There are two arriving processes: A and B. one arriving rate is lambda1, the second arriving rate is lambda2.
The serving rate is u.
All A's are in one FIFO queue. All B's are in the second FIFO queue. These customers are dequeued from the two queues in roundrobin way to be served.
but if one queue is empty, dequeue from the other queue. The arrivals are independent and Poisson. The services times are independent and exponential.
What is the average time spent in the whole system (from arrive to serving finished)? any hints or any references?
Thanks in advanced
UPDATE
Thanks for the comments. I added some assumptions to this question.
In fact, I am comparing the above scheduling algorithm with the following schedule algorithm in perspective of the total time spent in the system, and fairness.
The following scheduling has average time is 1/(u-lambda1-lambda2), I guess.
Fairness is not very clearly defined, tbh.
The service discipline in both the round-robin and FIFO cases is nonpreemptive and work-conserving.
Average delay $\overline{W}$: Define the aggregate average delay as a weighted sum of individual delays: \begin{align} \overline{W} &= \frac{\lambda_A}{\lambda_A+\lambda_B}\overline{W}_A + \frac{\lambda_B}{\lambda_A+\lambda_B} \overline{W}_B \quad (Eq. 1)\\ &=\frac{\overline{N}_A + \overline{N}_B}{\lambda_A+\lambda_B}\\ &=\frac{\overline{N}}{\lambda_A+\lambda_B} \quad (Eq. 2) \end{align} where the last two equalities use Little's theorem and these definitions: \begin{align} \overline{N}_A &= \mbox{avg number of jobs of type A} \\ \overline{N}_B &= \mbox{avg number of jobs of type B} \\ \overline{N} &= \overline{N}_A + \overline{N}_B \end{align}
Regardless of the arrival processes, if the service times are indistinguishable over job types, i.i.d., and independent of the arrivals, the total average number of jobs in the system, which is $\overline{N}$, is the same for the round-robin and FIFO policies. Thus, by (Eq. 2), the average delay $\overline{W}$ is also the same for round-robin and FIFO. So you can calculate $\overline{W}$ for the more complicated round-robin policy simply by calculating it for the FIFO policy. Under independent Poisson arrivals, this amounts to calculating average delay in a standard FIFO M/G/1 queue: $$ \mbox{RoundRobin or FIFO} \implies \overline{W} = \mbox{M/G/1 formula}$$
Individual average delays $\overline{W}_A, \overline{W}_B$: Again assume independent Poisson arrivals. Define $\overline{X}$ as the average service time and let $\overline{U}_{seen, A}$ and $\overline{U}_{seen,B}$ be the average unfinished work "seen" by an arrival of type A and B, respectively. Then $$ \mbox{FIFO} \implies \overline{W}_A = \overline{U}_{seen,A}+\overline{X} , \quad \overline{W}_B = \overline{U}_{seen,B} + \overline{X}$$ By PASTA, both types of jobs (A and B) "see" the same average unfinished work when they arrive: $$\overline{U}_{seen,A} = \overline{U}_{seen,B}$$ Thus $$ \mbox{FIFO} \implies \overline{W}_A=\overline{W}_B=\overline{W}$$ where the final equality holds by (Eq. 1). On the other hand, the round-robin system will likely have $\overline{W}_A\neq \overline{W}_B$ even though aggregate average delay $\overline{W}$ is the same as FIFO. Intuitively I would expect $\overline{W}_A<\overline{W}_B$ if $\lambda_A<\lambda_B$.