Component and System Reliability Modelling

Introduction

Reliability engineering asks a practical question: what is the probability that an item will perform its required function, for a specified time, under stated operating conditions? For a component, the answer may be obtained from test or field data. For a system containing many interacting components, reliability must also reflect the system architecture, redundancy, dependencies, maintenance policy, operating environment and failure consequences.

System reliability modelling provides a structured way to transform component behaviour into system-level performance measures. The principal tools introduced in this chapter are reliability block diagrams (RBDs), fault-tree analysis (FTA), event-tree analysis (ETA), and Markov models. These tools are related but answer different questions. An RBD generally models success paths; an FTA models combinations of failures leading to a defined top event; an ETA models the consequences of an initiating event; and a Markov model represents transitions among time-dependent operating and failed states.

The models are useful in mechanical engineering for pumps, compressors, powertrains, aircraft systems, production lines, safety systems and maintainable industrial assets. Their results support design decisions, redundancy selection, maintenance planning, safety justification and reliability allocation.

Learning objectives

After completing this chapter, the reader should be able to:

1. Define component and system reliability and distinguish series, parallel and standby structures.
2. Construct and evaluate reliability block diagrams.
3. Develop fault trees and event trees and identify minimal cut and path sets.
4. Formulate simple Markov models for repairable and standby systems.
5. Account for redundancy degradation, common-cause failure and imperfect switching.
6. Allocate a system reliability target among components.
7. Critically interpret model assumptions and communicate engineering conclusions.

1. Reliability foundations for systems

Reliability is denoted by R(t) and is the probability that an item performs its required function throughout the interval from time 0 to time t. The item, mission, environment and success criterion must be specified. If T is the random time to failure, then

where F(t) is the cumulative distribution function of failure time. The associated probability density is f(t) = dF(t)/dt, when it exists. The hazard rate is

with units of inverse time. It describes the instantaneous failure rate conditional on survival to time t. Reliability and hazard are not interchangeable: a constant hazard is an assumption, not a universal property of components.

For a constant failure rate λ, the exponential model gives

This model is appropriate mainly during a useful-life period in which failures occur randomly and the hazard is approximately constant. It is less suitable for dominant wear-out mechanisms such as fatigue, corrosion or bearing degradation. For a Weibull model with shape parameter β and scale parameter η,

where t and η have the same time units. β < 1 indicates decreasing hazard, β = 1 gives the exponential case, and β > 1 indicates increasing hazard.

A system model must define:

• the required function and mission duration;
• whether components are independent or dependent;
• whether failures are detected immediately;
• whether repair or replacement occurs during the mission;
• the effect of switching, load sharing and common stresses; and
• whether success means full performance or an acceptable degraded performance.

A mathematically precise model can still be operationally wrong if these definitions are unclear.

2. Reliability block diagrams

An RBD is a success-oriented representation. Blocks represent components or functional subsystems, and the arrangement specifies the combinations required for system success. A block is not necessarily a physical object; it may represent a function delivered by several physical parts.

2.1 Series systems

In a series system every component must operate. If component failures are statistically independent,

where R_i(t) is the reliability of component i. For two components, R_s = R₁R₂. If each component has the same reliability R_c, then R_s = R_cⁿ.

For independent exponential components,

Thus series failure rates add under the stated assumptions. This result is often used for preliminary allocation, but it should not be applied to dependent components without modification.

2.2 Parallel systems

A parallel system succeeds if at least one component operates. For statistically independent components,

For two identical components, R_p = 1 − (1 − R_c)². Redundancy improves reliability only if the redundant elements are genuinely available, can carry the required load, and are not defeated by a common cause or a shared support system.

2.3 Worked example: series and parallel structures

Two independent pumps each have reliability 0.90 over a mission. The delivery system requires both a pump and a controller in series. The controller reliability is 0.98.

If only one pump is installed, system reliability is

R_s = 0.90 × 0.98 = 0.882.

If two pumps are installed in active parallel and either can deliver the required flow, pump-subsystem reliability is

R_p = 1 − (1 − 0.90)² = 0.99.

