The Newsvendor Problem

A newspaper seller must decide, each morning, how many papers to buy for the day. Buy too few and she turns away paying customers — profit walking down the street. Buy too many and the leftovers are tomorrow's recycling — cash in the bin. She must commit before she knows the day's demand, and she gets exactly one shot: no reordering, no second chance. This little dilemma — stocking a perishable good once, under uncertain demand — is the newsvendor problem, and it governs far more than newspapers: fashion buys, bakery batches, flu-vaccine doses, concert staffing, cloud capacity for a launch.

The two ways to be wrong

Because demand D is random and the order Q is fixed in advance, exactly one of two mistakes happens each day. Each carries a cost.

Underage cost c_u — the cost of stocking one unit too few: a lost sale, so c_u is the profit forgone on that unit (plus any goodwill lost).
Overage cost c_o — the cost of stocking one unit too many: an unsold item, so c_o is what you paid for it less any salvage value.

Ordering one more unit is a gamble: it helps (saves c_u) if demand turns out to reach that unit, and hurts (costs c_o) if it does not. The optimal Q is where that gamble is exactly break-even.

The critical fractile

Consider raising the order from Q to Q+1. The extra unit sells only if demand is at least Q+1 — probability P(D > Q) — earning c_u; otherwise it is wasted, costing c_o. Keep adding units while the expected gain beats the expected loss, and stop when they balance:

c_u\,P(D > Q) = c_o\,P(D \le Q).

Using P(D > Q) = 1 - P(D \le Q) and solving gives the single most useful formula in single-period inventory — the critical fractile (or critical ratio):

Order up to the quantity Q^{*} at which \displaystyle P(D \le Q^{*}) = \frac{c_u}{c_u + c_o}.
That target probability is the optimal service level — the chance of not running out.
So Q^{*} is the corresponding quantile of the demand distribution.

The whole decision reduces to a ratio of two costs, read off as a point on the demand distribution.

Reading it off the demand curve

When demand follows a normal distribution with mean \mu and standard deviation \sigma, the optimal order is Q^{*} = \mu + z\,\sigma, where z is the z-value whose cumulative probability equals the critical fractile. Slide the cost ratio and watch the optimal quantile move along the demand distribution:

When the critical fractile is 0.5 (underage and overage equally painful), Q^{*} sits right at the mean. Make lost sales hurt more than leftovers and the fractile climbs above 0.5, pushing Q^{*} to the right — you deliberately over-stock. Make leftovers the bigger worry and the fractile drops, pulling Q^{*} below the mean.

Worked example

A bakery sells croissants that cost \pounds 1 to make and sell for \pounds 3; unsold ones are worthless at close. So the underage cost (a lost sale) is the forgone profit c_u = 3 - 1 = \pounds 2, and the overage cost (an unsold croissant) is the wasted production c_o = \pounds 1. The critical fractile is

\frac{c_u}{c_u + c_o} = \frac{2}{2 + 1} = 0.667.

Daily demand is roughly normal with mean \mu = 100 and standard deviation \sigma = 20. The z-value for a cumulative probability of 0.667 is about z = 0.43, so

Q^{*} = \mu + z\sigma = 100 + 0.43 \times 20 \approx 109 \text{ croissants.}

The bakery stocks above its average demand — not because it expects 109, but because a missed sale (£2) stings twice as much as a wasted croissant (£1). The cost ratio, not the average, sets the target.

Then c_u = c_o, the critical fractile is c_u/(c_u+c_o) = 0.5, and the optimal order is exactly the median demand — for a symmetric distribution, the mean. Only in this perfectly balanced case does "order the average" happen to be right. The moment the two costs differ — and they almost always do — the optimal order moves off the average toward whichever mistake is cheaper. High-margin goods with cheap production (newspapers, fast fashion) have a high fractile and are stocked aggressively; low-margin perishables with costly waste (sushi, blood products) have a low fractile and are stocked lean.

The most common newsvendor blunder is to order the mean and feel prudent. That is optimal only in the knife-edge case c_u = c_o. Whenever the costs are unequal — the usual situation — ordering the average leaves money on the table. If under-stocking is far more expensive than over-stocking (a critical vaccine, a flagship product launch), the right service level might be 95% or 99%, far above the mean. If leftovers are ruinous (a wedding cake, a bespoke print run), it might be well below. Let the cost ratio decide the service level — never the average alone.