The Central Idea
A system is reliable when it continues to function as required.
If
\[ R=P(\text{system works}), \]then the failure probability is
\[ \boxed{Q=1-R}. \]Reliability and failure probability are complements.
Reliability
Reliability is the probability that an item, component, or system performs successfully under defined conditions for a specified period.
Reliability is therefore always tied to:
- a function;
- a time period;
- operating conditions;
- a definition of success.
Failure Probability
If reliability is
\[ R, \]then failure probability is
\[ \boxed{Q=1-R}. \]Likewise,
\[ \boxed{R=1-Q}. \]Example: Component Reliability
Suppose a component has reliability
\[ R=0.97. \]Then its failure probability is
\[ Q=1-0.97. \] \[ \boxed{Q=0.03}. \]Reliability as a Function of Time
Reliability often changes with time.
We write
\[ \boxed{R(t)=P(T>t)}, \]where \(T\) is the time to failure.
This is also called the survival function.
Failure Distribution
The cumulative failure probability by time \(t\) is
\[ F(t)=P(T\le t). \]Since success beyond time \(t\) and failure by time \(t\) are complementary,
\[ \boxed{R(t)=1-F(t)}. \]Example: Failure by a Deadline
Suppose
\[ F(100)=0.08. \]Then the probability of surviving beyond time \(100\) is
\[ R(100)=1-0.08 \] \[ \boxed{0.92}. \]Series Systems
In a series system, every required component must work for the system to work.
If independent components have reliabilities
\[ R_1,R_2,\ldots,R_n, \]then
\[ \boxed{ R_{\text{series}} = \prod_{i=1}^{n}R_i }. \]Why Series Reliability Multiplies
The system succeeds only if
\[ C_1\cap C_2\cap\cdots\cap C_n \]occurs.
Under independence,
\[ P(C_1\cap\cdots\cap C_n) = P(C_1)\cdots P(C_n). \]Example: Two Components in Series
Suppose
\[ R_1=0.95 \]and
\[ R_2=0.90. \]Then
\[ R_{\text{series}} =0.95(0.90) \] \[ \boxed{0.855}. \]Series Systems Become Less Reliable as Required Components Accumulate
If every component has reliability below \(1\), multiplying several reliabilities generally produces a smaller system reliability.
A long chain of individually reliable components can therefore produce a noticeably less reliable overall system.
Parallel Systems
In a simple parallel system, the system works as long as at least one component works.
The system fails only if every parallel component fails.
Parallel Reliability
If independent components have failure probabilities
\[ Q_i=1-R_i, \]then
\[ P(\text{all fail}) = \prod_{i=1}^{n}Q_i. \]Therefore,
\[ \boxed{ R_{\text{parallel}} = 1-\prod_{i=1}^{n}(1-R_i) }. \]Example: Two Components in Parallel
Suppose both components have reliability
\[ 0.90. \]Each has failure probability
\[ 0.10. \]Both fail with probability
\[ 0.10(0.10)=0.01. \]Therefore,
\[ R_{\text{parallel}} =1-0.01 \] \[ \boxed{0.99}. \]Redundancy
Redundancy means providing additional components or paths so that one failure does not necessarily cause total system failure.
Parallel structure is one common form of redundancy.
Redundancy Can Increase Reliability
Compare a single component with reliability \(0.9\) to two independent parallel components, each with reliability \(0.9\).
Single component:
\[ R=0.9. \]Parallel pair:
\[ R=1-(0.1)^2 \] \[ =\boxed{0.99}. \]Redundancy Has Costs
Redundant components may increase:
- purchase cost;
- maintenance;
- weight;
- energy use;
- complexity;
- space requirements.
Reliability improvement must therefore be compared with implementation cost.
Mixed Series-Parallel Systems
Real systems often contain both series and parallel structures.
A useful strategy is to simplify the system one block at a time.
- Combine simple parallel groups.
- Combine simple series groups.
- Repeat until one equivalent reliability remains.
Example: Parallel Pair Followed by a Series Component
Suppose components A and B are in parallel, each with reliability \(0.9\). Their block reliability is
\[ R_{AB}=1-(0.1)(0.1)=0.99. \]Suppose this block is in series with component C of reliability \(0.95\).
Then
\[ R_{\text{system}} =0.99(0.95) \] \[ \boxed{0.9405}. \]Independence Is a Modeling Assumption
The simple multiplication formulas for series and parallel systems depend on independence.
