A telecommunications network has two types of failure: short lasting and long la

Question

A telecommunications network has two types of failure: short lasting and long lasting. The short lasting failures occur with the probability of 0.4 per day. The restoration after the failure occurs next day, with probability 0.9. The long lasting failures occur not as frequently, with probability of 0.1 per day, but the restoration after this type of failure occurs next day with probability 0.2. Two types of failure do not communicate with each other. Consider the network status as a three-state Markov chain, with the state space, X = {N,S,L). Here N denotes the normal functioning, while S and L represent the short and long lasting failures 1. Show the transition probability matrix per day for this network. Normal Short Long Normal Short ong 2. Find the probability that the long lasting failure is not resolved within three consecutive days, PIX (1) = X(2)-x(3) = L|X(0) = N) 3. Find the probability that at least one failure occurs after normal functioning, PIX (I) = {S or L} IX(0) = N]

Explanation / Answer

1.

The short lasting failures occur with probability of 0.4 per day. So, the transition fom state N to S is 0.4.

The long lasting failures occur with probability of 0.1 per day. So, the transition fom state N to L is 0.1.

The transition from state N to N is 1 - (0.4 + 0.1) = 0.5

The restoration after the short failure occurs next day with probability 0.9. So, the transition fom state S to N is 0.9.

The transition from state S to S is 1 - 0.9 = 0.1

The restoration after the long failure occurs next day with probability 0.2. So, the transition fom state L to N is 0.2.

The transition from state N to N is 1 - 0.2 = 0.8

So, the transition probability matrix per day for this network is,

2.

As, the Markov chain transition dependeds only on the current state,

P[X(1) = X(2) = X(3) = L | X(0) = N) = P(X(1) = L | X(0) = N) * P(X(2) = L | X(1) = L) * P(X(3) = L | X(2) = L)

= 0.1 * 0.8 * 0.8 = 0.064

3.

As, State S and State L are disjoint,

P[X(1) = {S or L} | X(0) = N] = P[X(1) = S | X(0) = N] + P[X(1) = L | X(0) = N]

= 0.4 + 0.1 = 0.5

Normal Short Long Normal 0.5 0.4 0.1 Short 0.9 0.1 0 Long 0.2 0 0.8

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