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IE383_Assignment6_sol

# IE383_Assignment6_sol - IE 383 Assignment 6 solution...

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IE 383 Assignment 6 solution Problem 1 a) Linear model = + + Y a bt ε t Y Y*t t^2 1 8025 8025 1 2 7200 14400 4 3 7145 21435 9 4 6775 27100 16 5 6952 34760 25 6 6542 39252 36 7 6525 45675 49 8 6121 48968 64 Sum 36 55285 239615 204 According to method of least squares, it follows that So the fitted regression equation is: From this equation we can predict that b) = . α 0 4 = = a1 Y1 8025 = - - =- b1 Y8 Y18 1 272 = - - = Y1 a1 b11 αα 8433 = - * Y1 a1 2 b1 - = 1 αα 8841

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t ( ) Y t * ( ) α Y t (1- )* α ( - ) Y t 1 ( ) Y t * α ( ) Y t (1- )* - α Yt 1 Yt 1 8025 8433 8841 2 7200 2880 5059.800 7939.80 0 3175.920 5304.600 8480.520 3 7145 2858 4763.880 7621.88 0 3048.752 5088.312 8137.064 4 6775 2710 4573.128 7283.12 8 2913.251 4882.238 7795.490 5 6952 2781 4369.877 7150.67 7 2860.271 4677.294 7537.564 6 6542 2617 4290.406 6907.20 6 2762.882 4522.539 7285.421 7 6525 2610 4144.324 6754.32 4 2701.729 4371.253 7072.982 8 6121 2448 4052.594 6500.99 4 2600.398 4243.789 6844.187 = - - = . Y8 a8 b81 αα 6500 994 = - * Y8 a8 2 b8 - = . 1 αα 6844 187 = . =- . a8 6157 801 b8 228 795 = + ( - )= . Y9 a8 b8 9 8 5929 01 = + ( - )= . Y10 a8 b8 10 8 5700 21 = + ( - )= . Y11 a8 b8 11 8 5471 42 c) Table : Forecast with linear regression t ( ) Y t ( ) Y t e2 ( ) abs e 9 6000 5928.393 5127.583 71.607 10 5500 5710.119 44150.014 210.119 11 5000 5491.845 241911.738 491.845 MSE 97063.11 MAD 257.86 Table : Forecast with exponential smoothing t ( ) Y t ( ) Y t e2 ( ) abs e 9 6000 5929.006 5040.113 70.994 10 5500 5700.211 40084.472 200.211 11 5000 5471.416 222232.944 471.416 MSE 89119.18 MAD 247.54 Based on the comparison of MSE and MAD for the two forecasting approach, exponential smoothing is the more desirable. Note: we are comparing the approaches of linear programming and exponential smoothing, not the methods used for evaluation.
d) We can transform the exponential model = Y aebt + ε into linear model

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