63874-Ch15

# Work element k t eak preceded by t ebk preceded by 1

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Work element k T eAk Preceded by: T eBk Preceded by 1 1 min - 1 min - 2 3 min 1 3 min 1 3 4 min 1 4 min 1, 8 4 2 min - 2 min 8 5 1 min 2 - - 6 2 min 2, 3, 4 2 min 2, 3, 4 7 3 min 5, 6 3 min 6, 9 8 - - 4 min - 9 - - 2 min 4 T wc 16 min 21 min Solution : (a) Precedence diagrams: 118

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Assembly Lines-3e-S 07-05/06, 06/04/07 1 2 3 4 5 6 7 Model A 1 3 4 2 1 2 3 1 2 3 4 6 7 1 3 4 2 2 3 Model B 9 8 2 4 1 2 3 4 6 7 9 8 Combined: A and B 5 AB AB A AB AB AB B AB B (b) Preliminary computations for line balancing solution. We will use the Kilbridge & Wester method. Given R pA = 7 units/hr, R pB = 5 units/hr. Element k T eA T eB R pA T eAk R pB T eBk Σ R pj T ejk 1 1 min 1 min 7 min 5 min 12 min 2 3 min 3 min 21 min 15 min 36 min 3 4 min 4 min 28 min 20 min 48 min 4 2 min 2 min 14 min 10 min 24 min 5 1 min - 7 min 0 7 min 6 2 min 2 min 14 min 10 min 24 min 7 3 min 3 min 21 min 15 min 36 min 8 - 4 min 0 20 min 20 min 9 - 2 min 0 10 min 10 min 217 min Given M = 1, E = 1, E r = 1, and therefore, AT = 60 min, 119
Assembly Lines-3e-S 07-05/06, 06/04/07 n = w = Minimum Integer 217 60 = 3.62 4 stations . 4 stations given. (c) Kilbridge & Wester method to solve line balancing problem ( TT s = 60 min): List of elements by column Allocation of elements to workstations Element TT k Column Station Element TT k TT si 1 12 min I 1 1 12 min 8 20 min I 8 20 min 2 36 min II 4 24 min 56 min 3 48 min II 2 3 48 min 4 24 min II 5 7 min 55 min 5 7 min III 3 2 36 min 6 24 min III 6 24 min 60 min 9 10 min III 4 9 10 min 7 36 min IV 7 36 min 46 min 217 min 217 min 217 min Line balance efficiency. Using maximum{ TT si } = 60 min E b = 217 4 60 ( ) = 0.904 = 90.4% (c) Fixed rate launching schedule. R p = 7 + 5 = 12/hr. T cf = 1 12 7 16 5 21 4 0 904 ( ) ( . ) x x + = 5.00 min If product A launched, T cjh = T cAh = 16 4 0 904 ( . ) = 4.424 min If product B launched, T cjh = T cBh = 21 4 0 904 ( . ) = 5.806 min The following table indicates the solution for one cycle (one hour). The cycle repeats every hour. ( Σ T cAh -mT cf ) 2 ( Σ T cBh -mT cf ) 2 A or B Σ T cjh Σ T cf T cAh T cBh T cf 0.332 0.650 A 4.424 5 4.424 5.806 5 1.327 0.053 B 10.23 10 4.424 5.806 5 0.120 1.073 A 14.654 15 4.424 5.806 5 0.850 0.212 B 20.46 20 4.424 5.806 5 0.013 1.603 A 24.884 25 4.424 5.806 5 0.479 0.476 B 30.69 30 4.424 5.806 5 0.013 2.238 A 35.114 35 4.424 5.806 5 0.213 0.846 A 39.538 40 4.424 5.806 5 1.077 0.118 B 45.344 45 4.424 5.806 5 0.054 1.323 A 49.768 50 4.424 5.806 5 0.653 0.329 B 55.574 55 4.424 5.806 5 0.000 1.904 A 59.998 60 4.424 5.806 5 Workstation Considerations 15.30 An overhead continuous conveyor is used to carry dishwasher base parts along a manual assembly line. The spacing between appliances = 2.2 m and the speed of the conveyor = 1.2 m/min. The length of each workstation is 3.5 m. There are a total of 25 stations and 30 workers on the line. Determine (a) elapsed time a dishwasher base part spends on the line, (b) feed rate, and (c) tolerance time. Solution : (a) L = 25(3.5) = 87.5 m, ET = 87.5 1.2 = 72.917 min 120

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Assembly Lines-3e-S 07-05/06, 06/04/07 (b) f p = 1.2 2.2 = 0.5455 parts/min (c) T t = 3.5 1.2 = 2.917 min/station 15.31 A moving belt line is used to assemble a product whose work content = 20 min. Production rate = 48 units/hour, and the proportion uptime = 0.96. The length of each station = 5 ft and manning level = 1.0. The belt speed can be set at any value between 1.0 and 6.0 ft/min. It is expected that the balance delay will be about 0.08 or slightly higher. Time lost for repositioning each cycle is 3 seconds. (a) Determine the number of stations needed on the line. (b) Using a tolerance time that is 50% greater than the cycle time, what would be
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