63876-Ch17

# Determine a yield of assemblies that are free of

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occurrence for defective parts is 2 min. Determine (a) yield of assemblies that are free of defective components, (b) proportion of assemblies that contain at least one defective component, (c) average production rate of good product, and (d) uptime efficiency. Solution : (a) P ap = (1 - 0.02 + 0.5x0.02) 6 = (0.99) 6 = 0.9415 (b) P qp = 1 - 0.9415 = 0.0585 (c) T p = 12/60 + 0.02(3) + 6(0.5)(0.02)(2) = 0.38 min R p = 60/0.38 = 157.6 cycles/hr R ap = (0.9415)(157.9) = 148.66 good asbys/hr 138

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Automated Asby-3e-S 7-12, 12/06, 06/04/07 (d) E = 0.2/0.38 = 0.526 = 52.6% 17.9 An eight-station automatic assembly machine has an ideal cycle time of 10 sec. Downtime is caused by defective parts jamming at the individual assembly stations. The average downtime per occurrence is 3.0 min. The fraction defect rate is 1.0% and the probability that a defective part will jam at a given station is 0.6 for all stations. The cost to operate the assembly machine is \$90.00 per hour and the cost of components being assembled is \$.60 per unit assembly. Ignore other costs. Determine (a) yield of good assemblies, (b) average production rate of good assemblies, (c) proportion of assemblies with at least one defective component, and (d) unit cost of the assembled product. Solution : (a) P ap = (1 - 0.01 + 0.6x0.01) 8 = (0.996) 8 = 0.9684 (b) T p = 0.1667 + 8(0.6)(0.01)(3) = 0.310667 min/asby R p = 60/0.310667 = 193.1 asbys/hr R ap = (0.9684)(193.1) = 187.04 good asbys/hr (c) P qp = 1 - 0.9684 = 0.0316 (d) C pc = [0.60+1.50(0.310667)]/0.9684 = 1.066/0.9684 = \$1.1007/asby 17.10 An automated assembly machine has four workstations. The first station presents the base part, and the other three stations add parts to the base. The ideal cycle time for the machine is 3 sec, and the average downtime when a jam results from a defective part is 1.5 min. The fraction defective rates ( q ) and probabilities that a defective part will jam the station ( m ) are given in the table below. Quantities of 100,000 for each of the bases, brackets, pins, and retainers are used to stock the assembly line for operation. Determine (a) proportion of good product to total product coming off the line, (b) production rate of good product coming off the line, (c) total number of final assemblies produced, given the starting component quantities. Of the total, how many are good product, and how many are products that contain at least one defective component? (d) Of the number of defective assemblies determined in above part (c), how many will have defective base parts? How many will have defective brackets? How many will have defective pins? How many will have defective retainers? Station Part identification q m 1 Base 0.01 1.0 2 Bracket 0.02 1.0 3 Pin 0.03 1.0 4 Retainer 0.04 0.5 Solution : (a) P ap = (1-0.01+1x0.01)(1-0.02+1x0.02)(1-0.03+1x0.03)(1-0.04+0.5x0.04) = (1.0)(1.0)(1.0)(0.98) = 0.98 (b) T p = 3/60 + (0.01+0.02+0.03+0.04x0.5)(1.5) = 0.17 min/cycle. R ap = 0.98(60/0.17) = 345.9 good asbys/hr (c) The diagram below shows quantities of components at the four workstations in the assembly machine: base 100,000 bracket 100,000 pin 100,000 retainer 100,000 1 2 3 4 95060 good asbys + 1940 with defective 1000 def 2000 def 3000 def 1940 def retainers Total number produced = 95,060 + 1,940 = 97,000 Number of units of good product = 95,060 Number of units containing at least one defect = 1,940 (d) Number of products containing defective base parts = 0 Number of products containing defective brackets = 0 Number of products containing defective pins = 0 Number of products containing defective retainers = 1,940 17.11 A six-station automatic assembly machine has an ideal cycle time of 6 sec. At stations 2 through 6, parts
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