Solve the previous problem using the Kilbridge and Wester method in part c

# Solve the previous problem using the kilbridge and

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7. Solve the previous problem using the Kilbridge and Wester method in part (c). Solution: (a)Precedence diagram same as in Problem 15.11. (b)Same as in Problem 15.11: n= 5 stations. (c)Line balancing solution using the Kilbridge & Wester method: (d)Same as in Problem 15.11: d= 0.14 = 14% List of elements by precedence columns Allocation of elements to stations Element Te(min) Column Station Element TeΣTe1 0.5 I 1 1 0.5 min 2 0.3 II 2 0.3 min 3 0.8 II 4 0.2 min 1.0 min 4 0.2 III 2 3 0.8 min 5 0.1 III 5 0.1 min 0.9 min 6 0.6 III 3 6 0.6 min 7 0.4 IV 7 0.4 min 1.0 min 8 0.5 IV 4 8 0.5 min 9 0.3 V 9 0.3 min 0.8 min 10 0.6 VI 5 10 0.6 min 0.6 min 4.3 min total 8. Solve the previous problem using the ranked positional weights method in part (c). Solution: Elements by ranked positional weights Allocation of elements to stations Element Te(min) RPW Station Element TeΣTe1 0.5 4.3 1 1 0.5 min 3 0.8 2.8 2 0.3 min 2 0.3 2.4 5 0.1 min 0.9 min 5 0.1 1.9 2 3 0.8 min 4 0.2 1.5 4 0.2 min 1.0 min 8 0.5 1.4 3 8 0.5 min 7 0.4 1.3 7 0.4 min 0.9 min 6 0.6 1.2 4 6 0.6 min 9 0.3 0.9 9 0.3 min 0.9 min 10 0.6 0.6 5 10 0.6 min 0.6 min 4.3 min total
59. 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. Element TePreceded by: Element TePreceded by: 1 0.2 min - 8 0.2 min 5 2 0.5 min - 9 0.4 min 5 3 0.2 min 1 10 0.3 min 6, 7 4 0.6 min 1 11 0.1 min 9 5 0.1 min 2 12 0.2 min 8, 10 6 0.2 min 3, 4 13 0.1 min 11 7 0.3 min 4 14 0.3 min 12, 13 Solution: (a)Assume Tr= 0. fp= 54ftftasby/ min./= 1.25 asbys/min, Tc= 0.8 min/asby(b)Tt= 65ftstationft// min.= 1.2 min/station(c)Twc= ΣTek= 0.2 + 0.5 + . . + 0.3 = 3.7 min w= Minimum Integer 3708. min.

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• Fall '14
• Chemical element, Cycle Time, Production line