Problems in RBD modelling.

[Gintama]

Hi all,

I have some problems to modelling the following system by RBD scheme.
I have a sequence of N equal devices, N/2 belong to system A and N/2 belong to system B, which are installed in the following alternative order: device A, device B, device A, device B and so on. The fault happens only when two consecutive devices fail, so I think that I have to model a series of N-1 parallel of two devices.
Problem: all the N/2 devices of system A are connected between them and to another system "1", the same is for N/2 devices of system B which are connected to system "2". So the fault globally happens when:

1)two consecutive devices (obviously one of system A and one system B) fail
or
2)both two system "1" and "2" fail
or
3)only one devices of system A and "2" fail
or
4) only one devices of system B and "1" fail.

Now I can’t model the first series of N-1 devices parallels by adding in series the block parallel of “1” and “2”. How can I resolve? I need to calculate MTBF of global system.

Thanks to everybody

 

[IRstuff]

A drawing would help

TTFN
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[Gintama]

Hi,

I hope this drawing will be useful.

http://we.tl/Guan4Ek454

Tks

 

[IRstuff]

1> I don't get what your yellow connections are doing. Are they representing parallel or serial connections? Your middle A shows connections to the other As, but no connection to 1, while the other 2 As are connected to 1.

2> (1-P1)(1-P2)
3> (1-PA)(1-P2)
4> (1-PB)(1-P1)

Assuming I understood what you described. You seemed to imply that the failure of any single A results in total system failure.


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[Gintama]

No, these connection are not RBD (indeed my problem is to find the RBD model to describe this issue) but communication links (could be fiber links for example). The failure of service on each zone happens when both adjacent A and B devices are in fault or can't communicate with system "1" and "2" respectively, so when events described in my post happen.

Middle A is connected by a ring topology to system “1”. The same for middle B to system “2”

Thanks for your help IRstuff.

 

[IRstuff]

OK
2> no change
3> (1-PA^3)(1-P2) 1 minus probability that all 3 PAs are working
4> (1-PB^3)(1-P1) 1 minus probability that all 3 PBs are working

As for 1> I would just brute force it; there are 64 possible states for the 6 A/B devices, i.e., 2^6. Create a table with those states and mark them as good or bad, and count up the bad ones.



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[Gintama]

I fully agree your approach. My problem is that I have a number of devices that is very bigger than the example in drawing. I don't have a specific line guide from the client, so I would ask to you if it is reasonable modelling an RBD only for the worst case. I mean: to calculate the availability in the point of global system that is the worst, in my example (see the new drawing) it will be the green service (because it is the service provided by the two devices depends ever also of their near devices of the same network to reach system "1" and 2" respectively).

http://we.tl/Cp05Ki8k4J

 

[IRstuff]

But, that's not the worst case, based on your OP, which implies that any A/B pair fail is a system fail. Your diagram only has the middle pair, which is 1/43 of the possible fail mode cases. I suggest that you convert the truth table into a Karnaugh map to find the simplified logic equations, which you can then write as probability equations. This is, by the way, already a gross simplification. To do it correctly, you would need to use the hot redundancy approach, since this approach ignores time.

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[IRstuff]

Now that I think about, the tree approach was the correct way, since we have to assume that the first instance of any A/B pair failure causes a repair cycle, so whatever would have happened in serial time would be a new calculation. That means that there are only 10/64 cases to consider.

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[Gintama]

Hi IRstuff,
the fault of service in one area covered by two consecutive devices, does not cause fault of the services in the other places of system. For this reason I proposed to you the middle couple fault as worst case of analysis. Consider for example that A e B devices are access point of a wireless network or CCTV cameras: if we lost the service in one area we don't lost the service in the other one.

I agree with you that in my OP I was talking about that any A/B faults cause a global fault of the system but now I'm reworking the analysis.

In this condition, I again would ask to you if it is reasonable to make an analysis of the worst case to demonstrate availability of design.

Thank you

 

[IRstuff]

You seem to be arguing both sides of the argument. If a single set of A/B does not cause an availability incident, how can it be the "worst" case? If is an availability incident, then, they all have to be.

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[Gintama]

With "worst" case I mean the point of analysis that depends of the biggest number of devices: the middle couple depends by all the other devices A and B to reach "1" and "2". So in my drawing the point of analysis could be the green one and so the point for which i would show the results.
Please note that drawing is only an example i should have a number very big of devices.


Sorry if I am repeating again but i want to be sure that your opinion is not based on a wrong english sentence about the question.

 

[IRstuff]

I don't think that you have sufficiently detailed the system, particularly the ring nodes, to say. In order to do what you are now describing, you would have to separately analyze the ring structure, unless there is some assurance that if the A or B fails, that includes the ring node, and vice-versa. A ring has built-in redundancy, and is generally a small fraction of the failure rate attributable to a given module. The weak links are 1 and 2, since they are a single point of failure. Given that, it seems to me that the reliability of the ring ought to be so high that you should not be able to tell the difference between different positions in the ring, but more detail would be required to determine that.



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