Chapter 9 - Immediate Task Time(Sec Predecessors A 12 B 15...

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Task Time (Sec) A 12 B 15 A C 8 A D 5 B,C E 20 D Total 60 A.) Cycle Time 20 sec/unit B.) Total Task Time 60 sec Minimum # of workstations 3 stations C.) We need 3 workstations to produce 180 PLA/hr Work Station Activity Time (Sec) 1 A,C 20 2 B,D 20 3 E 20 Immediate Predecessors
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Task Time (Sec) A 10 B 12 A C 8 A,B D 6 B,C E 6 C F 6 D,E Total 48 A.) Cycle Time 12 min/unit Total Task Time 48 min Minimum # of workstations 4 stations B.) We need 5 workstations to produce 5 Micro Popcorn per hour C.) 80% efficient Immediate Predecessors
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Question A Question B Hint: The sum of each element time in each workstation should be less than or equal the desired cycle tim Workstation 1: A, B, G Efficiency = (5+1.5+3)/10=95% Idle time = 1-95% = 5% Workstation 2: D, E, F Efficiency = (4+3+2)/10=90% Idle time = 1-90% = 10%
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Unformatted text preview: Worksta±on 3: C, H, I, J EFciency = (3+3.5+2+2)/10=105% Idle ±me = 1-105% = no idle ±me b) Assign tasks to worksta±ons and determine how much idle ±me is present each cycle. Queston C It is diFcult to a³ain a 100% eFciency in prac±ce, but for discussion’s sake, to improve the eFciency to 1 specially for the elements in worksta±on 3 should be shorten. Queston D The network diagram with the ac±vity ±me Theore±cal minimum number of worksta±ons = Total number of minutes in the whole diagram/10 minut minutes/10 minutes = 2.9 worksta±ons OR 3 worksta±ons me of 10 minutes 100%, the ±me per ac±vity tes cycle ±me = 29...
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  • Fall '14
  • Harshad number, Cycle Time, Workstation, Sun Microsystems

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