The complete system reliability is therefore

R_system = 0.99 × 0.98 = 0.9702.

The improvement is substantial, but it depends on independent pump failures and a controller that can support either pump. A shared power supply, common suction blockage or incorrect switching could reduce the actual value.

2.4 Mixed and bridge structures

Many systems cannot be reduced directly by series and parallel rules. A useful method is decomposition: condition on the state of a selected component, calculate the reliability of each conditional subnetwork, and combine the results using the law of total probability. Another method is enumeration of minimal path sets or use of Boolean algebra.

A minimal path set is a smallest set of functioning components sufficient for system success. A minimal cut set is a smallest set of component failures sufficient for system failure. For rare, independent failures, the probability of a top event can sometimes be approximated by summing the probabilities of minimal cut sets, but overlapping cut sets require correction; careless addition can double-count events.

2.5 Standby systems

In a standby system, one unit operates while another is reserved. The standby unit may be cold, warm or hot:

• A cold standby is unpowered or inactive and generally has negligible active failure during standby.
• A warm standby is partially energized and can fail while waiting, usually at a lower rate.
• A hot standby operates or is fully ready and may fail at nearly the active rate.

For two independent, identical units with constant active failure rate λ, perfect failure detection and instantaneous, perfect switching, a cold-standby system has reliability

The first term corresponds to the primary surviving to time t; the second term includes operation of the primary until time u followed by operation of the reserve for the remaining interval. More generally, for non-identical independent lifetimes with densities f₁ and reliabilities R₁ and R₂,

This assumes the reserve does not fail while dormant and switching is perfect. If switching succeeds with probability c, the second term is multiplied by c. Standby models must also consider coverage: the probability that a failure is detected and the reserve is successfully connected.

3. Fault-tree analysis

FTA is a top-down, failure-oriented method. The analyst begins with a defined top event, such as “loss of cooling,” and logically combines lower-level events using gates.

An OR gate represents failure if any input fails. For independent input events,

An AND gate represents failure only when all inputs fail. Under independence,

The AND formula is especially important for redundant systems: two independent failures may be required to produce system failure. Independence must be justified, not assumed merely because components have different names.

For rare events, the OR-gate expression is often approximated by P(A OR B) ≈ P(A) + P(B), but this approximation is accurate only when the probabilities are small and overlap is negligible.

3.1 Building a fault tree

A disciplined FTA proceeds as follows:

1. Define the top event, operating state, mission time and boundary of analysis.
2. Identify immediate causes using AND and OR logic.
3. Continue decomposition until basic events can be quantified or controlled.
4. Check that events are mutually meaningful and not duplicated under different names.
5. Identify minimal cut sets.
6. Quantify the tree using exact Boolean methods, approximations or simulation.
7. Test sensitivity to dominant basic events and common-cause assumptions.

FTA can expose design weaknesses that a simple RBD conceals, particularly shared power, control, environment or maintenance errors. It can also represent human actions and external events, although their probabilities require careful evidence.

3.2 Common-cause failures

A common-cause failure (CCF) is a single cause that defeats two or more nominally redundant elements. Examples include a shared electrical bus, a common lubrication system, a fire, a software defect, an environmental excursion or a maintenance error.

For two redundant channels, a simple illustrative model separates independent channel failures and a CCF event. If each channel has reliability R_c and the CCF probability is q, one approximate model is

Here the factor (1 − q) represents survival of the common-cause challenge. This is a simplified model: in real analyses the CCF event may overlap with individual failures, vary with time, and have several severity classes. CCF modelling should therefore be based on explicit causal analysis, not an arbitrary “redundancy discount.” Physical separation, diverse designs, independent utilities, testing and controlled maintenance are common risk controls.

4. Event-tree analysis

An event tree begins with an initiating event and follows the success or failure of subsequent safety functions. Each branch represents a conditional event, and each end state has a probability obtained by multiplying conditional branch probabilities along that path and summing mutually exclusive paths leading to the same outcome.

