Consider a system with two components. We observe the state...

Consider a system with two components. We observe the state of the system every hour. A given component operating at time n has probability p of failing before the next observation at time n + 1. A component that was in a failed condition at time n has a probability r of being repaired by time n + 1, independent of how long the component has been in a failed state. The component failures and repairs are mutually independent events. Let Xn be the number of components in operation at time n. The process {Xn, n = 0,1,...} is a discrete time homogeneous Markov chain with state space I = 0, 1, 2. a) Determine its transition probability matrix, and draw the state diagram. b) Obtain the steady state probability vector, if it exists.

Transcript

00:01 In part a, we are asked to determine the transition probability matrix and draw the state diagram.

00:09 Here we have three states, 0, 1, and 2.

00:23 First we consider starting from 0.

00:26 So when we are at 0, the probability of going to 2, which means that both machines become failed next time.

00:39 So one machine becoming failed, it has probability p and two machines becoming failed is p squared.

00:52 The probability of staying at zero, which means that both machines stay operating.

01:01 So, becoming f -p is the probability of becoming fail, so staying operating, it has probability 1 minus p, and we want both machines to stay operating, so we square.

01:18 Next is the probability of going to 1.

01:23 It means that one machine stays operating, the other becoming failed.

01:30 So we choose which one to stay operating.

01:35 So there's a 2, and the one staying, the one becoming failed has probability p, the one staying operating, so 1 minus p here.

01:50 Then use similar argument to deal with the, starting point two so when we are at two both machines failed so the probability of going to zero which means that next time both machines are repaired it is r square because r is the probability that one machine is repaired square means both machines are repaired and the next time that both machines are not repaired you similar i one machine is not repaired, it has probability 1 minus r and the square for the case that it both are not repaired.

02:48 And the probability of going to 1, we use similar argument, choose 1 to be repaired, and then are the probability of the one being repaired and the other one not repaired 1 minus r.

03:11 Next we deal with starting from 1.

03:16 So starting from 1 we can go to 0.

03:23 So the probability of 1 going to 0, we deal with one machine is operating the other, it's failed.

03:36 And going to 0 means the failed one is repaired and the operating one stay operating.

03:45 So we have 1 minus p for the operating one staying operating and the operating.

03:51 And r for the failed one being repaired.

04:03 Probability 1 go into 2.

04:07 It is used similar argument from 1 to 2.

04:13 It means that the failed one state failed, so it is not repaired, 1 minus r, and the operating one becomes failed.

04:27 So p here.

04:29 Finally, we determine this one.

04:33 Going from one to one.

04:37 This has two cases.

04:39 The one is the failed big and the operating one staying failed.

04:52 Operating one staying operating or it can be like this.

04:55 The failed one become operating, operating one become failed.

04:59 So we need to consider two cases.

05:02 Or we can just use this method.

05:05 One minus go to zero.

05:08 Probably going to 0 is 1 minus pr and then minus the probability going to 2.

05:17 Then we have this is 1 minus p minus r plus 2 pr.

05:35 So having this diagram then we write down the transition probability matrix...

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