With demand random and stock watched continuously, the order quantity and the reorder point are separate decisions, and only the second one carries any risk: the only interval in which a stockout can happen is the lead time between placing an order and receiving it. So the reorder point is a quantile of the demand over L periods, built by convolution. Then the sting: cycle service and fill rate are two different measurements of one policy, and on the worked instance they are 68.75% and 99.625%.
Core ideas
Three ideas. The first says where the risk is, the second says how to build the distribution it lives on, and the third is the one this course would choose if it could only keep one.
While stock is above the reorder point nothing can go wrong: an order has not been placed and none is needed. Once one is placed, the question is whether what remains covers demand until it arrives. Everything before that moment and everything after it is safe, so the reorder point is a decision about the distribution of demand over L periods and about nothing else.
Two periods of demand is the distribution of a sum, so the probabilities are convolved rather than the values multiplied. The mean does scale: L periods of a mean-1 demand has mean L. The shape does not. On the worked distribution one period runs 0 to 2 and two periods run 0 to 4 with 3/8 of the mass in the middle, so the tail grows more slowly than the support does — which is why doubling the lead time needs less than double the safety stock.
Cycle service is the probability that a cycle ends without a stockout. Fill rate is the fraction of demand met from stock. They are computed from the same policy and they are not the same number: the first counts cycles and the second counts units, and the order quantity appears in one and not in the other. Both get called “the service level”, and quoting one where a contract means the other is the most expensive confusion in this subject.
The material
The exposure window, the convolution that describes it, the reorder point, and then the two service levels side by side.
Stock is watched continuously. When it falls to the reorder point r, an order for Q is placed and arrives after a lead time of L periods. Demand in each period is an independent random variable with a known distribution.
Safety stock is r − μ, where μ is the expected demand over the lead time. It is the amount held against variation rather than against the average, and it can be zero or negative.
Q and r are separate decisions. Q is a cost question and the earlier lessons of this course answer it; r is a risk question and this lesson answers it. Treating them together is possible and it is not what makes this model useful: almost all of the value is in noticing that the risk window is the lead time and nothing else.
That distribution is what the reorder point is a quantile of. Set r = 2 — exactly the mean, so zero safety stock — and a cycle survives whenever lead-time demand is 0, 1 or 2, which is 1/16 + 1/4 + 3/8 = 11/16. As a percentage that is 68.75%, and it is an exact fraction printed to two places rather than a rounded decimal.
Cycle service is P(lead-time demand ≤ r), the probability that a cycle ends without running out. It does not mention Q.
Fill rate is 1 − E[shortage per cycle]/Q, the fraction of demand met from stock. The expected shortage is E[(lead-time demand − r)⁺], and dividing by the order quantity turns a per-cycle expectation into a per-unit one.
At r = 2 the expected shortage is 3/8 of a unit per cycle: with probability 1/4 you are one short and with probability 1/16 you are two short. Spread that over an order quantity of 100 and the fill rate is 1 − (3/8)/100 = 797/800, which is 99.625%. The same policy, at the same instant, is 68.75% reliable per cycle and 99.625% reliable per unit of demand.
The top row is the one to stare at. A reorder point of zero means ordering only when the shelf is empty, and fifteen cycles in sixteen will run short. It is also, measured by units, a 98% service: almost all demand is still met, because the shortages are small and the order quantity is large. Neither number is wrong. They are answers to different questions, and a supplier who promised one and delivered the other would be in court.
The gap between them is the order quantity. A larger Q spreads the same expected shortage over more demand, so the fill rate rises while the cycle service does not move at all — Q does not appear in it. Shrink Q to 1 on the worked instance and the fill rate falls to 5/8, which is 62.5%, now BELOW the cycle service of 68.75%. That inversion is the proof that one is not the other rescaled.
