Lead Time

412. Lead Time

412.1. Lead Time

Order today, receive after the lead time $𝐿$ . What drives safety stock is the distribution of demand over those $𝐿$ periods. Central question: why does its standard deviation scale with $\sqrt{𝐿}$ , not $𝐿$ ?

412.1.1. Setup

Daily demand $𝐷_{𝑡}$ is i.i.d. with mean $𝜇$ and standard deviation $𝜎$ . Take $𝜇 = 100$ , $𝜎 = 20$ .

412.1.2. Two days

Demand over two days is $𝐷_{1} + 𝐷_{2}$ (independent draws).

Means add:

𝔼 [𝐷_{1} + 𝐷_{2}] = 𝜇 + 𝜇 = 2 𝜇 = 200

Standard deviations do not add — variances do:

Var (𝐷_{1} + 𝐷_{2}) = Var (𝐷_{1}) + Var (𝐷_{2}) = 𝜎^{2} + 𝜎^{2} = 2 𝜎^{2}

So the spread grows by only $\sqrt{2}$ :

Std (𝐷_{1} + 𝐷_{2}) = \sqrt{2 𝜎^{2}} = 𝜎 \sqrt{2}

412.1.3. Generalize to $𝐿$ days

Var (\sum_{𝑡 = 1}^{𝐿} 𝐷_{𝑡}) = 𝐿 𝜎^{2} \Rightarrow Std (\sum_{𝑡 = 1}^{𝐿} 𝐷_{𝑡}) = 𝜎 \sqrt{𝐿}

Mean of lead-time demand scales with $𝐿$ : $𝜇_{𝐿} = 𝐿 𝜇$
Spread scales with $\sqrt{𝐿}$ : $𝜎_{𝐿} = 𝜎 \sqrt{𝐿}$

This $\sqrt{𝐿}$ is why safety stock $SS = 𝑧 \cdot 𝜎 \sqrt{𝐿}$ grows sub-linearly in lead time — halving $𝐿$ cuts safety stock by only $\sqrt{2}$ , not $2$ .

412. Lead Time

412.1. Lead Time

412.1.1. Setup

412.1.2. Two days

412.1.3. Generalize to 𝐿 days

412.1.3. Generalize to $𝐿$ days