“What does a stockout cost you?”
“I don’t know. Just give me 95 percent.”
Setting a service target does not remove \(b\) from the model.
It fixes \(b\) at whatever value would have produced the same policy.
You can recover that value, and you should.
Minimize the ordering and holding cost subject to a service constraint.
At the optimum, the Lagrange multiplier on the constraint is the shortage cost.
\[ \pi^{*} = \frac{Q h}{\lambda\, G^{0}(r)} \]
That number is not a curiosity. It is what the organization has just decided, without saying so.
A 95 percent fill rate target.
Implied shortage cost:
\[ \pi^{*} = \$919 \text{ per unit short} \]
An emergency transformer actually costs the utility about $5,700.
The usual accusation against service targets is that they are over-cautious.
Here the target is holding far too little: it implies a shortage cost less than a fifth of the real one.
About any service level, ask: what shortage cost would make it optimal?
Then ask whether that number is remotely like the real one.
Raising the target from 95 to 98 percent closes less than a third of the gap on this item.
Apply the same fill rate across a storeroom and you set a different implied shortage cost on every item, because the implied cost depends on the item’s own unit cost and demand rate.
Before raising service on everything, find out which items are carrying the average.
That is cheap to do and almost never done.
A fuse cutout moves 800 units a year, but not one at a time.
Demand epochs arrive 400 times a year, and each carries a lot: 1 unit with probability 0.4, 2 with 0.4, 4 with 0.2.
\[ E[Y] = 2.0, \qquad E[Y^{2}] = 5.2 \]
Lead time demand is now compound Poisson.
The variance-to-mean ratio rises, so the family changes.
The fill rate and the ready rate come apart. Lots still arrive as a Poisson process, so the ready rate is the share of lots that find stock; a lot of four can be partly filled, so the share of units filled is lower.
For the cutout at \((25, 75)\): ready rate 0.9060, unit fill rate 0.8978.
It does not change the formulas for \(\bar{B}\), \(\bar{I}\) or the ready rate, because those never asked how the demand arrived.
They only ever asked for the distribution of \(D(L)\). The fill rate is the one measure that also needs the lot size.
Order up to \(S\) whenever the position reaches or falls below \(s\).
With unit demand this is \((r, Q)\) with \(Q = S - s\).
With lots, it is not. A lot of four can carry the position from \(s+1\) to \(s-3\), so the order size varies.
The amount by which a demand carries the position below \(s\).
Its equilibrium distribution comes from the lot size distribution:
\[ P\{U = j\} = \frac{P\{Y > j\}}{E[Y]} \]
For the cutout, \(E[U] = 0.8\) units.
Ignore it and the model is biased in the direction that flatters the policy: it reports orders placed less often and stock held more steadily than either really is.
Scarf’s theorem. For a stock point reviewed at fixed intervals over a finite horizon, with independent random demands, a fixed ordering charge and convex holding and shortage costs, there exist \(s \le S\) in every period such that ordering up to \(S\) when the position is at or below \(s\) minimises expected total cost.
The proof turns on \(K\)-convexity, which is what survives of convexity once a fixed charge is added.
It says the form is right: nothing is gained by searching over more complicated rules.
It does not say what \(s\) and \(S\) are, and it does not make the cost convex in them. Computing the best pair is a separate problem.
And note the setting: fixed review intervals, not continuous review. Which brings us to the last topic.
Look every \(R\) time units and order up to \(S\). The \((R, S)\) policy.
Stock committed at a review must last until the order after next arrives.
So the exposure is the protection interval
\[ \tau = R + L \]
not the lead time alone.
\[ \mathit{IN} = S - D(\tau) \]
That is the dual of the \((r,Q)\) result.
One policy randomises the position over \(Q\) levels. The other randomises the exposure over \(R\) of time.
Both are the base-stock model averaged over a uniform quantity.
So the four measures, the critical ratio, and the newsvendor argument all carry over, with an averaged distribution in place of \(G\).
The calendar creates cycle stock: on average \(\lambda R/2\) units sitting because you did not look sooner.
That is the EOQ’s \(Q/2\) in different units.
And it is what makes the optimal review interval the EOQ time supply:
\[ R^{*} \approx \frac{Q_{\mathit{EOQ}}}{\lambda} \]
\(S^{*} = 14\) units.
Use the protection-interval distribution naively, without averaging over the review cycle, and you get \(S = 16\), at about 10.7 percent more cost.
The averaging is not a refinement. It is the model.
At the same fill rate, periodic review on the transformer costs $970 a year more, or 13.4 percent.
And the premium is not all safety stock. A longer exposure raises safety stock; a shorter review interval lowers cycle stock. The difference lands across all three cost terms.
So the choice between continuous and periodic review is a question about what the record-keeping discipline costs, not about the item.
Every one of those decisions was made for one item, from parameters that arrived as givens.
Next: a storeroom holds thousands of items, and not one of those parameters stands still.