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Age replacement of a wear-out item: optimum replacement age, cost against run-to-failure and the NAVAIR cost-benefit ratio

ToolMaintenance Interval Optimization
SourceNaval Air Systems Command (2005). Guidelines for the Naval Aviation Reliability-Centered Maintenance Process. NAVAIR 00-25-403, 1 July 2005. Appendix B, section 1.3.4, p. B-7 (cost-benefit ratio of a hard-time task). Age-replacement model: Barlow, R. E., and Hunter, L. C. (1960). "Optimum Preventive Maintenance Policies." Operations Research, Vol. 8, No. 1, pp. 90–100.
LevelProfessional

Problem

An item wears out with a Weibull life of shape 2.5 and characteristic life 1,000 h. A planned replacement costs 100 and a failure in service 1,000; a planned replacement takes 2 h and a failure replacement 24 h; the item runs 4,000 h per year. Find the replacement age that minimises the cost per hour, the saving over running to failure, the NAVAIR cost-benefit ratio, the age of least downtime and the failures and planned replacements per year.

Inputs to Enter

  • Analysis: age replacement; time unit hours; use per year 4,000 h
  • Life distribution: Weibull, shape β = 2.5, characteristic life η = 1,000 h (MTTF = η Γ(1.4) = 887.26 h)
  • Planned cost C_p = 100, failure cost C_f = 1,000; planned downtime T_p = 2 h, failure downtime T_f = 24 h
Open the Maintenance Interval Optimization tool (loaded with this example)

Answers

  • Run to failure: C_f / MTTF = 1,000 / 887.26 = 1.1271 per hour (4,508.24 per year, 4.508 failures per year)
  • Optimum age from h(t) ∫R − F(t) = C_p / (C_f − C_p) = 0.11111: t = 354.57 h, where R = 92.79 %, F = 7.21 % and the mean cycle ∫R = 347.15 h
  • Minimum cost C(t) = (100 × 0.92787 + 1,000 × 0.07213) / 347.15 = 0.4751 per hour, a saving of 57.9 %; 1,900.22 per year with 0.831 failures and 10.69 planned replacements per year
  • NAVAIR cost-benefit ratio CBR = 0.421 (N_S = 0.9279, MTTF_P = 251.68 h; below 1, so the hard-time task is cost-effective)
  • Age of least downtime 326.97 h: downtime 1.018 % (availability 98.982 %) against 2.634 % when run to failure
  • Sensitivity: cost ratio 5 gives 493.0 h, ratio 20 gives 262.4 h; replacing 10 % early or late costs under 1 % more

Notes

The tool's default inputs are this case; press Submit. It is an illustrative case, not a printed example: NAVAIR 00-25-403 prints the cost-benefit formula without numbers, and no public document with a worked age-replacement example was found. The numbers are checked against an independent scipy calculation (optimum 354.5744 h, 0.475055 per hour) in test_maintenance_interval.php, which also reproduces a vendor's printed example (Weibull β 2.847494, η 108.420135, costs 20 and 500: 28.657 h, 1.079 per hour).