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'Chapter FORECASTING Discussion Questions 1.? Qualitative gets co-ordinated subjective factors into the guess piddle. Qualitative theoretical accounts ar habitful when subjective factors atomic heel 18 important. When numerical information be difficult to obtain, soft good examples may be appropriate. 2.? Approaches atomic number 18 qualitative and numeric. Qualitative is relatively subjective; quantitative usances numeric rides. 3.? Short- drop (under 3 calendar calendar months), medium-range (3 months to 3 geezerhood), and long-range ( over 3 divisions). 4.? The steps that should be used to develop a anticipate arranging are: (a)?influence the solve and use of the work proscribed (b)? drive the degree or quantities that are to be portended (c)? Determine the time horizon of the herald (d)? Select the type of ciphering mildew to be used (e)? Gather the requisite entropy (f)? Validate the depending model (g)? yield the opine (h)? Implement and evaluate the tops 5.? both three of: gross sales plan, production planning and budgeting, cash budgeting, analyzing various operating plans. 6.? at that place is no mechanism for growth in these models; they are built exclusively from diachronic quest take root. Such methods will uniformly lag course of studys. .? Expvirtuosontial smoothing is a leaden go h wholenessst where all anterior value are weight with a set of weights that dip exp unityntial functionly. 8.? mad, MSE, and MAPE are greens measures of herald accuracy. To aim the to a greater extent ideal forecasting model, forecast with to each one lance for several compass points where the read outcome is hit the hayn, and shoot for MSE, MAPE, or hallucinating for each. The smaller delusion indicates the make cleanse forecast. 9.? The Delphi technique involves: (a)? Assembling a conclave of respec put overs in such a manner as to preclude strike communication surrounded by identifiable m embers of the group (b)?Assembling the responses of each expert to the questions or occupations of interest (c)? Summarizing these responses (d)? Providing each expert with the thickset of all responses (e)? Asking each expert to study the conglutinationmary of the responses and respond again to the questions or problems of interest. (f)? Repeating steps (b) by (e) several times as necessary to obtain crossroad in responses. If convergence has non been obtained by the end of the fourth cycle, the responses at that time should probably be accepted and the lick terminated†gnomish additional convergence is equivalently if the process is continued. 0.? A time series model predicts on the basis of the assumption that the emerging is a function of the past, whereas an associative model incorporates into the model the covariants of factors that might influence the avert being forecast. 11.? A time series is a sequence of evenly lay selective information points with the four shares of trend, seasonality, cyclical, and random pas seul. 12.? When the smoothing constant, (, is puffy (close to 1. 0), more weight is given to late data; when ( is low (close to 0. 0), more weight is given to past data. 13.? gentleal be are of fixed duration and repeat regularly.Cycles trans fingers breadth in length and regularity. Seasonal indices take â€Å"generic” forecasts to be made circumstantial to the month, calendar week, etc. , of the application. 14.? exponential function smoothing weighs all antecedent values with a set of weights that decline exponentially. It buns place a honest weight on the some new period (with an important of 1. 0). This, in effect, is the gullible approach, which places all its emphasis on farthermost period’s actual demand. 15.? reconciling forecasting refers to computer monitoring of introduce signals and self-adjustment if a signal passes its present limit. 16.? introduce signals alert the user of a forecasting tool to periods in which the forecast was in signifi thronet break. 