QUESTION 1 (15pts): In class we mentioned about the external and internal work e

Question

QUESTION 1 (15pts): In class we mentioned about the external and internal work elements in a worker-machine cycle; What are differences between internal and external work elements? QUESTION 2 (35pts): A repetitive work cycle takes 4.25 min (normal time). Then, an irregular work element (1.75 min) is performed every 8 cycles. Please note that two work-units are produced each cycle (hint: Tstd/2). The PFD allowance factor is 16%. Based on the given data calculate (a) the standard time per piece and (b) determine the number of work units produced in a given shift (an 8- hour shift) when performance is standard. (c) calculate the amount of time lost and the anticipated amount of time worked per 8-hour shift (hint: PFD allowance factor is still 16%) QUESTION 3 (25pts): The standard time for a manual work cycle is given as 2.58 min allowance factor is 13%. During a 8 hour shift, it is recorded that the employee lost a total of 53 min due to delays and rest breaks. The worker completed 214 work units on the same day. Calculate (a) the number of std (standard) hours accomplished, (b) worker efficiency per piece. The PFD QUESTION 4 (25pts): For a work cycle, the normal time in a worker-machine system is 6.27 min. the PFD allowance factor is 12%, and the machine allowance factor is 25%. The work cycle includes manual elements totaling a normal time of 5.92 min, all but 0.65 min of which are performed as internal elements. Determine (a) the standard time for the cycle and (b) calculate Qstd

Explanation / Answer

(a) Machine time per cycle Tm = 6.27 – 0.65 = 5.62 min

Internal normal time Tnwi = 5.92 – 0.65 = 5.27 min

Tstd = 0.65(1.12) + Max{5.27(1.12), 5.62(1.25)} = 0.728 + 7.025 = 7.753 min

(b) Qstd = 8(60)/7.753 = 61.9 pc (if rounded, 62 pc)

(c) Time worked = 480 – 39 = 441 min

Given that Q = 72 pc, then total machine time = 72(5.62) = 404.64 min

Total worker-controlled time = 441 – 404.64 = 36.36 min

Given Tnw = 0.65 min, total worker-controlled time at normal pace = 72(0.65) = 46.8 min

Pw = 46.8/36.36 = 1.287 = 128.7%

Solve it by using this example

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