Periodic review changes when the decision is taken. Under \((r, Q)\) the decision rides on the demand event. Under periodic review it rides on a review, an event the policy schedules for itself every \(R\) time units from the start of the replication.
A demand still changes \(I\), \(B\) and \(\mathit{IP}\), but takes no decision. That is the only change to the model, so it can be validated against the exact periodic formulas before it answers questions those formulas reach only by comparison or approximation.
The transformer, reviewed monthly, ordering up to \(S = 14\): the level chapter 8 found optimal. Two-month lead time, 45 a year, the usual costs. At each review it orders \(S - \mathit{IP}\) when that is positive.
A month with no demand at all leaves the position at \(S\) and nothing to order. With Poisson demand that happens with probability \(e^{-\lambda R} = e^{-45/12} = e^{-3.75} = 0.0235\), so the order rate is \(12(1 - 0.0235) = 11.72\) a year, not twelve.
| Measure | Exact | Simulated |
|---|---|---|
| \(\bar{I}\), units | 4.7751 | 4.7696 \(\pm\) 0.0143 |
| \(\bar{B}\), units | 0.1501 | 0.1507 \(\pm\) 0.0025 |
| \(\overline{\mathit{RR}}\) | 0.8928 | 0.8925 \(\pm\) 0.0012 |
| \(\overline{\mathit{FR}}\) | 0.8928 | 0.8929 \(\pm\) 0.0013 |
| Orders a year | 11.7178 | 11.7194 \(\pm\) 0.0068 |
| Ordering cost | $2,577.91 | $2,578.28 \(\pm\) 1.50 |
| Total cost | $8,397.22 | $8,397.75 \(\pm\) 19.49 |
The check: the fraction of reviews that place an order is 0.9767 \(\pm\) 0.0006 against \(1 - 0.0235 = 0.9765\).
Every measure is covered. The review event behaves as the periodic analysis assumes, down to the one review in forty-three that finds nothing to order.
Chapter 8 priced continuous review on the transformer by comparing two exact costs: \((r, Q) = (8, 7)\) at $7,225.15 against the best periodic policy, \(R = 0.120\) year with \(S = 15\), at $8,186.92. That is \(\Delta = \$961.77\) a year, 13.3%, at fill rates 0.8781 and 0.8819.
Let \(C_c\) and \(C_p\) represent the simulated costs of the two in one replication, and \(\Delta = C_p - C_c\). Driven by the same demand times, the \(n\) values of \(\Delta\) are paired.
| Common random numbers | Independent streams | |
|---|---|---|
| \(C_c\) | $7,231.50 \(\pm\) 15.42 | $7,231.50 \(\pm\) 15.42 |
| \(C_p\) | $8,191.96 \(\pm\) 24.05 | $8,182.21 \(\pm\) 19.06 |
| \(\Delta = C_p - C_c\) | $960.47 \(\pm\) 12.34 | $950.71 \(\pm\) 27.61 |
| Variance of \(\Delta\) | 1,092.73 | 5,468.41 |
The paired interval puts the price of not watching the transformer between $948 and $973 a year. The two separate intervals in the first column, read on their own, would only place it between about $920 and $1,000.
Against independent demands the difference is estimated with five times the variance and more than twice the half-width, from the same amount of simulation. That factor is what common random numbers buy, and it costs nothing but giving each source of randomness its own stream.
Under periodic review the position passes below \(s\) even with single-unit demand. Chapter 8 put the mean undershoot at \(\lambda R/2 = 45/24 = 1.875\) units when the spread \(S - s\) is wide, and at exactly 2.8403 at the narrow extreme, \((R, S)\) itself. Between the two, it said only that the spread decides. Let \(U = s - \mathit{IP}\) at every review that orders; \(S = 14\), monthly review, 3.75 units of demand a month:
| \(s\) | \(S - s\) | Prediction | Mean undershoot | Reviews that order |
|---|---|---|---|---|
| 13 | 1 | 2.8403, exact | 2.8422 \(\pm\) 0.0062 | 0.9767 \(\pm\) 0.0006 |
| 10 | 4 | 1.875 | 1.7981 \(\pm\) 0.0081 | 0.6472 \(\pm\) 0.0009 |
| 6 | 8 | 1.875 | 1.8796 \(\pm\) 0.0077 | 0.3798 \(\pm\) 0.0006 |
The check: at \(s = 13\) the prediction is exact, so the interval must cover 2.8403. It does, which validates the recording of \(U\) before it is used where the prediction is only a limit.