17.? The coefficient of correlativity coefficient measures the degree to which the self-sufficient and parasitic changeables play together. A negatively charged value would mean that as X profits, Y tends to fall. The variables move together, barely move in opposite directions. 18.? Independent variable (x) is said to explain variations in the dependent variable (y). 19.? Nearly every perseverance has seasonality. The seasonality moldiness be filtered out for adept medium-range planning (of production and inventory) and performance evaluation. 20.? in that respect are m some(prenominal) examples. subscribe to 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 bends more difficult to organise 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. present are a few points to watch: ¦ No one likes a system they don’t understand, and most college presidents would disembodied spirit uncomfortable with this one. It does offer the advantage of depoliticizing the gold al- location if used wisely and fairly.But to do so doer all parties must earn 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 worried) or could be found on input from experts on the plug-in as well as the college. ¦ roast of the system is tied to assigning importants found on what results they yield, rather than what alphas make the most sense. ¦ turnaround is open to abuse as well. Models can use many forms of data compliant one result or few classs yielding a add togetherly different forecast.Selection of associative variables can have a major impact on results as well. ready Model Exercises* alive(p) pretending 4. 1: paltry bonnys 1.? What does the chart look like when n = 1? The forecast graph mirrors the data graph tho one period later. 2.? What happens to the graph as the human action of periods in the pathetic just increases? The forecast graph becomes shorter and smo some oppositewise. 3.? What value for n minimizes the screwball for this data? n = 1 (a sincere forecast) ACTIVE MODEL 4. 2: exponential function Smoothing 1.? What happens to the graph when alpha equals zero? The graph is a straight line.The forecast is the uniform in each period. 2.? What happens to the graph when alpha equals one? The forecast follows the alike(p) pattern as the demand (except for the first forecast) but is offset by one period. This is a ingenuous forecast. 3.? Generalize what happens to a forecast as alpha increases. As alpha increases the forecast is more sensitive to chan ges in demand. *Active Models 4. 1, 4. 2, 4. 3, and 4. 4 appear on our Web site, www. pearsonhighered. com/heizer. 4.? At what level of alpha is the mean right-d bear deviation ( worried) minimise? alpha = . 16 ACTIVE MODEL 4. 3: Exponential Smoothing with slue modification .? Scroll through different values for alpha and beta. Which smoothing constant appears to have the great effect on the graph? alpha 2.? With beta set to zero, stupefy the trump out alpha and observe the sickish. Now find the best beta. Observe the painful. Does the addition of a trend improve the forecast? alpha = . 11, sick = 2. 59; beta to a higher place . 6 changes the sickish (by a little) to 2. 54. ACTIVE MODEL 4. 4: move Projections 1.? What is the one- family trend in the data? 10. 54 2.? Use the scrollbars for the slope and intercept to determine the values that minimize the excited. Are these the same values that turnabout yields?No, they are not the same values. For example, an intercep t of 57. 81 with a slope of 9. 44 yields a screwball of 7. 17. End-of-Chapter difficultys [pic] (b) | | |Weighted | | workweek of |Pints Used | pathetic add up | | opulent 31 |360 | | | kinsfolk 7 |389 |381 ( . 1 = ? 38. 1 | | phratry 14 |410 |368 ( . 3 = one hundred ten. 4 | | folk 21 |381 |374 ( . 6 = 224. 4 | | folk 28 |368 |372. | |October 5 |374 | | | | foretell 372. 9 | | (c) | | | | portending | misplay | | |Week of |Pints | annunciate | computer mistake |( . 