 1104. 6 ? 585 ? 3. 00 ? 342225 9. 00 ? 1755. 0 ? 690 ? 3. 45 ? 476100 ? 11. 90 ? 2380. 5 ? 608 ? 3. 66 ? 369664 ? 13. 40 ? 2225. 3 ? 390 ? 2. 88 ? 52100 8. 29 ? 1123. 2 ? 415 ? 2. 15 ? 172225 4. 62 892. 3 ? 481 ? 2. 53 ? 231361 6. 40 ? 1216. 9 ? 729 ? 3. 22 ? 531441 ? 10. 37 ? 2347. 4 ? 501 ? 1. 99 ? 251001 3. 96 997. 0 ? 613 ? 2. 75 ? 375769 7. 56 ? 1685. 8 ? 709 ? 3. 90 ? 502681 ? 15. 21 ? 2765. 1 ? 366 ? 1. 60 ? 133956 2. 56 585. 6 Column 6885 36. 6 totals January 400 February 380 400 20. 0 March 410 398 12. 0 April 375 399. 2 396. 67 24. 2 21. 67 May 405 396. 8 388. 33 8. 22 16. 67 MAD = 16. 11 19. 17 (d)Note that Amit has more forecast observations, while Barbaras moving average does not start until month 4. Also note that the MAD for Amit is an average of 4 numbers, while Barbaras is only 2. Amits MAD for exponential smoothing (16. 1) is lower than that of Barbaras moving average (19. 17). So his forecast seems to be better. 4. 48? (a) Quarter Contracts X Sales Y X2 Y2 XY 1 ? 153 ? 8 ? 23,409 ? 64 ? 1,224 2 ? 172 10 ? 29,584 100 ? 1,720 3 ? 197 15 ? 38,809 225 ? 2,955 4 ? 178 ? 9 ? 31,684 ? 81 ? 1,602 5 ? 185 12 ? 34,225 144 ? 2,220 6 ? 199 13 ? 39,601 169 ? 2,587 7 ? 205 12 ? 42,025 144 ? ,460 8 ? 226 16 ? 51,076 256 ? 3,616 Totals 1,515 95 b = (18384 8 ( 189. 375 ( 11. 875)/(290,413 8 ( 189. 375 ( 189. 375) = 0. 1121 a = 11. 875 0. 1121 ( 189. 375 = 9. 3495 Sales ( y) = 9. 349 + 0. 1121 (Contracts) (b) pic 4. 49? (a) Method ( Exponential Smoothing 0. 6 = ( Year Deposits (Y) Forecast Error Error2 1 ? 0. 25 0. 25 0. 00 ? 0. 00 2 ? . 24 0. 25 0. 01 ? 0. 0001 3 ? 0. 24 0. 244 0. 004 ? 0. 0000 4 ? 0. 26 0. 241 0. 018 ? 0. 0003 5 ? 0. 25 0. 252 0. 002 ? 0. 00 6 ? 0. 30 0. 251 0. 048 ? 0. 0023 7 ? 0. 31 0. 280 0. 029 ? 0. 0008 8 ? 0. 32 0. 298 0. 021 ? 0. 0004 9 ? 0. 24 0. 311 0. 071 ? 0. 0051 10 ? 0. 26 0. 68 0. 008 ? 0. 0000 11 ? 0. 25 0. 263 0. 013 ? 0. 0002 12 ? 0. 33 0. 255 0. 074 ? 0. 0055 13 ? 0. 50 0. 300 0. 199 ? 0. 0399 14 ? 0. 95 0. 420 0. 529 ? 0. 2808 15 ? 1. 70 0. 738 0. 961 ? 0. 925 16 ? 2. 3 0 1. 315 0. 984 ? 0. 9698 17 ? 2. 80 1. 906 0. 893 ? 0. 7990 18 ? 2. 80 2. 442 0. 357 ? 0. 278 19 ? 2. 70 2. 656 0. 043 ? 0. 0018 20 ? 3. 90 2. 682 1. 217 ? 1. 4816 21 ? 4. 90 3. 413 1. 486 ? 2. 2108 22 ? 5. 30 4. 305 0. 994 ? 0. 9895 23 ? 6. 20 4. 90 1. 297 ? 1. 6845 24 ? 4. 10 5. 680 1. 580 ? 2. 499 25 ? 4. 50 4. 732 0. 232 ? 0. 0540 26 ? 6. 10 4. 592 1. 