Solution-HW3.pdf - Cycle Time Analysis in Worker-Machine...

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Cycle Time Analysis in Worker-Machine Systems 2.7 The normal time of the work cycle in a worker-machine system is 5.39 min. The operator-controlled portion of the cycle is 0.84 min. One work unit is produced each cycle. The machine cycle time is constant. (a) Using a PFD allowance factor of 16% and a machine allowance factor of 30%, determine the standard time for the work cycle. (b) If a worker assigned to this task completes 85 units during an 8- hour shift, what is the worker’s efficiency? (c) If it is known that a total of 42 min was lost during the 8-hour clock time due to personal needs and delays, what was the worker’s performance on the porti on of the cycle he controlled? Solution : Machine time per cycle T m = 5.39 0.84 = 4.55 min T std = 0.84(1 + 0.16) + 4.55(1 + 0.30) = 6.889 min (b) H std = 85(6.889)/60 = 585.6/60 = 9.76 hr E w = 9.76/8.0 = 1.22 = 122% (c) Time worked = 480 42 = 438 min Total machine time = 85(4.55) = 386.75 min Total operator time = 438 386.75 = 51.25 min Total operator time at normal pace = 85(0.84) = 71.4 min P w = 71.4/51.25 = 1.393 = 139.3% 2.9 The work cycle in a worker-machine system consists of (1) external manual work elements with a total normal time of 0.42 min, (2) a machine cycle with machine
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