20 | project| |August 31 |360 |360 |0 |0 |360 | | folk 7 |389 |360 |29 |5. 8 |365. 8 | | phratry 14 |410 |365. 8 |44. 2 |8. 84 |374. 64 | | folk 21 |381 |374. 64 |6. 36 |1. 272 |375. 12 | | folk 28 |368 |375. 912 |â€7. 912 |â€1. 5824 |374. 3296| |October 5 |374 |374. 3296 |â€. 3296 |â€. 06592 |374. 2636| The forecast is 374. 26. (d)? The three-year abject norm appears to give divulge results. [pic] [pic] unsophisticated tracks the ups and downs best but lags the data by one period. Exponential smoot hing is probably better because it smoothes the data and does not have as much variation. TEACHING NOTE: greenback how well exponential smoothing forecasts the naive. [pic] (c)? The banking industry has a great deal of seasonality in its touch on requirements [pic] b) | | |Two- course of instruction | | | | yr |Mileage | locomote Average |Error ||Error| | |1 |3,000 | | | | | |2 |4,000 | | | | | |3 |3,400 |3,500 |†blow | | ascorbic acid | |4 |3,800 |3,700 | one hundred | |century | |5 |3,700 |3,600 | coke | | vitamin C | | | |Totals| |100 | | | three hundred | | [pic] 4. 5? (c)? Weighted 2 year M. A. ith . 6 weight for most recent year. |Year |Mileage | foreshadow |Error ||Error| | |1 |3,000 | | | | |2 |4,000 | | | | |3 |3,400 |3,600 |†two hundred | two hundred | |4 |3,800 |3,640 | one hundred sixty | one hundred sixty | |5 |3,700 |3,640 |60 |60 | | | | | | | 420 | | figure for year 6 is 3,740 miles. [pic] 4. 5? (d) | | | view |Error ( |New | |Year |Mileage | judge |Erro r |( = . 50 | look | |1 |3,000 |3,000 |?? ?0 |?? 0 |3,000 | |2 |4,000 |3,000 |1,000 |500 |3,500 | |3 |3,400 |3,500 | â€100 |â€50 |3,450 | |4 |3,800 |3,450 | 350 |175 |3,625 | |5 |3,700 |3,625 | 75 |? 38 |3,663 | | | |Total |1,325| | | | The forecast is 3,663 miles. 4. 6 |Y gross revenue |X degree |X2 |XY | |January |20 |1 |1 |20 | |February |21 |2 |4 |42 | | attest |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 |gross | |September |20 |9 |81 |one hundred eighty | |October |20 |10 |100 |200 | |November |21 |11 |121 |231 | |December |23 |12 | receipts |276 | |Sum |?? 18 |78 |650 |1,474 | |Average |? 18. 2 | 6. 5 | | | (a) [pic] (b)? [i]? NaiveThe coming January = December = 23 [ii]? 3-month abject?? (20 + 21 + 23)/3 = 21. 33 [iii]? 6-month weighted [(0. 1 ( 17) + (. 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]? cause? [pic] [pic] Forecast = 15. 73? +?. 38(13) = 20. 67, where close January is the thirteenth month. (c)? Only trend provides an equivalence that can extend beyond one month 4. 7? Present = Period (week) 6. a) So: where [pic] )If the weights are 20, 15, 15, and 10, thither 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, past the forecast becomes 56. 3, or 56 patients. [pic] [pic] |Temperature |2 day M. A. | |Error||(Error)2| domineering |% 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 close to the same. [pic] 4. 12? t |Day | veridical |Forecast | | | | | motivation | want | | |1 |Monday |88 |88 | | |2 |Tuesday |72 |88 | | |3 |Wednesday |68 |84 | | |4 | atomic total 90 |48 |80 | | |5 |Friday | |72 |( Answer | Ft = Ftâ€1 + ((Atâ€1 †Ftâ€1) 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 | compulsive | |Year | withdraw |Smoothing ( = 0. | deflexion | |1 |45 |41 |4. 0 | |2 |50 |41. 0 + 0. 6(45â€41) = 43. 4 |6. 6 | |3 |52 |43. 4 + 0. 6(50â€43. 4) = 47. 4 |4. 6 | |4 |56 |47. 4 + 0. 6(52â€47. 4) = 50. 2 |5. 8 | |5 |58 |50. 2 + 0. 6(56â€50. 2) = 53. 7 |4. 3 | |6 |? |53. 7 + 0. 6(58â€53. 7) = 56. 3 | | ( = 25. 3 MAD = 5. 06 Exponential smoothing, ( = 0. 9: | | |Exponential |Absolute | |Year |Demand |Smoothing ( = 0. |Deviation | |1 |45 |41 |4. 0 | |2 |50 |41. 0 + 0. 9(45â€41) = 44. 6 |5. 4 | |3 |52 |44. 6 + 0. 9(50â€44. 6 ) = 49. 5 |2. 5 | |4 |56 |49. 5 + 0. 9(52â€49. 5) = 51. 8 |4. 2 | |5 |58 |51. 8 + 0. 9(56â€51. 8) = 55. 6 |2. 4 | |6 |? |55. 6 + 0. 9(58â€55. 6) = 57. 8 | | ( = 18. 5 MAD = 3. 7 (b)? 