Recall that everything that can go wrong happens during a lead time. Let \(D(L)\) represent the lead time demand, the demand that arrives between placing an order and receiving it, and let \(\theta = E[D(L)]\) represent its mean.
Everything in the rest of this chapter is a functional of the distribution of \(D(L)\), and nothing in it is a functional of anything else.
That is a stronger statement than it looks. Two systems with different demand patterns and lead times, whose lead time demand distributions happen to coincide, have identical performance under the same policy. Thus, the modeling effort belongs on \(D(L)\), and the choice of formula for \(\bar{B}\) is a detail by comparison.
Period based. Let \(D_{i}\) represent the demand in period \(i\), and count \(L\) in whole periods:
\[ D(L) = \sum_{i=1}^{L} D_{i} \]
The natural form when the data are a history of weekly or monthly totals.
Event based. Let \(N(L)\) represent the number of demand events in the lead time and \(D_{i}\) the size of the \(i\)th:
\[ D(L) = \sum_{i=1}^{N(L)} D_{i} \]
The natural form for a slow mover, where a bucket is mostly zeros and the information is in how often and how large.
Notice that the two are the same object reached from different data. You should choose between them by asking which one the organization’s records can actually support, not by which is more elegant.
Let \(\sigma_{D}\) represent the standard deviation of demand in one period. With a constant lead time of \(L\) periods and independent period demands:
\[ \begin{aligned} E[D(L)] &= E\Big[\textstyle\sum_{i=1}^{L} D_{i}\Big] = \lambda L &&\text{expectation is linear}\\ \mathit{Var}[D(L)] &= \textstyle\sum_{i=1}^{L}\mathit{Var}[D_{i}] = \sigma_{D}^{2}L &&\text{independent periods}\\ \sigma_{D(L)} &= \sigma_{D}\sqrt{L} &&\text{square root} \end{aligned} \]
A supplier’s lead time doubles. What happens to the expected requirement, and to its standard deviation?
The mean doubles; the standard deviation is multiplied by \(\sqrt{2} = 1.414\). Thus, a longer lead time raises the reorder point more than it raises the safety stock. The common error is to multiply the standard deviation by \(L\).
The random sum gives the moments directly:
\[ E[D(L)] = E[L]\,E[D], \qquad \mathit{Var}[D(L)] = \underbrace{E[L]\,\mathit{Var}[D]}_{\text{the customer}} + \underbrace{\mathit{Var}[L]\,\big(E[D]\big)^{2}}_{\text{the supplier}} \]
| Case | What is random | \(E[D(L)]\) | \(\mathit{Var}[D(L)]\) |
|---|---|---|---|
| 1 | Demand only, lead time constant at \(L\) | \(L\,E[D]\) | \(L\,\mathit{Var}[D]\) |
| 2 | Lead time only, demand constant at rate \(\lambda\) | \(\lambda E[L]\) | \(\lambda^{2}\mathit{Var}[L]\) |
| 3 | Both | \(E[L]E[D]\) | \(E[L]\mathit{Var}[D] + \mathit{Var}[L](E[D])^{2}\) |
Notice that cases 1 and 2 are case 3 with one variance set to zero. Compute the two terms of case 3 separately, every time: a total that hides the split cannot tell you whether to spend money on the customer or the supplier.
Demand is Poisson at \(\lambda = 45\) a year, one unit per event. Let \(L\) represent the lead time in years.
Case 1, a constant two months.
\[ \begin{aligned} \theta &= \frac{45 \text{ units}}{1 \text{ year}} \times \frac{2 \text{ months}}{1} \times \frac{1 \text{ year}}{12 \text{ months}}\\ &= 7.5 \text{ units} \end{aligned} \]
\(D(L)\) is Poisson, so \(\mathit{Var}[D(L)] = 7.5\) and the standard deviation is \(\sqrt{7.5} = 2.74\) units.
Case 3, two months on average, standard deviation one month. \(E[L] = 1/6\) year and \(\mathit{Var}[L] = (1/12)^{2} = 1/144\).
\[ \begin{aligned} \mathit{Var}[D(L)] &= \lambda E[L] + \lambda^{2}\mathit{Var}[L]\\ &= 7.5 + 2025\left(\tfrac{1}{144}\right)\\ &= 7.5 + 14.0625 = 21.5625 \end{aligned} \]
The mean is still 7.5; the standard deviation is 4.64 units.
