can someone help me please solving this?
Consider a switch with two input links and one outgoing transmission link. Data packets arrive at the first input link according to a Poisson process with mean $\lambda_1$ and voice packets arrive at the second input link also according to a Poisson process with mean $\lambda_2$. Determine the total transit time when a packet arrives at either input until its transmission completion if the service time of both the data and voice packets are exponentially distributed with mean rates $\mu_1$ and $\mu_2$, respectively.
Since there are any assuntions about which one of the queue is the higher priority queue, I can't go on and find formulas needed. Should I make my assumption that voice channel is preferred and then
$W_v = R / (1-\rho_v)$
and
$W_d = R / ((1 - \rho_v)(1 - \rho_v - \rho_d))$,
with
$R = \rho / \mu$, $\rho = \rho_1 + \rho_2$, $\mu = \mu_1 + \mu_2$,
and calculate
$T_i = W_i + \bar x_i$?
Or should I consider that, because no priority is specified, and due to memoryless propriety of exponential distribution for service times, we simply have
$N= N_1 + N_2 = \lambda_1/(\mu_1 - \lambda_1) + \lambda_2/(\mu_2-\lambda_2)$
and then, for Little's formula,
$T_1 = N/\lambda_1$, $T_2=N/\lambda_2$.
Am I wrong? What do you think? Thank you all