Components may fail together because of:
- shared power supply;
- common environment;
- shared software;
- common manufacturing defect;
- the same operator error;
- the same external shock.
Common-Cause Failure
A common-cause failure is one event that can disable multiple components at the same time.
Common-cause failures can sharply reduce the benefit expected from redundancy.
Apparent independence should therefore be checked carefully.
Conditional Reliability
Reliability can be updated when new information becomes available.
For example,
\[ P(T>t+s\mid T>t) \]is the probability that an item survives an additional interval \(s\), given that it has already survived to time \(t\).
Memoryless Reliability
The exponential distribution has the memoryless property:
\[ P(T>t+s\mid T>t)=P(T>s). \]Under this model, surviving to the present does not change the modeled probability of surviving an additional interval of the same length.
Constant Failure Rate Model
Under a constant failure rate \(\lambda>0\),
\[ \boxed{ R(t)=e^{-\lambda t} }. \]The cumulative failure probability is
\[ \boxed{ F(t)=1-e^{-\lambda t} }. \]Example: Constant Failure Rate
Suppose
\[ \lambda=0.02 \]per year.
After \(5\) years,
\[ R(5)=e^{-0.02(5)} \] \[ =e^{-0.1} \] \[ \boxed{\approx0.9048}. \]Failure Probability under the Exponential Model
Using the previous example,
\[ F(5)=1-e^{-0.1}. \] \[ \boxed{F(5)\approx0.0952}. \]Mean Time to Failure
For an exponential lifetime model,
\[ \boxed{ E[T]=\frac{1}{\lambda} }. \]This is often called the mean time to failure, or MTTF, for a nonrepairable item.
Example: MTTF
If
\[ \lambda=0.02, \]then
\[ E[T]=\frac{1}{0.02} \] \[ \boxed{50}. \]The units match the time units used for \(\lambda\).
Hazard Rate
The hazard rate describes the instantaneous failure tendency among items that have survived to time \(t\).
Formally,
\[ \boxed{ h(t)=\frac{f(t)}{R(t)} }. \]Under an exponential model,
\[ h(t)=\lambda. \]Failure Rates Need Not Be Constant
Real systems may have:
- high early-life failure;
- a relatively stable middle period;
- increasing failure risk as components age.
A constant hazard model is useful, but it should not be assumed automatically.
Reliability over Repeated Independent Demands
Suppose a component succeeds independently on each use with probability
\[ r. \]The probability it succeeds on all \(n\) uses is
\[ \boxed{r^n}. \]Example: Repeated Uses
If success probability per use is
\[ 0.99 \]and the component must succeed on \(10\) independent uses, then
\[ P(\text{all succeed})=(0.99)^{10}. \] \[ \boxed{\approx0.9044}. \]Probability of At Least One Failure
If all \(n\) uses must succeed,
\[ P(\text{at least one failure}) = 1-r^n. \]In the previous example,
\[ 1-(0.99)^{10} \] \[ \boxed{\approx0.0956}. \]k-out-of-n Systems
A \(k\)-out-of-\(n\) system works if at least \(k\) of its \(n\) components work.
Series and parallel systems are special cases:
- series: \(n\)-out-of-\(n\);
- parallel: \(1\)-out-of-\(n\).
Identical Independent Components in a k-out-of-n System
If each component works independently with probability \(r\), then
\[ \boxed{ R = \sum_{j=k}^{n} \binom{n}{j} r^j(1-r)^{n-j} }. \]This is a binomial probability.
Example: Two-out-of-Three System
Suppose three independent components each have reliability
\[ r=0.9. \]The system works if at least two work:
\[ R = \binom{3}{2}(0.9)^2(0.1) + \binom{3}{3}(0.9)^3. \] \[ =3(0.81)(0.1)+0.729. \] \[ =0.243+0.729. \] \[ \boxed{R=0.972}. \]Availability
Reliability and availability are related but different.
Reliability asks whether the system avoids failure over an interval.
Availability asks whether the system is operational at a particular time, allowing for repair and restoration.
Repairable Systems
For repairable systems, failure is not necessarily permanent.
Useful quantities may include:
- failure rate;
- repair rate;
- mean time between failures;
- mean time to repair;
- long-run availability.
Reliability versus Safety
A reliable system is not automatically a safe system.
