Monte Carlo simulation

Why it matters

Many shop-floor systems are too messy for a neat formula: arrivals that are not Poisson, machines that break down, demand and lead time that are both random, several interacting queues. Monte Carlo simulation imitates such a system on paper or a computer with random numbers, so you can test a policy — an extra fitter, a higher reorder point, a second loading bay — before spending money on it. It is the general tool when analytical OR models run out.

Key ideas

What simulation is. A model of the system's logic is run step by step, with each uncertain quantity (demand, inter-arrival time, service time, time to failure) drawn at random from its probability distribution. Averaging the results of many runs estimates the performance measures. Simulation does not optimise by itself; it evaluates the alternatives you feed it.

Monte Carlo procedure.

1. Define the problem, the performance measure and the policies to compare.
2. Get the probability distribution of each random variable (from past records or a fitted distribution).
3. Form the cumulative probability distribution.
4. Assign random-number ranges in proportion to the probabilities: with two-digit numbers 00–99, a probability of 0.20 gets 20 numbers. The ranges follow the cumulative probabilities.
5. Draw random numbers (from a table or a generator) and read off the simulated values.
6. Run the system logic for the required number of trials and record the results.
7. Compute averages and compare alternatives; increase the number of trials until the estimates settle.

Accuracy. A simulation result is a statistical estimate. Its standard error falls as 1/√n, so four times as many trials only halves the error. Short runs (like the ten-trial examples done by hand) show the method; real decisions need hundreds or thousands of trials, a warm-up period for queues that start empty, and several independent replications.

Continuous distributions. For a continuous variable, the inverse-transform method sets x = F⁻¹(r) with r uniform on (0, 1). For an exponential distribution with mean 1/λ: x = −(1/λ)·ln(1 − r), equivalently −(1/λ)·ln r.

Advantages and limits. Simulation handles any distribution, interactions and time-varying behaviour; it is easy to explain to managers. But it gives estimates, not exact optima, depends on correct input distributions ("garbage in, garbage out"), and can need many runs. Use an analytical model (e.g. M/M/1) when its assumptions hold — it is exact and instant — and also to validate a simulation.

Formulas

• Cumulative probability: F(xₖ) = p₁ + p₂ + … + pₖ
• Random-number range for value xₖ (two digits): from 100·F(xₖ₋₁) to 100·F(xₖ) − 1
• Estimated mean: x̄ = (1/n) · Σ xᵢ
• Expected value for comparison: E(X) = Σ xₖ · pₖ
• Standard error of the estimate: s / √n
• Exponential variate: x = −(1/λ) · ln(1 − r)
• Monte Carlo estimate of π: π ≈ 4 · (points inside quarter or inscribed circle) / (total points)

Symbols: xₖ = possible value of the random variable (units/day, min); pₖ = its probability; F = cumulative probability; r = uniform random number in (0, 1); n = number of trials; s = sample standard deviation of the simulated results; λ = rate of the exponential distribution (per min).

Worked examples

Example 1 (demand simulation). Daily demand for a spare part is 0, 1, 2, 3 or 4 units with probabilities 0.10, 0.20, 0.40, 0.20, 0.10. Simulate 10 days with random numbers 48, 78, 19, 51, 56, 77, 15, 14, 68, 09.

1. Cumulative probabilities: 0.10, 0.30, 0.70, 0.90, 1.00.
2. Random-number ranges: 0 → 00–09; 1 → 10–29; 2 → 30–69; 3 → 70–89; 4 → 90–99.
3. Simulated demands: 48 → 2, 78 → 3, 19 → 1, 51 → 2, 56 → 2, 77 → 3, 15 → 1, 14 → 1, 68 → 2, 09 → 0.
4. Total = 17 units in 10 days, so the simulated mean = 1.7 units/day.
5. Theoretical mean = 0(0.10) + 1(0.20) + 2(0.40) + 3(0.20) + 4(0.10) = 2.0 units/day. The gap is sampling error from only 10 trials.

Example 2 (GATE-type single-server queue). Inter-arrival times 1, 2, 3, 4 min have probabilities 0.2, 0.3, 0.3, 0.2 (ranges 00–19, 20–49, 50–79, 80–99). Service times 1, 2, 3 min have probabilities 0.3, 0.5, 0.2 (ranges 00–29, 30–79, 80–99). Random numbers for inter-arrivals of customers 1–6: 64, 18, 37, 92, 05, 51; for services: 82, 45, 27, 71, 93, 10. The clock starts at 0 with the server free. Find the average wait in queue and server idle time.

1. Inter-arrival times: 3, 1, 2, 4, 1, 3 min → arrival clock times 3, 4, 6, 10, 11, 14.
2. Service times: 3, 2, 1, 2, 3, 1 min.
3. C1: arrives 3, starts 3, waits 0, ends 6. C2: arrives 4, starts 6, waits 2, ends 8. C3: arrives 6, starts 8, waits 2, ends 9.
4. C4: arrives 10, starts 10, waits 0, ends 12 (server idle 9–10). C5: arrives 11, starts 12, waits 1, ends 15. C6: arrives 14, starts 15, waits 1, ends 16.
5. Total wait = 0 + 2 + 2 + 0 + 1 + 1 = 6 min → average wait = 1.0 min per customer.
6. Server idle = 3 (0–3) + 1 (9–10) = 4 min out of 16 min, utilisation = 12/16 = 75 %.

Common mistakes

• Making random-number ranges overlap or leave gaps (for example 00–10 and 10–29); each range must have exactly 100·p numbers.
• Using individual probabilities instead of cumulative ones to set the ranges.
• Starting a customer's service at the arrival time even though the server is still busy.
• Drawing conclusions from a handful of trials as if they were exact.
• Reusing the same random-number stream for two different variables, which correlates them.
• Comparing two policies with different random numbers when the aim is a fair comparison (use common random numbers).

For GATE PI

• Assign random-number intervals from a probability table and read simulated values (NAT).
• Short hand simulations of demand, inventory or a single-server queue: average demand, waiting time, idle time.
• Concept MCQs on the purpose, accuracy and limitations of simulation, and the π estimate.
• Practise building the arrival–start–end–wait table neatly; most errors come from the clock column.

Quick check

1. Probabilities 0.25, 0.45, 0.30. Write the two-digit random-number ranges.
2. 400 random points in a unit square give 316 inside the quarter circle. Estimate π.
3. To halve the standard error of a simulation estimate, by what factor must the trials increase?
4. Can Monte Carlo simulation find the optimal policy by itself?

Answers: 1. 00–24, 25–69, 70–99. 2. 4 × 316/400 = 3.16. 3. Four times. 4. No — it evaluates the policies you test.