Series and Parallel System Reliability

How Component Lifetimes Determine System Reliability

A system’s lifetime depends not only on the lifetimes of its components but also on how those components are connected. If every component must work for the system to operate, the first component failure ends the system’s life. If the system works whenever at least one component works, it fails only when the last working component fails. These are the basic series and parallel configurations.

Let $T_i$ be the lifetime of component $i$, measured from a common time origin. Assume throughout that component lifetimes are independent. Independence lets us multiply component survival probabilities or failure probabilities; the system configuration then determines which product is relevant.

Definition: For $m$ components, a series system has lifetime $T_{\mathrm{ser}}=\min_{1\leq i\leq m}T_i$: it operates only while every component operates. A parallel system has lifetime $T_{\mathrm{par}}=\max_{1\leq i\leq m}T_i$: it operates while at least one component operates. Component $i$'s survival function is $S_i(t)=P(T_i>t)$, and its failure distribution function is $F_i(t)=P(T_i\leq t)=1-S_i(t)$.

Survival and Failure Probabilities

At a mission time $t$, a series system survives precisely when all its components survive past $t$. A parallel system survives precisely when at least one component survives past $t$, or, equivalently, when the components have not all failed by $t$. For independent component lifetimes, these events give the following formulas.

Theorem: For independent component lifetimes and $t\geq 0$, $$ S_{\mathrm{ser}}(t)=\prod_{i=1}^{m}S_i(t), \qquad F_{\mathrm{ser}}(t)=1-\prod_{i=1}^{m}S_i(t), $$ and $$ F_{\mathrm{par}}(t)=\prod_{i=1}^{m}F_i(t), \qquad S_{\mathrm{par}}(t)=1-\prod_{i=1}^{m}F_i(t). $$

For example, a series system has failed by $t$ if at least one component has failed; a parallel system has failed by $t$ only if every component has failed. The formulas use the independence assumption: without it, the probability of a joint event is not generally the product of its component probabilities.

Worked Example: Three Exponential Components in Series

Suppose three independent components have exponential lifetimes with rates $0.2$, $0.3$, and $0.5$ failures per hour. Find the reliability, failure probability, and hazard of the series system at two hours.

For an exponential lifetime with rate $\lambda_i$, the component survival function is $S_i(t)=e^{-\lambda_i t}$. Thus the system survival is $$ S_{\mathrm{ser}}(2) =e^{-0.2(2)}e^{-0.3(2)}e^{-0.5(2)} =e^{-(0.2+0.3+0.5)2} =e^{-2} \approx 0.13534. $$ The system failure probability by two hours is $$ F_{\mathrm{ser}}(2)=1-e^{-2}\approx 1-0.13534=0.86466. $$ The series-system survival at any $t\geq0$ is $e^{-(0.2+0.3+0.5)t}=e^{-t}$. Its density is $e^{-t}$, so its hazard, density divided by survival, is $e^{-t}/e^{-t}=1$ failure per hour. At two hours the density is $e^{-2}\approx0.13534$, and the hazard remains $1$.

System Densities and Hazards

For continuous component lifetimes, densities provide another route to system hazards. Assume each component lifetime is absolutely continuous, with density $f_i$, and write $h_i(t)=f_i(t)/S_i(t)$ wherever $S_i(t)>0$. The system hazard is likewise $h(t)=f(t)/S(t)$ wherever the system survival is positive. The density formulas below hold at times where the component survival functions are differentiable and their densities are defined.

Theorem: For independent absolutely continuous component lifetimes, $$ f_{\mathrm{ser}}(t) =\sum_{i=1}^{m} f_i(t)\prod_{j\ne i}S_j(t), \qquad h_{\mathrm{ser}}(t)=\sum_{i=1}^{m}h_i(t), $$ where the hazard identity applies when all $S_i(t)>0$. For a parallel system, $$ f_{\mathrm{par}}(t) =\sum_{i=1}^{m} f_i(t)\prod_{j\ne i}F_j(t), \qquad h_{\mathrm{par}}(t)= \frac{\sum_{i=1}^{m}f_i(t)\prod_{j\ne i}F_j(t)} {1-\prod_{i=1}^{m}F_i(t)}, $$ where the parallel hazard applies when $S_{\mathrm{par}}(t)>0$.

Proof for a series system. Independence gives $S_{\mathrm{ser}}(t)=\prod_i S_i(t)$. Differentiating this finite product, using $S_i'(t)=-f_i(t)$, gives $$ f_{\mathrm{ser}}(t)=-S_{\mathrm{ser}}'(t) =\sum_{i=1}^{m} f_i(t)\prod_{j\ne i}S_j(t). $$ When every $S_i(t)>0$, divide by $S_{\mathrm{ser}}(t)=\prod_iS_i(t)$. Each summand becomes $f_i(t)/S_i(t)=h_i(t)$; hence $h_{\mathrm{ser}}(t)=\sum_i h_i(t)$. $\square$

Proof for a parallel system. Independence gives $F_{\mathrm{par}}(t)=\prod_i F_i(t)$, since the maximum is at most $t$ exactly when every component lifetime is at most $t$. Differentiating this finite product gives the density of the parallel system: $$ f_{\mathrm{par}}(t)=F_{\mathrm{par}}'(t) =\sum_{i=1}^{m} f_i(t)\prod_{j\ne i}F_j(t). $$ Its survival is $1-\prod_iF_i(t)$. Dividing the density by this survival wherever it is positive yields the stated hazard. Absolute continuity ensures these distributions have densities; for continuous component lifetimes, ties between component failure times have probability zero. $\square$

The series hazard is especially simple: while the system is alive, each component is still a possible cause of system failure, and its instantaneous hazard contributes to the total. For a parallel system, a component’s failure ends the system’s life only if all the other components have already failed. The factors $F_j(t)$ in the density formula express that requirement.

