13.3 Random demand: the newsvendor problem
A newspaper seller must decide how many copies to buy before the day’s demand \(X\), with distribution function \(F\), is known. Each copy costs \(c\), sells for \(p\), and unsold copies are salvaged for \(s < c\). Ordering one copy too few loses the margin \(c_u = p - c\) (the underage cost); ordering one too many loses \(c_o = c - s\) (the overage cost).
Theorem 13.3.1 (Critical fractile). The expected profit is maximised by the order \(Q^*\) with \[ F(Q^*) = \frac {c_u}{c_u + c_o} , \] the critical fractile, or by the smallest \(Q\) with \(F(Q) \geq c_u/(c_u + c_o)\) if demand is discrete.
Proof. Increase the order from \(Q\) to \(Q + 1\) (or by \(dQ\)). The extra unit is sold, gaining \(c_u\), if demand exceeds \(Q\), which has probability \(1 - F(Q)\), and is left over, losing \(c_o\), otherwise. The expected marginal gain \(c_u(1 - F(Q)) - c_oF(Q)\) is decreasing in \(Q\) and changes sign where \(F(Q) = c_u/(c_u + c_o)\), so the expected profit, which is concave, is maximised there. □
When the margin is large compared with the loss on leftovers, the ratio is near \(1\) and the seller orders well above the mean demand; when leftovers are costly, below it. For normal demand \(N(\mu , \sigma ^2)\), \(Q^* = \mu + z\sigma \) with \(\Phi (z) = c_u/(c_u + c_o)\). With \(\mu = 200\), \(\sigma = 40\), \(p = 10\), \(c = 6\), \(s = 1\): \(c_u = 4\), \(c_o = 5\), the critical fractile is \(\frac 49 = 0.444\), \(z = -0.14\) and \(Q^* = 194\), slightly below the mean (Software 13.5.2). The same logic sizes airline overbooking, fashion orders, blood stocks and hotel capacity.
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