A work operation consisting of three elements has been subjected to a stopwatch time study. The recorded observations are shown in the following table. By union contract, the allowance time for the operation is personal time 5%, delay 5%, and fatigue 10%. Determine the standard time for the work operation.

JOB
ELEMENT OBSERVATIONS (MINUTES) PERFORMANCE
RATING (%)

1 2 3 4 5 6

A .1 .3 .2 .9 .2 .1 90

B .8 .6 .8 .5 3.2 .7 110

C .5 .5 .4 .5 .6 .5 80

Question

A work operation consisting of three elements has been subjected to a stopwatch time study. The recorded observations are shown in the following table. By union contract, the allowance time for the operation is personal time 5%, delay 5%, and fatigue 10%. Determine the standard time for the work operation.

JOB ELEMENT	OBSERVATIONS (MINUTES)						PERFORMANCE RATING (%)
JOB ELEMENT	1	2	3	4	5	6	PERFORMANCE RATING (%)
A	.1	.3	.2	.9	.2	.1	90
B	.8	.6	.8	.5	3.2	.7	110
C	.5	.5	.4	.5	.6	.5	80

JOB ELEMENT	OBSERVATIONS (MINUTES)						PERFORMANCE RATING (%)
JOB ELEMENT	1	2	3	4	5	6	PERFORMANCE RATING (%)
A	.1	.3	.2	.9	.2	.1	90
B	.8	.6	.8	.5	3.2	.7	110
C	.5	.5	.4	.5	.6	.5	80

Accepted Answer

First, delete the two observations that appear to be very unusual (.9 minute for job element A and 3.2 minutes for job element B). Then:

A’s average observed time = [latex]\frac{.1 + .3 + .2 + .2 + .1}{5}[/latex] = 0.18 min
B’s average observed time = [latex]\frac{.8 + .6 + .8 + .5 + .7}{5}[/latex] = 0.68 min
C’s average observed time = [latex]\frac{.5 + .5 + .4 + .5 + .6 + .5}{6}[/latex] = 0.50 min

A’s normal time = (0.18)(0.90) = 0.16 min
B’s normal time = (0.68)(1.10) = 0.75 min
C’s normal time = (0.50)(0.80) = 0.40 min
Normal time for job = 0.16 + 0.75 + 0.40 = 1.31 min
Note, the total allowance factor = 0.05 + 0.05 + 0.10 = 0.20

Then: Standard time = [latex]\frac{1.31}{1 - 0.20}[/latex] = 1.64 min

Question 10.SP.2: A work operation consisting of three elements has been subje...

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