For example, suppose an initiating pump failure has probability 0.02 during a mission. An emergency pump succeeds with conditional probability 0.95, and if it fails, an operator recovery action succeeds with conditional probability 0.80. The probability of loss of cooling, assuming the operator action is attempted only after emergency-pump failure, is

P(loss) = 0.02 × (1 − 0.95) × (1 − 0.80) = 0.0002.

The conditional probabilities must reflect the initiating condition. They are not necessarily equal to stand-alone component reliabilities. Event trees are particularly useful for accident progression, protective systems and consequence analysis. They complement fault trees: an FTA asks how an unwanted event can occur, whereas an ETA asks what can happen after an initiating event.

5. Markov reliability models

RBDs and static fault trees are often insufficient when state changes over time matter. Markov models represent mutually exclusive states and transitions between them. The Markov property requires that future behaviour depends on the current state, not on the detailed history, usually because transition times are exponential or the state has been expanded sufficiently to retain relevant history.

Let p_i(t) be the probability of being in state i at time t, and let Q be the transition-rate matrix. With a row-vector convention,

The diagonal element q_ii is the negative sum of rates leaving state i, and each off-diagonal element q_ij is the transition rate from state i to state j. State probabilities satisfy ∑ p_i(t) = 1. System reliability at time t is the sum of probabilities of all states classified as successful.

5.1 Repairable two-state component

For an item that transitions from up to down at failure rate λ and from down to up at repair rate μ,

If the item starts operating, its availability at time t is

The steady-state availability is A_∞ = μ/(λ + μ), or, equivalently, MTTR/(MTBF + MTTR) when constant-rate assumptions apply and MTBF = 1/λ, MTTR = 1/μ. Availability is not the same as mission reliability: availability permits repair and later restoration, whereas reliability usually concerns failure-free operation during a defined mission.

5.2 Standby and repair models

A two-unit repairable standby system may contain states such as: both units available, one operating and one standby, one operating with the other failed, and both failed. Transitions include active failure, standby failure, switching, repair and restoration. The state diagram must specify whether repair capacity is one or multiple technicians, whether a failed unit is detected immediately, and whether the system can operate in degraded mode.

Markov models are powerful for repairable systems, load sharing and imperfect coverage, but state-space size grows rapidly. Non-exponential failure distributions require supplementary variables, phase-type approximations, semi-Markov methods or other techniques. A Markov result is only as credible as its state definitions and transition rates.

6. Reliability allocation

Reliability allocation distributes a system target among components so that design teams have measurable objectives. For a series system with independent component reliabilities R_i,

If equal allocation is appropriate and the target is R_T for n statistically similar components,

For example, a five-component series system requiring reliability 0.95 must give each identical component a reliability of approximately 0.95^(1/5) ≈ 0.9898. This is a starting allocation, not proof of feasibility.

For unequal components, allocation may account for complexity, stress, maturity, cost and improvement potential. A useful sensitivity measure for a series system is the reliability importance of component i, often represented by the partial derivative

This indicates how strongly a small change in R_i affects system reliability, holding other reliabilities fixed. Importance measures should be interpreted with failure probability, cost and feasibility; improving a mathematically sensitive component may be impractical or less valuable than eliminating a common cause.

7. Integrated illustrative case: cooling-water supply

Background

Consider an illustrative industrial compressor whose bearing system requires cooling-water flow for a 24-hour operating mission. The design includes two independent pumps, either of which can provide the required flow, and a common controller. Each pump has mission reliability 0.96 and the controller has reliability 0.99.

Problem

Determine the idealized system reliability, then examine the effect of a common-cause failure that defeats both pumps with probability 0.015. Finally, identify design and governance actions needed before accepting the result.

Analysis and solution

Ignoring common cause, the pump subsystem reliability is

R_p = 1 − (1 − 0.96)² = 0.9984.

Including the controller,

R_system,ideal = 0.9984 × 0.99 = 0.988416.

If the common-cause event is independent of the residual pump failures and defeats the pump subsystem, the approximate reliability becomes

R_system,CCF = (1 − 0.015) × 0.9984 × 0.99 = 0.97359.

