SYSTEM RELIABILITY EVALUATION AND ALLOCATION
4.8 RELIABILITY EVALUATION OF COMPLEX SYSTEMS
1+3.6× 10−5 2.8× 10−5
1− e−2.8×10−5×5000
= 0.9756.
The mean time to failure is obtained from (4.37) as
MTTF= 1
3.6× 10−5 + 1
2.8× 10−5 − 3.6× 10−5
2.8× 10−5(3.6× 10−5+ 2.8 × 10−5)
= 4.34 × 104 hours.
Comparing these results with those in Example 4.7, we note the adverse effects of the imperfect switching system.
4.8 RELIABILITY EVALUATION OF COMPLEX SYSTEMS
So far we have studied series, parallel, series–parallel, parallel–series, k-out-of-n, and redundant systems. In reality, these configurations are frequently combined
RELIABILITY EVALUATION OF COMPLEX SYSTEMS 85
and form more complex systems in order to fulfill functional requirements.
Some networks, such as power supply grids, telecommunication systems, and computing networks, are so complicated in structure that they cannot easily be decomposed into the said configurations. Reliability evaluation of complex sys-tems requires more advanced methods. In this section we present three simple yet powerful approaches. For large-scale complex systems, manual calculation of reliability is difficult, if not prohibitive. Various commercial software packages, such as Reliasoft, Relex, and Item, are capable of calculating reliability and other measures of reliability of complex systems through simulation.
4.8.1 Reduction Method
Some systems are made up of independent series, parallel, series–parallel, paral-lel–series, k-out-of-n, and redundant subsystems. The system reduction method is to collapse a system sequentially into the foregoing subsystems, each repre-sented by an equivalent reliability block. The reliability block diagram is further reduced until the entire system is represented by a single reliability block. The method is illustrated in the following example.
Example 4.9 The reliability block diagram of an engineering system is shown in Figure 4.17. The times to failures of the components are modeled with the exponential distribution with the failure rates shown by the corresponding blocks with a multiplication of 10−4 failures per hour. Compute the system reliability at 600 hours of mission time.
SOLUTION The steps for calculating the system reliability are as follows:
1. Decompose the system into blocks A, B, C, and D, which represent a parallel–series, parallel, series, and cold standby subsystem, respectively, as shown in Figure 4.17.
2. Calculate the reliabilities of blocks A, B, C, and D. From (4.19), the reli-ability of block A is
RA= 1 − (1 − R1R2)(1− R3R4)= 1 −
FIGURE 4.17 Engineering system of Example 4.9
From (4.10), the reliability of block B is
RB = 1 − (1 − R5)(1− R6)= 1 − (1 − e−4.8×10−4×600)(1− e−3.3×10−4×600)
= 0.955.
From (4.1) we obtain the reliability of block C as RC = e−(1.7+2.5)×10−4×600= 0.7772.
From (4.31) the reliability of block D is
RD= (1 + 4.3 × 10−6× 600) × e−4.3×10−4×600= 0.9719.
The equivalent reliability block diagram is shown in Figure 4.18.
3. The equivalent system in Figure 4.18 is further reduced to blocks E and F, which are series and parallel subsystems, respectively.
4. Calculate the reliabilities of blocks E and F.
RE= RARD= 0.9736 × 0.9719 = 0.9462.
RF = 1 − (1 − RB)(1− RC)= 1 − (1 − 0.955)(1 − 0.7772) = 0.99.
The reliability block diagram equivalent to Figure 4.18 is shown in Figure 4.19.
5. The equivalent system in Figure 4.19 consists of two units in series. It is reduced to one single block, G.
6. Calculate the reliability of block G.
RG= RERF = 0.9462 × 0.99 = 0.9367.
Now that the original system has been reduced to a single unit, as shown in Figure 4.20, the reduction process is exhausted. Then the system reliability is R= RG= 0.9367.
B
C
A D
E F
FIGURE 4.18 Reduced system equivalent to Figure 4.17
E F
G
FIGURE 4.19 Reduced system equivalent to Figure 4.18
RELIABILITY EVALUATION OF COMPLEX SYSTEMS 87
G
FIGURE 4.20 Reduced system equivalent to Figure 4.19
4.8.2 Decomposition Method
The system reduction method is effective when a complex system can be par-titioned into a number of simple subsystems whose reliabilities are directly obtainable. In some situations we encounter more complex systems, such as the well-known bridge system shown in Figure 4.21, which cannot be solved by the reduction method. In this subsection we present a decomposition method also known as the conditional probability approach or Bayes’ theorem method.
The decomposition method starts with choosing a keystone component, say A, from the system being studied. This component appears to bind the system together. In Figure 4.21, for instance, component 5 is a such keystone component.
The keystone component is assumed to be 100% reliable and is replaced with a line in system structure. Then the same component is supposed to have failed and is removed from the system. The system reliability is calculated under each assumption. According to the rule of total probability, the reliability of the original system can be written as
R= Pr(system good | A) Pr(A) + Pr(system good | A) Pr(A), (4.40) where A is the event that keystone component A is 100% reliable, A the event that keystone component A has failed, Pr(system good| A) the probability that the system is functionally successful given that component A never fails, and Pr(system good| A) the probability that the system is functionally successful given that component A has failed. The efficiency of the method depends on the selection of the keystone component. An appropriate choice of the component leads to an efficient calculation of the conditional probabilities.
Example 4.10 Consider the bridge system in Figure 4.21. Suppose that the reli-ability of componenti is Ri, i= 1, 2, . . . , 5. Calculate the system reliability.