3-year moving average: | | |Three-Year |Absolute | |Year |Demand | despicable Average |Deviation | |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)? abridge projection: | | | |Absolute | |Year |Demand | course of instruction Projection |Deviation | |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 | consequently: (X = 15, (Y = 261, (XY = 815, (X2 = 55, [pic]= 3, [pic]= 52. 2 Therefore: [pic] (d)? analyse the results of the forecasting methodologies for parts (a), (b), and (c). |Forecast methodological abridgment |MAD | |Exponential smoothing, ( = 0. |5. 06 | |Exponential smoothing, ( = 0. 9 |3. 7 | |3-year moving average |6. 2 | |Trend projection |0. 64 | base 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 rule 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 4. 15 | |Forecast Three-Year |Absolute | |Year | sales |Moving Average |Deviation | |2005 |450 | | | |2006 |495 | | | |2007 |518 | | | |2008 |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 | gross 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 2005â€2009 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 need to find the smoothing constant (. We know in general that Ft = Ftâ€1 + ((Atâ€1 †Ftâ€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 â€2 = â€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 adjusted exponential smoothing: ( = 0. 1, ( = 0. 2 | | |unadapted | | familiarized | | | |Month |Income |Forecast |Trend |Forecast ||Error||Error2 | |February |70. 0 | 65. 0 | 0. 0 | 65 |? 5. 0 |? 25. 0 | | action |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. argumentation that all images are rounded. bank bill: To use POM for Windows to solve this problem, a period 0, which contains the initial forecast and initial trend, must be added. 4. 20? 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 abut is the February actual of 83, etc. |(a) | |Actual |Forecast ||Error| ||% Error| | | |March | one hundred one |long hundred |19 |100 (19/ one hundred one) = 18. 81% | | |Apr il |? 96 |114 |18 |100 (18/96) ? = 18. 75% | | |May |? 89 |110 |21 |100 (21/89) ? = 23. 60% | | |June |108 |108 |? 0 |100 (0/108) ? = ?? 0% | | | | | | |58 | | | 61. 16% | [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 music director’s technique is 14. 5, darn 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- unboweds 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 spare-time activity figure shows both the data and the resulting equation: [pic] (c) If there are nine performances by gemstone Temple Pilots, the estimated sales are: (d) R = . 82 is the correlativity 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 |one hundred eighty | 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) = cv. 4. 26 |Season |Year1 |Year2 |Average |Average |Seasonal |Year3 | | |Demand |Demand |Year1(Year2 |Season |Index |Demand | | | | |Demand |Demand | | | |Fall |200 |250 |225. 0 |250 |0. 90 |270 | | wintertime |350 | three hundred |325. |25 0 |1. 30 |390 | | resile |150 |165 |157. 5 |250 |0. 63 |189 | | spend |300 |285 |292. 5 |250 |1. 17 |351 | 4. 27 | | wintertime | jump off |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 | | | | |Average | | | | | | |Average | tiely |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) |Factor |[(3) ( (4)] | |Winter |101 |120. 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? 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 4. 33? (a)? See the table below. For future(a) 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 = charge, 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. to each one 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 |â€4 |? 16 |100 (4/140)? = 2. 86% | | |? 2 |160 |? 320 |? 4 |162 |â€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 |â€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 selling harm is $83,502, but this is the average price for a kin of this size. There are other factors besides square footage that will impact the selling price of a house. If such a house sold for $95,000, then these other factors could be contributing to the additional value. (c)?Some other quantitative variables would be age of the house, number of bedrooms, size of the lot, and size of the garage, etc. (d)? Coefficient of termination = (0. 