Thus a seemingly small common-cause probability reduces reliability by about 0.0148 absolute probability points. The calculation is illustrative, not a validated safety claim. Engineers must check whether the controller, power supply, suction line, cooling-water reservoir and maintenance procedures are shared. They must also verify that the pumps can deliver flow under actual pressure, that failure detection is timely, and that the standby or parallel arrangement does not overload a surviving pump.

Approach and outcome

The appropriate engineering approach is to combine an RBD for the nominal architecture with an FTA containing shared utilities and common environmental causes. An ETA can then describe consequences following loss of cooling, including alarm success, operator response and emergency shutdown. If repairs and repeated demands matter, a Markov or dynamic simulation model should replace the static mission model.

Learning points

1. Nominal redundancy is not equivalent to independent protection.
2. Shared support functions may dominate system risk.
3. Reliability calculations require an explicit mission and success definition.
4. Different models should be used together when architecture, causes and consequences require different viewpoints.
5. Model results should be verified through inspection, testing, failure reporting and sensitivity analysis.

8. Practical limitations, governance and best practice

The main limitations of reliability models are uncertain data, unrealistic independence assumptions, incomplete failure definitions, neglected human actions, changing operating conditions and inappropriate treatment of repair. Field data may contain censored observations, inconsistent failure codes or maintenance-induced removals. Engineering judgement is useful but should be documented and subjected to sensitivity analysis.

Reliability and safety models also have governance implications. A model should be traceable: assumptions, data sources, version, analyst, review status and changes should be recorded. Safety-critical decisions should not rely solely on a convenient numerical result. Independent review, configuration control, hazard management and compliance with applicable sector standards are essential. Ethical practice requires communicating uncertainty and avoiding false precision.

Modern digital-twin and condition-monitoring systems can update reliability estimates using sensor data, prognostics and Bayesian methods. Artificial intelligence may help detect degradation patterns, but it does not remove the need for physically meaningful failure modes, representative data, validation, cybersecurity and human oversight. A predictive model that is accurate in one operating regime may fail when load, environment or maintenance practice changes.

Best practice is to:

• define function, mission and failure criteria before selecting a method;
• use RBDs for success architecture, FTA for failure causation, ETA for consequences and Markov models for state transitions;
• distinguish reliability, availability, maintainability and safety;
• model common causes, imperfect switching and shared utilities explicitly;
• use exact probability expressions where practical and label approximations;
• validate assumptions against test, inspection and field evidence;
• perform sensitivity and uncertainty analysis; and
• connect model outputs to design actions, inspection plans and measurable requirements.

Chapter Summary

System reliability modelling converts component behaviour and system architecture into quantitative estimates of successful operation. Series systems require every component to function, while parallel systems succeed when at least one independent path remains available. Standby systems require additional treatment of dormancy, detection, switching and coverage. Reliability block diagrams are success-oriented; fault trees analyse combinations of failures; event trees follow consequences from initiating events; and Markov models represent transitions among operating, failed and repaired states.

The basic series and parallel formulas depend strongly on independence and clearly defined success criteria. Common-cause failures can substantially reduce the value of redundancy and should be represented explicitly. Reliability allocation distributes system targets among components, but allocation must be balanced with cost, feasibility, failure physics and importance. Professional analysis requires transparent assumptions, reliable data, sensitivity studies, validation and careful communication of uncertainty.

Key Terms

Reliability: Probability that an item performs its required function for a specified time under stated conditions.

Mission time: Duration over which successful operation is evaluated.

Failure rate: Instantaneous conditional rate of failure, usually denoted by λ(t).

Hazard function: Failure rate conditional on survival to a specified time.

Series system: System in which every required element must operate.

Parallel system: System that succeeds when at least one required path operates.

Standby system: System in which a reserve becomes active after failure of an operating unit.

Cold standby: Reserve that is inactive and assumed not to fail while dormant.

Coverage: Probability that a failure is detected and the intended protective or switching action succeeds.

Reliability block diagram: Success-oriented representation of functional system architecture.

Fault-tree analysis: Top-down analysis of causes leading to a defined failure event.

Event-tree analysis: Forward analysis of possible consequences following an initiating event.

