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 |
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 = \frac{.1 + .3 + .2 + .2 + .1}{5} = 0.18 min
B’s average observed time = \frac{.8 + .6 + .8 + .5 + .7}{5} = 0.68 min
C’s average observed time = \frac{.5 + .5 + .4 + .5 + .6 + .5}{6} = 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 = \frac{1.31}{1 – 0.20} = 1.64 min