Lead-time demand is how much stock will be consumed between placing an order and receiving it. The expected value is simply daily demand multiplied by lead time, but the expected value is not the number that matters — demand during the lead time is a distribution, and it is the upper tail of that distribution that causes stockouts.
The spread of that distribution is σ × √L, because variances add over independent days. On the sample entries daily variability of 34 units over 9 days gives a lead-time standard deviation of 102 units, not 306 — the square root again. Expected demand is 1,890 units, but there is a one-in-ten chance demand exceeds 2021 units over the same window.
That gap is precisely what safety stock exists to cover, and it is why holding exactly the expected lead-time demand produces a stockout about half the time. This page inverts the question: given the stock you actually hold, what is the probability that lead-time demand exceeds it? At 2,100 units against expected demand of 1,890, the position is 2.06 standard deviations above the mean, a 2.0% chance of running out before the replenishment arrives.
The formula
d- Average demand per day
L- Lead time in days
σ- Standard deviation of daily demand
σ<sub>L</sub>- Standard deviation over the whole lead time = σ × √L
How it works, step by step
- Enter average daily demand and the lead time in days.
- Enter the standard deviation of daily demand.
- Enter the stock you currently hold.
- The gauge shows expected lead-time demand; the spread and stockout risk follow.
Worked examples
210 a day over 9 days
Expected lead-time demand is 1,890 units. The standard deviation is 34 × √9 = 102, so a one-in-ten bad period reaches 2021 units and a one-in-a-hundred reaches 2128.
Holding 2,100 units
That is 210 units above expected demand, 2.06 standard deviations, giving a 2.0% stockout probability. Holding exactly the expected 1,890 would give a 50% chance instead — which is the entire argument for safety stock.
How to read your score
Frequently asked questions
Why is the spread σ × √L rather than σ × L?
Because variances of independent days add, and standard deviation is the square root of variance. High and low days partly cancel over a longer window, so the spread grows more slowly than the mean.
What if the lead time itself varies?
Then use √(L × σ_d² + d² × σ_L²), which adds the variance from lead time. When suppliers are unreliable this second term usually dominates and the demand variability barely matters.
Does this assume normally distributed demand?
Yes, which is reasonable for fast-moving items where many small orders aggregate. For slow movers with occasional large orders a Poisson or compound distribution fits far better.
How does this relate to the reorder point?
The reorder point is exactly this figure plus safety stock. This page shows the distribution that the reorder point is trying to cover.
Lead-time demand reference
| Daily demand | 3 day lead | 7 day lead | 9 day lead | 14 day lead | 21 day lead |
|---|---|---|---|---|---|
| 50 a day | 150 | 350 | 450 | 700 | 1,050 |
| 120 a day | 360 | 840 | 1,080 | 1,680 | 2,520 |
| 210 a day | 630 | 1,470 | 1,890 | 2,940 | 4,410 |
| 400 a day | 1,200 | 2,800 | 3,600 | 5,600 | 8,400 |
| 800 a day | 2,400 | 5,600 | 7,200 | 11,200 | 16,800 |
Expected demand only — the spread around it is σ × √L and grows far more slowly.
| Stock on hand | Above expected demand | Z position | Stockout probability |
|---|---|---|---|
| 1,700 | -190 | -1.86 | 96.9% |
| 1,890 | 0 | 0.00 | 50.0% |
| 2,000 | 110 | 1.08 | 14.0% |
| 2,100 | 210 | 2.06 | 2.0% |
| 2,200 | 310 | 3.04 | 0.1% |
| 2,400 | 510 | 5.00 | 0.0% |
Expected lead-time demand is 1,890 units with a standard deviation of 102. Holding the expected figure exactly gives a 50% stockout risk.