A company can employ 3 workers. Each worker independently stays on the job for an exponentially distributed time with a mean of one year and then quits. When a worker quits, it takes the company an exponentially distributed time with a mean of 1/10 of a year to hire a replacement for that worker (independently of how long it takes to replace other workers). If the company takes in \$1000 per day in revenue when it has 3 workers, \$800 per day when it has 2 workers, and \$500 per day when it has 0 or 1 workers, what is the company’s long-run average daily revenue?
If we take every individual's working as a timeline, and let $X_1$ time span is the first worker worked then he was replaced by another worker after $Y_1$ days and then the replaced worker works $X_2$ days, in this way, $\sum_{i=1}^{N(t)} (X_i+Y_i)$, would be total working before time $t$ by replacing a single individual worker. But here we have $3$ of them, where I had gotten confused totally, because to find average revenue per day, we need to add them in parallel.
So, it's like parallel and series circuit thing (as much as I can understand it)
Can anyone please solve it problem? (Maybe I am over complicating it!)
Let $N(t)$ be the number of workers employed at time $t$, then $\{N(t):t\geqslant0\}$ is a continuous-time Markov chain on $\{0,1,2,3\}$ with transition rates $$ q_{ij} = \begin{cases} 10,& j=i+1\\ 1,& j=i-1\\ 0,& \text{otherwise}. \end{cases} $$ This is a birth-death chain, so we may find the stationary distribution by solving the detailed balance equations $\pi_iq_{ij}=\pi_jq_{ji}$; that is: \begin{align} 10\pi_0 &= \pi_1\\ 10\pi_1 &= \pi_2\\ 10\pi_2 &= \pi_3, \end{align} Together with $\sum_{i=0}^3 \pi_i=1$, this yields $$ \pi = \begin{pmatrix}\frac1{1111}&\frac{10}{1111}&\frac{100}{1111}&\frac{1000}{1111} \end{pmatrix}. $$ We compute the long-run average daily revenue by multiplying $\pi$ with the vector of revenues: $$ \begin{pmatrix}\frac1{1111}&\frac{10}{1111}&\frac{100}{1111}&\frac{1000}{1111} \end{pmatrix}\begin{pmatrix}500\\500\\800\\1000 \end{pmatrix} = \frac{1085500}{1111}\approx 977.0477. $$