Core Reliability Metrics
Production Line — 5 breakdowns in a month
Given
- Breakdown 1: 2.5 h
- Breakdown 2: 1.8 h
- Breakdown 3: 4.2 h
- Breakdown 4: 3.1 h
- Breakdown 5: 2.4 h
Calculation
MTTR = 2.8 hours
Breakdown 3 (4.2 h) is the outlier — finding out why it took longer (parts, diagnosis, access) is usually the quickest way to bring MTTR down.
IT Server — 12 incidents over 6 months
Given
- 8 incidents × 0.75 h = 6 h
- 3 incidents × 2 h = 6 h
- 1 major incident × 8 h = 8 h
Calculation
MTTR = 1.67 hours
Within the 1–4 h IT benchmark. The single major incident raises the average — resolve its root cause.
Fleet of 10 delivery trucks — 1 year operation
Given
- 10 trucks × 8 h/day × 250 days = 20,000 h
- 25 failures across fleet
Calculation
MTBF = 800 hours
MTBF alone does not say when failures happen. Before setting a PM interval, check whether failures cluster by age (a Weibull fit) — time-based PM only helps with wear-out.
Industrial pump operating 24/7 for 6 months
Given
- Operating time = (6×30×24) − 48 h downtime = 4,272 h
- 3 major failures
Calculation
MTBF = 1,424 hours
Track the figure as more months of data arrive; three failures is a small sample. Condition monitoring can catch degradation between PM visits.
100 LED bulbs tested — 20 fail over testing period
Given
- Total component-hours = 3,000,000 h
- 20 failures observed
Calculation
MTTF = 150,000 hours
Only 20 of 100 bulbs failed, so this is an estimate from partial data that assumes a constant failure rate. A Weibull fit would show whether failures are wear-out before planning group replacement.
Availability
Open calculator →Web server — one calendar month (744 h)
Given
- Total planned time: 744 h
- Planned maintenance: 4 h
- Unplanned downtime: 6 h
Calculation
Availability = 98.66%
Below the 99.9% target. 10 h downtime/month = 120 h/year. Focus on eliminating the 6 h unplanned outages.
Using MTBF/MTTR formula
Given
- MTBF = 500 h
- MTTR = 5 h
Calculation
Availability = 99.01%
Reducing MTTR from 5 h to 2 h would improve availability to 99.60%.
Injection moulding machine — 8 h shift
Given
- Planned: 480 min
- Breakdowns: 30 min
- Changeovers: 20 min
- Ideal cycle: 2 min, Actual: 200 units
- Defects: 10 units
Calculation
OEE = 79.2%
Good for a typical plant. Availability is the weakest factor — target the 50 min of stoppage time first.
Downtime Cost
Open calculator →Automotive assembly line — 3-hour unplanned stoppage
Given
- Lost revenue: $8,000/h
- Overtime labour: $2,000 total
- Customer penalty: $5,000
- Repair parts: $800
Calculation
Total Downtime Cost = $31,800
A $5,000 predictive maintenance programme that prevents one such event per year pays back about 6 times its cost.
Spare Parts
Open calculator →Critical pump seal — stockroom optimisation
Given
- Mean annual demand (λ) = 4 seals/yr
- Lead time: 3 weeks
- Target service level: 95%
Calculation
Stock 1 seal to achieve 95% SL
For 99% SL, stock 2 seals. For a $15 seal, the cost difference is trivial vs hours of production downtime.
Advanced Reliability Analysis
Reliability R(t)
Open calculator →Aircraft component — 500-hour mission
Given
- MTBF = 2,000 h
- Mission time t = 500 h
- λ = 1 / 2,000 = 0.0005 /h
Calculation
R(500) = 77.88%
Roughly one unit in five would fail within the mission — too high for a critical function. Redundancy or a more reliable component is needed.
System Reliability
Open calculator →Parallel pump system (1-of-2 redundancy)
Given
- Pump A: R = 0.90
- Pump B: R = 0.90
Calculation
System R = 99%
Adding one redundant pump improves system reliability from 90% to 99%, assuming the pumps fail independently — shared power or suction lines reduce the real gain.
3-component series: pump → valve → filter
Given
- R_pump = 0.95
- R_valve = 0.98
- R_filter = 0.99
Calculation
System R = 92.2%
The pump is the weakest link. A 1% reliability improvement in the pump yields more than improvements in other components.
Failure Rate
Open calculator →Electronic controller — fleet data
Given
- 200 units tested for 5,000 h
- 8 failures observed
- Total unit-hours = 200 × 5,000 = 1,000,000 h
Calculation
λ = 8,000 FIT | MTBF = 125,000 h
Automotive and other safety-related applications usually demand much lower failure rates per component, so this design would need work before use there.
