OPIM 101 - Spring 2012 Exam 1 Review Solutions

OPIM 101 - Spring 2012 Exam 1 Review Solutions - 6...

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Exam 1 Review, Spring 2012 Solutions to Sample Questions Multiple Process Flows 1) Minutes per hour 2) 67 minutes/hour 3) 67 60 = 111.6667% 4) 100% Batching 1) Consider that a full production run produces both Product A and Product B. Set capacity to equal demand, and add up the setup times for each product. Batch Size = Capacity × Setup Time 1 - Capacity × Processing Time = (175 units/hour) × ( 1 3 hour + 1 6 hour) 1 - (175 units/hour) × ( 1 500 units/hour) = 135 units 2) Utilization = Time Producing Time Producing + Idle Time = 1 500 (400) 1 500 (400) + 0 . 5 = 61 . 54% 1) 450 cans/month 2) Q ? = r 2 × K × R h = r 2(5)(450) 0 . 05 = 300 cans 1
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2 3) Solve for T in Little’s Formula. Recall that inventory in Little’s Law is average inventory. I = RT 300 2 = (450 cans/month) T T = 175 450 . Inventory Turns = 1 T = 450 150 = 3 4) Annual demand is 5400 cans. So the number of orders per year is 5400 300 , or 18. Queuing 1) I = RT = (1 1
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Unformatted text preview: 6 clients/minute)(2 minutes) = 7 3 clients 2) Note: Utilization is p am = 2 (0 . 85714)(3) = 0 . 7777778 I q = RT q = (1 1 6 clients/minute)( p m Util 2( m +1)-1 1-Util ( CV 2 a + CV 2 p ) 2 ) = (1 1 6 clients/minute)( 2 3 . 77778 8-1 . 22222 (3 2 + 1 4 2 ) 2 ) = (1 1 6 clients/minute)( 2 3 2 . 84216 4 . 53125) = (1 1 6 clients/minute)(8 . 5857 minutes) = 10 . 0166 clients Decision Analysis 1) The expected value of investing is (0 . 1)(45-15) + (0 . 9)(10-15) = $1 . 5 million 2) We want to nd the probability of P (success | labeled dud): P (success) = P (success | labeled promise) P (labeled promise) + P (success | labeled dud) P (labeled dud) . 10 = 0 . 32(0 . 20) + P (success | labeled dud)(0 . 80) So, P (success | labeled dud) = 0.045...
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This note was uploaded on 04/04/2012 for the course OPIM 101 taught by Professor Lee during the Spring '08 term at UPenn.

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OPIM 101 - Spring 2012 Exam 1 Review Solutions - 6...

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