Engineering Example 1

4 Engineering Example 1

4.1 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)

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.

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 $R_{1} + R_{2} - R_{1} R_{2}$ , 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

$1 - (1 - R_{1}) (1 - R_{2}) (1 - R_{3})$

Figure 6 :

{At least one of the three components must function for the system to function}

Problem in words

Show that the expression for the system reliability for three components in parallel is equal to $R_{1} + R_{2} + R_{3} - R_{1} R_{2} - R_{1} R_{3} - R_{2} R_{3} + R_{1} R_{2} R_{3}$
Find an expression for the reliability of the system when the reliability of each of the components is the same i.e. $R_{1} = R_{2} = R_{3} = R$
Find the system reliability when $R = 0.75$
Find the system reliability when there are two parallel components each with reliability $R = 0.75$ .

Mathematical statement of the problem

Show that $1 - (1 - R_{1}) (1 - R_{2}) (1 - R_{3}) \equiv R_{1} + R_{2} + R_{3} - R_{1} R_{2} - R_{1} R_{3} - R_{2} R_{3} + R_{1} R_{2} R_{3}$
Find $1 - (1 - R_{1}) (1 - R_{2}) (1 - R_{3})$ in terms of $R$ when $R_{1} = R_{2} = R_{3} = R$
Find the value of (b) when $R = 0.75$
Find $1 - (1 - R_{1}) (1 - R_{2})$ when $R_{1} = R_{2} = 0.75$ .

Mathematical analysis

$1 - (1 - R_{1}) (1 - R_{2}) (1 - R_{3}) \equiv 1 - (1 - R_{1} - R_{2} + R_{1} R_{2}) (1 - R_{3})$
$= 1 - ((1 - R_{1} - R_{2} + R_{1} R_{2}) \times 1 - (1 - R_{1} - R_{2} + R_{1} R_{2}) \times R_{3})$

$= 1 - (1 - R_{1} - R_{2} + R_{1} R_{2} - (R_{3} - R_{1} R_{3} - R_{2} R_{3} + R_{1} R_{2} R_{3}))$

$= 1 - (1 - R_{1} - R_{2} + R_{1} R_{2} - R_{3} + R_{1} R_{3} + R_{2} R_{3} - R_{1} R_{2} R_{3})$

$= 1 - 1 + R_{1} + R_{2} - R_{1} R_{2} + R_{3} - R_{1} R_{3} - R_{2} R_{3} + R_{1} R_{2} R_{3}$

$= R_{1} + R_{2} + R_{3} - R_{1} R_{2} - R_{1} R_{3} - R_{2} R_{3} + R_{1} R_{2} R_{3}$
When $R_{1} = R_{2} = R_{3} = R$ the reliability is
$1 - {(1 - R)}^{3}$ which is equivalent to $3 R - 3 R^{2} + R^{3}$
When $R_{1} = R_{2} = R_{3} = 0.75$ we get
$1 - {(1 - 0.75)}^{3} = 1 - 0.2 5^{3} = 1 - 0.015625 = 0.984375$
$1 - {(0.25)}^{2} = 0.9375$

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).