Operations Management, Maintenance and Reliability

Importance of Maintenance and Reliability

• Maintenance helps achieve reliability in a system.
• Pickering Nuclear Generation Station:
  • Supplies 13% of Ontario's electricity.
  • Requires major inspection every ten years.
  • Involves high costs and additional staff.
  • Preventive maintenance is crucial due to safety and potential electricity cutoff risks.
• Consequences of not conducting preventive maintenance:
  • Sudden breakdowns requiring costly repairs.
  • Impact on consumers (e.g., Canada's Wonderland needing to provide safe rides).
  • Impact on production units (e.g., Redpath Sugar Refinery aiming for high utilization).

Objectives of Maintenance and Reliability

• Maintaining the capability of the system.
• Good maintenance improves reliability by removing variability.
• Scheduled maintenance helps identify and resolve problems, increasing system reliability.

Definition of Maintenance

• Activities to keep equipment in working order, such as inspection and replacement of parts.
• Example: Home furnace maintenance before winter.

Definition of Reliability

• Probability that a machine part or product will function properly for a specified time period under stated conditions.
• Maintenance in a production company (e.g., Redpath Sugar Refinery):
  • Helps prevent breakdowns and equipment failure.
  • Minimizes production loss from failures.
  • Reduces the frequency of failures.
  • Lowers interruptions of the production line.
  • Avoids missing delivery deadlines.
  • Increases customer satisfaction.
  • Results in higher profit.
  • Extends the useful life of machinery and equipment.
  • Increases facility utilization.

Types of Maintenance

• Determined through cost-benefit analysis.
  • Consider the costs of preventive maintenance versus breakdown costs.

1. Corrective or Breakdown Maintenance

• Let the system run until it breaks down, then repair it.
• Justifiable when failure doesn't result in huge loss.
• Suitable for small factories where downtime isn't critical.

2. Scheduled Maintenance

• Frequent maintenance and inspection to ensure the system never fails.
• Example: Pickering Nuclear Station's maintenance plan.
• Addresses the issues of sudden failures:
  • Higher costs due to unpreparedness.
  • Lower quality of repair.
  • More frequent disruptions.
  • Excessive delay and lower utilization.
  • More spoilt material due to unfinished operations.
  • Need for recalibration after disruption.

3. Preventive Maintenance

• Scheduled at regular intervals, regardless of whether a problem is apparent.
• Example: Annual car service.

4. Scheduled Maintenance (as needed)

• Done based on asset conditions.
• Example: Scheduling a car maintenance upon hearing a strange noise.

Cost Benefit Trade-off in Maintenance

1. Costs of Doing Maintenance

• Preventive costs:
  • Training staff to avoid mistakes.
  • Running process control efforts.
• Detection costs:
  • Hiring experts to inspect the process.
  • Finding the source of the problem.
  • Resolving the problem.

2. Costs of Not Doing Maintenance

• Loss of output and profit due to sudden disruption.
• Costs related to expedited replacement and repair.
• Missed deadlines and expedited shipping.
• Costs related to redoing operations or discarding damaged parts.

3. Cost Analysis

• $Cost$ is on the Y axis.
• $Maintenance Commitment$ is on the X axis.
• Breakdown cost decreases with higher maintenance.
• Preventive maintenance cost increases with higher maintenance.
• Optimal point is where the summation of breakdown cost and preventive maintenance cost is minimized.

4. Example: XYZ Company

• Company rolls seal plates and experiences breakdowns.
• Data collected over 20 months shows different breakdown rates.
• Each breakdown costs $300.
• Preventive maintenance costs $150/month and reduces breakdowns to an average of one per month.
• Analysis:
  • Calculate expected number of breakdowns:

$Expected \space Number \space of \space Breakdowns = \sum (Number \space of \space Breakdowns \times Frequency)$

*   Expected Breakdown cost per month:

1.6 \times $300 = $480

• Cost of preventative maintenance is $450 per month.
• Conclusion: XYZ should contract preventive maintenance service.

Failure Rate and Mean Time Between Failures (MTBF)

1. Failure Rate

• Frequency with which a system or component fails.
• Function of time and life cycle.

