63874-Ch15

# Because of the variability in the time required to

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the products between stations. Because of the variability in the time required to perform the assembly operations, it has been determined that the tolerance time should be 1.5 times the cycle time of the line. (a) Determine the ideal minimum number of workers on the line. (b) Use the Kilbridge and Wester method to balance the line. (c) Compute the balance delay for your solution in part (b). Element Time T e Preceded by: Element Time T e Preceded by: 1 0.4 min - 6 0.2 min 3 2 0.7 min 1 7 0.3 min 4 3 0.5 min 1 8 0.9 min 4, 9 4 0.8 min 2 9 0.3 min 5, 6 5 1.0 min 2, 3 10 0.5 min 7, 8 Solution : (a) T c = 400 300 min./ / day asbys day = 1.333 min/asby. No T r value given. Assume T r = 0. T wc = ek k T = 0.4 + 0.7 + . . . + 0.5 = 5.6 min w = Minimum Integer 56 1333 . . = 4.2 w = 5 workers (b) Line balancing solution using Kilbridge & Wester method. Assume M = 1.0. 107

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Assembly Lines-3e-S 07-05/06, 06/04/07 1 2 3 4 5 6 7 9 8 10 0.4 0.7 0.5 0.8 1.0 0.2 0.3 0.3 0.9 0.5 I II III IV V VI List of elements by precedence columns Allocation of elements to stations Element T e (min) Column Station Element T e Σ T e 1 0.4 I 1 1 0.4 min 2 0.7 II 2 0.7 min 1.1 min 3 0.5 II 2 3 0.5 min 4 0.8 III 4 0.8 min 1.3 min 5 1.0 III 3 5 1.0 min 6 0.2 III 6 0.2 min 1.2 min 7 0.3 IV 4 7 0.3 min 9 0.3 IV 9 0.3 min 0.6 min 8 0.9 V 5 8 0.9 min 0.9 min 10 0.5 VI 6 10 0.5 min 0.5 min 5.6 min total Note : In the above solution, the Kilbridge & Wester method was followed very precisely, so that the solution consisted of a total of six stations. A better solution, but one that violates the K&W algorithm, would be to combine elements 9 and 8 at station 4 and elements 7 and 10 at station 5, for a total of only 5 stations. (c) Use cycle time of 1.3 min (station 2) rather than 1.333 min. Using the six stations in the K&W solution, balance delay d = 6 13 56 6 13 ( . ) . ( . ) - = 0.282 = 28.2% 15.15 Solve the previous problem using the ranked positional weights method in part (b). Solution : (a) Same solution as in Problem 15.14: ( T c = 1.333 min, T r = 0, T wc = 5.6 min) w = 5 workers. (b) Line balancing solution using ranked positional weights method. Assume M = 1.0. Elements by ranked positional weights Allocation of elements to stations Element T e (min) RPW Station Element T e Σ T e 1 0.4 5.6 1 1 0.4 min 2 0.7 4.5 2 0.7 min 1.1 min 3 0.5 3.4 2 3 0.5 min 5 1.0 2.7 4 0.8 min 1.3 min 4 0.8 2.5 3 5 1.0 min 6 0.2 1.9 6 0.2 min 1.2 min 9 0.3 1.7 4 9 0.3 min 8 0.9 1.4 8 0.9 min 1.2 min 7 0.3 0.8 5 7 0.3 min 10 0.5 0.5 10 0.5 min 0.8 min 108
Assembly Lines-3e-S 07-05/06, 06/04/07 5.6 min total (c) Use cycle time of 1.3 min (station 2) rather than 1.333 min. Using the five stations in the RPW solution, balance delay d = 5 13 56 5 13 ( . ) . ( . ) - = 0.138 = 13.8% 15.16 A manual assembly line operates with a mechanized conveyor. The conveyor moves at a speed of 5 ft/min, and the spacing between base parts launched onto the line is 4 ft. It has been determined that the line operates best when there is one worker per station and each station is 6 ft long. There are 14 work elements that must be accomplished to complete the assembly, and the element times and precedence requirements are listed in the table below. Determine (a) feed rate and corresponding cycle time, (b) tolerance time for each worker, and (c) ideal minimum number of workers on the line. (d) Draw the precedence diagram for the problem. (e) Determine an efficient line balancing solution. (f) For your solution, determine the balance delay.

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