Activity: evaluating a policy from the Poisson table
Chapter 8, Continuous and Periodic Review Systems
Work in pairs. 25 minutes.
Names: ______________________________________________
The item. The storeroom stocks a capacitor bank controller. Crews draw 24 a year, one at a time, as a Poisson process. The supplier quotes a constant lead time of two months. A controller costs $1,800, the carrying charge is 25% a year, and unmet demand is backordered. The storeroom runs \((r, Q) = (5, 3)\).
Notation. \(\theta\) is the mean lead time demand. \(G(x) = P\{D(L) \le x\}\) is the Poisson distribution function and \(G^{1}(x) = E[(D(L) - x)^{+}]\) its first order loss function. For a position \(j\), the ready rate is \(G(j-1)\) and the expected backorders are \(G^{1}(j)\). Then
\[ \overline{\mathit{RR}} = \frac{1}{Q}\sum_{j=r+1}^{r+Q} G(j-1), \qquad \bar{B} = \frac{1}{Q}\sum_{j=r+1}^{r+Q} G^{1}(j), \qquad \bar{I} = \frac{Q+1}{2} + r - \theta + \bar{B} \]
The table. Poisson lead time demand with mean 4.
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(G(x)\) | 0.0183 | 0.0916 | 0.2381 | 0.4335 | 0.6288 | 0.7851 | 0.8893 | 0.9489 | 0.9786 | 0.9919 | 0.9972 | 0.9991 | 0.9997 |
| \(G^{1}(x)\) | 4.0000 | 3.0183 | 2.1099 | 1.3480 | 0.7815 | 0.4103 | 0.1954 | 0.0848 | 0.0336 | 0.0123 | 0.0041 | 0.0013 | 0.0004 |
Report probabilities and averages to four decimal places.
Part A. Predict
A colleague says: “We reorder at 5, so we run out whenever lead time demand exceeds 5. The fill rate is \(G(5) = 0.7851\).” Is that too high, too low, or right? Why?
Part B. Evaluate \((5, 3)\)
- Compute \(\theta\), showing the unit conversion. Which values can the inventory position take, and with what probability each?
- For each position, read the ready rate and the expected backorders from the table. Average them to get \(\overline{\mathit{RR}}\) and \(\bar{B}\). Why is the fill rate equal to the ready rate here?
- Compute \(\bar{I}\).
- Check the ready rate by a second route: \(\overline{\mathit{RR}} = 1 - \left[G^{1}(r) - G^{1}(r+Q)\right]/Q\).
Part C. Meet a 95% fill rate
- Holding \(Q = 3\), find the smallest \(r\) whose fill rate is at least 0.95. Show each \(r\) you tried.
- Report \(\bar{B}\) and \(\bar{I}\) at that \(r\).
- Compute \(h\). How much more stock is on the shelf, on average, than at \(r = 5\), and what does that cost a year?
Part D, if you have time. What the planning package does
The storeroom’s planning package sets \(r = \theta + z\sigma\), with \(\sigma = \sqrt{\theta}\) and \(z = 1.645\) “for 95% service”, and rounds up.
- What \(r\) does it set? What fill rate does that policy actually deliver, at \(Q = 3\)?
- What does the extra stock cost a year against your answer to Part C?
- Read the \(G(x)\) row for the smallest \(x\) with \(G(x) \ge 0.95\). What does that tell you about whether the normal or the meaning of “95%” caused the gap?
Exit ticket (tear off and hand in)
Name: ____________________
- In one sentence: why is the fill rate of an \((r, Q)\) policy higher than \(G(r)\)?
- True or false, and why: holding \(r\) fixed, raising \(Q\) lowers the fill rate.