Loom breakdowns, repair cost and a Poisson shortcut
A textile unit in Coimbatore operates 10 identical power looms. Historical logs show that on any working day each loom independently has probability 0.20 of needing a mechanical repair, and loom failures are independent. A day with no more than one repair is considered normal. Each repair costs ₹1,500, and the unit also incurs a fixed daily maintenance cost of ₹2,000 regardless of the number of repairs. The production manager has a daily contingency budget of ₹6,500 for repair and maintenance and wants to know how often the budget will be exceeded. The accountant, who has studied only the Poisson approximation, proposes modelling the number of repairs as Poisson with mean 2, arguing that 10 is large enough. The manager is unsure because the daily repair pattern is not a rare event and the two models give different tail probabilities. Before renewing the maintenance contract, the manager asks for a binomial analysis and a check of the accountant's approximation.
1. Define the random variable, state the binomial assumptions in context, and compute the probability that a working day is normal, that is, has at most one repair.
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Let X be the number of looms needing repair on a working day. The assumptions are: a fixed number n = 10 of looms; each loom has two outcomes, repair or no repair; the repair probability p = 0.20 is the same for every loom; and the looms fail independently. Then X ~ Binomial(10, 0.20).
P(X = 0) = (0.8)^10 = 0.107374. P(X = 1) = C(10,1)(0.2)(0.8)^9 = 10 × 0.2 × 0.134218 = 0.268435.
P(X ≤ 1) = 0.107374 + 0.268435 = 0.375809.
So about 37.6% of working days are normal in this sense. The independence assumption is the one to check against logs: if one repair overloads the maintenance crew and raises the chance of a second breakdown, the binomial model would understate the tail.
2. Find the expected daily repair-and-maintenance cost and the probability that the daily cost exceeds ₹6,500.
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Let C be the daily cost. Then C = 2000 + 1500X. Since E(X) = np = 10 × 0.20 = 2,
E(C) = 2000 + 1500(2) = ₹5,000.
The budget is exceeded when 2000 + 1500X > 6500, i.e. 1500X > 4500, so X > 3, which means X ≥ 4.
From the binomial distribution: P(X = 2) = C(10,2)(0.2)^2(0.8)^8 = 45 × 0.04 × 0.167772 = 0.301990. P(X = 3) = C(10,3)(0.2)^3(0.8)^7 = 120 × 0.008 × 0.209715 = 0.201327.
P(X ≤ 3) = 0.107374 + 0.268435 + 0.301990 + 0.201327 = 0.879126. P(X ≥ 4) = 1 − 0.879126 = 0.120874.
The budget is exceeded on about 12.1% of working days, roughly 30 days in a 250-day year. The expected cost of ₹5,000 is below budget, but the tail risk is not negligible.
3. Compute the Poisson approximation with mean 2, compare it with the binomial tail, and advise whether the accountant's model is safe for the maintenance contract.
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With λ = np = 2, the Poisson probabilities are P(X = r) = e⁻²2^r/r!. Thus P(0) = 0.135335, P(1) = 0.270671, P(2) = 0.270671 and P(3) = 0.180447. Their sum is 0.857124, so
P(X ≥ 4) ≈ 1 − 0.857124 = 0.142876.
The exact binomial tail is 0.120874, so the Poisson approximation overstates the exceedance probability by about 0.0220, or 2.2 percentage points. In relative terms this is about 18%, not a small rounding error. The reason is that p = 0.20 is not small and n = 10 is not large; the Poisson model forces variance equal to mean (2), while the binomial variance is np(1 − p) = 1.6, so the Poisson model has a heavier right tail. The accountant's model should not be used for the contract decision. Use the binomial calculation, and check the independence assumption against the repair logs.