Read the split. The lead time contributes \(14.0625/21.5625 = 65\%\) of the variance the safety stock must cover, on an item whose demand nobody can control and whose lead time a purchasing department could negotiate.
Every model in this chapter needs a distribution. The procedure has three steps, in this order:
Case 3 for the transformer. The ratio is \(21.5625/7.5 = 2.875\), so the negative binomial. Matching moments:
\[ \begin{aligned} \hat{p} &= \frac{7.5}{21.5625} = 0.3478\\ \hat{r} &= \frac{7.5^{2}}{21.5625 - 7.5} = 4.00 \end{aligned} \]
The check, better than a moment match. Build the lead time from four independent stages averaging half a month each: Erlang, shape 4, rate \(\mu = 24\) a year. Poisson demand over an Erlang time is exactly negative binomial, with \(r = 4\) and
\[ p = \frac{\mu}{\lambda + \mu} = \frac{24}{69} = 0.3478 \]
Set \(Q = 1\). Every demand triggers an order for one unit, so the policy replenishes one for one and the inventory position never moves.
Let \(S\) represent the base-stock level, the value at which the position is held, and let \(r = S - 1\) represent the reorder point. Let
\[ \mathit{ss} = r - \theta \]
represent the safety stock, the reorder point in excess of expected lead time demand.
A definition, not an interpretation
“Expected stock on hand when an order arrives” is correct only when at most one order is outstanding. One for one, the expected number outstanding is \(\theta\), which for the transformer is 7.5.
Why \(k\) drops out. Under \(Q = 1\) the order rate is \(\lambda\) whatever \(S\) is, so \(k\lambda\) is a constant added to every alternative. It cannot change which \(S\) is least. It does not drop out of the comparison with \(Q > 1\).
An order is placed at the instant of every demand, so the orders outstanding now are exactly those triggered by the demands of the last \(L\) time units.
\[ \begin{aligned} \mathit{IO} &= D(L) &&\text{one order per demand in the last } L\\ \mathit{IP} &= I + \mathit{IO} - B = S &&\text{the position never moves}\\ S &= I + D(L) - B &&\text{substitute for } \mathit{IO}\\ \mathit{IN} &= I - B = S - D(L) &&\text{rearrange} \end{aligned} \]
\(\mathit{IN} = S - D(L)\) is the whole model. Everything else is read off it, using \(I = (\mathit{IN})^{+}\) and \(B = (-\mathit{IN})^{+}\).
\[ \begin{aligned} \bar{B}(S) &= E\big[(-\mathit{IN})^{+}\big] = E\big[(D(L) - S)^{+}\big] = G^{1}(S) &&\text{first order loss function}\\[4pt] \bar{I}(S) &= E\big[(S - D(L))^{+}\big] = E\big[(D(L) - S)^{+} + S - D(L)\big] &&(x)^{+} = (-x)^{+} + x\\ &= S - \theta + G^{1}(S) &&\text{take expectations}\\[4pt] \overline{\mathit{RR}}(S) &= P\{\mathit{IN} \ge 1\} = P\{D(L) \le S - 1\} = G(S - 1) &&\text{shelf not empty} \end{aligned} \]
Under Poisson demand, PASTA makes the ready rate the fill rate too.
Notice that the on-hand formula is not an approximation. It exceeds \(S - \theta\) by the expected backorder level: stock you are short of is stock you are not holding.
From a table of the distribution function, with \(G^{0}(x) = P\{D(L) > x\}\):
\[ \bar{B}(S) = \theta - \sum_{x=0}^{S-1} G^{0}(x) \]
For the transformer at \(S = 11\), from the \(\lambda = 7.5\) column:
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(G^{0}(x)\) | 0.9994 | 0.9953 | 0.9797 | 0.9409 | 0.8679 | 0.7586 | 0.6218 | 0.4754 | 0.3380 | 0.2236 | 0.1378 |
\[ \bar{B}(11) = 7.5 - 7.3384 = 0.1616 \]
Notice the upper limit is \(S - 1\), not \(S\). Running the sum to \(x = 11\) removes one term too many and reports \(G^{1}(12) = 0.0823\), half the true backorder level.
