Random processes: Repair time I have a question that is to do with qeueing theory and repair times:
Assume that a small office has 4 printers. Each printer breaks down independently of the
other printers and independently of the past history of the printer, on average once every 50
working days. When a printer breaks down, the repair time of that printer is exponentially
distributed with an average of 20 working days. Any number of printers can be repaired at
the same time, so that the repair time of a printer does not depend on how many printers are
broken down. Draw a transition diagram of the process $X$ when $X(t)$ denotes the number of
working printers on day $t$. Find the stationary distribution of the process. What percentage of time are none of the printers working?
I am not too sure where to start. I assume that it is a birth/death process?
 A: Let $\lambda_R$  = rate of printer repair, $\lambda_B$ = rate of printer break down. Note that these are Poisson processes (which are known to have exponential wait times and are "memory-less"). A poisson process is any process that has a constant probability over time of an event happening. In a Poisson process, on average $\lambda$ events happen in a given interval.
Let's start by looking at one printer. It follows the following propensity functions:
Broken $\xrightarrow{\lambda_R}$ Working, Working$\xrightarrow{\lambda_B}$ Broken
Written as a markov process,
$\frac{dP(X=0)}{\partial{t}}=\lambda_B P(X=1)-\lambda_R P(X=0)$, 
where X is the number of working printers (and we're only looking at 1 printer). Since we're looking at the stationary distribution, we set the derivative equal to 0. Then,
$0=\lambda_B P(X=1)-\lambda_R P(X=0)$
Dropping the $X=$ notation, so $P(X)$ for $X=1 = P(1)$, we find that
$\frac{P(1)}{P(0)}=\frac{\lambda_R}{\lambda_B}$. Then noting by conservation of probability that $P(0)+P(1)=1$, and substituting in $P(0)$, we find
$P(0)=\frac{1}{1+\frac{\lambda_R}{\lambda_B}}$
Therefore, 
$P(1)=\frac{\frac{\lambda_R}{\lambda_B}}{1+\frac{\lambda_R}{\lambda_B}}$
Testing our intuition, if the repair rate were very large and the breaking rate were small, you can see that the limit of $P(1)\rightarrow1$
Note that the solution can be thought of as a Bernoulli process with distribution Bernoulli($\frac{\frac{\lambda_R}{\lambda_B}}{1+\frac{\lambda_R}{\lambda_B}}$). What is the distribution of the sum of four of these distributions? Let $B_P$ be a bernoulli variable with "success" parameter $p_w=P(1)=\frac{\frac{\lambda_R}{\lambda_B}}{1+\frac{\lambda_R}{\lambda_B}}$
$Z=B_p+B_p^\prime+B_p^{\prime\prime}+B_p^{\prime\prime\prime}$
Where primes denote independent Bernoulli variables. Remember that sums of independent random variables correspond to their convolution. Fortunately, convolutions of independent Bernoulli variables have been worked out (for example, see case of sum of 2 Bernoulli RVs). Namely,
$\sum\limits_{i=1}^N Bernoulli(p_w) = Binomial(N,p_w)$. Where N is the total number of printers here, and $p$ is the success rate as defined above for this problem.
Then the question of what is the probability at any given time of 0 printers functioning is plug and chug into the Binomial PDF.
$P(X=0)=\binom{4}{0} p_w^0 (1-p_w)^4 = (1-p_w)^4$ So plugging in $\lambda_B=\frac{1}{50}$, $\lambda_R=\frac{1}{20}$, I get the .6663% of the time all printers are broken. Let me know if you have questions.
Also, note that you can solve this by writing out the full transition probabilities between the states 0 printers through 4 printers functioning, and you should get the same answer. Computing P(0) in that case is slightly more involved but the approach is the same.
