Overview

429. Overview

429.1. Newsvendor

Single-period inventory under uncertain demand: order $𝑄$ before seeing demand. After demand realizes, leftover stock is salvaged at $𝑆 < 𝐶$ , and unmet demand is lost.

The trade-off:

Order too much → leftovers, lose $𝐶 - 𝑆$ per unsold unit.
Order too little → stockouts, lose $𝑃 - 𝐶$ per missed sale.

429.1.1. Assumptions

Single period (no carryover to next period)
Single product
Demand $𝐷$ is random with known distribution $𝐹$
Lead time is zero — receive order before demand realizes
Linear underage and overage costs

429.1.2. Setup

$𝑃$ = sale price per unit
$𝐶$ = purchase cost per unit
$𝑆$ = salvage value per unsold unit (with $𝑆 < 𝐶$ )
$𝐷$ = random demand, with CDF $𝐹 (𝑑) = 𝑃 (𝐷 \leq 𝑑)$ and density $𝑓$
$𝑄$ = order quantity (decision variable)

Define:

Underage cost (lost margin per missed sale): $𝐶_{𝑢} = 𝑃 - 𝐶$
Overage cost (loss per unsold unit): $𝐶_{𝑜} = 𝐶 - 𝑆$

429.1.3. Derive $𝑄^{*}$ via marginal analysis

Imagine we’ve decided to order $𝑄$ units. Should we order one more?

The $(𝑄 + 1)$ -th unit either sells or doesn’t:

It sells if demand exceeds $𝑄$ , with probability $𝑃 (𝐷 > 𝑄) = 1 - 𝐹 (𝑄)$ . Profit: $𝐶_{𝑢}$ per unit (margin we’d otherwise lose).
It doesn’t sell if demand $\leq 𝑄$ , with probability $𝐹 (𝑄)$ . Cost: $𝐶_{𝑜}$ per unit (overage).

Expected marginal profit of ordering one more unit:

Δ (𝑄) = (1 - 𝐹 (𝑄)) 𝐶_{𝑢} - 𝐹 (𝑄) 𝐶_{𝑜}

Order more units while $Δ (𝑄) > 0$ , stop when $Δ (𝑄) = 0$ :

(1 - 𝐹 (𝑄^{*})) 𝐶_{𝑢} = 𝐹 (𝑄^{*}) 𝐶_{𝑜}

𝐶_{𝑢} - 𝐹 (𝑄^{*}) 𝐶_{𝑢} = 𝐹 (𝑄^{*}) 𝐶_{𝑜}

𝐶_{𝑢} = 𝐹 (𝑄^{*}) (𝐶_{𝑢} + 𝐶_{𝑜})

𝐹 (𝑄^{*}) = \frac{𝐶_{𝑢}}{𝐶_{𝑢} + 𝐶_{𝑜}}

This ratio is the critical ratio (CR). Read it as: “What probability of not stocking out balances the two costs?”

429.1.4. Inverting $𝐹$ to get $𝑄^{*}$

𝑄^{*} = 𝐹^{- 1} (CR) = 𝐹^{- 1} (\frac{𝐶_{𝑢}}{𝐶_{𝑢} + 𝐶_{𝑜}})

For any demand distribution, the recipe is the same: compute CR, look up the quantile.

429.1.5. Normal demand specialization

If $𝐷 \sim 𝒩︀ (𝜇, 𝜎^{2})$ , write $𝑄^{*} = 𝜇 + 𝑧^{*} 𝜎$ where $𝑧^{*} = Φ^{- 1} (CR)$ is the standard-normal quantile of CR.

Common $𝑧^{*}$ values:

$CR = 0.5$ → $𝑧^{*} = 0.00$ (order the median)
$CR = 0.67$ → $𝑧^{*} \approx 0.44$
$CR = 0.84$ → $𝑧^{*} \approx 1.00$
$CR = 0.95$ → $𝑧^{*} \approx 1.65$
$CR = 0.99$ → $𝑧^{*} \approx 2.33$

When $𝐶_{𝑢} ≫ 𝐶_{𝑜}$ (stockouts much worse than leftovers), CR $\to 1$ and $𝑧^{*}$ pushes you to over-stock. Conversely when $𝐶_{𝑜} ≫ 𝐶_{𝑢}$ , CR $\to 0$ and $𝑧^{*} < 0$ (under-stock the median).

