2 System reliability

It is reasonable to ask whether, in designing a system, an engineer should design a system using components in series or in parallel. The engineer may not have a choice of course! We may represent a system consisting of $n$ components say $C_1,C_2\dots ,C_n$ with reliabilities (these are just probability values) $R_1,R_2\dots ,R_n$ respectively as series and parallel systems as shown below.

Figure 2

With a series design, the system will fail if any component fails. With a parallel design, the system will work as long as any component works.

Assuming that the components are independent, we can express the reliability of the series design as

$R_{\text{Series}}=R_1\times R_2\times \cdots \times R_n$

simply by multiplying the probabilities.

Since each reliability value is less than one, we may conclude that a series design is less reliable than its least reliable component.

Similarly (although by no means as clearly!), we can express the reliability of the parallel design as

$R_{\text{Parallel}}=1-\left(1-R_1\right)\left(1-R_2\right)\dots \left(1-R_n\right)$

The derivation of this result is illustrated in Example 3 below for the case $n=3$ . In this case, the algebra involved in fairly straightforward. We can conclude that the parallel design is at least as reliable as the most reliable component.

Engineers will sometimes include ‘redundant’components in parallel to improve reliability. The spare wheel of a car is a well known example.

Example 2

Series design

Consider the three components $C_1,C_2$ and $C_3$ with reliabilities $R_1,R_2$ and $R_3$ connected in series as shown below

Find the reliability of the system where $R_1=0.2,R_2=0.3$ and $R_3=0.4$ .

Solution

Since the components are assumed to act independently, we may clearly write

$R_{\text{Series}}=R_1\times R_2\times R_3$

Taking $R_1=0.2,R_2=0.3$ and $R_3=0.4$ we obtain the value $R_{\text{Series}}=0.2\times 0.3\times 0.4=0.024$

Example 3

Parallel design

Consider the three components $C_1,C_2$ and $C_3$ with reliabilities $R_1,R_2$ and $R_3$ connected in parallel as shown below

Find the reliability of the system where $R_1=0.2,R_2=0.3$ and $R_3=0.4$ .

Solution

Observing that $F_i=1-R_i$ , where $F_i$ represents the failure of the $i$ th component and $R_i$ represents the reliability of the $i$ th component we may write

$F_{\text{System}}=F_1{F}_2{F}_3\to R_{\text{System}}=1-F_1{F}_2{F}_3=1-\left(1-R_1\right)\left(1-R_2\right)\left(1-R_3\right)$

Again taking $R_1=0.2,R_2=0.3$ and $R_3=0.4$ we obtain

$F_{\text{System}}=1-\left(1-0.2\right)\left(1-0.3\right)\left(1-0.4\right)=1-0.336=0.664$

Hence series system reliability is less than any of the component reliabilities and parallel system reliability is greater than any of the component reliabilities.

Task!

Consider the two components $C_1$ and $C_2$ with reliabilities $R_1$ and $R_2$ connected in series and in parallel as shown below. Assume that $R_1=0.3$ and $R_2=0.4$ .

Let $R_{\text{Series}}$ be the reliability of the series configuration and $R_{\text{Parallel}}$ be the reliability of the parallel configuration

1. Why would you expect that $R_{\text{Series}}<0.3$ and $R_{\text{Parallel}}>0.4$ ?
2. Calculate $R_{\text{Series}}$
3. Calculate $R_{\text{Parallel}}$

Answer