Worked Example: A Parallel System with Exponential and Weibull Components

Let two independent components form a parallel system. Component 1 has an exponential lifetime with rate $1$ per time unit. Component 2 has a Weibull lifetime with shape $2$ and scale $3$, so $S_2(t)=\exp[-(t/3)^2]$. Find system reliability, failure probability, and hazard at $t=2$.

For component 1, $S_1(2)=e^{-1}\approx0.36788$ and $F_1(2)=1-e^{-1}\approx0.63212$. For component 2, $$ S_2(2)=e^{-4/9}\approx0.64118, \qquad F_2(2)=1-e^{-4/9}\approx0.35882. $$ The parallel system fails only if both components have failed, so $$ F_{\mathrm{par}}(2)=F_1(2)F_2(2) =(1-e^{-1})(1-e^{-4/9}) \approx0.22682. $$ Therefore its reliability is $$ S_{\mathrm{par}}(2)=1-F_{\mathrm{par}}(2)\approx0.77318. $$

The component hazards are $h_1(t)=1$ and $h_2(t)=2t/9$. In particular, $h_2(2)=4/9\approx0.44444$, not $8/9$. The component densities at $t=2$ are $$ f_1(2)=h_1(2)S_1(2)=e^{-1}\approx0.36788, \qquad f_2(2)=h_2(2)S_2(2)=\frac{4}{9}e^{-4/9}\approx0.28497. $$ For two parallel components, the density is $f_1F_2+f_2F_1$. Thus $$ f_{\mathrm{par}}(2) \approx(0.36788)(0.35882)+(0.28497)(0.63212) \approx0.31214. $$ Dividing by system survival gives $$ h_{\mathrm{par}}(2) =\frac{f_{\mathrm{par}}(2)}{S_{\mathrm{par}}(2)} \approx\frac{0.31214}{0.77318} \approx0.4037. $$

Combining Series and Parallel Subsystems

Many systems are assembled in stages. When disjoint groups of independent components form subsystems, each subsystem can first be treated as a component with its own lifetime distribution. Independence between the groups then allows the subsystem distributions to be combined using the same formulas.

Worked Example: A Parallel Pair in Series with a Third Component

Components 1 and 2 are in parallel, and that subsystem is in series with component 3. Their independent lifetimes are exponential with rates $1$, $2$, and $0.5$, respectively. Find the system reliability and failure probability at $t=1$.

The parallel subsystem survives if at least one of components 1 and 2 survives. Its reliability is $$ S_{12,\mathrm{par}}(1) =1-(1-e^{-1})(1-e^{-2}) =e^{-1}+e^{-2}-e^{-3} \approx0.36788+0.13534-0.04979 =0.45343. $$ Component 3 survives to time 1 with probability $S_3(1)=e^{-0.5}\approx0.60653$. The final system is a series connection of this component and the parallel subsystem. The groups use disjoint independent component lifetimes, so their lifetimes are independent; consequently, $$ S_{\mathrm{system}}(1) =S_{12,\mathrm{par}}(1)S_3(1) \approx(0.45343)(0.60653) \approx0.27502. $$ The probability that the system has failed by time 1 is $$ F_{\mathrm{system}}(1)=1-S_{\mathrm{system}}(1) \approx1-0.27502=0.72498. $$

Common Mistakes and a Practical Check

Using the wrong event for system failure. A series system fails at the first component failure; a parallel system fails only when every component has failed. Write the system lifetime as a minimum or maximum before calculating.
Multiplying the wrong probabilities. For a series system, multiply component survivals to get system survival. For a parallel system, multiply component failure probabilities to get system failure probability.
Adding hazards in a parallel system. The sum-of-hazards formula applies to a series system under the stated conditions. A parallel-system hazard must account for the other components having failed, as in the density formula.
Confusing a Weibull hazard with its shape parameter. With survival $S(t)=\exp[-(t/\eta)^k]$, the hazard is $kt^{k-1}/\eta^k$. For shape $2$ and scale $3$, this is $2t/9$, which equals $4/9$ at $t=2$.
Assuming independence without justification. Products of component probabilities require independent lifetimes. If dependence is present, the joint distribution is needed instead.

Reliability calculations translate an engineering layout into a probability model. At a chosen mission time, the survival formulas give system reliability directly; differentiating them gives system densities, and dividing by survival gives hazards. These tools also make clear what a configuration change accomplishes: adding a required series component can reduce reliability, whereas adding a parallel backup can increase it, subject to the independence and lifetime assumptions.

Check Your Understanding

Use the system-lifetime definitions and independence formulas to answer the following.

Two independent components have survivals $S_1(t)$ and $S_2(t)$. Write the survival and failure probabilities for their series system.
Explain why the failure probability of a parallel system is a product of component failure probabilities.
For independent continuous component lifetimes, why is the hazard of a series system the sum of the component hazards when all component survivals are positive?
A parallel system has two components with $F_1(t)=0.2$ and $F_2(t)=0.4$. Find its failure probability and reliability at time $t$.
What assumption is needed to multiply the subsystem survival probabilities in a mixed series-parallel calculation?