An airline reservation system has two computers only one of which is in operation at any given time. A computer may break down on any given day with probability $p$. There is a single repair facility which takes $2$ days to restore a computer to normal. Form a markov chain by taking as states the pairs $(x, y)$ where $x$ is the number of machines in operating condition at the end of a day and $y$ is $1$ if a day's labor has been expended on a machine not yet repaired and $0$ otherwise.
$$ \begin{array}{c c} & \begin{array}{c c } (2,0) & (1,0) & (1,1) & (0,1) \\ \end{array} \\ \begin{array}{c c}(2,0)\\(1,0)\\(1,1)\\(0,1) \end{array} & \left[ \begin{array}{c c} q & p & 0 & 0 \\ 0 & 0 & q & p \\ q & p & 0 & 0 \\ 0 & 1 & 0 & 0\\ \end{array} \right] \end{array} $$
I don't understand what $y$ represents in the phrase, "If a day's labor has been expended on a machine not yet repaired". To me a machine not yet repaired is a broken machine and if the machine is broken then how can you expend a day of labor on it?
Can someone explain the intuition behind this more clearly? For instance, what does $(1,0) \rightarrow (1,1)$ mean in terms of the real physical thing this is modeling?
A machine can be broken, but the repair facility can fix it. However, it takes 2 days to fix. We need $y$ to keep track of the machines that have broken down and are in the process of being repaired, having been worked on for one day; one more day's repair will get it back in working order. Thus if the state is $(1,0)$, we have one working machine and one broken machine that the repair facility has not yet worked on. It will start now, so that the next day's state will have $y=1$. With probability $q$, the one working machine stays in working order so the next day's state is $(1,1)$. But with probability $p$, the working machine breaks down so the next day's state is $(0,1)$.