The lead-time standard deviation drives most of the safety stock, and it takes far more goods receipts to pin down than most planners use.
Use the standard error of a sample standard deviation, roughly σ ÷ √(2(n − 1)), and solve for n: receipts needed = 1 + ½ × (z × σ ÷ h)². For part AF-7, whose lead time varies by 1.9 weeks, knowing that spread to within 0.3 weeks at 95 per cent confidence takes 79 goods receipts, which at 21.8 orders a year is 3.6 years of history. Most planners set safety stock from a fraction of that.
Worked in full in Supply Chain Management by Julian R. Sterling, with every figure reproduced in a free workbook.See the book on Amazon →
On a part like AF-7 the supplier's irregularity is the largest single input to safety stock, larger than the forecast error. Yet the lead-time standard deviation is usually computed from whatever receipts happen to be in the system, and then treated as a known constant. The question is how many deliveries it takes before the number deserves that confidence, and what the uncertainty costs in stock while you wait.
| Input | Value |
|---|---|
| Mean weekly demand / standard deviation | 1,840 / 462 units |
| Mean lead time | 6 weeks |
| Lead-time standard deviation, measured | 1.9 weeks |
| Precision wanted on that figure, ± h | 0.3 weeks |
| Confidence / z, two-sided | 95% / 1.96 |
| Economic order quantity / orders a year | 4,384.5 / 21.8 |
| Cycle service level / z | 98% / 2.0537 |
| Purchase cost / holding cost per unit-year | 28.6 / 3.8324 EUR |
1. How precise a sample standard deviation is. For roughly normal data, the standard deviation estimated from n observations has a standard error of about σ ÷ √(2(n − 1)). Set the confidence half width, z times that standard error, equal to the precision you want, and solve:
h = z × σ ÷ √(2(n − 1)) ⇒ n = 1 + ½ × (z × σ ÷ h)²
z × σ ÷ h = 1.96 × 1.9 ÷ 0.3 = 12.413
n = 1 + ½ × 154.1 = 1 + 77.04 = 78.04, rounded up to 79 receipts
Excel: =ROUNDUP(1+0.5*(NORM.S.INV(0.975)*1.9/0.3)^2,0)
2. Convert receipts into time. A receipt is one replenishment. AF-7 is ordered in batches of 4,384.5 units against 95,680 a year, 21.82 orders a year, so 79 receipts take 3.62 years to accumulate. A precision of 0.3 weeks on 1.9 is 15.8 per cent of the figure, and it already needs the better part of four years.
3. Translate the uncertainty into stock. Safety stock is z × √(L × σd² + d² × σL²). At the measured 1.9 weeks that is 7,547 units. At the edges of the interval:
σL = 1.6 weeks: 2.0537 × √(6 × 462² + 1,840² × 1.6²) = 6,478 units
σL = 2.2 weeks: 8,632 units
Even with 79 receipts the right safety stock sits somewhere in a band 2,155 units wide, worth 61,627 EUR of stock. That is the irreducible fog of a well-measured spread, and it is why the safety stock should be recomputed as receipts arrive, not set once.
The receipts needed grow with the square of the precision. Halving the half width quadruples the history.
| Half width, weeks | Receipts | Years of AF-7 orders | Safety stock range, units | Range at cost, EUR |
|---|---|---|---|---|
| ± 1.0 | 8 | 0.4 | 4,119 to 11,203 | 202,581 |
| ± 0.5 | 29 | 1.3 | 5,778 to 9,362 | 102,501 |
| ± 0.3 | 79 | 3.6 | 6,478 to 8,632 | 61,627 |
| ± 0.2 | 175 | 8.0 | 6,832 to 8,269 | 41,110 |
| ± 0.1 | 695 | 31.8 | 7,188 to 7,907 | 20,562 |
Read the other way, the table says what a short history is worth. The interval below is the approximate 95 per cent range around a measured 1.9 weeks; for the smallest samples it flatters the precision, as the note further down shows.
| Receipts | Months of history | Spread could be, weeks | Safety stock could be, units |
|---|---|---|---|
| 6 | 3.3 | 0.72 to 3.08 | 3,585 to 11,860 |
| 12 | 6.6 | 1.11 to 2.69 | 4,782 to 10,442 |
| 22 | 12.1 | 1.33 to 2.47 | 5,521 to 9,636 |
| 44 | 24.2 | 1.50 to 2.30 | 6,121 to 9,003 |
With a year of receipts, about 22, safety stock could defensibly be anywhere from 5,521 to 9,636 units: 117,670 EUR of stock separates the two ends. With six receipts the number is barely information at all.
Two errors usually travel together. The first is computing σL from a handful of recent deliveries, say the last six, and feeding it into the safety stock formula with four significant figures. The second is reacting to each new receipt: one late shipment moves a six-point standard deviation by a large amount and the safety stock jumps with it, then falls back. Both come from forgetting that the spread is itself an estimate with an error bar.
Three practical remedies follow from the formula. Pool receipts across parts that share a supplier and a shipping lane, since the delay usually belongs to the lane: four parts ordered as often as AF-7 reach 79 receipts in 10.9 months rather than 3.6 years. Accept a lower confidence where the stakes are small: at 90 per cent, z = 1.645 and 56 receipts suffice. And until the history is long enough, set the safety stock towards the upper end of the interval, because on AF-7 a unit short loses 18.9 EUR of margin while a unit held costs 3.8324 EUR a year, and 90.5 per cent of the variance is the supplier's.
The √(2(n − 1)) rule is a large-sample approximation and gives a symmetric interval. On short histories the exact chi-square interval is wider and skewed upwards: with six receipts it runs from 1.19 to 4.66 weeks, safety stock 5,049 to 17,762 units, and with 22 receipts from 1.46 to 2.72 weeks. The formula also assumes lead times are roughly normal. Real lead times are often skewed, with a long tail of very late deliveries, and then the standard deviation converges more slowly still. If the receipts show a tail, look at the late deliveries directly before trusting any spread.
The receipt count and the variance split for AF-7 are in the free workbook for this case. The safety stock formula used in step 3 is set out in how to calculate safety stock when lead time varies, and its forecast-error counterpart in how to calculate safety stock from forecast error.
Take each goods receipt, measure the days from order to receipt, and compute the sample standard deviation with =STDEV.S over the history. Use delivered dates, not the supplier's quoted lead time, which has no spread. For part AF-7 the result is 1.9 weeks on a 6-week mean, and to know it to within 0.3 weeks at 95 per cent confidence needs about 79 receipts.
Usually not on its own. AF-7 is ordered 21.8 times a year, so a year gives about 22 receipts and an approximate 95 per cent interval of 1.33 to 2.47 weeks around a 1.9-week spread. Safety stock at a 98 per cent service level could then be anywhere from 5,521 to 9,636 units, a range worth 117,670 EUR of stock.
Pool receipts from parts that share the supplier and the shipping lane, since the delay usually comes from the lane rather than the part. Four AF-7-like parts on one lane reach 79 receipts in 10.9 months instead of 3.6 years. Accepting 90 per cent confidence instead of 95 also cuts the requirement from 79 to 56 receipts.
This article is one calculation from Supply Chain Management. The book takes the same case from first principles to the decision, chapter by chapter, and every figure it prints is a live formula in the free companion workbooks.
Get the book on Amazon →Free companion files
Also on Amazon UK · Amazon Germany · Amazon France · Amazon Canada
Reading guide: business operations: cost, contracts, pricing and supply chain → · All 1340 articles →
If this book helped, or didn’t, a few lines on Amazon are worth more than they look: they are what the next reader goes on. Write a review. The workbook stays free either way.