63876-Ch17

# Analyze the problem and determine the following a the

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assemblies is far below that designed production rate. Analyze the problem and determine the following: (a) The designed average production rate that the foreman alluded to. (b) What is the proportion of assemblies coming off the system that contain one or more defective components? (c) What seems to be the problem that limits the assembly system from achieving the expected production rate? (d) What is the production rate that the system is actually achieving? State any assumptions that you make in determining your answer. Station Assembly time Feed rate f Selector θ q m 2 4 sec 32/min 0.25 0.01 1.0 3 7 sec 20/min 0.50 0.005 0.6 4 5 sec 20/min 0.20 0.02 1.0 5 3 sec 15/min 1.0 0.01 0.7 140

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Automated Asby-3e-S 7-12, 12/06, 06/04/07 Solution : T d = 2 min, T c = 7 + 2 = 9 sec = 0.15 min Cycle rate R c = 1/0.15 = 6.667 cycles/min (a) Σ ( mq ) = 1(0.01) + 0.6(0.005) + 1(0.02) + 0.7(0.01) = 0.04 T p = 0.15 + 0.04(2) = 0.23 min R p = 60/0.23 = 260.9 asbys/hr (b) P ap = (1-0.01+1x0.01)(1-0.005+0.6x0.005)(1-0.02+1x0.02)(1-0.01+0.7x0.01) = (1)(0.998)(1)(0.997) = 0.995 P qp = 1 - 0.995 = 0.005 R ap = 260.9(0.995) = 259.6 good asbys/hr (c) Station 2: f θ = 32(.25) = 8 components/min Station 3: f θ = 20(.50) = 10 components/min Station 4: f θ = 20(.20) = 4 components/min Station 5: f θ = 15(1.0) = 15 components/min The problem is that the feeder for station 4 is slower than the machine's cycle rate of 6.667 cycles/min (d) If the machine operates at the cycle rate that is consistent with the feed rate of Station 4, then T c = 15 sec = 0.25 min T p = 0.25 + 0.04(2) = 0.33 min R p = 60/0.33 = 181.8 asbys/hr R ap = 181.8(0.995) = 180.9 good asbys/hr 17.13 For Example 17.4 in the text, dealing with a single-station assembly system, suppose that the sequence of assembly elements were to be accomplished on a seven-station assembly system with synchronous parts transfer. Each element is performed at a separate station (stations 2 through 6) and the assembly time at each respective station is the same as the element time given in Example 17.4. Assume that the handling time is divided evenly (3.5 sec each) between a load station (station 1) and an unload station (station 7). The transfer time is 2 sec, and the average downtime per downtime occurrence is 2.0 min. Determine (a) production rate of all completed units, (b) yield, (c) production rate of good quality completed units, and (d) uptime efficiency. Solution : (a) T c = 7 + 2 = 9.0 sec = 0.15 min F = 0.02(1.0) + 0.01(0.6) + 0.015(0.8) + 0.02(1.0) + 0.012 = 0.070 T p = 0.15 + 0.070(2.0) = 0.15 + 0.14 = 0.29 min R p = 1/0.29 = 3.45 asbys/min = 206.9 asbys/hr (b) P ap = (1.0)(0.996)(0.997)(1.0) = 0.993 (c) R ap = 206.9(0.993) = 205.5 good asbys/hr (d) E = 0.15/0.29 = 0.5172 = 51.7% Comment : Comparing the values with those in Example 17.4, production rate is more than double for the multi-station system, yield is the same, and line efficiency is greatly reduced because of the much faster cycle time. Single Station Assembly Systems 17.14 A single-station assembly machine is to be considered as an alternative to the dial-indexing machine in Problem 17.4. Use the data given in the table for that problem to determine (a) production rate, (b) yield of good product (final assemblies containing no defective components), and (c) proportion uptime of the system. Handling time to load the base part and unload the finished assembly is 7 sec and the downtime averages 1.5 min every time a component jams. Why is the proportion uptime so much higher than in the case of the dial- indexing machine in Problem 17.4?
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• Spring '10
• Hani
• Cycle Time, Production line, Stations of the Cross, Uptime

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