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Part of HELM Book 1.3

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Engineering Example 1: Reliability in a communication network

Introduction

The reliability of a communication network depends on the reliability of its component parts. The reliability of a component can be represented by a number between 0 and 1 which represents the probability that it will function over a given period of time.

A very simple system with only two components $C_1$ and $C_2$ can be configured in series or in parallel.

If the components are in series then the system will fail if one component fails (see Figure 4).

{figure4}

Figure 4: Both components 1 and 2 must function for the system to function

If the components are in parallel then only one component need function properly (see Figure 5) and we have built-in redundancy.

{figure5}

Figure 5: Either component 1 or 2 must function for the system to function

The reliability of a system with two units in parallel is given by $1 - (1 - R_1)(1 - R_2)$ which is the same as $R_1 + R_2 - R_1R_2$, where $R_i$ is the reliability of component $C_i$. The reliability of a system with $3$ units in parallel, as in Figure 6, is given by

\\[1 - (1 - R_1)(1 - R_2)(1 - R_3)\\]

{figure6}

Figure 6: At least one of the three components must function for the system to function

Problem in words

(a) Show that the expression for the system reliability for three components in parallel is equal to \\[R_1 + R_2 + R_3 - R_1R_2 - R_1R_3 - R_2R_3 + R_1R_2R_3\\]

(b) Find an expression for the reliability of the system when the reliability of each of the components is the same i.e. $R_1 = R_2 = R_3 = R$

(d) Find the system reliability when there are two parallel components each with reliability $R = 0.75$.

Mathematical statement of the problem

(a) Show that $1-(1-R_1)(1-R_2)(1-R_3) \\equiv R_1+R_2+R_3-R_1R_2-R_1R_3-R_2R_3+R_1R_2R_3$

(b) Find $1 - (1 - R_1)(1 - R_2)(1 - R_3)$ in terms of $R$ when $R_1 = R_2 = R_3 = R$

(d) Find $1 - (1 - R_1)(1 - R_2)$ when $R_1 = R_2 = 0.75$.

Mathematical analysis

(a)

\\[ \\begin{align*} 1 - &(1 - R_1)(1 - R_2)(1 - R3) \\equiv 1 - (1 - R_1 - R_2 + R_1R_2)(1 - R_3)\\\\
&= 1 - ((1 - R_1 - R_2 + R_1R_2) \\times 1 - (1 - R_1 - R_2 + R_1R_2) \\times R_3)\\\\
&= 1 - (1 - R_1 - R_2 + R_1R_2 - (R_3 - R_1R_3 - R_2R_3 + R_1R_2R_3))\\\\
&= 1 - (1 - R_1 - R_2 + R_1R_2 - R_3 + R_1R_3 + R_2R_3 - R_1R_2R_3)\\\\
&= 1 - 1 + R1 + R2 - R1R2 + R3 - R1R3 - R2R3 + R1R2R3\\\\
&= R_1 + R_2 + R_3 - R_1R_2 - R_1R_3 - R_2R_3 + R_1R_2R_3\\end{align*}\\]

(b) When $R_1 = R_2 = R_3 = R$ the reliability is

\\[ 1 - (1 - R)^3 \\textrm{ which is equivalent to } 3R - 3R^2 + R^3 \\]

\\[ 1 - (1 - 0.75)^3 = 1 - 0.25^3 = 1 - 0.015625 = 0.984375\\]

(d) $1 - (0.25)^2 = 0.9375$

Interpretation

The mathematical analysis confirms the expectation that the more components there are in parallel then the more reliable the system becomes ($1$ component: $0.75$; $2$ components: $0.9375$; $3$ components: $0.984375$). With three components in parallel, as in part (c), although each individual component is relatively unreliable ($R = 0.75$ implies a one in four chance of failure of an individual component) the system as a whole has an over $98\\%$ probability of functioning (under $1$ in $50$ chance of failure).

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