Given the two servers $X$ and $Y$ with buffer capacity $K_1$ and $K_2$. The calls come in according to a Poisson process with rate $\lambda$ and the call duration are i.i.d $exp(\mu)$. Assume any calls came in are routed to server $X$ with probability $p$ and to server $Y$ with probability $1-p$.
(a) Find the long-run probability of having a call blocked.
(b) Find $p$ so that it minimizes the long-run probability of having a call blocked.
My attempt
(a) Since the calls come in follow the splitting Poisson process, we have two independent $M/M/1/c$ queue with $c= K_1+1$ and $c = K_2+1$ for $X$ and $Y$, respectively. Since $X$ and $Y$ has a buffer capacity of $K_1$ and $K_2$ and follows a Poisson process with rates $\lambda_1 = p\lambda$ and $\lambda_2 = (1-p)\lambda$, respectively, the incoming call gets blocked if and only if the number of calls in the system is $(K_1+1)(K_2+1)$ (the $+1$ includes the call in service).
Now, the long-run probability of being full in $M/M/1/(K_1+1)$ queue is $(1-\rho)\rho^{K_1+1} = (1-\frac{p\lambda}{\mu})(\frac{p\lambda}{\mu})^{K_1+1}$ ($\frac{p\lambda}{\mu} < 1$). Similarly, for $M/M/1/(K_2+1)$ queue, the long-run probability of being full is $(1-\rho)\rho^{K_2+1} = (1-\frac{(1-p)\lambda}{\mu})(\frac{(1-p)\lambda}{\mu})^{K_2+1}$ ($\frac{(1-p)\lambda}{\mu} < 1$).
Thus, the long-run probability of having a call blocked is
$$(1-\frac{(1-p)\lambda}{\mu})(\frac{(1-p)\lambda}{\mu})^{K_2+1} (1-\frac{p\lambda}{\mu})(\frac{p\lambda}{\mu})^{K_1+1}$$.
(b) From (a), it is easy to see that the long-run probability of having a call blocked $\geq 0$, and the equality occurs at $p=0$ or $p=1$, we conclude that $\fbox{p=0, 1}$ is the minimizer of that long-run probability.
My question. Is my attempt above incorrect for part (b)? It just seems trivial, but I could not see where it went wrong.