ENGG*3070 – Assignment No. 4 (Due Nov. 24, 2011) Flexible Manufacturing systems (FMS) 1. A FMS consists of three stations plus a load/unload station. Station 1 loads and unloads parts using two...

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Answered Same DayNov 21, 2021

Answer To: ENGG*3070 – Assignment No. 4 (Due Nov. 24, 2011) Flexible Manufacturing systems (FMS) 1. A FMS...

Sonam answered on Nov 22 2021
127 Votes
2.
MLT1 = 7.0 + 8.7 + 10.5 + 11.25 + 12.6 = 50.05 minutes
Rp* = 0.1778 pc/min from Problem1
N*
= 0.1778(50.05) = 8.9
(a) For N = 5 < N* = 8.9,
MLT1 = 50.05 minutes, and Tw = 0 Rp = 5/50.05 = 0.0999 pc/min = 5.99 pc/hour
U1 = (7.0/2)*(0.0999) = 0.350 = 35.0%
U2 = (8.7/2)*(0.0999) = 0.435 = 43.5%
U3 = (10.5/3)*(0.0999) = 0.350 = 35.0%
U4 = (11.25/2)*(0.0999) = 0.562 = 56.2%
U5 = (12.6/3)*(0.0999) = 0.420 = 42.0%
Us = [(2* 0.35) + (2* 0.435) + (3*0.35) + (2*0.562)]/9 = 0.416 = 41.6%
(b) N = 8 < N* = 8.9,
Rp = 8/50.05 = 0.1598 pc/min = 9.59 pc/hour
MLT1 =50.05 minutes, and Tw = 0.
U1 = (7.0/2)(0.1598) = 0.559 = 55.9%
U2 = (8.7/2)(0.1598) = 0.695 = 69.5%
U3 = (10.5/3)(0.1598) = 0.559 = 55.9%
U4 = (11.25/2)(0.1598) = 0.899 = 89.9%
U5 = (12.6/3)(0.1598) = 0.671 = 67.1%
Us = [(2* 0.559) + (2* 0.695) + (3* 0.559) +...
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