Application of queueing theory Jake's Machine Shop contains a grinder for sharpening the machine cutting tools. A decision must now be made on the speed at which to set the grinder. 
The grinding time required by a machine operator to sharpen the cutting tool has an exponential distribution, where the mean 1/µ can be set at 0.5 min, 1 min, and 1.5 min, depending upon the speed of the grinder. The running and maintenance costs go up rapidly with the speed of grinder, so the estimated cost per minute is 1.60 dollars for providing a mean of 0.5 min, 0.40 dollars for a mean of 1 min, and 0.20 dollars for a mean of 1.5 min. 
The machine operates arrive randomly to sharpen their tools at a mean rate of 1 every 2 min. The estimated cost of an operator being away from his machine to the grinder is 0.80 dollars per min. 
I want to determine which grinder speed minimizes the expected total cost per min. 
Can anyone enplane to me how to approach problems like this one? I am self studying the Queueing system and Markov chain, just because it looks interesting to me. I am actuarial student, I do not need this, but I would really like to understand this bc it is really interesting..I choose this problem randomly from the book. Thank you in advance!    
 A: The grinder can be in two states -- either in use, or not in use. The period of being in use lasts for time $1/\mu$ per minute on average, so the rate here is $\mu$. When the grinder is not in use the operator arrives at rate 0.5 per minute (1 per 2 minutes). Therefore we have a continuous time Markov chain on the state space {in use, not in use} with transition rate matrix
$$Q=\begin{pmatrix}
-\mu & \mu \\
1 & -1 \end{pmatrix}.$$
We need to compute the stationary distribution of this matrix. Do you know how to do this? We seek a vector $\pi=\begin{pmatrix}\pi_1 &\pi_2\end{pmatrix}$ which satisfies $\pi Q=0$ and $\pi_1+\pi_2=1$. The solution is
$$ \pi_1 = \frac{1}{1+\mu} \text{ and } \pi_2 = \frac{\mu}{1+\mu}.$$
The costs of this system depend on the choice of $\mu$, the cost of being in the in use state 
are given in the question and depend on the choice of $\mu$, let's call them $c(\mu)$, and the cost of being in the not in use state is 0.8. Therefore the total cost is
$$\frac{c(\mu) + 0.8\mu}{1+\mu}$$
which you seek to minimise over your available choices of $\mu$.
