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<quiz>
  <question type="description">
    <name>
      <text>1.3.3.5</text>
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      <text><![CDATA[<h3> Engineering Example 1: Reliability in a communication network </h3><b>

Introduction</b>&nbsp;<p><br></p><p>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 <b>series</b> then the system will fail if one component fails (see the Figure below).&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</p><p><b>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<img src="@@PLUGINFILE@@/1-3-fig-4-series.svg" alt="" width="209" height="37" role="presentation" class="img-responsive atto_image_button_text-bottom"></b><br></p><p>If the components are in <b>parallel</b> then only one component need function properly and we have built-in redundancy (see the Figure below).&nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <img src="@@PLUGINFILE@@/1-3-fig-5-parallel1.svg" alt="" width="215" height="104" role="presentation" class="img-responsive atto_image_button_text-bottom">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<br></p><p><b>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</b><br></p><p>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, (see the Figure below), is given by</p><p>\[ 1 - (1 - R_1)(1 - R_2)(1 - R_3). \]</p><p><img src="@@PLUGINFILE@@/1-3-fig-6-parallel2.svg" alt="" width="231" height="151" role="presentation" class="img-responsive atto_image_button_text-bottom"><br></p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<br></p><p><b>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;</b><br></p><p><br></p><p><b>Problem in words</b></p><p>(a) Show that the expression for the system reliability for three components in parallel is equal to&nbsp;</p><p>\[ R_1 + R_2 + R_3 - R_1R_2 - R_1R_3 - R_2R_3 + R_1R_2R_3. \]</p><p>(b) Find an expression for the reliability of the system when the reliability of each of the components is the same i.e. \( R_1 = R_2 = R_3 = R. \)</p><p>(c) Find the system reliability when \( R = 0:75.\)</p><p>(d) Find the system reliability when there are two parallel components each with reliability \( R = 0:75.\)</p><p><br></p><p><b>Mathematical statement of the problem.</b></p><p>(a) Show that \(1-(1-R_1)(1-R_2)(1-R_3) - R_1+R_2+R_3-R_1R_2-R_1R_3-R_2R_3+R_1R_2R_3.\)</p><p>(b) Find \(1 - (1 - R_1)(1 - R_2)(1 - R_3)\) in terms of \(R\) when \(R_1 = R_2 = R_3 = R.\)</p><p>(c) Find the value of (b) when \( R = 0.75.\)</p><p>(d) Find \(1 - (1 - R_1)(1 - R_2)\) when \(R_1 = R_2 = 0.75.\)</p><p><br></p><p><b>Mathematical analysis</b></p><p>(a) \[ 1 - (1 - R_1)(1 - R_2)(1 - R_3) - 1 - (1 - R_1 - R_2 + R_1R_2)(1 - R_3) \]</p><p>\[ = 1 - ((1 - R_1 - R_2 + R_1R_2) - 1 - (1 - R_1 - R_2 + R_1R_2) - R_3) \]</p><p>\[= 1 - (1 - R_1 - R_2 + R_1R_2 - (R_3 - R_1R_3 - R_2R_3 + R_1R_2R_3))\]</p><p>\[= 1 - (1 - R_1 - R_2 + R_1R_2 - R_3 + R_1R_3 + R_2R_3 - R_1R_2R_3) \]</p><p>\[= 1 - 1 + R_1 + R_2 - R_1R_2 + R_3 - R_1R_3 - R_2R_3 + R_1R_2R_3 \]</p><p>\[= R_1 + R_2 + R_3 - R_1R_2 - R_1R_3 - R_2R_3 + R_1R_2R_3.\]</p><p>(b) When \( R_1 = R_2 = R_3 = R\) the reliability is</p><p>\[ 1 - (1 - R)^3 \text{&nbsp; &nbsp;which is equivalent to&nbsp; &nbsp;} 3R - 3R^2 + R^3. \]</p><p>(c) When \( R_1 = R_2 = R_3 = 0.75\) we get</p><p>\[ 1 - (1 - 0.75)^3 = 1 - 0.25^3 = 1 - 0.015625 = 0.984375.\]</p><p>(d) \(1 - (0.25)^2 = 0.9375.\)</p><p><br></p><p><b>Interpretation</b></p><p>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).</p>]]></text>
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