Skip to content
SILProof

Reliability · field data · IEC 61508 · IEC 61511

Chi-square failure rate estimator

Estimate a failure rate from your own field experience: failures observed and operating time give λ with a stated confidence.

Count only the failure mode you need, e.g. dangerous undetected failures for λDU.

Confidence level
Try an example:
λ at 90 % confidence (upper bound)One-sided · time-terminated

607.6 FIT

6.08 × 10⁻⁷ per hour · point estimate 228.3 FIT

Chi-square distribution, 6 degrees of freedom (2n + 2)
00.050.10.15051015χ²Densityχ² = 10.64 (90 %)

Shaded: 90 % of the area. Its right edge is the χ² value; λ ≤ χ² / 2T. More failures widen the curve.

Upper bound λ
6.08 × 10⁻⁷ /h · 0.00532 /yr
Upper bound MTBF
187.9 years
Point estimate n / T
228.3 FIT
Cumulative operating time
8.76 × 10⁶ h · 1,000 device-years
χ² (90 %, 6 degrees of freedom)
10.645

Working

  1. T = 200 devices × 5 years × 8,760 h = 8.76 × 10⁶ device-hours
  2. Point estimate λ = n / T = 2 / 8.76 × 10⁶ = 2.28 × 10⁻⁷ /h
  3. χ²(90 %; 2n + 2 = 6) = 10.645
  4. Upper bound λ = χ² / (2T) = 10.645 / (2 × 8.76 × 10⁶) = 6.08 × 10⁻⁷ /h = 607.6 FIT

What the calculator is for

Prior-use claims, Route 2H and plant-specific failure data all start from the same question: given the failures you have seen over the operating time you have, what failure rate can you defend? The number of failures n over a cumulative time T gives a point estimate n / T, but with few failures that estimate is uncertain. The chi-square distribution turns the same observation into an upper bound that the true rate stays below with a chosen confidence.

  • Point estimate: λ = n / T
  • Upper bound at confidence CL: λ = χ²(CL; 2n + 2) / (2T)
  • T is the cumulative operating time: number of devices × time each was in service.

Worked example

A site has 200 pressure transmitters of one type, each in service for 5 years: T = 200 × 5 × 8,760 = 8.76 × 10⁶ device-hours. Proof tests and trips revealed 2 dangerous undetected failures. The point estimate is 2 / 8.76 × 10⁶ = 228.3 FIT.

At 90 % confidence, χ²(0.9; 6) = 10.645, so λDU ≤ 10.645 / (2 × 8.76 × 10⁶) = 607.6 FIT. At 70 % confidence the bound is 412.7 FIT. Use the bound, not the point estimate, in the verification.

Multiplier table

Upper bound λ × T = χ²(CL; 2n + 2) / 2 for n failures
Failures n70 %90 %95 %
01.2042.3032.996
12.4393.8904.744
23.6165.3226.296
34.7626.6817.754
45.8907.9949.154
57.0069.27510.513
68.11110.53211.842
79.20911.77113.148
810.30112.99514.435
911.38714.20615.705
1012.47015.40716.962

Divide the multiplier by the cumulative operating hours to get the upper-bound failure rate per hour.

Common mistakes

  • Counting the wrong failures. For λDU count only dangerous undetected failures, found by proof tests, demands or inspection, not every repair order.
  • Reporting n / T. With few failures the point estimate is optimistic; IEC 61508 and IEC 61511 ask for an upper confidence bound.
  • Treating zero failures as zero rate. No failures in 10⁶ hours still means λ ≤ 1,204 FIT at 70 % confidence.
  • Pooling unlike devices. Combine operating time only for the same device type in similar service and environment.

Basis

For a constant failure rate, failures in a fixed (time-terminated) observation follow a Poisson distribution; the one-sided upper bound uses chi-square with 2n + 2 degrees of freedom, as in IEC 61508-6 and IEC 60605-4. IEC 61508-2 Route 2H calls for failure data with at least 90 % confidence; IEC 61511 expects reliability data at an upper confidence limit of at least 70 %. The chi-square values are computed exactly, not read from a table.

SILProof

Have a field-based failure rate?

SILProof keeps the risk basis, SRS and three-gate SIL verification of each safety instrumented function on one record.

Use it in a full SIL verification

Frequently asked questions

How do I calculate a failure rate from field data?

Add up the operating time of all devices of the type (devices × hours in service) and count the failures of the mode you need. The point estimate is failures ÷ hours; the upper bound at confidence CL is χ²(CL; 2n + 2) ÷ (2 × hours).

Why use the chi-square distribution?

It gives a confidence bound for a failure rate estimated from a limited number of failures, which is what functional safety standards ask for when field data supports a claim.

What confidence level should I use?

IEC 61508-2 Route 2H requires at least 90 %. IEC 61511 expects reliability data at an upper confidence limit of at least 70 %. Use the level your project or standard specifies.

What if no failures were observed?

The point estimate is zero, but the upper bound is not: χ²(CL; 2) / 2T. At 70 % that is 1.204 / T, at 90 % it is 2.303 / T.

All functional safety calculators →