63876-Ch17

The components fed into the selector can have any of

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eight stations is an automatic assembly operation that uses a feeder-selector. The components fed into the selector can have any of five possible orientations, each with equal probability, but only one of which is correct for passage into the feed track to the assembly workhead. Parts rejected by the selector are fed back into the hopper. What minimum rate must the feeder deliver components to the selector during system uptime in order to keep up with the assembly machine? Solution : T p = 60/400 = 0.15 min/asby T p = T c + FT d = T c + 1 50 2 5 ( . ) = T c + 0.05 T c = 0.15 - 0.05 = 0.10 min/asby R c = 1 1 0.10 c T = = 10 asbys/min Min f θ = 0.20 f = 10 asbys/min Feeder rate f = 10 0 20 . = 50 parts/min Multi-Station Assembly Systems 17.4 A dial indexing machine has six stations that perform assembly operations on a base part. The operations, element times, q and m values for components added are given in the table below (NA means q and m are not applicable to the operation). The indexing time for the dial table is 2 sec. When a jam occurs, it requires 1.5 137

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Automated Asby-3e-S 7-12, 12/06, 06/04/07 min to release the jam and put the machine back in operation. Determine (a) production rate for the assembly machine, (b) yield of good product (final assemblies containing no defective components), and (c) proportion uptime of the system. Station Operation Element time q m 1 Add part A 4 sec 0.015 0.6 2 Fasten part A 3 sec NA NA 3 Assemble part B 5 sec 0.01 0.8 4 Add part C 4 sec 0.02 1.0 5 Fasten part C 3 sec NA NA 6 Assemble part D 6 sec 0.01 0.5 Solution : (a) Σ ( mq ) = 0.6(0.015) + 0.8(0.01) + 1(0.02) + .5(0.01) = .042 T p = 0.1333 + 0.042(1.5) = 0.19633 min/asby R p = 60/0.19633 = 305.6 asbys/hr (b) P ap = (1-0.015+0.6x.015)(1-0.01+0.8x.01)(1-0.02+1x0.02)(1-0.01+0.5x0.01) = (0.994)(0.998)(1.0)(0.995) = 0.98705 (c) E = 0.1333/0.19633 = 0.679 = 67.9% 17.5 An eight-station assembly machine has an ideal cycle time of 6 sec. The fraction defect rate at each of the 8 stations is q = 0.015 and a defect always jams the affected station. When a breakdown occurs, it takes 1 minute, on average, for the system to be put back into operation. Determine the production rate for the assembly machine, the yield of good product (final assemblies containing no defective components), and proportion uptime of the system. Solution : T p = 0.1 + 8(1.0)(0.015)(1.0) = 0.22 min/asby. R p = 60/0.22 = 272.7 asbys/hr If defects always jam the affected station, then m = 1.0 P ap = (1 - 0.015 + 1x0.015) 8 = 1.0 = yield E = 0.1/0.22 = 0.4545 = 45.45% 17.6 Solve Problem 17.5 but assume that defects never jam the workstations. Other data are the same. Solution : T p = 0.1 + 8(0)(0.015)(1.0) = 0.10 min/asby. R p = 60/0.10 = 600 asbys/hr If defects never jam, then m = 0 P ap = (1 - 0.015 + 0x0.015) 8 = 0.8861 = yield E = 0.1/0.1 = 100% 17.7 Solve Problem 17.5 but assume that m = 0.6 for all stations. Other data are the same. Solution : T p = 0.1 + 8(0.6)(0.015)(1.0) = 0.172 min/asby R p = 60/0.172 = 348.8 asbys/hr P ap = (1 - 0.015 + 0.6x0.015) 8 = 0.953 = yield E = 0.1/0.172 = 0.5814 = 58.14% 17.8 A six-station automatic assembly line has an ideal cycle time of 12 sec. Downtime occurs for two reasons. First, mechanical and electrical failures cause line stops that occur with a frequency of once per 50 cycles. Average downtime for these causes is 3 min. Second, defective components also result in downtime. The fraction defect rate of each of the six components added to the base part at the six stations is 2%. The probability that a defective component will cause a station jam is 0.5 for all stations. Downtime per occurrence for defective parts is 2 min. Determine (a) yield of assemblies that are free of defective
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• Spring '10
• Hani
• Cycle Time, Production line, Stations of the Cross, Uptime

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