In: Mechanical Engineering
1) Find out the standard time using the following data:
Average time for machine elements = 6 min.
Average time for manual elements = 4 min.
Performance rating = 110%
Allowances = 10% of normal time
2) The actual observed time for an operation was 1 minute per piece. If the performance rating of the operator was 120 and a 5 per cent personal time is to be provided, the standard time in minute per piece is:
Q 1 | |
Input | |
1) | Avg time for machine elements = 6 min |
2) | Avg time for manual elements = 4 min |
3) | Performace rating = 110 % |
4) | Allowance - = 10 % normal time |
To find Standard time | |
Answer - | |
Let us find out Normal time for Manual elements | |
We know, | |
Normal time = (average element time ) X ( Peformance rating / 100) | |
'= ( 4 ) /x (110 /100) | |
Normal time for manual element = 4.4 min | |
Thus normal time for whole operation = Avg time for machine elements + | |
Normal time for manual elements | |
Normal time for whole operation = 6 +4.4 = 10.4 min | |
We know, | |
Standard time = normal time + allowance time | |
Here allowance time = 10 % normal time | |
'= 0.1 x 10.4 = 1.04 min | |
Hence standard time = 10.4 +1.04 = 11.44 min | |
Standard time for operation = 11.44 min | |
Q 2 | |
Input | |
1) | Avg time for machine elements = 0 min |
2) | Avg time for manual elements = 1 min |
3) | Performace rating = 120 % |
4) | Allowance - = 5 % normal time |
To find Standard time | |
Answer - | |
Let us find out Normal time for Manual elements | |
We know, | |
Normal time = (average element time ) X ( Peformance rating / 100) | |
'= ( 1 ) /x (120 /100) | |
Normal time for manual element = 1.2 min |