Worked Examples

Real-world calculations for all 20 reliability and quality tools

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

Total = 2.5 + 1.8 + 4.2 + 3.1 + 2.4 = 14.0 h
MTTR = 14.0 ÷ 5 repairs

MTTR = 2.8 hours

Slightly above the 2.0 h manufacturing benchmark. Review Breakdown 3 (4.2 h) for improvement.

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

Total = 6 + 6 + 8 = 20 h
MTTR = 20 ÷ 12 incidents

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 = 20,000 ÷ 25

MTBF = 800 hours

Schedule preventive maintenance every ~640 h (80% of MTBF) to catch failures before they occur.

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 = 4,272 ÷ 3

MTBF = 1,424 hours

Good reliability for industrial pumps. Consider condition monitoring to detect degradation between PM intervals.

100 LED bulbs tested — 20 fail over testing period

Given

  • Total component-hours = 3,000,000 h
  • 20 failures observed

Calculation

MTTF = 3,000,000 ÷ 20

MTTF = 150,000 hours

Suitable for applications requiring ≥ 50,000 h lamp life. Replace batch proactively at 80% of MTTF.

Availability

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Web server — one calendar month (744 h)

Given

  • Total planned time: 744 h
  • Planned maintenance: 4 h
  • Unplanned downtime: 6 h

Calculation

Uptime = 744 − 10 = 734 h
A = 734 ÷ 744 × 100%

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

A = 500 ÷ (500 + 5) = 500 ÷ 505

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

A = (480−50) ÷ 480 = 89.6%
P = (2×200) ÷ 430 = 93.0%
Q = (200−10) ÷ 200 = 95.0%
OEE = 89.6 × 93.0 × 95.0%

OEE = 79.2%

Good for a typical plant. Availability is the weakest factor — target the 50 min of stoppage time first.

Downtime Cost

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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

Lost revenue: 3 × $8,000 = $24,000
Total = $24,000 + $2,000 + $5,000 + $800

Total Downtime Cost = $31,800

A $5,000 predictive maintenance programme that prevents one such event per year delivers 6× ROI.

Spare Parts

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Critical pump seal — stockroom optimisation

Given

  • Mean annual demand (λ) = 4 seals/yr
  • Lead time: 3 weeks
  • Target service level: 95%

Calculation

Lead-time demand = 4 × (3/52) ≈ 0.23
Using Poisson CDF: P(X ≤ 1) = 97.5% → stock 1 seal

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)

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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) = e^(−0.0005 × 500) = e^(−0.25)

R(500) = 77.88%

Below the 90% threshold for aviation-critical components. Consider redundancy or shorter inspection interval.

System Reliability

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Parallel pump system (2-of-2 redundancy)

Given

  • Pump A: R = 0.90
  • Pump B: R = 0.90

Calculation

Rs = 1 − (1−0.90) × (1−0.90) = 1 − 0.01

System R = 99%

Adding one redundant pump improves system reliability from 90% to 99%. Worth the investment for critical services.

3-component series: pump → valve → filter

Given

  • R_pump = 0.95
  • R_valve = 0.98
  • R_filter = 0.99

Calculation

Rs = 0.95 × 0.98 × 0.99

System R = 92.1%

The pump is the weakest link. A 1% reliability improvement in the pump yields more than improvements in other components.

Failure Rate

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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 ÷ 1,000,000 = 8×10⁻⁶ /h
FIT = λ × 10⁹ = 8,000 FIT
MTBF = 1 ÷ λ = 125,000 h

λ = 8,000 FIT | MTBF = 125,000 h

Good for industrial electronics. Automotive targets are typically < 100 FIT — more work needed for that application.

Weibull Analysis

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Bearing wear-out failure analysis

Given

  • β = 2.8 (wear-out mode)
  • η = 8,000 h (characteristic life)

Calculation

R(6,000) = e^−(6000/8000)^2.8 = e^−(0.75)^2.8
(0.75)^2.8 ≈ 0.454
R = e^−0.454

R(6,000 h) = 63.5%

Only 63.5% survival at 6,000 h. Set replacement interval at 4,000–5,000 h for 85%+ reliability.

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 = 8 × 4 × 6

RPN = 192 (High Priority)

Improve detection with better test coverage. Reducing D from 6 to 3 drops RPN to 96 — below the action threshold.

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

σ_GRR = √(0.025² + 0.015²) = 0.0292 mm
σ_Total = √(0.100² + 0.0292²) = 0.1042 mm
%GRR = (0.0292 / 0.1042) × 100%

%GRR = 28% (Marginal)

Repeatability dominates. Investigate fixture consistency. Retrain operators and recalibrate caliper before accepting this MSA.

Quality & Statistical Tools

Process Capability

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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 = (205−195) ÷ (6×1.2) = 10 ÷ 7.2
Cpk = min[(205−201)/(3×1.2), (201−195)/(3×1.2)]
= min[1.11, 1.67]

Cp = 1.39 | Cpk = 1.11

Process is capable (Cpk > 1.0) but off-center. Shift mean from 201°C toward 200°C to raise Cpk to ≥ 1.33.

DPMO & Sigma Level

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Invoice processing — accounts payable team

Given

  • 1,000 invoices processed
  • 3 opportunities per invoice (amount, vendor, date)
  • 36 defects found in audit

Calculation

DPMO = (36 ÷ (1,000 × 3)) × 1,000,000
= 36 ÷ 3,000 × 1,000,000

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

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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

n = (1.96 × 12 / 2)² = (11.76)² = 138.3

Required sample n = 139 bolts

If sampling 139 is too costly, widen margin of error to 3 kN → n = 62. Use control charts to reduce required future sample sizes.

Control Chart

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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_X = 49.98 + 0.577×0.06 = 50.015 mm
LCL_X = 49.98 − 0.577×0.06 = 49.945 mm
UCL_R = 2.114 × 0.06 = 0.127 mm

UCL=50.015, LCL=49.945 mm

All points in control. Cpk = 1.67 — this is a world-class process. Continue monitoring at current frequency.

Pareto Chart

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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

Top 2 cumulative: 42 + 27 = 69%
Top 3 cumulative: 42 + 27 + 15 = 84%

Late delivery + defects = 69% of all complaints

Fix the delivery process and product quality. Addressing just these two issues eliminates 69% of customer complaints.

Fishbone Diagram

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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

Root cause weighting via team vote:
Worn tools: 40% | SOP gap: 30% | Material: 20% | Other: 10%

Primary cause: worn cutting tools + inadequate SOP

Implement daily tool wear inspection and launch a 2-week operator training programme on updated SOPs.

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

CV = (3.2 / 501.4) × 100% = 0.64%
Cp = (510−490) ÷ (6×3.2) = 20 ÷ 19.2

Cp = 1.04 — marginally capable

Right-skewed histogram: mean is above target. Reduce fill head valve opening slightly to center the distribution.

Scatter Diagram

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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 = Σ[(Tᵢ−T̄)(Sᵢ−S̄)] / √[Σ(Tᵢ−T̄)² × Σ(Sᵢ−S̄)²]
r = +0.87 (strong positive)

r = 0.87 — Strong positive correlation

Higher cure temperature → stronger bonds. Raise cure temperature to 175°C as the optimal operating point identified from the regression line.

Check Sheet

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Defect tally on finished assemblies — one shift

Given

  • Missing screw: 14
  • Wrong label: 8
  • Scratch/cosmetic: 22
  • Wrong colour: 3
  • Other: 2

Calculation

Total defects: 49
Scratch % = 22/49 × 100%
Missing screw % = 14/49 × 100%

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.

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