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A software program consists of two modules M1 and M2 that can fail independently, but never simultaneously. The program is considered to have failed if any of these modules fails. Both the modules are 'repairable' and so the program starts working again as soon as the repair is done. Assume that the mean time to failure (MTTF) of M1 is T1 with a mean time to repair (MTTR) of R1. The MTTF of M2 is T2 with an MTTR of R2. What is the availability of the overall program given that the failure and repair times are all exponentially distributed random variables?

  1. ((T1T2)/(T1R1 + T2R2))
  2. ((R1R2)/(T1R1 + T2R2))
  3. ((T1T2)/(T1T2 + T1R1 + T2R2))
  4. ((T1T2)/(T1T2 + T1R2 + T2R1))
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5 votes
5 votes

Answer should be (D) ,

Reliability can be defined as the probability that a system will produce correct outputs up to some given time t. Reliability is enhanced by features that help to avoid, detect and repair hardware faults. A reliable system does not silently continue and deliver results that include uncorrected corrupted data. 

Availability means the probability that a system is operational at a given time, i.e. the amount of time a device is actually operating as the percentage of total time it should be operating. [email protected]http://en.wikipedia.org/wiki/Reliability,_availability_and_serviceability_(computing)

Now ,

since "The program is considered to have failed if any of these modules fails" so both module in series then the reliability of two program module ,

Reliability = Reliability of M1* Reliability of M2 

Reliability of M1 = (MTTF1) /(MTTF1+MTTR1) = T1 /(T1+R1) ;

Reliability of M2 = (MTTF2) /(MTTF2+MTTR2) = T2 /(T2+R2) ,

so , 

Reliability = Reliability of M1* Reliability of M2  = (T1 /(T1+R1)) * (T2 /(T2+R2))

               = (T1*T2) / ((T1+R1)*(T2+R2)) = (T1*T2) / (T1*T2+T1*R2+R1*T2+R1*R2) ,

 { since "modules M1 and M2 that can fail independently, but never simultaneously" , so  R1*R2 can be ignore for high availability of the program} 

therefore availability of overall program =  (T1*T2) / (T1*T2+T1*R2+R1*T2)

[email protected]http://www.sars.org.uk/old-site-archive/BOK/Applied%20R&M%20Manual%20for%20Defence%20Systems%20(GR-77)/p4c06.pdf

correct me ..... ???

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4 votes
4 votes
Fraction of Time Available = Duration for which both are working / Duration in all 3 possible cases (pass-pass,fail-pass,pass-fail)

Duration for which both are working = both are in amidst of time period and going to fail = T1*T2

Duration for all 3 possible cases =  T1*T2 + T1*R2 + T1*R3
Answer:

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