Given a M/M/1 queueing system where vehicles are admitted an on-ramp at a rate of $2$ vehicles/s and serviced by highway admission rate of $2.5$ vehicles/s. What is the fraction of time that the on-ramp holds $10$ or more vehicles?
Not sure how to approach this homework problem.
The best I can come up with is that the fraction of the time that system has $N$ or more vehicles is the probability that the steady state system has $N$ or more vehicles, in this case $10$.
So for ${\Bbb P} (N>0) = U$, where $U$ is utilization and $U = \dfrac{\lambda}{\mu} = \dfrac{2}{2.5} = 0.8$. So for...
$$ \begin{aligned} {\Bbb P} (N>1) &= \left(\frac{\lambda}{\mu}\right)^2 \\ {\Bbb P} (N > 2) &= \left(\frac{\lambda}{\mu}\right)^3 \\ {\Bbb P} (N > 3) &= \left(\frac{\lambda}{\mu}\right)^4 \\ \vdots \\ {\Bbb P} (N > 9) &= \left(\frac{\lambda}{\mu}\right)^{10} \\ \vdots \\ {\Bbb P} (N > k-1) &= \left(\frac{\lambda}{\mu}\right)^k \end{aligned} $$
Does this seem like the approach?
It's right.
To derive the formula, first we have
$$P(N=k)=(1-\rho)\rho^k$$
Wikipedia
So, $P(N\ge0)=\sum_\limits{i=0}^\infty (1-\rho)\rho^i=(1-\rho)\sum_\limits{i=0}^\infty \rho^i=\frac{1-\rho}{1-\rho}=1$, by using the geometric series.
Now,
$P(N>k) = P(N\ge0)-P(0\le N\le k)$
$=1-(1-\rho)(1+\rho+\rho^2+\dots+\rho^k)$
$=1-(1-\rho)\frac{1-\rho^{k+1}}{1-\rho}$
$=\rho^{k+1}$