507 ? 2. 2712 27 ? 7. 0 5. 497 2. 202 ? 4. 8524 28 10. 10 6. 818 3. 281 10. 7658 29 15. 20 8. 787 6. 412 41. 1195 (Continued) 4. 49? (a)? (Continued) Method ( Exponential Smoothing 0. 6 = ( Year Deposits (Y) Forecast Error Error2 30 ? 18. 10 12. 6350 5. 46498 29. 8660 31 ? 24. 10 15. 9140 8. 19 67. 01 32 ? 25. 0 20. 8256 4. 774 22. 7949 33 ? 30. 30 23. 69 6. 60976 43. 69 34 ? 36. 00 27. 6561 8. 34390 69. 62 35 ? 31. 10 32. 6624 1. 56244 2. 44121 36 ? 31. 70 31. 72 0. 024975 0. 000624 37 ? 38. 50 31. 71 6. 79 ? 46. 1042 38 ? 47. 90 35. 784 12. 116 146. 798 39 ? 49. 10 43. 0536 6. 046 36. 56 40 ? 55. 80 46. 814 9. 11856 83. 1481 41 ? 70. 10 52. 1526 17. 9474 322. 11 42 ? 70. 90 62. 9210 7. 97897 63. 66 43 ? 79. 10 67. 7084 11. 3916 129. 768 44 ? 94. 00 74. 5434 19. 4566 378. 561 TOTALS 787. 30 150. 3 1,513. 22 AVERAGE 17. 8932 3. 416 34. 39 (MAD) (MSE) Next period forecast = 86. 2173 Standard error = 6. 07519 Method ( Linear Regression (Trend Analysis) Year Period (X) Deposits (Y) Forecast Error2 ? 1 ? 1 0. 25 17. 330 309. 061 ? 2 ? 2 0. 24 15. 692 253. 823 ? 3 ? 3 0. 24 14. 054 204. 31 ? 4 ? 4 0. 26 12. 415 160. 662 ? 5 ? 5 0. 25 10. 777 121. 594 ? 6 ? 6 0. 30 ? 9. 1387 89. 0883 ? 7 ? 7 0. 31 ? 7. 50 61. 0019 ? 8 ? 8 0. 32 ? 5. 8621 38. 2181 ? ? 9 0. 24 ? 4. 2238 19. 9254 10 10 0. 26 ? 2. 5855 8. 09681 11 11 0. 25 ? 0. 947 1. 43328 12 12 0. 33 ? 0. 691098 0. 130392 13 13 0. 50 ? 2. 329 3. 34667 14 14 0. 95 ? 3. 96769 9. 10642 15 15 1. 70 ? 5. 60598 15. 2567 16 16 2. 30 ? 7. 24427 24. 4458 17 17 2. 0 ? 8. 88257 36. 9976 18 18 2. 80 ? 10. 52 59. 6117 19 19 2. 70 ? 12. 1592 89. 4756 20 20 3. 90 ? 13. 7974 97. 9594 21 21 4. 90 ? 15. 4357 111. 0 22 22 5. 30 ? 17. 0740 138. 628 23 23 6. 20 ? 18. 7123 156. 558 24 24 4. 10 ? 20. 35 264. 083 25 25 4. 50 ? 21. 99 305. 62 26 26 6. 10 ? 23. 6272 307. 203 27 27 7. 70 ? 25. 2655 308. 547 28 28 10. 10 ? 26. 9038 282. 367 29 29 15. 20 ? 28. 5421 178. 011 30 30 18. 10 ? 30. 18 145. 936 31 31 24. 10 ? 31. 8187 59. 58 32 32 25. 60 ? 33. 46 61. 73 33 33 30. 30 ? 35. 0953 22. 9945 34 34 36. 0 ? 36. 7336 0. 5381 35 35 31. 10 ? 38. 3718 52. 8798 36 36 31. 70 ? 40. 01 69. 0585 37 37 38. 50 ? 41. 6484 9. 91266 38 38 47. 90 ? 43. 2867 21. 2823 39 39 49. 10 ? 44. 9250 17. 43 40 40 55. 80 ? 46. 5633 ? ? 85. 3163 41 41 70. 10 ? 48. 2016 ? 479. 54 42 42 70. 90 ? 49. 84 ? 443. 28 43 43 79. 10 ? 51. 4782 ? 762. 964 44 44 94. 00 ? 53. 1165 1,671. 