# M/M/1 priority queueing system (?), an exercise

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