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A software program consists of two modules M_{1} and M_{2} 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 M_{1} is T_{1} with a mean time to repair (MTTR) of R_{1}. The MTTF of M_{2} is T_{2} with an MTTR of R_{2}. What is the availability of the overall program given that the failure and repair times are all exponentially distributed random variables?

- ((T
_{1}T_{2})/(T_{1}R_{1}+ T_{2}R_{2})) - ((R
_{1}R_{2})/(T_{1}R_{1}+ T_{2}R_{2})) - ((T
_{1}T_{2})/(T_{1}T_{2}+ T_{1}R_{1 + }T_{2}R_{2})) - ((T
_{1}T_{2})/(T_{1}T_{2}+ T_{1}R_{2 + }T_{2}R_{1}))

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 ..... ???