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Reliability NetworksTable of Contents:
An engineering system can form various configurations/networks in conducting reliability analysis. This section is concerned with the reliability evaluation of such commonly occurring networks/configurations. Series NetworkThis network is the simplest reliability network/configuration, and its block diagram is shown in Figure 3.2. FIGURE 3.2 Block diagram of a /.unit series system (network). The diagram denotes a kunit series system (network), and each block in the diagram represents a unit. For the successful operation of the series system, all к units must function normally. In other words, if any one of the к units malfunctions/fails, the series system fails. The series system, shown in Figure 3.2, reliability is expressed by where R_{s} is the series system reliability. Ej is the successful operation (i.e., success event) of unit i, for / = 1, 2, 3,... ., k. P[E_{l}E_{2}E_{2}....E_{k}) is the occurrence probability of events E_{x},E_{2},E_{2},....,E_{k}. For independently failing all units, Equation (3.18) becomes where /^{5}(£,)is the probability of occurrence of event E,, for i = 1,2, 3 ,...,k. If we let /?, = P(Ej), for / = 1, 2, 3.....k. Equation (3.19) becomes
where /?, is the unit i reliability for i= 1,2, 3,..., k. For constant failure rate A, of unit i from Equation (3.11) (A, (/) = А,), we get where Rj (t) is the reliability of unit i at time t. By substituting Equation (3.21) into Equation (3.20), we get
where R_{s} (r) is the series system reliability at time t. By substituting Equation (3.22) into Equation (3.12), we obtain the following expression for the series system mean time to failure:
where MTTF_{S} is the series system mean time to failure. By substituting Equation (3.22) into Equation (3.6), we obtain the following expression for the series system hazard rate:
where A_{s} (t) is the series system hazard rate. Here, it is to be noted that the righthand side of Equation (3.24) is independent of time 1. Thus, the lefthand side of this equation is simply X_{s}, the failure rate of the series system. It means that whenever we add up the failure rates of items/units, we automatically assume that these items/units form a series network/configuration, a worstcase design scenario in regard to reliability. Example 3.5 Assume that a system has four independent and identical subsystems, and the constant failure rate of a subsystem is 0.0006 failures per hour. All four subsystems must operate normally for the system to function successfully. Calculate the following:
In this case, the subsystems of the system form a series configuration/network. Thus, by substituting the given data values into Equation (3.22), we get
Substituting the given data values into Equation (3.23) yields
Using the specified data values in Equation (3.24) yields
Thus, the system reliability, mean time to failure, and failure rate are 0.9809, 416.66 hours, and 0.0024 failures per hour, respectively. Parallel NetworkIn this case, the system has к simultaneously operating units/items, and at least one of these units/items must work normally for the successful operation of the system. The A:unit parallel system/network block diagram is shown in Figure 3.3, and each block in the diagram represents a unit. FIGURE 3.3 Block diagram of a parallel system/network with к units. The failure probability of the parallel system/network shown in Figure 3.3 is expressed by
where F_{p} is the failure probability of the parallel system. £,• is the failure (i.e., failure event) of unit /, for /=1,2, 3,..., k. Р(ЕЕЕгЕ2....Ек) is the probability of occurrence of events E ,E ,E ,...,and E_{k}. For independently failing parallel units, Equation (3.25) is written as where P(E_{i}) is the occurrence probability of failure event Ё, for / = 1, 2, 3,..., k. If we let Fj = P(Ei), for / = 1,2, 3,..., к, then Equation (3.26) becomes
where Fj is the failure probability of unit /, for /=1,2, 3,..., k. By subtracting Equation (3.27) from unity, we obtain where R_{p} is the reliability of the parallel system/network. For constant failure rate of A, of unit /, subtracting Equation (3.21) from unity and then inserting it into Equation (3.28) yields
where R_{p} (/) is the parallel system/network reliability at time t. For identical units, by substituting Equation (3.29) into Equation (3.12), we obtain the following expression for the parallel system/network mean time to failure:
where MTTF_{p} is the identical units parallel system/network mean time to failure. A is the unit constant failure rate. Example 3.6 Assume that a system has four independent, identical, and active units. At least one of these units must operate normally for the system to operate successfully. Calculate the system’s reliability if each unit’s failure probability is 0.15. By inserting the given values into Equation (3.28), we get
Thus, the reliability of the system is 0.9994. Example 3.7 Assume that a system has four independent and identical units in parallel. The constant failure rate of a unit is 0.006 failures per hour. Calculate the system’s mean time to failure. By substituting the given data values into Equation (3.30), we get
Thus, the system’s mean time to failure is 347.22 hours. 
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