2. Mean Time Between Failures (MTBF)

• Measures reliability.
• Longer MTBF = lower failure rate = higher reliability.
• $failure\space rate \space percentage ={\frac{number \space of \space failures}{number \space of \space all \space units \space tested}}*100$
• $MTBF={\frac{1}{failure\space rate}}$

3. Example: Air Conditioning Systems

• 20 AC systems tested for 1000 hours.
• Two systems failed (one at 200 hours, one at 600 hours).
• Calculations:

$Failure \space Rate \space$

$Failure \space Rate \space (per \space operating \space hour) = {\frac{2}{(18 \times 1000) + 200 + 600}} = 0.000106$

$MTBF = {\frac{1}{0.000106}} = 9434 \space hours$

• Expected failures during a 6-day space shuttle trip:

$0.000106 \times 24 \times 6 = 0.0153 \space failures$

Series System Reliability

• All components must function for the system to work.
• System fails if any component fails.
• Example: Extension cords connected in series to mow a lawn.
• Applications: Computer networks, chains, multi-cell batteries, decorative tree lights.

1. Reliability Calculation

• For a series system with components having reliabilities R1, R2, …, Rn, the system reliability is:

$R<em>{system} = R</em>1 \times R<em>2 \times … \times R</em>n$

• Example:
  • Four components with reliabilities 0.95, 0.85, 0.90, 0.85.
  • $R_{system} = 0.95 \times 0.85 \times 0.90 \times 0.85 = 0.617 \approx 62\%$.

2. Impact of Increasing Components

• As the number of components in a series system increases, the overall reliability declines significantly.
• System reliability cannot be greater than the smallest component reliability.

3. Example: Truck Cab Assembly Line

• Five welding robots (each with 96% reliability) in series.
• $R_{system} = 0.96^5 \approx 0.815 (81.5\%)$.
• To achieve 96% system reliability, each robot's reliability must be:

$R_{robot} = \sqrt[5]{0.96} \approx 0.992 (99.2\%)$.

Increasing System Reliability Through Redundancy

• Adding backup components to ensure system function even if one component fails.
• $System \space Reliability = P(first \space component \space works) + P(second \space component \space works \space if \space first \space doesn't)$
• $R<em>{system} = R</em>A + (1 - R<em>A) \times R</em>B$

1. Example: Bank of Montreal Loan Processing Center

• Three clerks in series with reliabilities 90%, 80%, 99%.
• System reliability: $0.90 \times 0.80 \times 0.99 \approx 0.71 (71\%)$.
• Bank adds redundancy for the two least reliable clerks.

2. Reliability with Redundancy

• Reliability of Clerk 1 (with redundancy): $0.90 + (1 - 0.90) \times 0.90 = 0.99 (99\%)$.
• Reliability of Clerk 2 (with redundancy): $0.80 + (1 - 0.80) \times 0.80 = 0.96 (96\%)$.
• New system reliability: $0.99 \times 0.96 \times 0.99 \approx 0.94 (94\%)$.

Parallel System Reliability

• System works as long as at least one component works.
• Applications: Jet engines, braking systems, projector light bulbs.
• Reliability Calculation:

$R_{parallel} = 1 - P(none \space of \space the \space components \space work)$

$R<em>{parallel} = 1 - ((1 - R</em>1) \times (1 - R<em>2) \times (1 - R</em>3) \times (1 - R_4))$

1. Example System

• Four components in parallel with reliabilities 95%, 85%, 90%, 85%.
• $R_{parallel} = 1 - ((1 - 0.95) \times (1 - 0.85) \times (1 - 0.90) \times (1 - 0.85))$
• $R_{parallel} = 1 - (0.05 \times 0.15 \times 0.10 \times 0.15) \approx 0.9998 (99.98\%)$.

Combining Series and Parallel Systems

1. Example: Medical Control System

• Three components in series with reliabilities 99%, 98%, 90%.
• System reliability: $0.99 \times 0.98 \times 0.90 \approx 0.873 (87.3\%)$.
• System changes to a redundant parallel configuration.
• New Configuration Analysis:
  • Subsystem S1 (series): $R_{S1} = 0.99 \times 0.98 \times 0.90 \approx 0.873$.
  • Subsystem S2 (series): $R_{S2} = 0.99 \times 0.98 \times 0.90 \approx 0.873$.

2. Parallel Redundancy

• The new system has S1 and S2 in parallel:

$R<em>{system} = 1 - ((1 - R</em>{S1}) \times (1 - R_{S2}))$

$R_{system} = 1 - ((1 - 0.873) \times (1 - 0.873)) \approx 0.984 (98.4\%)$.

• Redundancy improves reliability by more than 11% (from 87.3% to 98.4%).