Weibull Analysis
Open calculator →Bearing wear-out failure analysis
Given
- β = 2.8 (wear-out mode)
- η = 8,000 h (characteristic life)
Calculation
R(6,000 h) = 64.0%
Only 64% survival at 6,000 h. With β = 2.8 (wear-out), R(4,000 h) = 86.6% — a replacement interval near 4,000 h keeps survival above 85%.
Automotive fuel injector failure mode analysis
Given
- Failure Mode: Clogged injector
- Severity (S) = 8 — engine stall
- Occurrence (O) = 4 — occasional
- Detection (D) = 6 — not easily caught in test
Calculation
RPN = 192 (High Priority)
Improve detection with better test coverage. Reducing D from 6 to 3 drops RPN to 96 (medium) — but with severity 8, keep the failure mode on the action list regardless of RPN.
Gage R&R
Open calculator →Caliper measurement system validation
Given
- 2 operators, 10 parts, 2 replicates each
- σ_repeatability = 0.025 mm
- σ_reproducibility = 0.015 mm
- σ_Part = 0.100 mm
Calculation
%GRR = 28% (Marginal)
Repeatability dominates. Investigate fixture consistency. Retrain operators and recalibrate caliper before accepting this MSA.
Quality & Statistical Tools
Process Capability
Open calculator →Chemical reactor temperature control
Given
- USL = 205°C, LSL = 195°C
- Process mean μ = 201°C
- Process std dev σ = 1.2°C
Calculation
Cp = 1.39 | Cpk = 1.11
Cpk 1.11 is below the usual 1.33 requirement because the process runs off-centre. Re-centring at 200°C would lift Cpk to the Cp value of 1.39.
DPMO & Sigma Level
Open calculator →Invoice processing — accounts payable team
Given
- 1,000 invoices processed
- 3 opportunities per invoice (amount, vendor, date)
- 36 defects found in audit
Calculation
DPMO = 12,000 → ~3.8 Sigma
Below the 4σ (6,210 DPMO) target. Top defects: incorrect amounts (55%) and wrong vendor codes (30%) — use Pareto to prioritize.
Sample Size
Open calculator →Incoming inspection — bolt tensile strength
Given
- Confidence level: 95% (Z = 1.96)
- Process std dev σ = 12 kN
- Acceptable margin of error E = 2 kN
Calculation
Required sample n = 139 bolts
If sampling 139 is too costly, widen the margin of error to 3 kN → n = 62, at the cost of a less precise estimate.
Control Chart
Open calculator →Automotive shaft diameter — X̄-R chart (n=5)
Given
- Grand mean X̄̄ = 49.98 mm
- Average range R̄ = 0.06 mm
- Constants: A₂=0.577, D₃=0, D₄=2.114
Calculation
UCL=50.015, LCL=49.945 mm
All points within limits, so the process is stable. Capability against the drawing tolerance is a separate check (Cp/Cpk) — control limits are not specification limits.
Pareto Chart
Open calculator →Customer complaint analysis — 6 months
Given
- Late delivery: 245 (42%)
- Product defects: 156 (27%)
- Poor service: 89 (15%)
- Wrong product: 58 (10%)
- Other: 35 (6%)
Calculation
Late delivery + defects = 69% of all complaints
Fix the delivery process and product quality first — these two categories account for 69% of customer complaints.
Fishbone Diagram
Open calculator →High scrap rate on CNC machined parts (target 2%, actual 8%)
Given
- Man: New operators (3 hired last month)
- Machine: Worn cutting tools, vibration
- Material: Hardness variation in new supplier batch
- Method: Inadequate SOP for new operators
Calculation
Primary cause: worn cutting tools + inadequate SOP
Implement daily tool wear inspection and launch a 2-week operator training programme on updated SOPs.
Histogram
Open calculator →Fill volume analysis — 50 bottles sampled
Given
- Target: 500 mL
- Sample mean x̄ = 501.4 mL
- Sample std dev σ = 3.2 mL
- USL = 510, LSL = 490 mL
Calculation
Cp = 1.04 — marginally capable
The mean (501.4 mL) sits 1.4 mL above target, so Cpk ≈ 0.90 is lower than Cp. Re-centring the fill helps, but with Cp only 1.04 the spread itself also needs reducing.
Scatter Diagram
Open calculator →Cure temperature vs. bond strength (n=30 samples)
Given
- Temperature range: 150–180°C
- Bond strength: 12–28 MPa
- Pearson r calculated from data
Calculation
r = 0.87 — Strong positive correlation
Higher cure temperature is associated with stronger bonds across 150–180°C. Confirm with a designed experiment before changing the process — correlation alone does not prove cause, and a straight-line fit cannot locate an optimum.
Check Sheet
Open calculator →Defect tally on finished assemblies — one shift
Given
- Missing screw: 14
- Wrong label: 8
- Scratch/cosmetic: 22
- Wrong colour: 3
- Other: 2
Calculation
Scratch (44.9%) | Missing screw (28.6%) are dominant
Transfer these two categories to a Pareto chart. Together they account for 73.5% of defects — focus quality action here.