Minimal cut set: Smallest combination of failures sufficient to cause system failure.

Minimal path set: Smallest combination of functioning elements sufficient for system success.

Common-cause failure: One cause that defeats multiple nominally redundant elements.

Markov model: State-transition model satisfying the Markov property.

Availability: Probability that an item is operating or ready when required, including repair effects.

Reliability allocation: Distribution of a system reliability requirement among subsystems or components.

Transition-rate matrix: Matrix containing rates of movement between Markov states.

Review Questions

Short Answer Questions

1. Define reliability and explain why mission time must be specified.
2. State the independent-component reliability formulas for a series and a parallel system.
3. Distinguish a cold-standby system from a hot-standby system.
4. What is a common-cause failure, and why does it threaten redundancy?
5. Explain one important difference between reliability and availability.

Long Answer Questions

1. Explain how to construct and quantify a fault tree, including OR gates, AND gates and minimal cut sets.
2. Compare RBDs, fault trees, event trees and Markov models in terms of purpose, assumptions and suitable applications.
3. Derive the reliability expression for two identical units in cold standby under perfect switching and constant failure rate.
4. Discuss the assumptions behind reliability allocation in an independent series system and explain why equal allocation may be inappropriate.
5. Describe how common-cause failures, imperfect switching and shared utilities should be incorporated into a redundant mechanical system model.

Analytical / Application Questions

1. Three independent components have mission reliabilities 0.98, 0.97 and 0.99 in series. Calculate system reliability and identify which component has the greatest direct reliability sensitivity.
2. A two-pump parallel system has individual reliability 0.95, but a common-cause event defeats both pumps with probability 0.02. Using the simplified independent CCF model in this chapter, calculate system reliability before and after the CCF adjustment.
3. Design an event-tree structure for a loss-of-cooling initiating event that includes emergency-pump success, alarm success and operator recovery. State the conditional probabilities that would be required to quantify each end state.

Multiple Choice Questions with Answers

Question 1

For two independent components in series, each with reliability 0.95 over the mission, what is the system reliability?

A. 0.9025

B. 0.9500

C. 0.9750

D. 1.9025

Correct Answer: A. 0.9025

Question 2

A parallel system succeeds when:

A. All components fail simultaneously

B. At least one required component operates

C. Every component operates

D. The repair rate exceeds the failure rate

Correct Answer: B. At least one required component operates

Question 3

Which model is primarily success-oriented?

A. Reliability block diagram

B. Event tree

C. Fault tree

D. Failure-mode table

Correct Answer: A. Reliability block diagram

Question 4

For independent events A and B, the probability of an AND combination is:

A. P(A) + P(B)

B. 1 − P(A)P(B)

C. P(A)P(B)

D. P(A)/P(B)

Correct Answer: C. P(A)P(B)

Question 5

A common-cause failure is best described as:

A. A failure caused only by normal wear of one component

B. A single cause that defeats multiple redundant elements

C. A repair completed before failure occurs

D. A failure that occurs only in a cold standby unit

Correct Answer: B. A single cause that defeats multiple redundant elements

Question 6

For a constant failure rate λ, component reliability at time t is:

A. λt

B. e^(−λt)

C. 1 − e^(−λt)

D. e^(λ/t)

Correct Answer: B. e^(−λt)

Question 7

Which method is most suitable for modelling transitions among operating, failed and repaired states?

A. Markov model

B. Static series RBD only

C. Pareto chart

D. Material balance

Correct Answer: A. Markov model

Question 8

Two identical independent components each have reliability 0.90 in active parallel. Their subsystem reliability is:

A. 0.81

B. 0.90

C. 0.99

D. 1.80

Correct Answer: C. 0.99

Question 9

An event tree normally begins with:

A. A reliability allocation target

B. An initiating event

C. A repair invoice

D. A component material specification

Correct Answer: B. An initiating event

Question 10

For five identical independent components in series with system target reliability 0.95, equal allocation requires each component reliability approximately:

A. 0.9500

B. 0.9700

C. 0.9898

D. 0.9999

Correct Answer: C. 0.9898