Tools › Reliability › Maintainability and Availability › Maintainability and Availability › Example
Five-subsystem series system of MIL-HDBK-338B Figure 5.8-1: MTTR, maximum repair time, the availability family, a repaired 1-of-2 pair and the MIL-HDBK-470A demonstration plan
| Tool | Maintainability and Availability |
| Source | U.S. Department of Defense (1998). Electronic Reliability Design Handbook. MIL-HDBK-338B, 1 October 1998. Section 5.8.1, Figure 5.8-1 (series system of five subsystems), eqs. 5.75, 5.115–5.118, and the subsystem-3 redundancy case; Section 10.4.1 eqs. 10.21–10.22; U.S. Department of Defense (1997). Designing and Developing Maintainable Products and Systems, Volume I. MIL-HDBK-470A, 4 August 1997. Appendix B, Test Method 1 example (pp. B-28 to B-29). |
| Level | Professional |
Problem
A system of five subsystems in series has MTBF 400 / 50 / 100 / 200 / 25 h and MTTR 2 / 5 / 2 / 1 / 5 h (lognormal repair times, sigma of ln t 0.55, crews of one). Find the system MTBF, the failure-rate-weighted MTTR, the maintenance time ratio and the inherent availability; with preventive maintenance every 500 h lasting 1 h and a 0.7 h administrative and logistic delay per event, the achieved and operational availabilities; the median and 90 / 95 % maximum repair times of the mixture; the steady-state availability and mean time to first failure of a 1-of-2 redundant pair (λ 0.001 /h, μ 0.5 /h, one repair crew); and the sample sizes of a demonstration of a 0.5 h MTTR against a 0.75 h maximum with 5 % risks.
Inputs to Enter
- Tasks Sub1–Sub5: MTBF 400 / 50 / 100 / 200 / 25 h; repair time 2 / 5 / 2 / 1 / 5 h entered as the mean, lognormal, σ 0.55, crew 1, delay 0
- Mission 100 h; PM interval 500 h, PM duration 1 h, PM crew 1; admin + logistic delay 0.7 h; ready time 0
- Redundant group 1-of-2, 1 channel, λ 0.001 /h, μ 0.5 /h; spares lead time 168 h, 1 system, fill rate 90 %
- Demonstration μ0 0.5 h, μ1 0.75 h, α = β = 5 %, prior σ² of ln t 0.6, prior s.d. 0.5 h, M_max at 95 %
Answers
- Σλ = 0.0775 /h, MTBF = 12.90 h (handbook 12.9); MTTR = Σλ_i M_i/Σλ_i = 0.33/0.0775 = 4.258 h (handbook 4.26); α = 0.33; A_i = 1/(1 + 0.33) = 0.7519 (handbook 0.752); product of the subsystem availabilities 0.7353
- Mixture: median 3.598 h, M_max 90 % 8.015 h, M_max 95 % 9.909 h (numerical inversion, scipy brentq agrees to 1e-9); MMH/OH = 0.332
- MTBM = 1/(0.0775 + 0.002) = 12.58 h, M̄ = 4.176 h, A_a = 0.7508; MDT = 4.876 h, A_o = 0.7206
- 1-of-2 pair, one crew: A = (μ² + 2λμ)/(μ² + 2λμ + 2λ²) = 0.99999203, MTTF = (3λ + μ)/(2λ²) = 251,500 h (single unit 1,000 h); the handbook's subsystem-3 case λ 0.04, μ 0.2 gives 0.946 as printed
- Demonstration: n = 55.6 → 56 (Plan A, eq. B-4), n = 43.3 → 43 (Plan B, eq. B-5), as the handbook's 30 / 45 minute example prints; P(accept) at a true mean of 40 min = 0.203 (Figure B-2 reads 0.21)
Notes
The tool's default inputs are this case; press Submit. Figure 5.8-1 prints only MTBF and MTTR per subsystem: the lognormal spread is the handbook's suggested 0.55 (§5.6.2.4), the preventive maintenance, delay and spares inputs are illustrative. Checked by test_maintainability_availability.php (256 checks), which also reproduces the handbook's lognormal, normal and exponential repair-time examples of §5.6.2 and the availability Example 3 of §10.4.