SOLUTION Component 5 is chosen as the keystone component, denoted A.
Assume that it never fails and is replaced with a line in the system configuration.
1 2
3 4
5
FIGURE 4.21 Bridge system
1 2
3 4
FIGURE 4.22 Bridge system when component 5 never fails
1 2
3 4
FIGURE 4.23 Bridge system when component 5 has failed
Then the system is reduced as shown in Figure 4.22. The reduced system is a series–parallel structure, and the conditional reliability is
Pr(system good| A) = [1 − (1 − R1)(1− R3)][1− (1 − R2)(1− R4)].
The next step is to assume that component 5 has failed and is removed from the system structure. Figure 4.23 shows the new configuration, which is a paral-lel–series system. The conditional reliability is
Pr(system good| A) = 1 − (1 − R1R2)(1− R3R4).
The reliability and unreliability of component 5 are Pr(A)= R5and Pr(A)= 1 − R5, respectively. Substituting the equations above into (4.40) yields the reliability of the original system as
R= [1 − (1 − R1)(1− R3)][1− (1 − R2)(1− R4)]R5
+ [1 − (1 − R1R2)(1− R3R4)](1− R5)
= R1R2+ R3R4+ R1R4R5+ R2R3R5− R1R2R4R5
− R2R3R4R5− R1R2R3R5
− R1R3R4R5− R1R2R3R4+ 2R1R2R3R4R5. (4.41) As illustrated in Example 4.10, calculating the reliability of the bridge system needs the selection of only one keystone component, and (4.40) is applied once.
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For some complex systems, the reliabilities of decomposed systems cannot be written out directly. In these situations we may select additional keystone com-ponents and apply (4.40) successively until each term in the equation is easily obtainable. For example, if Pr(system good| A) cannot be worked out immedi-ately, the decomposed system with A functional may be further decomposed by selecting an additional keystone component, say B. Applying (4.40) to keystone component B, we may write the reliability of the original system as
R= Pr(system good | A · B) Pr(A) Pr(B) + Pr(system good | A · B) Pr(A) Pr(B)
+ Pr(system good | A) Pr(A). (4.42)
The decomposition method discussed above selects one keystone component at a time. W. Wang and Jiang (2004) suggest that several such components be chosen simultaneously for some complex networks. For example, if two keystone components, say A and B, are selected, the original system will be decomposed into four subsystems with conditionsA· B, A · B, A · B, and A · B, respectively, where A· B is the event that both A and B are functioning, A · B is the event that A is not functioning and B is,A· B is the event that A is functioning and B is not, andA· B is the event that both A and B are not functioning. By applying the rule of total probability, the reliability of the original system can be written as R= Pr(system good | A · B) Pr(A) Pr(B) + Pr(system good | A · B) Pr(A) Pr(B)
+ Pr(system good | A · B) Pr(A) Pr(B)
+ Pr(system good | A · B) Pr(A) Pr(B). (4.43)
Equation (4.43) has four terms. In general, for binary components, if m key-stone components are selected simultaneously, the reliability equation contains 2m terms. Each term is the product of the reliability of one of the decomposed subsystems and that of the condition on which the subsystem is formed.
4.8.3 Minimal Cut Set Method
The decomposition method studied earlier is based on the rule of total probability.
In this subsection we present an approach to system reliability evaluation by using a minimal cut set and the inclusion–exclusion rule. First let’s discuss cut sets. A cut set is a set of components whose failure interrupts all connections between input and output ends and thus causes an entire system to fail. In Figure 4.21, for example,{1, 3, 5} and {2, 4} are cut sets. Some cut sets may contain unnecessary components. If removed, failure of the remaining components still results in system failure. In the example above, cut set {1, 3, 5} contains component 5, which can be eliminated from the cut set without changing the failure state of the system. Such cut sets can be further reduced to form minimal cut sets. A minimal cut set is the smallest combination of components which if they all fail will cause the system to fail. A minimal cut set represents the smallest collection of components whose failures are necessary and sufficient to result in system
failure. If any component is removed from the set, the remaining components collectively are no longer a cut set. The definitions of cut set and minimal cut set are similar to those defined in Chapter 6 for fault tree analysis.
Since every minimal cut set causes the system to fail, the event that the system breaks is the union of all minimal cut sets. Then the system reliability can be written as
R= 1 − Pr(C1+ C2+ · · · + Cn), (4.44) where Ci (i= 1, 2, . . . , n) represents the event that components in minimal cut set i are all in a failure state and n is the total number of minimal cut sets. Equation (4.44) can be evaluated by applying the inclusion–exclusion rule, which is Example 4.11 Refer to Example 4.10. If the five components are identical and have a common reliabilityR0, calculate the reliability of the bridge system shown in Figure 4.21 using the minimal cut set method.
SOLUTION The minimal cut sets of the bridge system are{1, 3}, {2, 4}, {1, 4, 5}, and{2, 3, 5}. Let Aidenote the event that componenti has failed, i= 1, 2, . . . , 5.
Then the events described by the minimal cut sets can be written asC1 = A1· A3, C2= A2· A4,C3= A1· A4· A5, andC4= A2· A3· A5. From (4.44) and (4.45) and using the rules of Boolean algebra (Chapter 6), the system reliability can be written as
Note that (4.41) gives the same result when all components have equal relia-bility R0.
CONFIDENCE INTERVALS FOR SYSTEM RELIABILITY 91