hw9_Sp09_solutions

# hw9_Sp09_solutions - IEOR 130 Methods of Manufacturing...

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IEOR 130 Methods of Manufacturing Improvement Spring, 2009, Prof. Leachman Solutions to HW 9 1. (a) WIP = 200 lots, CT = 50 days, Production rate = WIP/CT = 4 lots per day (b) 5 visits, 5 hours per visit, 5 machines, 90% availability Workload = (4 lots/day)(5 visits/lot)(5 hours per visit) = 100 hours per day Total machine time = (5 machines)(24 hours per day) = 120 machine hours per day Available machine time = (.9)(total machine time) = (.9)(120) = 108 hours per day Utilization of total time = 100/120 = 0.833 Utilization of available time = 100/108 = 0.926 (c) If no changes are made in scheduling policy or operation, then increasing the production rate will require an increase in the WIP. How much to increase the target WIP level depends on the variability in the factory. The management can incrementally increase the WIP level and see what throughput results until the desired throughput is achieved. They also can try some off-line simulation analysis. (d) To decrease cycle time by a certain amount, they should decrease the WIP limit proportionally (200 lots in WIP corresponds to 50 days CT.) The production rate will be automatically decreased. 2. (a) Total workload is computed based on number of visits to bottleneck remaining and processing time per visit: { 21(5) + 34(4) + 39(3) + 45(2) + 41(1) + 20(0) }{5} = 2445 hours (b) If workload target is 2500 hours, then we need 55 hours more workload. Each lot contributes (5 visits)(5 hours/visit) = 25 hours of workload, so we should release 3 lots. (Two lots would only provide 50 hours workload, not quite enough.) (c) For the usual distribution of WIP, the workload is { 20(5) + 40( 4 + 3 + 2 +1 ) + 20(0) } {5} = 2500 hours Thus the 200-lot constant WIP level would seem to be equivalent to the 2500-hour target workload level. (d) Even though the total WIP was 200 lots, the Workload Regulating method released 2 lots because the workload was shifted towards the second half of the fab process, where fewer visits to the bottleneck remain. 3. (a) The load limit is (5 machines)(20 hours lead time)(1.5 safety factor) = 150 hours The virtual inventory is 130 hours -> amount of workload needed is 20 hours. Dividing by 5 hours per first visit of a new lot, this translates into 4 lots to be released.

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## This note was uploaded on 03/16/2010 for the course ORMS 130 taught by Professor Leachman during the Spring '10 term at Berkeley.

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hw9_Sp09_solutions - IEOR 130 Methods of Manufacturing...

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