Exam2-sol - F R 1 = 0 6985 ⇒ Q 1 = n R 1 1-F R 1 q 2 × k × d h n R 1 1-F R 1 2 = 861 c Imputed shortage cost = Qh(1-F R d = frac 861 × 3 ×

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Midterm 2 Solutions 1. a) Incremental b) C ( Q ) Q = ± 1 . 65 , Q 300 1 . 65 × 300+( Q - 300) × 1 . 45 Q = 60 Q + 1 . 45 , 300 < Q Q (0) = q 2 × 20 × 75 × 52 0 . 2 × 1 . 65 = 688 , not realizable Q (1) = q 2 × (20+60) × 75 × 52 0 . 2 × 1 . 45 = 1467 , realizable Q * = 1467 c) d Q = 75 × 52 1467 = 2 . 66 d) 6(1467) = 1 . 45 × 75 × 52+(20+60) × 52 × 75 1467 +0 . 2 × 1 . 45 × 1467 2 +0 . 2 × 60 2 = 6086 . 4 2. d (400 , 85 2 ) C = $150 $ K = 2500 Z = 5 months P = $75 / unit I = 30% a) μ LT = 400 × 5 = 2000 σ LT = 85 × 5 = 190 . 066 b) Q 0 = q 2 × 2500 × 400 × 12 0 . 3 × 150 = 730 1 - F ( R ) = 730 × 0 . 3 × 150 75 × 4800 F ( R ) = 0 . 90875 R - 2000 190 . 066 = 1 . 33 R 1 = 2253 n ( R 1 ) = 190 . 066 × 0 . 0427 = 8 . 116 Q 1 = q 2 × 4800 × (2500+75 × 8 . 116) 0 . 3 × 150 = 814 3. a) F ( R ) = α = 0 . 95 z = 1 . 65 n ( R ) = 190 . 066 × 0 . 0206 = 3 . 915 R = 190 . 066 × 1 . 65 + 2000 = 2314 Q = q 2 × 2500 × 4800 0 . 3 × 150 = 730 ( Q,R ) = (730 , 2314) 1
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b) β = 0 . 95 n ( R ) = EOQ (1 - β ) = 730 × 0 . 05 = 36 . 5 L ( Z ) = 36 . 55 190 . 066 = 0 . 1923 z = 0 . 52 R 1 = 190 . 066 × 0 . 52 + 2000 = 2099
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Unformatted text preview: F ( R 1 ) = 0 . 6985 ⇒ Q 1 = n ( R 1 ) 1-F ( R 1 ) + q 2 × k × d h + ( n ( R 1 ) 1-F ( R 1 ) ) 2 = 861 c) Imputed shortage cost = Qh (1-F ( R )) d = frac 861 × . 3 × 150(1-. 6985) × 4800 = $26 . 77 d) In general it is more difficult to meet an x% type I service level than an x% type II service level, since in a type I service regime even for very small amount of shortage in a cycle, that cycle is counted as having a ”stockout”. 2...
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This note was uploaded on 12/08/2010 for the course MSCI 311 taught by Professor Dumeiring during the Fall '09 term at Waterloo.

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Exam2-sol - F R 1 = 0 6985 ⇒ Q 1 = n R 1 1-F R 1 q 2 × k × d h n R 1 1-F R 1 2 = 861 c Imputed shortage cost = Qh(1-F R d = frac 861 × 3 ×

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