Reliability concerns whether required functions continue to operate.
Safety concerns whether unacceptable harm is avoided.
A system can be highly reliable at performing a function that is itself unsafe.
Reliability versus Quality
Quality and reliability also differ.
Quality may describe whether a product meets specifications when delivered.
Reliability asks whether performance persists over time and use.
Expected Failure Cost
Reliability can be connected to decision analysis by assigning a loss to failure.
If failure probability is \(Q\) and failure cost is \(C\), a simple expected failure cost is
\[ \boxed{E[L]=QC}. \]Example: Expected Failure Cost
Suppose
\[ Q=0.02 \]and a failure costs
\[ \$500{,}000. \]Then
\[ E[L]=0.02(500{,}000) \] \[ \boxed{\$10{,}000}. \]Reliability Improvement as a Decision
Suppose an upgrade increases reliability from \(0.96\) to \(0.99\).
Failure probability falls from
\[ 0.04 \]to
\[ 0.01. \]Whether the upgrade is worthwhile depends on the reduction in expected loss relative to the upgrade cost.
Failure Probability Reduction
The absolute reduction in failure probability is
\[ 0.04-0.01 \] \[ =\boxed{0.03}. \]The relative reduction is
\[ \frac{0.04-0.01}{0.04} \] \[ =0.75. \]This is a
\[ \boxed{75\%\text{ relative reduction}}. \]Reliability Allocation
In system design, a target reliability may be allocated across components.
For a simple independent series system with equal component reliabilities \(r\),
\[ r^n=R_{\text{target}}. \]Therefore,
\[ \boxed{ r=R_{\text{target}}^{1/n} }. \]Example: Required Component Reliability
A two-component series system must have reliability
\[ 0.96. \]If both components have equal reliability \(r\),
\[ r^2=0.96. \]Therefore,
\[ r=\sqrt{0.96} \] \[ \boxed{\approx0.9799}. \]Testing Reliability
Reliability is often estimated from observed successes and failures.
If \(x\) successes occur in \(n\) comparable trials, a simple empirical estimate is
\[ \hat R=\frac{x}{n}. \]The estimate itself is uncertain and should not be treated as exact.
Example: Empirical Reliability
Suppose \(985\) of \(1000\) units successfully complete a test.
Then
\[ \hat R=\frac{985}{1000} \] \[ \boxed{0.985}. \]Reliability Data Can Be Censored
In lifetime studies, some items may still be functioning when observation ends.
Their exact failure times are unknown.
Such observations are called censored, and specialized survival-analysis methods account for them.
Reliability and Sensitivity Analysis
Reliability models can be sensitive to:
- component failure probabilities;
- assumed independence;
- repair times;
- common-cause events;
- usage intensity;
- environmental conditions.
Sensitivity analysis helps identify which assumptions matter most.
Reliability and Value of Information
Additional inspection or testing may reduce uncertainty about system reliability.
The value of that information depends on whether it changes maintenance, replacement, redesign, or operating decisions.
Common Error: Adding Reliabilities in Series
For independent required components in series, reliability is multiplied, not added.
\[ R_{\text{series}}=R_1R_2. \]Common Error: Multiplying Reliabilities in Parallel
In a simple parallel system, the system fails only if all components fail.
Therefore,
\[ R_{\text{parallel}} = 1-(1-R_1)(1-R_2), \]not \(R_1R_2\).
Common Error: Assuming Independence Automatically
Multiplication rules require an independence assumption.
Shared dependencies can make the true system reliability materially lower than an independence model suggests.
Common Error: Confusing Reliability with Failure Rate
Reliability \(R(t)\) is a probability of surviving beyond time \(t\).
A failure or hazard rate describes failure intensity over time.
These quantities are related but are not interchangeable.
Common Error: Treating MTTF as a Guaranteed Lifetime
Mean time to failure is an average under a model.
It does not mean each unit will last exactly that long.
Common Error: Assuming Redundancy Eliminates Failure
Redundancy can greatly reduce failure probability, but common-cause events and shared dependencies may still cause complete failure.
Redundancy reduces risk; it does not automatically make risk zero.
A Reliable Procedure for Simple Reliability Problems
- Define what counts as system success.
- Identify component reliabilities or failure probabilities.
- Determine whether the structure is series, parallel, or mixed.
- Check whether independence is a reasonable assumption.
- Combine series components by multiplication.
