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Nahmias Chapter Selected Question Solutions Essay

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SOLUTIONS TO SELECTED PROBLEMS FROM NAHMIAS’ BOOK CHAPTER 2 FORECASTING 2. 13 Fcst 1 Fcst 2 Demand Err 1 Err 2 Er1^2 Er2^2 |Err1| 223 210 256 33 46 1089 2116 33 289 320 340 51 20 2601 400 51 430 390 375 -55 -15 3025 225 55 134 112 110 -24 -2 576 4 24 190 150 225 35 75 1225 5625 35 550 490 525 -25 35 625 1225 25 1523. 5 1599. 166 37. 16666 (MSE1 (MSE2) (MAD1) (Err2( (e1/D(*100 (e2/D((100 6 12. 89062 17. 96875 20 15. 00005. 88253 15 14. 66667 4. 00000 2 21. 81818 1. 81818 75 15. 55556 33. 33333 35 4. 761905 6. 66667 32. 16666 14. 11549 11. 61155 (MAD2) (MAPE1) (MAPE2) 2. 14It means that E(ei)  (  0. This will show up by considering [pic] A bias is indicated when this sum deviates too far from zero. 2. 16 MA (3) forecast:    258. 33 MA (6) forecast:    249. 33 MA (12) forecast:    205. 33 2. 17, 2. 18, and 2. 19. One-step-ahead Two-step-ahead

Month Forecast Forecast Demand    e1   e2 July 205. 50 149. 75 223 -17. 50 -73. 25 August 225. 25 205. 50 286 -60. 75 -80. 50 September 241. 50 225. 25 212 29. 50 13. 25 October 250. 25 241. 50 275 -24.

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75 -33. 50 November 249. 00 250. 25 188 61. 00 62. 25 December 240. 25 249. 00 312 -71. 75 -63. 00 MAD =  44. 2 54. 3

The one step ahead forecasts gave better results (and should have according to the theory). 2. 20Month Demand MA(3) MA(6) July 223 226. 00 161. 33 August 286 226. 67 183. 67 September 212 263. 00 221. 83 October 275 240. 33 233. 17 November 188 257. 67 242. 17 December 312 225. 00 244. 00 MA (6) Forecasts exhibit less variation from period to period. 2. 21An MA(1) forecast means that the forecast for next period is simply the current period’s demand.

Month Demand MA(4) MA(1)  Error Month Demand MA(4) MA(1)  Error July 223 205. 50 280 57 August 286 225. 25 223 -63 September 212 241. 50 286 74 October 275 250. 25 212 -63 November 188 249. 00 275 87 December 312 240. 25 188 -124 MAD = 78. 0 (Much worse than MA(4)) 2.

35a)V1 = (16 + 32 + 71 + 62)/4 = 45. 25 V2 = (14 + 45 + 84 + 47)/4 = 47. 5 1. G0 = (V2 – V1)/N = 0. 5625 2.

S0 = V2 + G0 (N-1/2) = 47. 5 + (0. 5625)(3/2) = 48. 34 3. ct = [pic]-2N+1 = ( t ( 0 c-7 = [pic] = 0. 36 c-6 = [pic] = 0. 71 c-5 = [pic] = 1. 56 c-4 = [pic] = 1. 35 c-3 = [pic] = 0. 30 c-2 = [pic] = 0. 95 c-1 = [pic] = 1. 76 c0 = [pic] = 0. 97 (c7 + c3)/2 = . 33 (c6 + c2)/2 = . 83 (c5 + c1)/2 = 1. 66 (c4 + c0)/2 = 1. 16 Sum = 3. 98 Norming factor  =  4/3. 9   =   1. 01 Hence the initial seasonal factors are: c-3 = . 33 c-1 = 1. 67 c-2 = . 83 c-0 = 1. 17 b)( = 0. 2, ( = 0. 15, ( = 0. 1, D1 = 18 S1 = ((D1/c-3) + (1-()(S0 + G0) = 0. 2(18/0. 33) + 0. 8(48. 34 + 0. 6) = 50. 03 G1 = ((S1 – S0) + (1 – () = G0 = 0. 1(50. 03 – 48. 34) + 0. 9(0. 56) = 0. 70 c1 = ((D1/S1) + (1-()c3 = 0. 15(18/50. 03) + 0. 85(0. 33) = . 3345 c)Forecasts for 2nd, 3rd and 4th quarters of 1993 F1,2 = [S1 + G1]c2 = (50 + . 70)0. 83 = 42. 08 F1,3 = [S1 + 2G1]c3 = (50 + 2(. 70))1. 67 = 85. 84 F1,4 = [S1 + 3G1]c4 = (50 + 3(. 70))1. 17 = 60. 96 2. 36Forecast Forecast fromfrom Period    Dt    30(d)   (et (  31(c)    ( et ( 1 2 5135. 815. 242. 08 8. 92 3 8682. 4 3. 685. 84 0. 16 4 66 56. 5 9. 5 60. 96 5. 04 MAD = 9. 43MAD = 4. 71 MSE = 111. 42MSE = 35. 0 Hence we conclude that Winter’s method is more accurate. 2. 37S1 = 50. 03 ( = 0. 2 ( = 0. 15 ( = 0. 1D1 = 18 G1 = 0. 67D2 = 51 D3 = 85 D4 = 66 S2 = 0. 2(51/0. 83) + 0. 8(50. 03 + 0. 70) = 52. 87 G2 = 0. 1(52. 87 – 50. 03) + 0. 9(0. 70) = 0. 914 S3 = 0. 2(86/1. 67) + 0. 8(52. 87 + 0. 914) = 53. 33 G3 = 0. 1(53. 33 – 52. 85) + 0. 9(0. 885) = 0. 8445 S4 = 0. 2(66/1. 17) + 0. 8(53. 33 + 0. 8445) = 54. 62 G4 = 0. 1(54. 62 – 53. 33) + 0. 9(0. 8445) = 0. 8891 c1 = (. 15)[18/50] + (0. 85)(. 33) = . 3345 ( . 34 c2 = (. 15)[51/52. 85] + 0. 85(0. 83) = . 8502  ( . 85 c3 = (. 15)(86/53. 9) + 0. 85(1. 67) = 1. 6616 ( 1. 66 c4 = (. 15)(66/54. 59) + 0. 85(1. 17) = 1. 1758  ( 1. 18 The sum of the factors is 4. 02. Norming each of the factors by multiplying by 4/4. 02  =  . 995 gives the final factors as: c1 = . 34 c2 = . 84 c3 = 1. 65 c4 = 1. 17 The forecasts for all of 1995 made at the end of 1993 are: F4,9 = [S4 + 5G4]c1 = [54. 62 + 5(0. 89)]0. 34 = 20. 08 F4,10 = [S4 + 6G4]c2 = [54. 62 + 6(0. 89)]0. 84 = 50. 37 F4,11 = [S4 + 7G4]c3 = [54. 62 + 7(0. 89)]1. 65 = 100. 40 F4,12 = [S4 + 8G4]c4 = [54. 62 + 8(0. 89)]1. 17 = 72. 24 2. 40 c1 = 0. 7 c2 = 0. 8 c3 = 1. 0 c4 = 1. 5

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Nahmias Chapter Selected Question Solutions Essay. (2018, Feb 05). Retrieved from https://graduateway.com/nahmias-chapter-selected-question-solutions/

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