Question 10.EX.2: USING TIME STUDIES TO COMPUTE STANDARD TIME. Management Scie...

USING TIME STUDIES TO COMPUTE STANDARD TIME. Management Science Associates promotes its management development seminars by mailing thousands of individually composed and typed letters to various firms. A time study has been conducted on the task of preparing letters for mailing. On the basis of the following observations, Management Science Associates wants to develop a time standard for this task. The firm’s personal, delay, and fatigue allowance factor is 15%.

OBSERVATIONS (MINUTES)
JOB ELEMENT 1 2 3 4 5 PERFORMANCE RATING
(A) Compose and type letter 8 10 9 21^* 11 120%
(B) Type envelope address 2 3 2 1 3 105%
(C) Stuff, stamp, seal, and sort envelopes 2 1 5^* 2 1 110%

APPROACH \blacktriangleright Once the data have been collected, the procedure is to:
1. Delete unusual or nonrecurring observations.
2. Compute the average time for each element, using Equation (10-1).
3. Compute the normal time for each element, using Equation (10-2).
4. Find the total normal time.
5. Compute the standard time, using Equation (10-3).

Average observed time = \frac{Sum  of  the  times  recorded  to  perform  each  element}{Number  of  observations}                  (10-1)

Normal time = Average observed time × Performance rating factor              (10-2)

Standard time = \frac{Total  normal  time}{1 - Allowance  factor}                  (10-3)

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SOLUTION \blacktriangleright
1. Delete observations such as those marked with an asterisk (*). (These may be due to business interruptions, conferences with the boss, or mistakes of an unusual nature; they are not part of the job element, but may be personal or delay time.)

2. Average time for each job element:
Average time for A = \frac{8 + 10 + 9 + 11}{4} = 9.5 min
Average time for B = \frac{2 + 3 + 2 + 1 + 3}{5} = 2.2 min
Average time for C = \frac{2 + 1 + 2 + 1}{4} = 1.5 min

3. Normal time for each job element:
Normal time for A = (Average observed time) × (Performance rating)
= (9.5)(1.2) = 11.4 min
Normal time for B = (2.2)(1.05) = 2.31 min
Normal time for C = (1.5)(1.10) = 1.65 min

Note: Normal times are computed for each element because the performance rating factor (work pace) may vary for each element, as it did in this case.

4. Add the normal times for each element to find the total normal time (the normal time for the whole job):
Total normal time = 11.40 + 2.31 + 1.65 = 15.36 min

5. Standard time for the job:
Standard time = \frac{Total  normal  time}{1 – Allowance  factor} = \frac{15.36}{1 – 0.15} = 18.07 min

Thus, 18.07 minutes is the time standard for this job.

INSIGHT \blacktriangleright When observed times are not consistent they need to be reviewed. Abnormally short times may be the result of an observational error and are usually discarded. Abnormally long times need to be analyzed to determine if they, too, are an error. However, they may include a seldom occurring but legitimate activity for the element (such as a machine adjustment) or may be personal, delay, or fatigue time.

LEARNING EXERCISE \blacktriangleright If the two observations marked with an asterisk were not deleted, what would be the total normal time and the standard time? [Answer: 18.89 min, 22.22 min.]

RELATED PROBLEMS \blacktriangleright 10.22–10.25, 10.28a,b, 10.29a, 10.30a (10.41–10.43 are available in MyOMLab)

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