How to find the intensity matrix $Q$ in a reliability process (Markovs chains)?

Question

I'm trying to understand how the intensity matrix can be set up in this problem:

Question A system has two components such that if one of them breaks, it will stop working. Each of the components' lifespan is exponentially distributed, i.e. $Exp(1/400)$ (and both of them are independent of each other). A broken component can be repaired in $Exp(1/20)$ time and there are two people who can repair them, they can be repaired simultaneously. All times are independent of each other. Find the availability of the system, i.e. probability that the system works after some time.

In the solution below, we will use $\lambda=1/400$ and $\mu=1/20$.

Solution

My initial thought was to set up four states, i.e.

1. Both components work
2. The first component does not work
3. The second component does not work
4. Both components don't work

however, it seems like this can be simplified into three states as shown below. This does sound reasonable, however, what does not feel intuitive is the intensity matrix $Q$, in particular row 1 and 3. Are they correct?

$$\begin{array}{l} {S_0} = \text{both components work}\\ {S_1} = \text{one component works the other doesn't}\\ {S_2} = \text{both components don't work}\\ \\ Q = \left( {\begin{array}{*{20}{c}} { - 2\lambda }&{2\lambda }&0\\ \mu &{ - \mu - \lambda }&\lambda \\ 0&{2\mu }&{ - 2\mu } \end{array}} \right) \end{array}$$

Row 1 case In my mind, if we are in $S_0$, i.e. both components work, it means three things can happen: either one of the components breaks or both of them (or nothing happens). In this case, the first row should be $(-2\lambda, \lambda, \lambda)$.

However, in the solution, they say two things can happen: either the first component breaks or the second, and this means that the transition to $S_1$ is $\lambda +\lambda$. In one way, this makes sense, because we can think of the addition as "either or, or both", i.e. it does take into account that both break. However, it still sounds strange why we cannot transition into $S_2$ from $S_0$, i.e. that both components fail simultaneously.

Question Is there a fault in my reasoning or in the solution from the book?

Answer 1

The two Markov formulations (the 4-state and the 3-state) yield indeed the same result. The reason we cannot transition into $S_2$ from $S_0$ is that the probability of two changes happening at the same time is assumed to be $0$.