Demand arriving while stock is on the shelf is met from the shelf and costs nothing to have been uncertain about.
Only demand arriving during a replenishment lead time can find the shelf empty.
So every measure and every cost in this chapter is a functional of the distribution of \(D(L)\) and of nothing else.
Two systems with different demand patterns and different lead times perform identically under the same policy whenever their lead time demand distributions coincide.
The modelling effort belongs on \(D(L)\), not on the formula built over it.
Get the distribution right and the rest is arithmetic.
Get it wrong and no amount of exactness downstream will save you.
Demand random, lead time constant. Lead time random, demand constant. Both random.
\[ E[D(L)] = E[L]\,E[D] \]
\[ \mathit{Var}[D(L)] = E[L]\mathit{Var}[D] + \mathit{Var}[L]\left(E[D]\right)^{2} \]
The three cases are one formula with terms switched off. A constant lead time kills the second term; constant demand kills the first.
\(\lambda = 45\) a year, one unit at a time, lead time two months exactly.
\[ \theta = E[D(L)] = 45\left(\frac{2}{12}\right) = 7.5 \text{ units} \]
Unit demand plus a constant lead time plus Poisson arrivals gives a Poisson lead time demand, with variance 7.5 as well.
Let the lead time be random and that stops being true. The variance rises and the family changes.
Two moments do not determine a distribution.
The variance-to-mean ratio chooses among the candidates:
\[ \mathit{VMR} = \frac{\mathit{Var}[D(L)]}{E[D(L)]} \]
\(\mathit{VMR} = 1\): Poisson. \(\mathit{VMR} > 1\): negative binomial. \(\mathit{VMR} < 1\): rarer, and worth checking your data before believing it.
Lead time demand is a non-negative count.
A normal puts mass below zero, and the slower and dearer the item, the more of its mass sits there.
The approximation is worst exactly where it matters: slow-moving, expensive items, which is where the money is.
This book models lead time demand with a normal nowhere.
Set \(Q = 1\).
Every demand of one unit triggers an order for one unit, so the position never moves.
\(S\) is the base-stock level, and \(r = S - 1\): the position reaches \(r\) exactly when one unit has been demanded.
This book states base-stock results in \(S\) and \((r,Q)\) results in \(r\). That relation is where the two meet.
The key insight: with one-for-one replenishment, an order is placed at the instant of every demand.
So the orders outstanding now are exactly those triggered in the last \(L\) time units:
\[ \mathit{IO} = D(L) \]
Substituting into \(\mathit{IP} = I + \mathit{IO} - B\) with \(\mathit{IP} = S\):
\[ \mathit{IN} = S - D(L) \]
That is the whole model.
\[ \bar{B}(S) = E\left[(D(L)-S)^{+}\right] = G^{1}(S) \]
\[ \bar{I}(S) = S - \theta + G^{1}(S) \]
\[ \overline{\mathit{RR}}(S) = P\{D(L) \le S-1\} = G(S-1) \]
The on-hand formula is an identity, not an approximation. Expected on-hand exceeds \(S - \theta\) by exactly the expected backorder level.
Stock you are short of is stock you are not holding.
Safety stock is defined as \(\mathit{ss} = r - \theta\).
It is often glossed as “expected stock on hand when a replenishment arrives”.
That reading is correct only when at most one order is outstanding.
Under one-for-one the expected number outstanding is \(\theta\) itself, which here is 7.5. There are almost always several.
Treat it as a definition, not an interpretation.
Under \(Q = 1\) the order rate is \(\lambda\) whatever \(S\) is.
So \(k\lambda\) is a constant added to every alternative, and a constant cannot change which alternative is least.
\(k\) drops out of the choice of \(S\), exactly as it dropped out of the newsvendor.
It does not drop out of the comparison between \(Q = 1\) and \(Q > 1\). That bill arrives later.
\[ C(S) = h(S - \theta) + (h + b)\,G^{1}(S) \]
That is the newsvendor’s expected cost, with the holding rate playing the overage cost and the backorder rate playing the underage cost.
\[ S^{*} = \min\left\{S : G(S) \ge \frac{b}{b+h}\right\} \]
The critical ratio, and the discrete rule, unchanged.
The newsvendor solved a problem with one period and no future.
This one has an infinite horizon and a permanent stock.
They reach the same rule because both trade one unit too many against one unit too few, and in a one-for-one system that trade is made afresh at every demand.
\(h = \$950\), \(b = \$8{,}550\), so the ratio is
\[ \frac{8550}{9500} = 0.90 \]
Reading down the Poisson distribution function at \(\theta = 7.5\), the first value at or above 0.90 gives
\[ S^{*} = 11 \text{ units} \]
Now let \(Q > 1\). The position is no longer pinned.
In steady state the inventory position is uniform over the \(Q\) integers \(r+1, \ldots, r+Q\), and independent of the lead time demand.
That one fact carries the whole section.
Every measure at a general \(Q\) is the base-stock measure averaged over the band:
\[ \bar{B}(r,Q) = \frac{1}{Q}\sum_{s=r+1}^{r+Q}\bar{B}(s) \]
And every such average telescopes into a difference of two loss functions:
\[ \bar{B}(r,Q) = \frac{G^{2}(r) - G^{2}(r+Q)}{Q} \]
One order higher each time. Backorders need \(G^{2}\); their variance needs \(G^{3}\).
\[ G^{1}(s) = G^{1}(s-1) - G^{0}(s-1), \qquad G^{2}(s) = G^{2}(s-1) - G^{1}(s) \]
started from \(G^{1}(0) = \theta\) and \(G^{2}(0) = \theta^{2}/2\).
Watch the index on the second one. \(G^{2}\) steps down by \(G^{1}(s)\), not by \(G^{1}(s-1)\), because \(G^{2}\) is the tail sum of \(G^{1}\) from \(s+1\) upward.
Getting it wrong is the most common error in a hand calculation.
Nothing above is an approximation.
No assumption that at most one order is outstanding. No continuous family fitted to a count. No shortage priced per cycle.
Given \(D(L)\), the four measures at any \((r,Q)\) are exact, and they cost a short table of loss functions to compute.
Which raises an obvious question about the approximation everybody uses.
Next: the classical approximation, what it assumes, and the two errors that partly cancel.