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In the early days of its development the ARPA network looked something like the diagram shown below. Assume the probability of failure of any link in the network is p, and the link failures are independent. Let A, B, C, D, E be the events that the corresponding link fails. (a) Write an algebraic expression for the failure to connect between subscribers in Los Angeles and Boston. (b) Determine the probability of establishing a connection between these users. (c) Let p = 30.01. Compute the probability of establishing the above connection.

Explanation / Answer

In given network diagram it can be said that if at least 1 of the connections in a given path of 2 fails, the entire network will fail in that path. There are 2 alternatives:

1. ABE

2. CD

(a) probability of failure:

= [failure in path 1] + [failure in path 2]

=[failure in any 1 activity + failure in any 2 activities + failure in all 3 activities] + [failure in any 1 activity + failure in all 2 activities]

P= [ (3C1)*p+(3C2)*p^2 + (3C3)*p^3 ] + [(2C1)*p + (2C2)*p^2]

where C stands for Comibations

(b) prob of connection= 1-prob of failure (since both are mutually exclusive events)

Prob of connection = 1- [ (3C1)*p+(3C2)*p^2 + (3C3)*p^3 ] + [(2C1)*p + (2C2)*p^2]

(c) if p=0.01

Prob of connection = 1- [ (3C1)*p+(3C2)*p^2 + (3C3)*p^3 ] + [(2C1)*p + (2C2)*p^2]

=1- [COMBIN(3,1)*0.01 + COMBIN(3,2)*0.01*0.01 + COMBIN(3,3)*0.01^3 +COMBIN(2,1)*0.01 +COMBIN(2,2)*0.01 ^2 ]

(in excel you can paste this and get ans)

= 0.9496

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