Substitute into the cost rate and drop the constant \(k\lambda\):
\[ \begin{aligned} C(S) &= h\,\bar{I}(S) + b\,\bar{B}(S) &&\text{holding and backorder terms}\\ &= h\big(S - \theta + G^{1}(S)\big) + b\,G^{1}(S) &&\text{substitute both}\\ &= h(S - \theta) + (h + b)\,G^{1}(S) &&\text{collect } G^{1}(S) \end{aligned} \]
This is the newsvendor’s expected cost with \(c_{o} = h\) and \(c_{u} = b\). Thus, the answer is the critical ratio, with the discrete rule since \(S\) is a whole number:
\[ S^{*} = \min\left\{S : G(S) \ge \frac{b}{b + h}\right\} \]
\(h = 0.25(3{,}800) = \$950\) and \(b = \$8{,}550\) per unit per year; \(D(L)\) is Poisson with \(\theta = 7.5\).
\[ \begin{aligned} \frac{b}{b + h} &= \frac{8550}{9500} = 0.90 &&\text{critical ratio}\\ G(10) &= 0.8622 < 0.90 \quad\text{and}\quad G(11) = 0.9208 \ge 0.90 &&\text{so } S^{*} = 11\\ \bar{B} &= G^{1}(11) = 0.1616, \quad \bar{I} = 11 - 7.5 + 0.1616 = 3.6616 &&\text{units owed, units held}\\ C(11) &= 950(3.6616) + 8550(0.1616) = 3478.48 + 1381.35 = \$4{,}859.83 &&\text{a year} \end{aligned} \]
The fill rate is \(G(10) = 0.8622\), so the storeroom is out of stock about 14% of the time and holds about three and a half transformers.
The average demand waits
\[ \frac{0.1616}{45} \text{ years} \times \frac{365 \text{ days}}{1 \text{ year}} = 1.3 \text{ days} \]
The critical ratio is a shortcut, so we cost the neighbors directly.
| \(S\) | 9 | 10 | 11 | 12 | 13 |
|---|---|---|---|---|---|
| \(\bar{I}\) | 2.0229 | 2.7993 | 3.6616 | 4.5823 | 5.5397 |
| \(\bar{B}\) | 0.5229 | 0.2993 | 0.1616 | 0.0823 | 0.0397 |
| \(C(S)\) | $6,392.69 | $5,218.57 | $4,859.83 | $5,057.03 | $5,601.71 |
Notice that the minimum is at 11, where the critical ratio pointed, and that the curve is flat near it: levels 10 and 12 cost 7.4% and 4.1% more. Two routes, one answer.
We dropped \(k\) because \(Q = 1\) made it a constant. What is that constant for the transformer, at \(k = \$220\) an order?
\[ k\,\overline{\mathit{OF}} = k\lambda = 220(45) = \$9{,}900 \text{ a year} \]
That is more than twice the $4,859.83 we just minimized. Thus, the policy is well chosen given \(Q = 1\), and \(Q = 1\) has not been justified at all.
One-for-one replenishment is right when an order is nearly free or a unit is nearly priceless, and the transformer is neither.
Lead time demand is the one random quantity. Every measure in the chapter is a functional of its distribution.
Its moments are one formula with terms switched off. For the transformer, 65% of the variance under a variable lead time comes from the supplier.
A family is a choice. The variance to mean ratio picks it; an Erlang lead time shows the negative binomial is exact, not assumed.
Base stock is \(\mathit{IN} = S - D(L)\). \(\bar{B} = G^{1}(S)\), \(\bar{I} = S - \theta + G^{1}(S)\), \(\overline{\mathit{RR}} = G(S-1)\).
Its cost is the newsvendor’s. \(S^{*} = 11\) transformers at $4,859.83 a year, and an ordering bill of $9,900 the derivation left out.
Read before next session: General Batch Sizes and The Classical Approximation.
Today’s material was Lead Time Demand and The Base-Stock Model.
Next session:
Bring the base-stock formulas. Next session averages them.