63874-Ch15

# A or b σ t cjh σ t cf t c a h t c b h t cf 0212

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A or B Σ T cjh Σ T cf T c A h T c B h T cf 0.212 0.077 B 1.407 1.684 2.144 1.407 1.684 0.033 0.307 A 3.551 3.368 2.144 1.407 1.684 0.413 0.009 B 4.958 5.052 2.144 1.407 1.684 0.134 0.138 A 7.102 6.736 2.144 1.407 1.684 0.682 0.008 B 8.509 8.42 2.144 1.407 1.684 0.301 0.035 B 9.916 10.104 2.144 1.407 1.684 0.074 0.216 A 12.06 11.788 2.144 1.407 1.684 0.536 0.000 B 13.467 13.472 2.144 1.407 1.684 0.207 0.080 B 14.874 15.156 2.144 1.407 1.684 0.032 0.312 A 17.018 16.84 2.144 1.407 1.684 0.407 0.010 B 18.425 18.524 2.144 1.407 1.684 0.130 0.141 A 20.569 20.208 2.144 1.407 1.684 0.674 0.007 B 21.976 21.892 2.144 1.407 1.684 0.296 0.037 B 23.383 23.576 2.144 1.407 1.684 0.071 0.221 A 25.527 25.26 2.144 1.407 1.684 0.529 0.000 B 26.934 26.944 2.144 1.407 1.684 15.23 For Problem 15.19, determine (a) the fixed rate launching interval, and (b) the launch sequence of models A, B, and C during one hour or production. 112

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Assembly Lines-3e-S 07-05/06, 06/04/07 Solution : R p = 10 + 20 + 30 = 60/hr. T cf = 1 60 10 45 20 35 30 25 39 0 93 0 94 ( ) ( . )( . ) x x x + + = 0.929 min If product A launched, T cjh = T c A h = 45 39 0 94 0 93 ( . )( . ) = 1.320 min If product B launched, T cjh = T c B h = 35 39 0 94 0 93 ( . )( . ) = 1.027 min If product C launched, T cjh = T c C h = 25 39 0 94 0 93 ( . )( . ) = 0.733 min We note that the hourly production rates of the three models (10/hr for A and 20/hr for B and 30/hr for C) are all divisible by 10 (1 per 6 min for A, 2 per 6 min for B, and 3 per 6 min for C). Thus the sequence should repeat every 6 launches. The following table indicates the solution for one cycle (see equations for first three columns following launching sequence): Eq. A Eq. B Eq. C A,B,or C Σ T cjh Σ T cf T c A h T c B h T c C h T cf 1.15 1.01 1.04 B 1.03 0.93 1.320 1.027 0.733 0.929 1.24 2.04 1.01 C 1.76 1.86 1.320 1.027 0.733 0.929 1.09 2.00 1.59 A 3.08 2.79 1.320 1.027 0.733 0.929 100 2.15 1.51 C 3.81 3.72 1.320 1.027 0.733 0.929 100 2.04 3.01 B 4.84 4.65 1.320 1.027 0.733 0.929 100 100 3.00 C 5.57 5.57 1.320 1.027 0.733 0.929 Eq. A: ( 29 pA cAh cf Am R T mT Q Σ - + Eq. B: ( 29 pB cBh cf Bm R T mT Q Σ - + Eq. C: ( 29 pC cCh cf Cm R T mT Q Σ - + 15.24 Two models A and B are to be assembled on a mixed model line. Hourly production rates for the two models are: A, 25 units/hr; and B, 18 units/hr. The work elements, element times, and precedence requirements are given in the table below. Elements 6 and 8 are not required for model A, and elements 4 and 7 are not required for model B. Assume E = 1.0, E r = 1.0, and M i = 1. (a) Construct the precedence diagram for each model and for both models combined into one diagram. (b) Find the theoretical minimum number of workstations required to achieve the required production rate. (c) Use the Kilbridge and Wester method to solve the line balancing problem. (d) Determine the balance efficiency for your solution in (c). Work element k T eAk Preceded by: T eBk Preceded by 1 0.5 min - 0.5 min - 2 0.3 min 1 0.3 min 1 3 0.7 min 1 0.8 min 1 4 0.4 min 2 - - 5 1.2 min 2, 3 1.3 min 2, 3 6 - - 0.4 min 3 7 0.6 min 4, 5 - - 8 - - 0.7 min 5, 6 9 0.5 min 7 0.5 min 8 T wc 4.2 min 4.5 min Solution : (a) Precedence diagrams. 113
Assembly Lines-3e-S 07-05/06, 06/04/07 1 2 3 4 5 7 9 0.5 0.3 0.7 0.4 1.2 0.6 0.5 Model A: 25 units/hr 1 2 3 5 6 8 9 0.5 0.3 0.8 1.3 0.4 0.7 0.5 Model B: 18 units/hr 1 2 3 4 5 6 7 8 9 AB AB AB A AB B A B AB Combined: A and B (b) Minimum number of workstations to achieve production rates for A and B. Element k R pA T eAk R pB T eBk Σ R pj T ejk 1 12.5 min 9.0 min 21.5 min 2 7.5 min 5.4 min 12.9 min 3 17.5 min 14.4 min 31.9 min 4 10.0 min - 10.0 min 5 30.0 min 23.4 min 53.4 min 6 - 7.2 min 7.2 min 7 15.0 min - 15.0 min 8 - 12.6 min 12.6 min 9 12.5 min 9.0 min 21.5 min Total = 186.0 min Given M = 1, E = 1, E r = 1, therefore, AT = 60 min, n = w = Min Int 186 0 60 .

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• Spring '10
• Hani
• Chemical element, Cycle Time, Production line

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