Example

Given (newsstand selling newspapers):

Sale price: $𝑃$ = $3 / unit
Purchase cost: $𝐶$ = $1 / unit
Salvage value: $𝑆$ = $0 / unit (unsold papers worthless)
Demand: $𝐷 \sim 𝒩︀ (𝜇 = 100, 𝜎 = 20)$

Step 1 — underage and overage

𝐶_{𝑢} = 𝑃 - 𝐶 = 3 - 1 = 2 (\$/unsold-demand-unit)

𝐶_{𝑜} = 𝐶 - 𝑆 = 1 - 0 = 1 (\$/leftover)

Step 2 — critical ratio

CR = \frac{𝐶_{𝑢}}{𝐶_{𝑢} + 𝐶_{𝑜}} = \frac{2}{2 + 1} = \frac{2}{3} \approx 0.667

Each unit ordered should have a $\approx 67 %$ chance of selling — overstocking is twice as cheap as stocking out.

Step 3 — quantile lookup

For normal demand, $𝑧^{*} = Φ^{- 1} (0.667) \approx 0.44$ :

𝑄^{*} = 𝜇 + 𝑧^{*} 𝜎 = 100 + 0.44 \cdot 20 \approx 109 newspapers

Step 4 — interpret

At $𝑄^{*} \approx 109$ : about a $67 %$ chance of selling out (matching CR), $33 %$ chance of leftovers. Sensible because the underage cost ( $𝐶_{𝑢} = 2$ ) is twice the overage cost ( $𝐶_{𝑜} = 1$ ) — the optimum biases toward stocking more than the mean (100).

newsvendor.py

import scipy.stats as stats

P, C, S = 3, 1, 0  # sale price, purchase cost, salvage value
mu, sigma = 100, 20  # demand mean and standard deviation

C_u = P - C  # underage cost
C_o = C - S  # overage cost
CR = C_u / (C_u + C_o)  # critical ratio

z_star = stats.norm.ppf(CR)  # quantile
Q_star = mu + z_star * sigma

429.1.6. Derivation via calculus (integral form)

The marginal argument above has a continuous twin: write expected cost as a function of $𝑄$ and minimize directly. With overage $𝐶_{𝑜}$ and underage $𝐶_{𝑢}$ :

𝒞︀ (𝑄) = 𝐶_{𝑜} \cdot 𝔼 [{(𝑄 - 𝐷)}^{+}] + 𝐶_{𝑢} \cdot 𝔼 [{(𝐷 - 𝑄)}^{+}]

Rewrite the expectations as integrals. Each ${(\cdot)}^{+}$ clips the integrand to one side of $𝑄$ :

𝔼 [{(𝑄 - 𝐷)}^{+}] = \int_{0}^{𝑄} (𝑄 - 𝑥) 𝑓 (𝑥) d 𝑥, 𝔼 [{(𝐷 - 𝑄)}^{+}] = \int_{𝑄}^{\infty} (𝑥 - 𝑄) 𝑓 (𝑥) d 𝑥

𝒞︀ (𝑄) = 𝐶_{𝑜} \int_{0}^{𝑄} (𝑄 - 𝑥) 𝑓 (𝑥) d 𝑥 + 𝐶_{𝑢} \int_{𝑄}^{\infty} (𝑥 - 𝑄) 𝑓 (𝑥) d 𝑥

Differentiate (Leibniz’s rule — the boundary terms vanish because the integrand is $0$ at $𝑥 = 𝑄$ ):

\frac{d 𝒞︀}{d 𝑄} = 𝐶_{𝑜} 𝐹 (𝑄) - 𝐶_{𝑢} (1 - 𝐹 (𝑄))

Set to zero and solve:

𝐶_{𝑜} 𝐹 (𝑄^{*}) = 𝐶_{𝑢} (1 - 𝐹 (𝑄^{*})) ⟹ 𝐹 (𝑄^{*}) = \frac{𝐶_{𝑢}}{𝐶_{𝑢} + 𝐶_{𝑜}}

— the same critical ratio. Convexity check: $(d^{2} 𝒞︀) / (d 𝑄^{2}) = (𝐶_{𝑜} + 𝐶_{𝑢}) 𝑓 (𝑄) \geq 0$ , so the stationary point is the global minimum.