Classifying a System, and Describing Its State

All three are managing inventory. Almost nothing carries from one to the next.

What they share is the pair of questions: how much, and when. They do not share the machinery that answers them.

Choosing to treat it as constant is an approximation you make deliberately, having judged that the decision at hand is insensitive to the difference.

It is not a discovery that the lead time is constant.

Dependent demand is implied. Demand for a component follows from the production schedule for the assembly that consumes it, and that schedule is already known.

Such demand should be calculated, not forecast. Forecasting it independently throws away information the organization already has.

This question comes first, because answering it “dependent” sends you to requirements planning regardless of how everything else falls.

Deterministic demand is known in advance. Known does not mean constant: demand may vary period to period and still be deterministic, provided the variation is known before the decisions are made.

Stochastic demand must be described by a distribution. The question changes character, from “what will demand be” to “how much protection against what it might be,” and every result acquires a service dimension it did not have.

1. When variability is small relative to the decision, so ignoring it changes nothing that matters.

2. When the model sits inside a larger deterministic plan, as lot sizing does inside a material requirements plan.

Locations. A multi-echelon system is not several single-location problems. The demand a warehouse sees is not customer demand; it is the replenishment orders of the locations it serves, which are lumpier than the demand underneath them and are shaped by those locations’ own policies.

Simplicity repays real effort: a simpler classification yields a model whose behavior can be understood and whose parameters can be estimated.

But simplicity that discards what the answer depends on is not simplification. It is an error with a smaller model attached.

And the criterion is testable. When it is unclear whether a distinction matters, model it both ways and compare the recommendations. If they agree, the comparison is your justification. If they disagree, the question has answered itself.

Three counts describe the system at time \(t\):

• \(I(t)\), the inventory on hand: units physically present and available
• \(\mathit{IO}(t)\), the inventory on order: ordered but not yet arrived
• \(B(t)\), the backorders: demanded but not yet supplied

All three are counts of units. None can be negative.

Inventory position adds what is already coming:

\[ \mathit{IP}(t) = I(t) + \mathit{IO}(t) - B(t) \]

That second one is what every policy in this course actually reads.

A system holding stock while owing units would be one that had received inventory and declined to ship it to a waiting customer.

Under the ordinary assumption that arriving replenishments fill outstanding backorders first, that never happens. So

\[ I(t) = \left[\mathit{IN}(t)\right]^{+}, \qquad B(t) = \left[\mathit{IN}(t)\right]^{-} \]

This is also what makes the cost model coherent: \(h\) charges \(I(t)\), \(b\) charges \(B(t)\), and no unit is ever charged both ways.

Stock runs down, the level is reached, an order is placed.

The lead time has not elapsed, so on-hand inventory keeps falling, and the trigger level is still breached.

So the rule places another order. And another. Each for a shortfall the first order is already on its way to cover.

The system then orders again only when what is already outstanding is no longer enough.

The position is what the system will eventually have.

On-hand inventory is only what it has now.

Early in each cycle they lie on top of each other. No order is outstanding, so \(\mathit{IO}(t) = 0\) and position equals net inventory.

The dashed line reaches \(r\), an order is placed, and the position jumps by \(Q\) while net inventory does not move at all, because nothing has arrived.

The two run apart for exactly one lead time \(L\).

Net inventory continues down, crosses zero, and the shaded region below the axis is backorders accumulating while the order is in transit.

Net inventory does.

These integrals are where the two halves of the course meet.

In a deterministic model \(I(t)\) is a known piecewise-linear function and the integral is the area of a triangle.

In a stochastic model the same integral is an expectation.

The quantity being computed does not change. Only the method does.

Choosing what counts as “inside” is what generates the results.

We apply it three times, to three different flows of units.

Notice that no ordering policy appears. So no ordering policy can reduce it.

Average backorders and average customer waiting are two views of one quantity.

This matters practically. An organization that cannot price a backorder can usually state a tolerable wait, and that statement is enough to pin down what the backorder cost would have multiplied.

Its reciprocal is a quantity every operations manager already knows:

\[ \mathit{TO} = \frac{\lambda}{\bar{I}} = \frac{1}{\overline{T}_{s}} \]

Inventory turnover is nothing more than the reciprocal of the average time a unit sits on a shelf.

These are the same statement.

Turnover is reported by almost every organization and understood by people who would not sit through a derivation.

It also supplies a sanity check available before any modeling begins: an item whose computed \(\bar{I}\) implies a turnover far from what the business reports has been mis-specified somewhere.

Three counts and two composites describe any inventory system, and on hand and backordered are never both positive.

Policies read the position, not the shelf, because the position counts what is already coming.

Little’s law three times gives pipeline stock, the backorder-wait identity, and turnover.