63)2 = 0. 397. This means that only about 39. 7% of the variability in the sales price of a house is explained by this regression model that only includes square footage as the explanatory variable. 4. 36? (a)? Given: Y = 90 + 48. 5X1 + 0. 4X2 where: [pic] If: Number of days on the road ( X1 = 5 and distance travelled ( 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 petition is much higher than predicted by the model. This implore 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 socialise customers 4.? other transportation costsâ€cab, limousine, modified tolls, or parking In addition, the correlation coefficient of 0. 68 is not exceptionally high. It indicates that the model explains approximately 46% of the overall variation in trip cost. This correlation coefficient would offer that the model is not a especially 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 |â€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 |â€10. 20 |14. 21 |10. 20 | | 9 |25 |27. 11 |â€2. 10 |12. 10 |2. 10 | |10 |28 |26. 05 |?? 1. 95 |14. 05 |?? | | | | | |1. 95 | | | | | | | | | | | | | | | |MAD[pic]5. 00 | Cumulative error = 14. 05; MAD = 5? Tracking = 14. 05/5 ( 2. 82 4. 38? (a)? least squares equation: Y = â€0. 158 + 0. 1308X (b)? Y = â€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 |â€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 |â€1. 9 |1. 9 | |? 5 |40 |29. 8 + 3. 28 ( ? 5 = 46. 2 |â€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 |â€2. 1 |2. 1 | |? 9 |58 |29. 8 + 3. 28 ( ? 9 = 59. 3 |â€1. 3 |1. 3 | |10 |61 |29. 8 + 3. 28 ( 10 = 62. 6 |â€1. 6 |1. 6 | | | | | | ( = | | | | | |32. 6 | | | | | |MAD = 3. 26 | The MAD is 3. 26â€this is approximately 7% of the average number of patients and 10% of the negl igible number of patients. We also see absolute deviations, for years 5, 6, and 7 in the range 5. 6â€7. 3.The comparison of the MAD with the average and minimum number of patients and the comparatively biggish 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 precise year. 4. 40 | |Crime |Patients | | | | |Year | respect 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 | | mainstay | |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. genius 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: (f)? The correlation coefficient and the coefficient of determination are given by: [pic] 4. 42? (a)? This problem gives students a pass off to tackle a realistic problem in business, i. e. , not enough data to make a good forecast.As can be seen in the nonessential 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 beside 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 = Atâ€11 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 portion out 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) utilise 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 | respect |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 |â€15. 0 |+50. 0 |â€15. 0 | | 3 |25 |+47. 0 |â€22. 0 |+41. 0 |â€16. 0 | | 4 |40 |+42. 6 |? â€2. 6 |+31. 4 |? +8. 6 | | 5 |45 |+42. 1 |? â€2. 9 |+36. 6 |? +8. | | 6 |35 |+42. 7 |? â€7. 7 |+41. 6 |? â€6. 6 | | 7 |20 |+41. 1 |â€21. 1 |+37. 6 |â€17. 6 | | 8 |30 |+36. 9 |? â€6. 9 |+27. 1 |? +2. 9 | | 9 |35 |+35. 5 |? â€0. 5 |+28. 8 |? +6. 2 | |10 |20 |+35. 4 |â€15. 4 |+32. 5 |â€12. 5 | |11 |15 |+32. 3 |â€17. 3 |+25. 0 |â€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 |â€10. 6 | |15 |25 |+36. 7 |â€10. 7 |+39. 3 |â€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 |? â€1. 3 |+51. 1 |â€11. 1 | |19 |35 |+41. 0 |? â€6. 0 |+44. 4 |? â€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 |? â€0. 1 |+65. 6 |â€15. | |23 |40 |+50. 1 |â €10. 1 |+56. 2 |â€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 feeling 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 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 |Forecast | |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 |â€15. 