Solution nt s in the first expression 1 n s i t in

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Solution : nT s in the first expression = 1 n s i T = in the second expression, since T s is a constant ( T s = Max{ T si } in Eq. (15.11)) And T wc in the first expression = 1 n si i T = in the second expression, according to Eq. (15.15). Therefore nT s - T wc = ( 29 1 n s si i T T = - Q.E.D. 15.11 The table below defines the precedence relationships and element times for a new model toy. (a) Construct the precedence diagram for this job. (b) If the ideal cycle time = 1.1 min. repositioning time = 0.1 min, and uptime proportion is assumed to be 1.0, what is the theoretical minimum number of workstations required to minimize the balance delay under the assumption that there will be one worker per station? (c) Use the largest candidate rule to assign work elements to stations. (d) Compute the balance delay for your solution. Work element T e (min) Immediate predecessors 1 0.5 - 2 0.3 1 105
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Assembly Lines-3e-S 07-05/06, 06/04/07 3 0.8 1 4 0.2 2 5 0.1 2 6 0.6 3 7 0.4 4,5 8 0.5 3,5 9 0.3 7,8 10 0.6 6,9 Solution : (a) Precedence diagram: 1 2 3 4 5 6 7 8 9 10 0.5 0.3 0.8 0.2 0.1 0.6 0.4 0.5 0.3 0.6 (b) T s = T c - T r = 1.1 - 0.1 = 1.0 min With M = 1.0, n = w = Minimum Integer 4.3 1.0 wc s T T = = 4.3, Use n = 5 stations (c) Line balancing solution using the largest candidate rule. List of elements by T e value Allocation of elements to stations Element T e (min) Predecessors Station Element T e Σ T e 3 0.8 1 1 1 0.5 min 6 0.6 3 2 0.3 min 10 0.6 6, 9 4 0.2 min 1.0 min 1 0.5 - 2 3 0.8 min 8 0.5 3, 5 5 0.1 min 0.9 min 7 0.4 4, 5 3 6 0.6 min 2 0.3 1 7 0.4 min 1.0 min 9 0.3 7, 8 4 8 0.5 min 4 0.2 2 9 0.3 min 0.8 min 5 0.1 2 5 10 0.6 min 0.6 min 4.3 min total (d) Balance delay d = 5 10 4 3 5 10 ( . ) . ( . ) - = 0.14 = 14% 15.12 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: List of elements by precedence columns Allocation of elements to stations Element T e (min) Column Station Element T e Σ T e 1 0.5 I 1 1 0.5 min 2 0.3 II 2 0.3 min 106
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Assembly Lines-3e-S 07-05/06, 06/04/07 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 (d) Same as in Problem 15.11: d = 0.14 = 14% 15.13 Solve the previous problem using the ranked positional weights 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: Elements by ranked positional weights Allocation of elements to stations Element T e (min) RPW Station Element T e Σ T e 1 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 (d) Same as in Problem 15.11: d = 0.14 = 14% 15.14 A manual assembly line is to be designed to make a small consumer product. The work elements, their times, and precedence constraints are given in the table below. The workers will operate the line for 400 min per day and must produce 300 products per day. A mechanized belt, moving at a speed of 1.25 m/min, will transport the products between stations. Because of the variability in the time required to perform the assembly
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  • Spring '10
  • Hani
  • Chemical element, Cycle Time, Production line

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