- Combine parallel components through complementary failure probabilities.
- Simplify mixed systems one block at a time.
- Convert between reliability and failure probability as needed.
- Interpret the result in the system context.
A Reliable Procedure for Time-Based Reliability
- Define the lifetime variable \(T\).
- Choose or estimate an appropriate failure-time model.
- Compute \(R(t)=P(T>t)\).
- Compute \(F(t)=1-R(t)\) when failure probability is needed.
- Evaluate hazard or mean lifetime when relevant.
- Check whether model assumptions such as constant failure rate are reasonable.
- Use sensitivity analysis when key parameters are uncertain.
Practice 1: Complement
A component has reliability
\[ 0.94. \]Find its failure probability.
Practice 2: Two Components in Series
Two independent components have reliabilities
\[ 0.90 \]and
\[ 0.95. \]Find the series-system reliability.
Practice 3: Two Components in Parallel
Two independent components each have reliability
\[ 0.80. \]Find the parallel-system reliability.
Practice 4: Mixed System
Two independent components with reliability \(0.9\) are placed in parallel. That block is in series with a component of reliability \(0.8\).
Find the total system reliability.
Practice 5: Exponential Reliability
Suppose
\[ R(t)=e^{-0.05t}. \]Find \(R(4)\).
Practice 6: MTTF
Under an exponential model,
\[ \lambda=0.04. \]Find the mean time to failure.
Practice 7: Repeated Uses
A device succeeds independently on each use with probability \(0.98\).
Find the probability it succeeds on all \(5\) uses.
Practice 8: At Least One Failure
Using Practice 7, find the probability of at least one failure in the five uses.
Practice 9: Two-out-of-Three System
Three independent components each have reliability \(0.8\).
Find the reliability of a system that works when at least two components work.
Practice 10: Expected Failure Cost
A system has failure probability
\[ 0.015 \]and each failure causes a loss of
\[ \$200{,}000. \]Find the expected failure cost.
Practice Answers
Practice 1
\[ Q=1-0.94 \] \[ \boxed{0.06}. \]Practice 2
\[ R=0.90(0.95) \] \[ \boxed{0.855}. \]Practice 3
Each component fails with probability \(0.20\).
\[ R=1-(0.20)(0.20) \] \[ =1-0.04 \] \[ \boxed{0.96}. \]Practice 4
Parallel block:
\[ R_p=1-(0.1)^2=0.99. \]Series with the \(0.8\) component:
\[ R=0.99(0.8) \] \[ \boxed{0.792}. \]Practice 5
\[ R(4)=e^{-0.05(4)} \] \[ =e^{-0.2} \] \[ \boxed{\approx0.8187}. \]Practice 6
\[ E[T]=\frac{1}{0.04} \] \[ \boxed{25}. \]Practice 7
\[ (0.98)^5 \] \[ \boxed{\approx0.9039}. \]Practice 8
\[ 1-(0.98)^5 \] \[ \boxed{\approx0.0961}. \]Practice 9
\[ R = \binom32(0.8)^2(0.2) + \binom33(0.8)^3. \] \[ =3(0.64)(0.2)+0.512. \] \[ =0.384+0.512. \] \[ \boxed{0.896}. \]Practice 10
\[ E[L]=0.015(200{,}000) \] \[ \boxed{\$3{,}000}. \]What to Remember
Reliability converts system success and failure into probability models that can be analyzed and improved.
- Reliability and failure probability are complements.
- Time-based reliability is a survival probability.
- Independent series reliabilities multiply.
- Independent parallel reliability is one minus the probability that all components fail.
- Redundancy can improve reliability.
- Common-cause failures can undermine redundancy.
- The exponential model gives \(R(t)=e^{-\lambda t}\) under constant hazard.
- MTTF is an average lifetime, not a guaranteed lifetime.
- \(k\)-out-of-\(n\) systems connect reliability to binomial probability.
- Reliability should be evaluated together with failure consequences and system costs.
The most important modeling principle is: system reliability depends not only on component reliability, but also on how the components are connected and how their failures depend on one another.
Where We Go Next
Reliability analysis studies uncertain success and failure within systems. The next lesson moves from system components to multiple decision makers, where each participant's outcome can depend on what others choose.
The next lesson, Games and Strategic Decisions, develops payoff matrices, dominant strategies, best responses, equilibrium, and the mathematics of interdependent choices.