Cycle service rises with r and never falls, so “the smallest reorder point reaching 95%” is well defined and can be found by walking up. On the worked instance no reorder point gives between 93.75% and 100%: the distribution is discrete, r = 3 gives 15/16 and r = 4 gives 1, so the smallest reorder point meeting a 95% target is 4 — total coverage, two units of safety stock, and an achieved service of 100% rather than 95%.
Overshooting a target is what discrete distributions do, and it is a fact about the distribution rather than a rounding. It is also the reason a service target should be stated with the measurement attached: a 95% target on cycle service costs two units of safety stock here, while a 95% target on fill rate is already met at r = 0, where the safety stock is −2.
Raise the lead time to three periods. The mean rises from 2 to 3 and the support from 0…4 to 0…6, and a reorder point of 2 — unchanged — now delivers a cycle service of 11/32, which is 34.38%.
Holding the reorder point at the old mean was already marginal at 68.75%; at the longer lead time it is worse than a coin. Restoring the old service level takes more than the one extra unit of mean demand, because the distribution has spread as well as shifted — which is the practical content of the word “convolution”.
Everything on this page is exact. The per-period probabilities are fractions, the convolution is a finite sum of products of fractions, the expected shortage is a finite sum, and the two service levels are fractions printed as percentages to a stated number of places. Nothing here is rounded, and the distinction from the earlier lessons — where the economic order quantity and its cost genuinely are irrational — is one a reader should be able to state without looking.
Interactive
the convolution, the safety stock, and two service levels that are not the same number
Method
Five steps, and the first one is the question that decides all the others.
Cycle service counts cycles; fill rate counts units. They differ by an order of magnitude in the gap on the worked instance, and a target quoted without its definition is not a number. If nobody can say which is meant, that is the finding, and it is worth more than any calculation that follows.
Convolve the per-period distribution with itself L times. The mean is L times one period's mean, but the shape is not one period's shape stretched, and using a scaled single-period distribution understates the middle and overstates the tails.
For a cycle-service target, the smallest r with P(lead-time demand ≤ r) ≥ α. Cycle service is monotone in r, so walking up from zero finds it and the first one that clears is the answer.
E[(lead-time demand − r)⁺] is a short sum over the tail. Divide by the order quantity for the fill rate. Print both, always: the pair is the description of the policy and either one alone is an invitation to be misread.
A discrete distribution overshoots. On the worked instance a 95% cycle-service target is met at r = 4 and delivers 100%, which costs two units of safety stock that nobody asked for. Reporting “95% achieved” there would be false in the cheap direction and would hide the overshoot.
Demand of 0, 1 or 2 a period with probabilities 1/4, 1/2, 1/4; a lead time of two periods; an order quantity of 100. Every figure is the panel's own and every one of them is exact.
The two right-hand columns are the lesson. One policy, one instant, two numbers that differ by thirty percentage points at r = 2 and by ninety at r = 0. Both are called the service level in ordinary use, and the only defence is to print the definition beside the figure every single time — which is what the panel does and why it does it.
The last block is the part that proves they are genuinely different quantities rather than one rescaled. With Q = 1 the fill rate drops below the cycle service, so no fixed relationship holds between them. The order quantity is in one definition and not in the other, and that single structural difference is enough to make the ordering depend on the data.
For a rehearsal, switch the panel to the lumpy example: demand 0, 1 or 4 with probabilities 3/5, 1/5, 1/5, a lead time of three periods, a reorder point of 4 and an order quantity of 60. The supplied first move is that the mean lead-time demand is 3. Predict which of the two service figures will be the larger, then check the sizes: they come out 72.80% and 98.893%, and the smallest reorder point reaching a 95% cycle service is 8.
You should be able to say why the lead time is the only exposure window, build a lead-time distribution by convolution, compute safety stock, cycle service and expected shortage from it, turn the shortage into a fill rate with the order quantity, find the smallest reorder point reaching a cycle-service target, and report what it actually achieves rather than what was asked for.
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