The snapback timing method in direct time study was used to obtain the times for

Question

The snapback timing method in direct time study was used to obtain the times for a worker-machine task. The recorded times are listed in the table below. Element d is a machine-controlled element and the time is constant. Elements a, b, c, e, and f are operator-controlled and were performance rated at 90%. Element f is an irregular element, performed every five cycles. The operator-controlled elements are all external to machine-controlled element d. The machine allowance is zero, and the PFD allowance is 13%. Determine (a) the normalized time for the cycle and (b) the standard time for the cycle. The worker's actual time spent working during an 8-hour shift was 7.08 hours, and he produced 400 units of output during this time. Determine (c) the worker's performance during the operator-controlled portions of the cycle and (d) the worker's efficiency during this shift.

Explanation / Answer

Ans(a):
Given that performance rated at 90% or 0.90
Element f is performed every five cycle so time for 1 cycle = 0.62/5 = 0.124
so normalized time for elements a, b, c, e, f is given by
Normal time =((0.14 +0.25 +0.18 +0.20 + 0. 124)*0.90 = 0.8046 minutes
Cycle time is given by
Cycle time = performance rated operator controlled time + machine controlled element = 0.8046 +0.45 =1.2546 minutes

Ans(b):

Standard time for the cycle is given by
Standard time = operator controlled elements *(1+ PFD) + machine controlled element time (1+ machine allowance)
=0.8046*(1.13) + 0.45 (1+0) =1.359 minutes

Ans(c):
Given that worker produced 400 units of output during 8 hour shift while time spent was only 7.08 hours.
so machine time taken =Time taken for d * no of units = 0.45*400 =180 minutes
Operator time = Time spent by operator - machine time = 7.08*60 - 180 =424.8 - 180 = 244.8 minutes

Operator cycle time = (operator time) / (no of units) = 244.8 /400 = 0.612 minutes/cycle

Operator performance = (normal cycle time of operator)/ (operator cycle time taken) =0.8046/0.612 = 1.31470588235
which is approx 131.47%

Ans(d):
Standard time = Number of units * standard cycle time = 400* 1.359 =543.6 minutes = 543.6/60 hours=9.06 hours
Allowed time is 8 hours shift.
so Worker effeciency is given by
Worker effeciency =Standard time / allowed time = 9.06/8 =1.1325
which is approx 113.25%

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