Time study and standard time

Why it matters

Standard time is the number behind almost every plant decision: how many workers and machines a line needs, what a part costs, what the delivery promise is and what incentive a worker earns. Stopwatch time study is the most widely used way to set it for repetitive manual and machine-paced jobs, so you must know how to get from raw readings to a defensible standard.

Key ideas

Time study is a work-measurement technique for recording the times and rates of working for the elements of a specified job under specified conditions, and analysing the data to obtain the time needed to do the job at a defined level of performance. It should be done after method study, on the improved and standardised method.

Qualified worker and standard performance. The standard is set for a qualified worker (trained, with the physical attributes and skill for the job) working at standard performance: the rate a qualified worker naturally achieves without over-exertion over a working day, given motivation. On the British Standard scale this is 100; on the older 60–80 scale it is 80.

Procedure

1. Select the job and the worker (an average, qualified, cooperative operator); record the method, tools, materials and conditions.
2. Break the job into elements: short, distinct parts with clear break points. Separate manual from machine elements, constant from variable, repetitive from occasional, and record foreign elements (not part of the job) separately so they can be excluded.
3. Time each element with a stopwatch (continuous or snap-back/fly-back method) or electronic timer, and rate the worker's pace at the same time.
4. Decide the number of cycles needed for the required accuracy.
5. Compute the normal (basic) time, then add allowances to get the standard time.

Performance rating adjusts observed time to what a qualified worker at standard pace would take. Methods: speed (pace) rating, where the observer judges pace against the 100 standard; the Westinghouse (levelling) system, which adds factors for skill, effort, conditions and consistency to 1; synthetic rating, which compares observed times of some elements with predetermined (PMTS) times for them; and objective rating. A rating of 120 % means the worker was faster than standard, so normal time is longer than observed time.

Allowances are extra time added to normal time: personal needs; basic and variable fatigue (posture, force, heat, noise, monotony); unavoidable delays; and policy or special allowances (contingency, set-up, machine interference). Total allowances usually fall in the range of about 10–20 %, but the exact values come from the company's or ILO allowance tables.

Allowances defined two ways. If allowance is a percentage of normal time, multiply. If it is a percentage of the total working time (shift), divide by (1 − A). Read the question carefully; the second gives a larger standard time.

Formulas

Observed (selected) time OT = Σx / n

• x = element time from each cycle (min); n = number of cycles; foreign elements and abnormal readings removed.

Normal (basic) time NT = OT × R / 100

• R = performance rating (%), standard pace = 100.

Standard time ST = NT × (1 + A) when A is a fraction of normal time.

Standard time ST = NT / (1 − A) when A is a fraction of the total working time.

Number of cycles N′ = [ 40 √(nΣx² − (Σx)²) / Σx ]²

• For 95 % confidence and ±5 % accuracy from n trial readings. General form N′ = (z·s / (k·x̄))², where z = 1.96 for 95 % confidence, s = sample standard deviation, k = accuracy fraction, x̄ = mean.

Output per shift = Available time / ST

Worked examples

Example 1 (standard). A job has three elements with average observed times 0.25, 0.60 and 0.15 min, rated 120 %, 90 % and 100 %. Allowances are 15 % of normal time. Find the standard time and the standard output in a 480-min shift.

1. Element normal times: 0.25 × 1.20 = 0.30, 0.60 × 0.90 = 0.54, 0.15 × 1.00 = 0.15 min.
2. NT = 0.30 + 0.54 + 0.15 = 0.99 min.
3. ST = 0.99 × 1.15 = 1.139 min per unit.
4. Output = 480 / 1.1385 = 421 units per shift.

Example 2 (GATE level). Ten trial readings of an element (min): 0.40, 0.48, 0.36, 0.50, 0.42, 0.46, 0.38, 0.52, 0.44, 0.40. (a) How many readings are needed for 95 % confidence and ±5 % accuracy? (b) With a rating of 110 % and allowances of 12 %, find the standard time if the allowance is (i) 12 % of normal time and (ii) 12 % of total working time.

1. Σx = 4.36, Σx² = 1.9264, n = 10.
2. nΣx² − (Σx)² = 19.264 − 19.0096 = 0.2544; √0.2544 = 0.5044.
3. N′ = (40 × 0.5044 / 4.36)² = (4.627)² = 21.4, so 22 readings; 12 more are needed.
4. OT = 4.36 / 10 = 0.436 min; NT = 0.436 × 1.10 = 0.4796 min.
5. (i) ST = 0.4796 × 1.12 = 0.537 min.
6. (ii) ST = 0.4796 / (1 − 0.12) = 0.4796 / 0.88 = 0.545 min.

Common mistakes

• Dividing by the rating instead of multiplying (a fast worker, R > 100 %, gives NT > OT).
• Mixing the two allowance definitions; check whether the percentage is of normal time or of the shift.
• Adding allowances before rating; rating applies to observed time only, allowances to normal time.
• Rating machine-controlled elements: an automatic machine time is not rated (R = 100 %).
• Including foreign elements or freak readings in the average.
• Rounding N′ down; always round the number of cycles up.

For GATE PI

Very common: normal and standard time from observed time, rating and allowances (both allowance definitions), output per shift, and the number of observations for a given confidence and accuracy. Also one-mark questions on rating systems, types of elements and allowances. Practise reading which basis the allowance uses and carrying three significant figures.

Quick check

1. Observed time 2.0 min, rating 90 %, allowance 10 % of normal time. Standard time?
2. A worker rated 125 % takes 0.8 min. What is the normal time?
3. Normal time 1.0 min, allowance 20 % of the working day. Standard time?
4. Which rating system uses skill, effort, conditions and consistency?

Answers: 1. 2.0 × 0.9 × 1.1 = 1.98 min; 2. 1.0 min; 3. 1.0 / 0.8 = 1.25 min; 4. Westinghouse (levelling) system.