At a spread of eight, about two months of demand, the formula is right to the second decimal. At four it is 4% high, not low: the undershoot is smaller than the long-run figure. The formula serves any \((R, s, S)\) policy whose spread holds two review intervals of demand or more.
Let \(C_1\) and \(C_2\) represent two policies’ simulated costs in one replication. The quantity of interest is the difference:
\[ \begin{aligned} \mathit{Var}[C_1 - C_2] &= \mathit{Var}[C_1] + \mathit{Var}[C_2] - 2\,\mathit{Cov}[C_1, C_2] &&\text{variance of a difference}\\ \mathit{Cov}[C_1, C_2] &= 0 &&\text{independent runs: noisier than either}\\ \mathit{Cov}[C_1, C_2] &> 0 &&\text{same demands: heavy demand is dear for both} \end{aligned} \]
Common random numbers do not make either estimate more precise; they make the comparison more precise, which is the thing being asked.
Give each source of randomness its own stream. Two policies place different numbers of orders, so if demand times and lead times share one stream, the draws drift apart after the first order one places and the other does not, and the demands are no longer common.
With more than two policies the question is which is best, and how sure we can be. Multiple comparison with the best (MCB) compares each policy with the best of the others and returns the subset that cannot be ruled out. Let \(\delta\) represent the indifference zone, the smallest cost difference worth distinguishing: a decision about the item, not about statistics. Here \(\delta = \$75\), about 1% of the transformer’s cost.
| \(r\) at \(Q = 5\) | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|
| Exact cost | $8,016.43 | $7,405.97 | $7,396.80 | $7,792.96 | $8,442.11 |
| Above the best | $619.63 | $9.17 | $396.16 | $1,045.31 | |
| Simulated | $8,033.53 | $7,419.78 | $7,405.71 | $7,798.73 | $8,445.30 |
| Half-width | $29.03 | $19.41 | $13.38 | $9.93 | $9.32 |
| MCB interval | \([0, 702.82]\) | \([-60.92, 89.08]\) | \([-89.08, 60.92]\) | \([0, 468.02]\) | \([0, 1114.59]\) |
The check: every interval covers its exact cost, the smallest average is at \(r = 9\), and the MCB intervals containing zero, the policies not ruled out, are \(r = 8\) and \(r = 9\).
With \(\delta = \$75\), MCB keeps \(r = 8\) and \(r = 9\) and rules out the other three. That is the right answer to the question asked, which policies are worth considering: the two differ by $9.17 a year, far inside the $75 the planner said was not worth distinguishing.
The paired difference says something sharper:
\[ \begin{aligned} C_8 - C_9 &= \$14.08 \pm \$9.45 &&\text{excludes zero: the two can be told apart}\\ n &\approx 30\left(\frac{9.45}{9.17}\right)^{2} \approx 32 &&\text{replications to resolve the true \$9.17} \end{aligned} \]
Separating the two policies is cheap. It is also pointless. A statistically significant difference of nine dollars a year on a $7,400 item is not a reason to prefer one reorder point to the other, and the indifference zone is how the planner says so before the run rather than after it.
Comparing a handful of policies is enumeration. With hundreds of candidates, or a cost surface with no formula to guide the search, the problem is simulation optimization: a search in which every evaluation is a noisy estimate. Cross-entropy, simulated annealing and R-SPLINE are among the solvers, and the KSL textbook runs each on the \((r, Q)\) problem.
Two pieces of advice carry over from the earlier chapters:
Periodic review is one more event, and it reproduces the exact periodic results, down to the reviews that find nothing to order: \(1 - e^{-3.75} = 0.9765\).
Continuous review is worth $960 a year on the transformer, measured with five times less variance under common random numbers.
The undershoot formula \(\lambda R/2\) holds once the spread covers two review intervals of demand.
Common random numbers make the comparison precise, through the covariance term, provided each source of randomness has its own stream.
An indifference zone decides what is worth distinguishing before the run; MCB then keeps \(r = 8\) and \(r = 9\).
Read before next session: Multi-Echelon Systems, The KSL Supply Chain Package and Designing the Software.
Today’s material was Periodic Review and Comparing and Choosing Policies.
Next session:
Bring METRIC’s 0.4055 and VARI-METRIC’s 0.4815. Next session measures which is right.