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Reliability Phase Modeling

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meverett85

Aerospace
Sep 23, 2013
5
US
Question on how to find reliability when you have two different time periods rolled up into the overall reliability requirement. I'm having a doozy of a time trying to create a reliability model for a range safety system that has multiple independent strings.

For illustrative purposes, let's say there is a requirement to be .999 reliable over a 3 hour period. Let's say the system has 3 independent strings (i.e. broadcasting transmitters). We will call them Transmitter 1 (T1), Transmitter 2 (T2), and Transmitter 3 (T3).

During phase 1 (0 to 1 hour), there is no hit against reliability as long as T3 is working along with either T1 and/or T2.

So a conditional table has 3 passing possibilities for phase 1 (0 to 1 hour):

1)T1=T2=T3 are working
2)T1 not working; T2=T3 are working
3)T1=T3 are working; T2 not working

Adding up each probability gives you R(1) for phase 1 reliability. Easy enough so far...

During phase 2 (1 to 3 hours), T2 and T3 are required however T1 is not. Here is the kicker that confuses me: What if you could repair failed transmitters during phase 1 before they entered phase 2?

I believe I need to multiply the transmitter dependability variables (MTBF/MTBF+MTTRF) so that you don't encounter a failure condition as soon as you transition from phase 1 to phase 2. There are two acceptable entry conditions into Phase 2:

1)T1(D)*T2(D)*T3(D)
2)(1-T1(D))*T2(D)*T3(D)

Adding these two values would give Dependability (D).

Now, how do I create the conditional table for phase 2?

Do you create a separate Phase 2 conditional table for each of the two passable dependability conditions above?

If so, how would you roll all of these variables up into your overall reliability for the 3 hour period?

Phase 1 Reliability (R1), Dependability (D), and Phase 2 Reliability from two different conditional tables R2 and R3....

Do you just multiply all of these together? R1*D*R2*R3? Or would it be R1*D*(R2+R3)

If this doesn't make any sense...I apologize. It's late and my head hurts!

 
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Reliability of repairable items is the probability that failure will not occur within the time period of interest = 1 - P(failure)

Reliability of two transmitters is the probability of one transmitter not failing during the total time of phase 1 and phase 2, squared.

If a transmitter is repaired before entering phase 2, then it essentially becomes a new unit again at the start of phase 2

Say the probability of failure of one transmitter during it's first 1 hour of service is 0.05, then the probability of 2 failing is .05^2 = 0.025; reliability = 1-0.025 = 97.5% With three units, 0.05^3 = .000125

If the probability of a transmitter failing during its first 2 hours of service is 0.1, then the probability of two failing during that time is 0.1^2 = 0.01, reliability of two units for 2 hours of service is 90%
If one is repaired before entering the 2nd hour, then it starts anew with its P(f) = 0.05 and P(other failing)= 0.1 P(of only 1 failing) is 0.05 * 0.01 = .005; reliability is 99.5%

Independent events are seldomly independent.
 
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