Penalty Clauses and the Poisson Habit
A Bengaluru grocery-delivery platform fills about 500 orders a day. Over several months the operations team has found that 3 per cent of delivered orders come back as returns (wrong item, damaged pack). The service-level agreement with a large client imposes a penalty of ₹20,000 for every day on which more than 22 orders are returned. The platform works 250 days a year. The analytics intern, told that returns are 'rare events', proposes to model daily returns with a Poisson distribution with λ = 15 and to read P(returns > 22) straight from it. A second proposal is to spend ₹1,80,000 a year on an extra packing-station supervisor who is expected to bring the return rate down to 2 per cent. Management wants to know how often the penalty will bite, what it costs per year, and whether the supervisor pays for herself on penalty savings alone.
1. Identify the exact distribution of the daily number of returns and its parameters. Is the intern's Poisson model defensible? State the mean and standard deviation under the binomial model.
Read the worked answer
Let X = returns among the 500 orders in a day. Each order is a Bernoulli trial (return / no return) with p = 0.03, the orders are independent and n is fixed, so X ~ Binomial(n = 500, p = 0.03). This is the exact model; the Poisson is an approximation, not the truth.
Mean = np = 500 × 0.03 = 15. Variance = npq = 500 × 0.03 × 0.97 = 14.55, so SD = 3.81.
The Poisson with λ = np = 15 is defensible because n is large, p is small and failures are counted per unit of exposure. The cost is that it forces variance = mean = 15, whereas the binomial gives 14.55; the Poisson is slightly more spread out and therefore slightly conservative in the upper tail, which is acceptable when the tail drives a penalty. The intern's λ value is right; only the label 'Poisson' needs the qualifier 'approximation'.
2. Using a normal approximation with continuity correction, estimate P(more than 22 returns on a day), the expected number of penalty days a year, and the expected annual penalty.
Read the worked answer
Under the binomial, μ = 15, σ = √14.55 = 3.814.
The event is X ≥ 23. With continuity correction the boundary is 22.5: z = (22.5 − 15)/3.814 = 7.5/3.814 = 1.97 P(Z > 1.97) = 1 − 0.9756 = 0.0244.
Expected penalty days = 250 × 0.0244 ≈ 6.1 days a year. Expected penalty = 6.1 × ₹20,000 ≈ ₹1,22,000 a year - with sampling variation around it, so a budget of ₹1.4–1.5 lakh would be prudent.
Why the approximation is allowed: np = 15 and nq = 485 are both well above 5, so the binomial is symmetric enough for the normal curve; the continuity correction matters because a discrete count is being fitted by a continuous curve (without it, z = (23 − 15)/3.814 = 2.10 and the tail would be understated at 0.018).
3. At a 2 per cent return rate, recompute the penalty exposure and advise management on the supervisor proposal, considering only penalty savings.
Read the worked answer
At p = 0.02: mean = 500 × 0.02 = 10; variance = 500 × 0.02 × 0.98 = 9.8; σ = 3.13.
P(X ≥ 23): z = (22.5 − 10)/3.13 = 3.99, tail ≈ 0.00003. Expected penalty days ≈ 250 × 0.00003 ≈ 0.008 per year, i.e. essentially never; expected penalty ≈ ₹150.
Penalty saving from the supervisor ≈ ₹1,22,000 − ₹150 ≈ ₹1,22,000 a year, against a cost of ₹1,80,000 a year. On penalty savings alone the proposal does not pay for itself: it is short by about ₹58,000.
Judgement for management: the honest conclusion is 'reject on penalty grounds alone', but record the other effects that could change the decision - fewer replacements and reshipments, less staff time on complaints, customer retention, and the option of negotiating a higher trigger (say more than 25) with the client. Also flag that a threshold far out in the tail is sensitive: moving the trigger by two or three returns would alter the whole calculation, so the SLA should be read before the supervisor is rejected outright.