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2009 Midterm Solutions

# 2009 Midterm Solutions - MIE363-2009 Midterm Solutions...

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MIE363-2009 Midterm Solutions Question 1 a) Since the series is assumed to be stationary the forecast for February 2009 equals the forecast for January 2009. The forecast for January & February 2009 using MA(3) is: 250 + 300 + 280 3 = 277 b) Same assumption of stationary series holds here as well. Forecast for January & February 2009 using EXP(0.2), assuming that forecast for January 2008 was equal to the demand: Equation: F t +1 = α D t + (1 - α ) F t = 0 . 2 · D t + 0 . 8 · F t Month Demand Forecast Jan ’08 200 200 Feb ’08 300 200 Mar ’08 200 220 Apr ’08 250 216 May ’08 320 223 Jun ’08 370 242 Jul ’08 280 268 Aug ’08 200 270 Sep ’08 310 256 Oct ’08 250 267 Nov ’08 300 264 Dec ’08 280 271 Jan ’09 273 Feb ’09 273 1

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c) Month Demand MA(3) Forecast e for MA(3) EXP(0.2) e for EXP(0.2) Jan ’09 330 273 -53 273 -57 Feb ’09 300 273 -23 273 -27 MSE = 1 n n t - 1 e 2 t MAD = 1 n n t - 1 | e t | MA(3) EXP(0.2) MSE 38.2 42.2 MAD 1686.7 2007.4 d) Sum of the weights of all observations must equal to 1. 0 . 5 + 2 x = 1 x = 0 . 25 Therefore, the weight for demand observed in December ’08 is 0.5 and for November and October ’08 is 0.25. F January = 0 . 5 · F December + 0 . 25 · ( F Novembet + F October ) = 0 . 5 · 280 + 0 . 25(250 + 300) = 277 . 5 278 units e) The α used for exponential smoothing should result in the same average age for the observations as the forecast method described in part d.
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2009 Midterm Solutions - MIE363-2009 Midterm Solutions...

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