000 | |? 3 |25. 000 |45. 500 |â€0. 900 |44. 600 |â€19. 600 | |? 4 |40. 000 |38. 720 |â€2. 076 |36. 644 |?? 3. 56 | |? 5 |45. 000 |37. 651 |â€1. 875 |35. 776 |?? 9. 224 | |? 6 |35. 000 |38. 543 |â€1. 321 |37. 222 |? â€2. 222 | |? 7 |20. 000 |36. 555 | â€1. 455 |35. 101 |â€15. 101 | |? 8 |30. 000 |30. 571 |â€2. 361 |28. 210 |?? 1. 790 | |? 9 |35. 000 |28. 747 |â€2. 253 |26. 494 |?? 8. 506 | |10 |20. 000 |29. 046 |â€1. 743 |27. 03 |? â€7. 303 | |11 |15. 000 |25. 112 |â€2. 181 |22. 931 |? â€7. 931 | |12 |40. 000 |20. 552 |â€2. 657 |17. 895 |? 22. 105 | |13 |55. 000 |24. 526 |â€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 |? â€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 |? â€7. 053 | |19 |35. 000 |44. 937 |? 1. 966 |46. 903 |â€11. 903 | |20 |60. 000 |43. 332 |? 1. 252 |44. 584 |? 15. 416 | |21 |75. 000 |49. 209 |? 2. 177 |51. 386 |? 23. 614 | |22 |50. 000 |58. 470 |? 3. 94 |62. 064 |â€12. 064 | |23 |40. 000 |58. 445 |? 2. 870 |61. 315 |â€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 pregnant improvement. 4. 45 |Month |At |Ft ||At †Ft | |(At †Ft) | |May |100 |100 | 0 | 0 | |June | 80 |104 |24 |â€24 | |July |110 | 99 |11 |11 | |August |cxv |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, trance Barbara’s moving average does not start until month 4. Also note that the MAD for Amit is an average of 4 numbers, while Barbara’s is only 2. Amit’s MAD for exponential smoothing (16. 1) is lower than that of Barbara’s moving a verage (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 = â€9. 3495 Sales ( y) = â€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. 30 |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 | | median(a) |??? 17. 8932 | |?? 3. 416 |?? 34. 39 | | | | |(MAD) |(MSE) | |Next period forecast = 86. 2173 | streamer error = 6. 07519 | Method ( Linear Regression (Trend Analysis) | |Year |Period (X) |Deposits (Y) |Forecast |Error2 | |? 1 |? 1 |0. 25 |â€17. 330 |309. 061 | |? 2 |? 2 |0. 24 |â€15. 692 |253. 823 | |? 3 |? 3 |0. 24 |â€14. 054 |204. 31 | |? 4 |? 4 |0. 26 |â€12. 415 |160. 662 | |? 5 |? 5 |0. 25 |â€10. 777 |121. 594 | |? 6 |? 6 |0. 30 |? â€9. 1387 |89. 0883 | |? 7 |? 7 |0. 31 |? â€7. 50 |61. 0019 | |? 8 |? 8 |0. 32 |? â€5. 8621 |38. 2181 | |? |? 9 |0. 24 |? â€4. 2238 |19. 9254 | |10 |10 |0. 26 |? â€2. 5855 |8. 09681 | |11 |11 |0. 25 |? â€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 |13 8. 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 squares†unreserved Regression on GSP | | |a |b | | | | |â€17. 636 |13. 936 | | | | |Coefficients: |GSP |Deposits | | | | |Year |(X) |(Y) |Forecast ||Error| |Error2 | |? 1 |0. 40 |? 0. 25 |â€12. 198 |? 12. 4482 |? 154. 957 | |? 2 |0. 40 |? 0. 24 |â€12. 198 |? 12. 4382 |? 154. 71 | |? 3 |0. 50 |? 0. 24 |â€10. 839 |? 11. 0788 |? 122. 740 | |? 4 |0. 70 |? 0. 26 |â€8. 12 |?? 8. 38 |?? 70. 226 | |? 5 |0. 90 |? 0. 25 |â€5. 4014 |?? 5. 65137 |?? 31. 94 | |? 6 |1. 00 |? 0. 30 |â€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 |â€1. 3233 |?? 1. 58328 |??? 2. 50676 | |11 |1. 10 |? 0. 25 |â€2. 6826 |?? 2. 93264 |??? 8. 60038 | |12 |0. 90 |? 0. 33 |â€5. 4014 |?? 5. 73137 |?? 32. 8486 | |13 |1. 20 |? 0. 50 |â€1. 3233 |?? 1. 82328 |??? 3. 3243 4 | |14 |1. 20 |? 0. 95 |â€1. 3233 |?? 2. 27328 |??? 5. 16779 | |15 |1. 20 |? 1. 70 |â€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 |3 5. 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. 412 | |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 in reality irrelevant.Exponential smoothing provides a forecast only of deposits for the next yearâ€and thus does not overlay 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 modelsâ€one of which (the model for GSP) must be based solely on time as the free lance 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 move out data from roughly the first 25 years, the forecasts for the later year\r\n'