Reliability Engineering & Quality Management Examples
Practical examples and real-world scenarios for reliability engineering calculations and quality management tools to help you understand and apply these metrics effectively
Reliability Engineering Examples
MTTR Examples
Example 1: Manufacturing Equipment
Scenario
A production line had 5 breakdowns in the past month:
- • Breakdown 1: 2.5 hours
- • Breakdown 2: 1.8 hours
- • Breakdown 3: 4.2 hours
- • Breakdown 4: 3.1 hours
- • Breakdown 5: 2.4 hours
Calculation
Total Repair Time = 2.5 + 1.8 + 4.2 + 3.1 + 2.4 = 14.0 hours
Number of Repairs = 5
MTTR = 14.0 ÷ 5 = 2.8 hours
Example 2: IT Server Maintenance
Scenario
Server incidents over 6 months:
- • 8 incidents requiring 45 minutes each
- • 3 incidents requiring 2 hours each
- • 1 major incident requiring 8 hours
Calculation
Total Time = (8 × 0.75) + (3 × 2) + (1 × 8) = 20 hours
Number of Repairs = 8 + 3 + 1 = 12
MTTR = 20 ÷ 12 = 1.67 hours
MTBF Examples
Example 1: Fleet of Delivery Trucks
Scenario
Fleet performance over 1 year:
- • 10 trucks in operation
- • Each truck operates 8 hours/day, 250 days/year
- • Total failures across fleet: 25
Calculation
Total Operating Time = 10 × 8 × 250 = 20,000 hours
Number of Failures = 25
MTBF = 20,000 ÷ 25 = 800 hours
Example 2: Manufacturing Pump System
Scenario
Pump system data:
- • Operates 24/7 for 6 months
- • 3 major failures occurred
- • Total downtime: 48 hours
Calculation
Total Operating Time = (6 × 30 × 24) - 48 = 4,272 hours
Number of Failures = 3
MTBF = 4,272 ÷ 3 = 1,424 hours
Availability Examples
Example 1: Web Server Availability
Scenario
Web server performance in one month:
- • Total time in month: 744 hours (31 days)
- • Planned maintenance: 4 hours
- • Unplanned downtime: 6 hours
Calculation
Total Downtime = 4 + 6 = 10 hours
Uptime = 744 - 10 = 734 hours
Availability = (734 ÷ 744) × 100% = 98.7%
Example 2: Using MTBF and MTTR
Scenario
Equipment reliability data:
- • MTBF = 500 hours
- • MTTR = 5 hours
Calculation
Availability = MTBF ÷ (MTBF + MTTR)
Availability = 500 ÷ (500 + 5)
Availability = 500 ÷ 505 = 99.0%
OEE Examples
Example 1: Packaging Line
Availability Data
- • Planned: 8 hours
- • Actual: 7.5 hours
- • Availability: 93.8%
Performance Data
- • Ideal cycle: 0.5 min/unit
- • Actual count: 800 units
- • Performance: 88.9%
Quality Data
- • Total count: 800 units
- • Good count: 760 units
- • Quality: 95.0%
OEE Calculation
OEE = Availability × Performance × Quality
OEE = 93.8% × 88.9% × 95.0%
OEE = 79.2%
Example 2: Injection Molding Machine
Shift Data
- • Shift duration: 8 hours (480 minutes)
- • Breakdowns: 30 minutes
- • Changeovers: 20 minutes
- • Ideal cycle time: 2 minutes
- • Parts produced: 200
- • Defective parts: 10
Detailed Calculation
Availability:
(480 - 30 - 20) ÷ 480 = 89.6%
Performance:
(2 × 200) ÷ 430 = 93.0%
Quality:
(200 - 10) ÷ 200 = 95.0%
OEE = 89.6% × 93.0% × 95.0% = 79.2%
Quality Management Tool Examples
Pareto Analysis Examples
Example 1: Customer Complaint Analysis
Complaint Data (6 months)
- • Late delivery: 245 complaints (42%)
- • Product defects: 156 complaints (27%)
- • Poor service: 89 complaints (15%)
- • Wrong product: 58 complaints (10%)
- • Billing errors: 23 complaints (4%)
- • Others: 12 complaints (2%)
Pareto Analysis Results
Top 2 issues: 69% of complaints
Top 3 issues: 84% of complaints
Action: Focus on delivery process and quality control
Example 2: Manufacturing Defect Analysis
Defect Types (Monthly)
- • Dimensional errors: 187 units (35%)
- • Surface finish: 134 units (25%)
- • Material flaws: 98 units (18%)
- • Assembly errors: 67 units (12%)
- • Contamination: 34 units (6%)
- • Others: 21 units (4%)
Cost Impact Analysis
Total defects: 541 units
Cost per defect: $45 average
Monthly cost: $24,345
Fixing top 3 issues could save 78% of costs
Fishbone Diagram Examples
Example: High Scrap Rate Problem
Root Cause Categories
Man (People)
- • Insufficient training
- • Operator fatigue
- • New employees
Machine
- • Worn cutting tools
- • Machine vibration
- • Calibration drift
Material
- • Supplier quality issues
- • Material hardness variation
- • Storage conditions
Analysis Results
Problem: Scrap rate increased from 2% to 8%
Primary cause: Worn cutting tools (40% contribution)
Secondary cause: New operator training (30%)
Actions: Implement preventive tool maintenance + enhanced training program
Control Chart Examples
Example: Automotive Parts Diameter Control
Process Data
- • Target diameter: 50.0 mm
- • Tolerance: ±0.1 mm
- • Sample size: 5 parts/hour
- • Process average: 49.98 mm
- • Average range: 0.06 mm
Control Limits
- • UCL (X-bar): 50.01 mm
- • LCL (X-bar): 49.95 mm
- • UCL (R): 0.13 mm
Process Analysis
Process Status: In statistical control
Capability: Cpk = 1.67 (Very capable)
Defect Rate: < 0.1%
Result: Process meets all quality requirements
Process Capability Examples
Example: Chemical Process Temperature Control
Process Parameters
- • Specification: 200°C ± 5°C
- • USL: 205°C
- • LSL: 195°C
- • Process mean: 201°C
- • Process std dev: 1.2°C
Capability Calculations
Cp: (205-195)/(6×1.2) = 1.39
Cpk: min[(205-201)/(3×1.2), (201-195)/(3×1.2)]
= min[1.11, 1.67] = 1.11
Result: Process is capable but slightly off-center
Industry Benchmarks
Typical Values by Industry
Manufacturing
MTBF: 200-2000 hours | MTTR: 1-8 hours | OEE: 60-85%
IT Systems
MTBF: 8000+ hours | MTTR: 0.5-4 hours | Availability: 99.9%+
Automotive
MTBF: 500-5000 hours | MTTR: 0.5-2 hours | OEE: 70-90%
Improvement Targets
World Class MTTR
< 1 hour for critical equipment
World Class Availability
99.9% (8.77 hours downtime/year)
World Class OEE
≥ 85% overall effectiveness