Solving the Complex RBD-Series Parallel System with a Keystone Component

The general form of this type of reliability block diagram has two series circuits in a parallel arrangement combined through a center or “keystone” unit. You are often given the reliability of each block and asked to calculate the reliability of the circuit.

For example let’s say we are given the following five block arrangement pictured in Figure 1. Each of the blocks is identical and each has a failure rate of 10 failures per million hours (FPMH). For a 1,000 hour mission, what is the reliability of the circuit?

 

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Figure 1 – Five Element RBD

 

The most commonly used method of solving this problem is by way of using a Bayesian approach. According to Bayes Theorem, the probability that the system is working (reliability) can be summed up from two components; first, the reliability of the system when Block C is functioning properly multiplied by the probability that C is working properly, and; second, the reliability of the system when C is not functioning properly multiplied by the probability that C is not working properly.

Mathmatically:

RSystem = RSystemC working * Probability C working + RSystem C failed * Probability C failed

Figures 2 and 3 show the block diagrams of the two scenarios:

 

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Figure 2 – Scenario A:
The Circuit with Block C Working

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Figure 3 – Scenario B:
The Circuit with Block C Failed

 

Based on these two scenarios, the following equation expresses the system reliability:

Rs = C * ( ( A + D – AD ) * ( B + E – BE ) ) + ( 1 – C ) * ( AB + DE – ABDE )

 We begin by calculating the reliability of Scenario A. Since each block in parallel circuit AD and BE is identical we can use the Redundancy Calculator tool in QuART PRO, as seen in Figure 4. Using a failure rate of 10FPMH and 1000 hours, we get a reliability of 99.99% for each. Since the product of these two parallel groups reflects the circuit when C is functioning, we then multiply this product by the reliability of Block C.

 

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Figure 4 – QuART PRO Redundancy Calculator

 

We can do the same for Scenario B (the parallel arrangements of AB and DE). The result is then multiplied by the probability that C is in the failed state, 1 – Reliability of C.

Combining the two results yields a System reliability of 97.85%

Had the blocks not been identical, we could have just as easily used the Failure Rate calculator in QuART PRO, as seen in Figure 5, to calculate the failure rate for the two scenarios. Given the stated mission time, we can translate this failure rate to a system reliability number.

 

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Figure 5 – QuART PRO Failure Rate Calculator

 

Another approach is to use a technique familiar with Fault Tree practitioners, the “cut-set” method. Here you simply examine the circuit in terms of all the combinations that would result in system failure.

If we look at the original circuit we notice that there are four block combinations that can cause the system to fail. These are AD, BE, ACE, and BCD. These four “cut-sets” are shown in Figure 6.

 

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Figure 6 – Original Circuit Cut-Sets

 

We then generate a list of all possible combinations for the five blocks in the circuit. Since each block has a 90% probability of working we can calculate the probability of each potential system state for each outcome combination. Table 1 lists all the possible combinations.

 Table 1 – System States

A

B

C

D

E

System Status

Probability

Failed Failed Failed Failed Failed Failure

0.0000

Failed Failed Failed Failed Good Failure

0.0001

Failed Failed Failed Good Failed Failure

0.0001

Failed Failed Failed Good Good Success

0.0008

Failed Failed Good Failed Failed Failure

0.0001

Failed Failed Good Failed Good Failure

0.0008

Failed Failed Good Good Failed Failure

0.0008

Failed Failed Good Good Good Success

0.0073

Failed Good Failed Failed Failed Failure

0.0001

Failed Good Failed Failed Good Failure

0.0008

Failed Good Failed Good Failed Failure

0.0008

Failed Good Failed Good Good Success

0.0073

Failed Good Good Failed Failed Failure

0.0008

Failed Good Good Failed Good Failure

0.0073

Failed Good Good Good Failed Success

0.0073

Failed Good Good Good Good Success

0.0656

Good Failed Failed Failed Failed Failure

0.0001

Good Failed Failed Failed Good Failure

0.0008

Good Failed Failed Good Failed Failure

0.0008

Good Failed Failed Good Good Success

0.0073

Good Failed Good Failed Failed Failure

0.0008

Good Failed Good Failed Good Success

0.0073

Good Failed Good Good Failed Failure

0.0073

Good Failed Good Good Good Success

0.0656

Good Good Failed Failed Failed Success

0.0008

Good Good Failed Failed Good Success

0.0073

Good Good Failed Good Failed Success

0.0073

Good Good Failed Good Good Success

0.0656

Good Good Good Failed Failed Success

0.0073

Good Good Good Failed Good Success

0.0656

Good Good Good Good Failed Success

0.0656

Good Good Good Good Good Success

0.5905

 

Lastly, the probabilities for the successful outcomes are added, as follows:

 

DE 0.0008 +
CDE 0.0073 +
BDE 0.0073 +
BCD 0.0073 +
BCDE 0.0656 +
ADE 0.0073 +
ACE 0.0073 +
ACDE 0.0656 +
AB 0.0008 +
ABE 0.0073 +
ABD 0.0073 +
ABDE 0.0656 +
ABC 0.0073 +
ABCE 0.0656 +
ABCD 0.0656 +
ABCDE 0.5905 = 0.9785

 

The probability of system success is calculated to be 97.85%, which agrees, as it should, with the results from the QuART PRO Redundancy Calculator.