Virginia College promotes a wide variety of executive training courses for firms
ID: 347406 • Letter: V
Question
Virginia College promotes a wide variety of executive training courses for firms in the Arlington, Virginia, region. Director Wendy Tate believes that individually typed letters add a personal touch to marketing. To prepare letters for mailing, she conducts a time study of her secretaries. On the basis of the observations shown in the following table, she wishes to develop a time standard for the whole job.
*Disregard—secretary stopped to answer the phone
^Disregard—interruption by supervisor
The college uses a total allowance factor of 12%.
Tate decides to delete all unusual observations from the time study.
The standard time for this process = ___ minutes
Observation (minutes per cycle) Performance Element Typing letter Typing envelope Stuffing envelope Sealing, sorting Rating 90% 100% 90 % 120% 2 3.6 0.8 0.5 2.8A 5 2.7 3.2 0.7 4.5A 3.3 0.7 0.4 0.9 2.6 0.7 0.5 1.0 2.8 0.5 1.9 1.0 2.1 0.9 0.3Explanation / Answer
According to Time study method the standard cycle time is calculated as follows:
1. Obtain the elemental times for each process element
2. Determine the performance rating (PR) of worker for each process element.
3. Calculate average time for each element. (disregard times of all unusual observations)
4. Compute normal time for each element as follows:
Normal Time = NT = Element average time x PR x frequency
5. Determine the allowance factor (A) for the complete process.
Fatigue factor = 12%
6. Calculate Standard time as follows:
Standard time = (Total normal time)/ (1 - A)
ST = (NT)/(1 - 0.12)
For given problem, frequency = 1
Observation (minutes per cycle)
Average time (t)
NT = t x RF
Element
Performance Rating
1
2
3
4
5
6
Typing letter
90%
2.6
3.6
2.8
2.1
2.7
3.3
2.85
0.9*2.85
= 2.565
Typing envelope
100%
0.7
0.8
0.5
0.9
3.2
0.7
0.72
0.72
stuffing envelope
90%
0.5
0.5
1.9
0.3
0.7
0.4
0.48
0.48*0.9
= 0.432
Sealing, sorting
120%
1
2.8
1
1
4.5
0.9
0.975
0.975*1.2
= 1.17
2.568+0.72+0.432+1.17
= 4.887
ST = (NT)/(1 - 0.12) = 4.887/(1 – 0.12) = 5.55
The standard time for this process is 5.55 minutes
Observation (minutes per cycle)
Average time (t)
NT = t x RF
Element
Performance Rating
1
2
3
4
5
6
Typing letter
90%
2.6
3.6
2.8
2.1
2.7
3.3
2.85
0.9*2.85
= 2.565
Typing envelope
100%
0.7
0.8
0.5
0.9
3.2
0.7
0.72
0.72
stuffing envelope
90%
0.5
0.5
1.9
0.3
0.7
0.4
0.48
0.48*0.9
= 0.432
Sealing, sorting
120%
1
2.8
1
1
4.5
0.9
0.975
0.975*1.2
= 1.17
2.568+0.72+0.432+1.17
= 4.887
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