1. 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 failure is , a repeater failure is , and a bridge failure is , derive an expression for parts (a) to (d).
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) to (d) for
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To derive expressions for parts (a) to (d), we need to understand how the failures of individual components affect the overall system reliability. Let's break it down step by step:
a) Probability of failure of system A:
Since System A consists of a single ring with 100 stations, we need to calculate the probability that at least one station or repeater fails. If the probability of a link failure is P(link), and the probability of a repeater failure is P(repeater), we can use the principle of complementary probability. The reliability of a single station or repeater would be (1 - P(link)) * (1 - P(repeater)), assuming failure events are independent. For the entire system, we have 100 independent stations or repeaters, so the probability of failure of system A is:
(P(failure of system A)) = 1 - (1 - P(link))^(100) * (1 - P(repeater))^(100)
b) Probability of complete failure of system B:
System B consists of four 25-station rings linked by a bridge. In order for the entire system to fail, either all four rings must fail or the bridge must fail. Assuming these events are independent, we can calculate the probability of complete failure of system B as:
(P(complete failure of system B)) = P(all rings fail) + P(bridge fails)
P(all rings fail) = (P(failure of a single ring))^4 = (1 - (1 - P(link))^(25))^4
P(bridge fails) = P(failure of bridge) = P(bridge)
c) Probability that a particular station will find the network unavailable, for systems A and B:
For system A, the network is unavailable to a specific station if either the station itself fails or any of the repeaters or links leading to that station fail. Using the principle of complementary probability, we can express this as:
(P(network unavailable for system A)) = 1 - (1 - P(repeater)) * (1 - P(link))^99
For system B, a station will find the network unavailable if either the station itself fails, any of the repeaters, links, or bridges leading to that station fail. Again using the principle of complementary probability, we have:
(P(network unavailable for system B)) = 1 - (1 - P(repeater)) * (1 - P(link))^24 * (1 - P(bridge))
d) Probability that any two stations selected at random will be unable to communicate, for systems A and B:
For system A, the probability that two randomly selected stations cannot communicate can be calculated using the probability that the network is unavailable for both stations:
(P(cannot communicate for system A)) = (P(network unavailable for system A))^2
For system B, the probability that two randomly selected stations cannot communicate can be calculated similarly as:
(P(cannot communicate for system B)) = (P(network unavailable for system B))^2
e) To compare the values of parts (a) to (d) for systems A and B, you would need specific values for P(link), P(repeater), and P(bridge) provided in the question. With those values, you can substitute them into the respective expressions for system A and system B to compare the probabilities.