Computation of Normal Time and Standard
Time for a Job
∑ (Normal
times for all the elements of the job)
[(μ of Recorded
Times) × (Performance Rating expressed in %-age with 100% as the ‘accepted’ performance)]
÷ 100
Note: ‘μ’ is Arithmetical Average or Mean.
(Normal time
for the job) ÷ (1 – Total allowance expressed as a fraction)
Important note:
Required number of readings to be recorded for
an element (i.e. sample size of readings), for calculating standard time for a
job, can be calculated as follows:
| Nreqd
  = | [[40.
  {Ntaken∑X^2 – (∑X) ^2}^ (1/2)] ÷ ∑X] ^2 | 
Where,
| Nreqd
  = | Number
  of readings required to be recorded for an element for calculating standard
  time for a job | 
| Ntaken
  = | Number
  of readings for the element taken, based on which Nreqd is being
  determined | 
| X
  = | Value
  of the individual readings of the element | 
Illustration:
Continuous Stopwatch Time Study Figures for a job are
given below. Calculate the Standard Time for the job assuming that the sample
size is adequate, and total allowances are 15%.
| Cycle | ||||
| 1 | 2 | 3 | 4 | |
| (Continuous Stopwatch Time Study Figures in Minutes) | ||||
| 1 | 0.09 | 0.16 | 0.28 | 0.41 | 
| 2 | 0.49 | 0.56 | 0.67 | 0.80 | 
| 3 | 0.89 | 0.95 | 1.07 | 1.21 | 
| 4 | 1.31 | 1.38 | 1.49 | 1.61 | 
| 5 | 1.70 | 1.76 | 1.88 | 2.00 | 
| 6 | 2.09 | 2.16 | 2.28 | 2.41 | 
| 7 | 2.50 | 2.57 | 2.68 | 2.80 | 
| 8 | 2.88 | 2.95 | 3.07 | 3.20 | 
| 9 | 3.29 | 3.36 | 3.40 | 3.62 | 
| 10 | 3.71 | 3.78 | 3.90 | 4.03 | 
| Performance Rating | 90 | 110 | 120 | 100 | 
Solution:
| Cycle | Elements | |||
| 1 | 2 | 3 | 4 | |
| (Stopwatch Time Study Figures in Minutes) | ||||
| 1 | 0.09 | 0.07 | 0.12 | 0.13 | 
| 2 | 0.08 | 0.07 | 0.11 | 0.13 | 
| 3 | 0.09 | 0.06 | 0.12 | 0.14 | 
| 4 | 0.10 | 0.07 | 0.11 | 0.12 | 
| 5 | 0.09 | 0.06 | 0.12 | 0.12 | 
| 6 | 0.09 | 0.07 | 0.12 | 0.13 | 
| 7 | 0.09 | 0.07 | O.11 | 0.12 | 
| 8 | 0.08 | 0.07 | 0.12 | 0.13 | 
| 9 | 0.09 | 0.07 | 0.04 | 0.22 | 
| 10 | 0.09 | 0.07 | 0.12 | 0.13 | 
| μ | 0.089 | 0.068 | 0.109 | 0.137 | 
| Performance Rating | 90 | 110 | 120 | 100 | 
| Normal Time for Element in Minutes | 0.0801 | 0.0748 | 0.1308 | 0.137 | 
Normal time for the
job
= 0.0801 + 0.0748 + 0.1308 + 0.137 = 0.4227 Minutes
Standard time for the job
= (Normal time
for the job) ÷ (1 – Total allowance expressed as a fraction)
= 0.4227 ÷ (1 – 0.15) = 0.4973 Minutes
 
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