46 TOTALS 990. 00 787. 30 7,559. 95 AVERAGE 22. 50 17. 893 171. 817 (MSE) Method ( Least squaresSimple Regression on GSP a b 17. 636 13. 936 Coefficients GSP Deposits Year (X) (Y) Forecast Error Error2 ? 1 0. 40 ? 0. 25 12. 198 ? 12. 4482 ? 154. 957 ? 2 0. 40 ? 0. 24 12. 198 ? 12. 4382 ? 154. 71 ? 3 0. 50 ? 0. 24 10. 839 ? 11. 0788 ? 122. 740 ? 4 0. 70 ? 0. 26 8. 12 8. 38 70. 226 ? 5 0. 90 ? 0. 25 5. 4014 5. 65137 31. 94 ? 6 1. 00 ? 0. 30 4. 0420 4. 342 18. 8530 ? 7 1. 40 ? 0. 31 ? 1. 39545 1. 08545 1. 17820 ? 8 1. 70 ? 0. 32 ? 5. 47354 5. 5354 26. 56 ? 9 1. 30 ? 0. 24 ? 0. 036086 0. 203914 0. 041581 10 1. 20 ? 0. 26 1. 3233 1. 58328 2. 50676 11 1. 10 ? 0. 25 2. 6826 2. 93264 8. 60038 12 0. 90 ? 0. 33 5. 4014 5. 73137 32. 8486 13 1. 20 ? 0. 50 1. 3233 1. 82328 3. 32434 14 1. 20 ? 0. 95 1. 3233 2. 27328 5. 16779 15 1. 20 ? 1. 70 1. 3233 3. 02328 9. 14020 16 1. 60 ? 2. 30 ? 4. 11418 1. 81418 3. 9124 17 1. 50 ? 2. 80 ? 2. 75481 0. 045186 0. 002042 18 1. 60 ? 2. 80 ? 4. 11418 1. 31418 1. 727 19 1. 70 ? 2. 70 ? 5. 47354 2. 77354 7. 69253 20 1. 90 ? 3. 90 ? 8. 19227 4. 29227 18. 4236 21 1. 90 ? 4. 90 ? 8. 19227 3. 29227 10. 8390 22 2. 30 ? 5. 30 13. 6297 8. 32972 69. 3843 23 2. 50 ? 6. 20 16. 3484 ? 10. 1484 ? 102. 991 24 2. 80 ? 4. 10 20. 4265 ? 16. 3265 ? 266. 56 25 2. 90 ? 4. 50 21. 79 ? 17. 29 ? 298. 80 26 3. 40 ? 6. 10 28. 5827 ? 22. 4827 ? 505. 473 27 3. 80 ? 7. 70 34. 02 ? 26. 32 ? 692. 752 28 4. 10 10. 10 38. 0983 ? 27. 9983 ? 783. 90 29 4. 00 15. 20 36. 74 ? 21. 54 ? 463. 924 30 4. 00 18. 10 36. 74 ? 18. 64 ? 347. 41 31 3. 90 24. 10 35. 3795 ? 11. 2795 ? 127. 228 32 3. 80 25. 60 34. 02 8. 42018 70. 8994 33 3. 0 30. 30 34. 02 3. 72018 13. 8397 34 3. 70 36. 00 32. 66 3. 33918 11. 15 35 4. 10 31. 10 38. 0983 6. 99827 48. 9757 36 4. 10 31. 70 38. 0983 6. 39827 ? 40. 9378 37 4. 00 38. 50 36. 74 1. 76 3. 10146 38 4. 50 47. 90 43. 5357 4. 36428 19. 05 39 4. 60 49. 10 44. 8951 4. 20491 17. 6813 40 4. 50 55. 80 43. 5357 ? 12. 2643 ? 150. 4 12 41 4. 60 70. 10 44. 951 ? 25. 20 ? 635. 288 42 4. 60 70. 90 44. 8951 ? 26. 00 ? 676. 256 43 4. 70 79. 10 46. 2544 ? 32. 8456 1,078. 83 44 5. 00 94. 00 50. 3325 ? 43. 6675 1,906. 85 TOTALS 451. 223 9,016. 45 AVERAGE ? 10. 2551 ? 204. 92 ? (MAD) ? (MSE) 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 yearand 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 modelsone 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 year