Solution a r p 29 29 100000 50 5 75 5333 pchr b t c

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Solution : (a) R p = ( 29 ( 29 100,000 50 5 7.5 = 53.33 pc/hr (b) T c = ( 29 60 0.96 53.33 = 1.08 min, w * = Minimum Integer 42.0 1.08 = 38.89 39 workers (c) T s = 1.08 - 0.1 = 0.98 min w = Minimum Integer ( 29 42.0 0.92 0.98 = 46.58 47 workers 15.4 A single model assembly line is being planned to produce a consumer appliance at the rate of 200,000 units per year. The line will be operated 8 hours per shift, two shifts per day, five days per week, 50 weeks per year. Work content time = 35.0 min. For planning purposes, it is anticipated that the proportion uptime on the line will be 95%. Determine (a) average hourly production rate, (b) cycle time, and (c) theoretical minimum number of workers required on the line. (d) If the balance efficiency is 0.93 and the repositioning time = 6 sec, how many workers will actually be required? Solution : (a) R p = 200 000 10 8 50 , ( )( ) = 50 pc/hr (b) T c = 60 0 95 50 ( . ) = 1.14 min (c) w = Minimum Integer 350 114 . . = 30.7 31 workers . (d) T s = 1.14 - 0.10 = 1.04 min E r = 1.04/1.14 = 0.9123 w = Minimum Integer 350 50 60 0 95 0 93 0 9123 . ( ) ( . )( . )( . ) = 36.2 37 workers . 15.5 The required production rate = 50 units per hour for a certain product whose assembly work content time = 1.2 hours. It is to be produced on a production line that includes four workstations that are automated. Because the automated stations are not completely reliable, the line will have an expected uptime efficiency = 90%. The remaining manual stations will each have one worker. It is anticipated that 8% of the cycle time 103
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Assembly Lines-3e-S 07-05/06, 06/04/07 will be lost due to repositioning at the bottleneck station. If the balance delay is expected to be 0.07, determine (a) the cycle time, (b) number of workers, (c) number of workstations needed for the line, (d) average manning level on the line, including the automated stations, and (e) labor efficiency on the line. Solution : (a) T c = 60 0 90 50 ( . ) = 1.08 min (b) T r = 0.08 T c , therefore, T s = 0.92 T c , therefore, E r = 0.92 E b = 1 - d = 1 - 0.07 = 0.93 w = Minimum Integer 12 60 0 93 0 92 108 . ( ) ( . )( . )( . ) = 77.9 78 workers . (c) n = 78 + 4 = 82 workstations . (d) M = 78/82 = 0.951 (e) Labor efficiency = EE b E r = 0.90(0.93)(0.92) = 0.77 = 77% 15.6 A final assembly plant for a certain automobile model is to have a capacity of 225,000 units annually. The plant will operate 50 weeks/yr, two shifts/day, 5 days/week, and 7.5 hours/shift. It will be divided into three departments: (1) body shop, (2) paint shop, (3) general assembly department. The body shop welds the car bodies using robots, and the paint shop coats the bodies. Both of these departments are highly automated. General assembly has no automation. There are 15.0 hours of work content time on each car in this third department, where cars are moved by a continuous conveyor. Determine (a) hourly production rate of the plant, (b) number of workers and workstations required in trim-chassis-final if no automated stations are used, the average manning level is 2.5, balancing efficiency = 90%, proportion uptime = 95%, and a repositioning time of 0.15 min is allowed for each worker. Solution : (a) R p = 225 000 50 2 5 7 5 , ( )( )( . ) = 60 cars/hr.
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  • Spring '10
  • Hani
  • Chemical element, Cycle Time, Production line

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