A traffic network is shown in the figure below. The reliability (probability of

Question

A traffic network is shown in the figure below. The reliability (probability of free-flowing without congestion) of link B is 0.75, link D is 0.8, and link E is 0.85; these are all statistically independent of each other and of other links in the network. The overall reliability of Link A is 0.65. However, if A is congested, then the reliability of C is 0.4, whereas if A is not congested, C has a reliability of 0.7.

Please answer the following questions using the terminology "A = A works; B = B works" and so on.

A) How do you write the event “A vehicle can travel from point 1 to point 2 without encountering congestion." in terms of A, B, C, D, and E? You may use "U" (capital u) to indicate unions, and "^" (caret, above the 6 on most keyboards) to indicate intersections.

B) Continuing with the traffic question, find the probability of vehicles being able to travel freely from point 1 to point 2. (3 decimal places, 0.XXX)

C) Continuing with the traffic question, if vehicles can flow freely, what is the probability that A is congested? (3 decimal places, 0.XXX)

Explanation / Answer

A) There are two paths of reaching 2 from 1. Path 1 is through A and B Path 2 is through C, D and E Hence we write the event “A vehicle can travel from point 1 to point 2 without encountering congestion." as (A^B) U (C^D^E) B) P(Vehicles being able to travel freely from point 1 to point 2) = P(Path 1) + P(Path 2) = P(A^B) + P(C^D^E) = P(A) x P(B) + P[(A) x P(C/A )] x P(D) x P(E ) + [P(A') x P(C/A' )] x P(D) x P(E ) Note that the free-flow probability of C depends on whether A is congested or not Hence we consider P(C/A) as well as P(C/A') Given P(A) = 0.65, P(B) = 0.75, P(C/A) = 0.7, P(C/A') = 0.4 , P(D) = 0.8, P(E ) = 0.85 Hence P(A') = 0.35 = 0.65 * 0.75 + 0.65 * 0.7 * 0.8 * 0.85 + 0.35 * 0.4 * 0.8 * 0.85 = 0.8921 P(Vehicles being able to travel freely from point 1 to point 2) = 0.892 C) If vehicles can flow freely, to find the probability that A is congested. We consider here the path 2 , since the free-flow of path 2 especially C depends on the congestion/non-congestion of A We have to find P(A is congested/path C is free-flowing) From Baye's theorem P(A'/C) = P(C/A')* P(A') / [P(C/A')P(A') + P(C/A)P(A)] = (0.4) * (0.35) / (0.4 * 0.35 + 0.7 * 0.65) = 0.2353 P(A is congested if vehicles can flow freely) = 0.2353

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