Question

Answer question in RED box:

I. A computer has five processing units (PUs). The lifetimes of the PUs are iid Exp(A) random variables. When a PU fails, the computer tries to isolate it automatically and reconfigure the system using the remaining PUs. However, this process succeeds with probability c, called the coverage factor. If the reconfiguring succeeds, the system continues with one less PU. If the process fails, the entire system crashes. Assume that the reconfiguring process is instantaneous and that once the system crashes it stays down forever. Let X(t) be 0 if the system is down at time t; otherwise it equals the number of working PUs at timet . Model X (t) t2 0 as a CTMC, and show its rate matrix. 2. Consider the five-processor system described in Problem 1. Suppose that the system mainte- nance starts as soon as the system crashes. The amount of time needed to repair the system is an Exp(A) random variable, at the end of which all five processors are functioning. Model this system as a CTMC and show its rate matrix 3. Consider Problem 1 again. Suppose the mean life time of a processor is 2 years and the coverage factor is .94. What is the probability that the five-processor system functions for 5 vears without fail? Assume all s are operating at time 0

Answer 1

ANSWER:

Given that,

• For a exponential distr,

P(X<x) = 1 - e^(-x/u)

For a single PU,

P(PU fails before 5years) = P(x<5) = 1 - e^(-5/2) = 0.918

=> p = 0.918

For the computer,

P(5 yrs w/o fail) = P(no PU fails) + P(1 PU fails but reconfigures) + P(2 PU fails but reconfigures) + P(3 PU fails but reconfigures) + P(4 PU fails but reconfigures)

= (1-p)^5 + 5C1*cp*(1-p)^4 + 5C2*(cp)^2*(1-p)^3 + 5C3*(cp)^3*(1-p)^2 + 5C4*(cp)^4*(1-p)

= 0.082^5 + 5*(0.94*0.918)*0.082^4 + 10*(0.94*0.918)^2*0.082^3 + 10*(0.94*0.918)^3*0.082^2 + 5*(0.94*0.918)^4*0.082

=0.2748448

= 0.275 or 27.5% approximately

according to HOMEWORKLIB RULES 1 an 2 can be answered upto my knowledge