Question

A worker-machine operation was found to involve 3.3 minutes of machine time per cycle in the course of 40 cycles of stopwatch study. The worker’s time averaged 1.9 minutes per cycle, and the worker was given a rating of 120 percent (machine rating is 100 percent). Midway through the study, the worker took a 10-minute rest break. Assuming an allowance factor of 12 percent of work time, determine the standard time for this job.

Answer 1

The standard time for this job will be "5.89 minutes". A further solution is provided below.

Given values are:

• Machine time per cycle = 3.3 minutes
• Performance rating = 100%
• Average cycle time = 1.9 minutes
• Performance rating = 120%
• Allowance factor = 12% or 0.12

For machine,

→  The normal time will be:

= $Average \ observed \ time\times Performance \ rating$

By substituting the values, we get

= $3.3\times \frac{100}{100}$

= $3.3 \ minutes$

When there is no allowance then the allowance factor will be zero (0).

→  The standard time will be:

= $\frac{Normal \ time}{1-Allowance \ factor}$

= $\frac{3.3}{1-0}$

= $3.3 \ minutes$

For workers,

→  The normal time will be:

= $Average \ observed \ time\times Performance \ rating$

= $1.9\times \frac{120}{100}$

= $2.28 \ minutes$

→  The standard time will be:

= $\frac{Normal \ time}{1-Allowance \ factor}$

= $\frac{2.28}{1-0.12}$

= $2.59 \ minutes$

hence,

→  The standard time for this job will be:

= $Standard \ time \ of \ machine+Standard \ time \ of \ worker$

= $3.3+2.59$

= $5.89 \ minutes$

Thus the above answer is correct.

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