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Reliability case scenarios

Four situations reliability engineers meet in practice, worked through from raw data to a decision. Each one shows the inputs, the formula, and the result — and links to the calculator that does the same job.
These are illustrative scenarios, not client case studies. The organisations and input figures are hypothetical but typical. Every result is computed from the stated inputs with the same formulas the calculators use, so you can reproduce each one.

1 · Welding line: what is unplanned downtime costing?

A robotic welding cell runs two shifts. The maintenance lead has six months of breakdown records and needs to put a number in front of the plant manager.

Input (6 months)Value
Planned production time1,920 h
Breakdowns24
Total repair time96 h
Cost of a stopped line$8,000 / h
MTBF (uptime ÷ failures)
(1,920 − 96) ÷ 24 = 76.0 h
MTTR (repair time ÷ failures)
96 ÷ 24 = 4.0 h
Availability
76.0 ÷ (76.0 + 4.0) = 95.00%

Downtime cost over the period: 96 h × $8,000 = $768,000.

What-if: if a preventive programme cut breakdowns to 15 and faster diagnosis brought MTTR to 3 h, MTBF becomes 125.0 h, availability 97.66%, and downtime cost $360,000 — a difference of $408,000 to weigh against the programme's cost.

ToolsMTBFMTTRAvailabilityCost of Downtime

2 · Aircraft component: mission reliability and spares

An operator flies a fleet with a line-replaceable unit whose field MTBF is known. Two questions: how likely is the unit to survive a typical flight, and how many spares should be held while repairs are away?

InputValue
Unit MTBF4,000 flight hours
Typical mission10 h
Fleet20 aircraft × 250 FH / month
Repair turnaround (lead time)2 months
Spares fill-rate target95%
Mission reliability, constant failure rate
R(10 h) = e^(−10 ÷ 4,000) = 99.75%
Expected removals
5,000 FH / month ÷ 4,000 = 1.25 per month
Demand during lead time (Poisson mean)
1.25 × 2 = 2.50 units

The smallest stock s with P(demand ≤ s) ≥ 95% is 5 spares, covering 95.8% of lead-time demand.

ToolsReliability R(t)Spare PartsFailure Rate

3 · Hospital imaging: turning an availability target into a repair deadline

A radiology department commits to 98% availability for a CT scanner. Its failure history gives an MTBF of 600 scheduled hours. The biomedical team needs to know how fast repairs must be to hit the target.

Rearranging A = MTBF ÷ (MTBF + MTTR)
MTTR ≤ MTBF × (1 − A) ÷ A = 600 × 0.02 ÷ 0.98 = 12.2 h

The current service contract averages 18 h to restore, giving 97.09% — below target, and about 2.4 lost hours in an 84-hour scanning week. The requirement to put in the contract is a restore time of no more than 12 hours.

ToolsAvailabilityMTTR

4 · Data centre: what does a redundant UPS module buy?

A single UPS module has an availability of 99.9%. The facility is deciding whether to add a second module in parallel so either one can carry the load.

ConfigurationAvailabilityExpected downtime / year
Single module99.900%8.76 h
Two in parallel (1-of-2)99.9999%0.53 min
Parallel availability
1 − (1 − 0.999)² = 99.9999%

The calculation assumes the two modules fail independently. In practice common-cause failures — shared switchgear, batteries of the same age, the same maintenance error — usually dominate, so the real gain is smaller than the arithmetic suggests. Treat the parallel figure as an upper bound.

ToolsSystem ReliabilityAvailability

What these show

  • Put downtime in money. An availability percentage rarely moves a budget; a cost per period and a what-if comparison usually does.
  • Work backwards from targets. An availability commitment implies a maximum repair time — that is the number to write into service contracts.
  • Size spares from demand, not habit. Removals per month and repair lead time give a Poisson demand you can stock against a fill-rate target.
  • Question redundancy maths. Parallel formulas assume independence; common-cause failures decide how much of the theoretical gain you actually get.