System A consists of a single ring with 100 stations, one per repeater. System B
ID: 643391 • Letter: S
Question
System A consists of a single ring with 100 stations, one per repeater. System B consists of four 25 stations rings linked by a bridge. If the probability of a link failureis Pl, a repeater failure is Pr , and a bridge failure is Pb , derive an expression for parts (a) to (e).
a) Probability of failure of system A.
b) Probability of complete failure of system B.
c) Probability that a particular station will find the network unavailable, for
systems A and B.
d) Probability that any two stations selected at random will be unable to
communicate for systems A and B.
e) Compare values of parts (a) and (b) for Pl = Pb = Pr = 10^-2 .
Explanation / Answer
a) Probability of failure of system A.
The probability of failing system A if any link fails or any repeater fails.
P(A fails) = 1 – (1 – Pl)100(1-Pr)100.
b) Probability of complete failure of system B.
The probability of failing system A if any link fails or any repeater fails in system B.
P[B fails] = [1 – (1-Pl)25(1-Pr)25]3
C) Probability that a particular station will find the network unavailable, for systems A and B.
For System A:
P[x in A; and A is unavailable] = 1 – (1-Pl)100(1-Pr)100.
For System B:
P[x in B; and ‘s ring in unavailable] = 1 – (1-Pl)25(1-Pr)25 .
d) Probability that any two stations selected at random will be unable to
communicate for systems A and B.
System A:
Pr[X and Y have no communication] = 1 – (1-Pl)100(1-Pr)100;
System B:
P1: Pr[X and Y on the same ring] = 1/3;
P2 = Pr [X and Y not on same ring] = 2/3;
P3 = Pr[X and Y on same ring and no communication] = 1 – (1-Pl)25(1-Pr)25;
P4 = Pr[X and Y have no communication, X and Y are not on the same ring]
= P[X’s ring doesn’t fail]* P[Y’s ring does not fail] * P[bridge does not fail]
= [(1-Pl)25(1-Pr)25]2(10Pb)
= 1 – (1-Pl)50(1-Pr)50(1-Pb);
Pr[No communication between X and Y] = P1P3 + P2P4
= (1/3)( 1 – (1-Pl)25(1-Pr)25) + (1/3)( 1 – (1-Pl)50(1-Pr)50(1-Pb))
e) Compare values of parts (a) and (b) for Pl = Pb = Pr = 10-2
P[A fails] = 1 – (1 – 10-2)100(1-10-2)100
= 1 – (0.99)100(0.99)100
= 1 – (0.366032)(0.366032)
P[A fails]= 0.86602.
P[B fails] = [1 – (1-10-2)25(1-10-2)25]3
= [1 – (0.99)25(0.99)25]3
= [1 – (0.777821)(0.777821)]3
= (0.394994)3
P[B fails]= 0.061627
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