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Problem 4. Two communication systems (shown below) are composed from several links, where, for proper operation,...

Problem 4. Two communication systems (shown below) are composed from several links, where, for proper operation, a connection must be available between two end points of each system. For a link to work properly, it must provide a connection across its end points. The links are assumed to fail independently of each other. Each link is numbered with i, and we assume that the probability that link fails is equal to p Find the probability of failure of each system, where Pi = 0.1,Pa-Pa-0.05, p.-ps-0.15,Pe-P7-02. 2 4 System (a) 4 2 System (b) Figure 1: Communication Systems

Problem 4. Two communication systems (shown below) are composed from several links, where, for proper operation, a connection must be available between two end points of each system. For a link to work properly, it must provide a connection across its end points. The links are assumed to fail independently of each other. Each link is numbered with i, and we assume that the probability that link fails is equal to p Find the probability of failure of each system, where Pi = 0.1,Pa-Pa-0.05, p.-ps-0.15,Pe-P7-02. 2 4 System (a) 4 2 System (b) Figure 1: Communication Systems
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Answer #1

system a:

P(subsystem 1-2-3 works) =(1-p1)*(1-p2)*(1-p3)=0.9*0.95*0.95=0.81225

P(subsystem 4-5 works)=(1-p4)*(1-p5)=0.85*0.85=0.7225

hence system (a) fails =P(subsystem 1-2-3 fails and subsystem 4-5 fails)

=(1-0.81225)*(1-0.7225)=0.0521

system b.

P(subsystem 4-5 works)=1-p4*p5 =1-0.15*0.15=0.9775

hence P(subsystem 1-4-5) works =P(1 works)*P(subsystem 4-5 works)=0.9*0.9775=0.87975

P(subsystem 6-7 works)=1-p6*p7 =1-0.2*0.2=0.96

P(subsystem 2-3-6-7 works)=0.95*0.95*0.96=0.8664

hence P(system (b) fails)=P(subsystem 6-7 fails and subsystem 2-3-6-7 fails)

=(1-0.87975)*(1-0.8664)=0.016065

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