Reliability Networks

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 Network

This network is the simplest reliability network/configuration, and its block diagram is shown in Figure 3.2.

Block diagram of a /.-unit series system (network)

FIGURE 3.2 Block diagram of a /.-unit series system (network).

The diagram denotes a k-unit 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

Rs is the series system reliability.

Ej is the successful operation (i.e., success event) of unit i, for / = 1, 2, 3,... ., k. P[ElE2E2....Ek) is the occurrence probability of events Ex,E2,E2,....,Ek.

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

Rs (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

MTTFS 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

As (t) is the series system hazard rate.

Here, it is to be noted that the right-hand side of Equation (3.24) is independent of time 1. Thus, the left-hand side of this equation is simply Xs, 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 worst-case 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:

  • • System reliability for an eight-hour mission.
  • • System mean time to failure.
  • • System failure rate.

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 Network

In 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.

Block diagram of a parallel system/network with к units

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

Fp 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 Ek.

For independently failing parallel units, Equation (3.25) is written as where

P(Ei) 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

Rp 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

Rp (/) 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

MTTFp 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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