Problem_solutions - g.2.4.2 1.2 1.0 1.2.4 d Task Number of...

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Ch 6 Problems 1. OT = 450 minutes a. Minimum cycle time = length of longest task, which is 2.4 minutes. Maximum cycle time = Σ task times = 18 minutes. b. Range of output: units 25 18 450 : . min 18 @ units 5 . 187 4 . 2 450 : . min 4 . 2 @ = = c. 8 to rounds which , 5 . 7 450 ) 18 ( 5 . 187 OT t Dx N = = = ; OR, Σ t/CT = 18/2.4 = 7.5 ≈ 8 d. cycle per minutes 6 . 3 125 450 CT CT, for Solving CT OT Output = = = e. Potential output: (1) units 50 9 450 CT OT : . min 9 CT = = = (2) units 30 15 450 : . min 15 CT = = 5. a. b. The minimum cycle time = maximum task time =1.2 minutes The maximum cycle time = .2 +.4 +.2 +.4 +1.2 +1.2 + 1.0 = 4.6 minutes CT) d (calculate minutes 2 day / nits u 40 2 min./day 480 output OT CT = = = c. stations 3 . 2 0 . 2 4.6 = = = CT t N a b d c f e
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Unformatted text preview: g .2 .4 .2 1.2 1.0 1.2 .4 d. Task Number of following tasks A 4 B 3 C 2 D 2 E 1 F 1 G Line Balancing Table (CT = 2 minutes) W. S. Time Remaining Eligible Tasks Task(s) Assigned Idle Time 1 2.0 a, d a 1.8 b, d b 1.4 c, d d 1.0 c c 0.8 0.8 2 2.0 0.8 e, f e 0.8 3 2.0 0.8 f f 0.8 4 2.0 1.0 g g 1.0 3.4 e. % 5 . 42 . 8 4 . 3 ) 2 )( 4 ( . 1 8 . 8 . 8 . percent Idle = = + + + = % 5 . 57 ) 2 )( 4 ( 6 . 4 E = = OR, Efficiency = 2.3/4 = 57.5% % idle time = 100% - 57.5 = 42.5% 14. See EXCEL file...
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