Work sampling

Why it matters

Many jobs cannot sensibly be stopwatched: maintenance crews, store-keepers, office staff, a bank of machines or a whole department. Work sampling answers "what fraction of the time are they working, idle or waiting?" from a few hundred random snapshots, at a fraction of the cost of continuous study, and it can even yield a standard time for long-cycle or non-repetitive work.

Key ideas

Work sampling (activity sampling, ratio-delay study) is a statistical technique in which a large number of instantaneous observations are made at random times of a group of machines, processes or workers. Each observation records what is happening at that instant (working, idle, waiting for material, etc.). The proportion of observations in which an activity occurs estimates the proportion of total time spent on it. L. H. C. Tippett introduced it in the British textile industry in the 1930s.

Why it works. Each observation is a Bernoulli trial: the activity is either occurring or not. With N random observations, the count follows a binomial distribution, which for large N is close to normal with mean p and standard deviation √(p(1 − p)/N). So the accuracy improves with the square root of N.

Procedure

1. Define the objective and the activity categories (clear and mutually exclusive).
2. Take a pilot study (say 50–100 observations) to get a first estimate of p.
3. Choose the confidence level and accuracy; compute N.
4. Schedule observations at random times (random-number tables or a random timer) spread over enough days to cover normal variation; avoid any fixed pattern workers could anticipate.
5. Observe and record; on a control chart, check that daily proportions stay within p ± 3√(p(1 − p)/n_day).
6. Compute p and the achieved accuracy; recompute N with the updated p if needed.

Absolute vs relative accuracy. Absolute accuracy E is in proportion units: p = 0.25 ± 0.03 means 22–28 %. Relative accuracy s is a fraction of p: ±5 % of 0.25 is ±0.0125. Questions use both; read carefully.

Advantages: no stopwatch, many workers or machines observed by one person, less observer fatigue and less disturbance to workers, results unaffected by short interruptions, cheap for long or irregular work. Limitations: no element-level detail, not suited to short repetitive cycles (use time study), needs many observations, results only valid for the period studied, and workers may change behaviour when seen.

Uses: machine and labour utilisation, setting allowances (delay proportions), estimating standard time for long-cycle or indirect work, and justifying staffing or equipment decisions.

Formulas

p = n / N

• n = observations in which the activity was seen; N = total observations.

Absolute accuracy: N = z² · p(1 − p) / E²; equivalently E = z · √(p(1 − p)/N)

• z = standard normal value (1.96 for 95 %, often rounded to 2; 1.645 for 90 %; 3 for 99.7 %); E = absolute error as a proportion.

Relative accuracy: N = z² · (1 − p) / (s² · p)

• s = accuracy as a fraction of p.

Normal time NT = (T × p_w × R) / Q

• T = total study time (min); p_w = working proportion; R = average performance rating (fraction); Q = units produced in T.

Standard time ST = NT × (1 + A); A = allowance fraction of normal time.

Worked examples

Example 1 (standard). A pilot study of 100 random observations of a crane found it idle 25 times. How many observations are needed at 95 % confidence (z = 1.96) for (a) relative accuracy ±5 % and (b) absolute accuracy ±3 %?

1. p = 25/100 = 0.25.
2. (a) N = 1.96² × (1 − 0.25) / (0.05² × 0.25) = 3.8416 × 0.75 / 0.000625 = 4,610 observations.
3. (b) N = 1.96² × 0.25 × 0.75 / 0.03² = 0.7203 / 0.0009 = 800.3, round up to 801 observations.

Relative accuracy on a small proportion is far more demanding than absolute accuracy.

Example 2 (GATE level). A worker was studied over 5 days of 480 min (T = 2,400 min). Of 400 random observations, 340 showed the worker working; the average rating was 110 %. 1,200 units were produced. Allowances are 15 % of normal time. Find the normal and standard time per unit, and check the absolute accuracy of p at 95 % confidence.

1. p_w = 340/400 = 0.85.
2. NT = 2,400 × 0.85 × 1.10 / 1,200 = 2,244 / 1,200 = 1.87 min per unit.
3. ST = 1.87 × 1.15 = 2.15 min per unit.
4. E = 1.96 × √(0.85 × 0.15 / 400) = 1.96 × 0.01785 = 0.035, i.e. p_w = 85 ± 3.5 %.

Common mistakes

• Confusing absolute and relative accuracy (the two formulas differ by a factor of p²).
• Using p as a percentage in one place and a fraction in another; keep p, E and s all as fractions or all as percentages.
• Taking observations at fixed intervals, which can lock onto a work cycle and bias p.
• Rounding N down instead of up.
• Forgetting the rating in the standard-time calculation, or dividing by total time instead of units produced.

For GATE PI

Expect NAT questions on the number of observations for a given confidence and accuracy, the accuracy achieved from given counts, utilisation from sample data, and standard time from a work-sampling study. One-mark questions compare work sampling with time study. Practise both accuracy formulas with z = 1.96 and z = 2.

Quick check

1. 600 observations, idle in 90. What is the idle proportion?
2. p = 0.2, z = 2, absolute accuracy ±0.04. How many observations?
3. If N is quadrupled, what happens to the absolute error?
4. Why must observation times be random?

Answers: 1. 0.15 (15 %); 2. N = 4 × 0.2 × 0.8 / 0.0016 = 400; 3. It halves; 4. To avoid bias from cycles or workers anticipating the observer, so every instant has an equal chance of being sampled.