# Jackson's theorem to optimize mean queue length of a traffic model

I am working on traffic signals for a city transport system. I modeled the city transport using a queuing network as shown in the following image Arrival rate of "A" cars from outside is S1 and arrival rate of "B" cars is S2. "A" and "B" cars can not enter into traffic-3 and traffic-1 respectively.
Traffic 1,2,3 are modeled as M/M/1-FCFS queue with infinite queue capacity.$(1-P_1)$ and $(1-P_3)$ are exit probabilities for "A" and "B" cars. $\mu_i$ is the service rate of $i_{th}$ traffic. It is given that $\sum_{i=1}^{3}\mu_i=Constant$.

So,overall arrival rate can be written as

$\lambda1=S1 + P_1.\lambda1$

$\lambda2=S2 + P_2.\lambda2$

Is it possible to find $\mu=\{\mu_1,\mu_2,\mu_3\}$ which will minimize the mean queue length of all three traffics. Can I use Jackson's theorem in this model?

• I'm not sure that you can use Jackson's theorem here (it's not one of the BCMP categories), but I'm also not sure that you need to. You don't need the queue length distribution to be product form; you just need the mean queue length at each node. My concern is that you do want to be able to treat each node in isolation as though it were M/M/1 (whether or not they are independent of the other nodes). I'm not immediately sure you can do that. If you can, you should be able to use some Lagrange multiplier technique to obtain the optimal allocation for constant $\sum_i \mu_i$. – Brian Tung Jun 1 '15 at 3:50
• (Not immediately sure because node 2 mixes customer classes that have distinct orbits.) – Brian Tung Jun 1 '15 at 3:51
• @Brian,What condition(s) a network should satisfy in order to treat each node in the network as isolated nodes? I thought node 3 is process shearing (PS) node which belongs to the BCMP categories. – marcella Jun 1 '15 at 4:55
• It is, but I don't think this counts as processor sharing. Each customer still has to be served in sequence; they're only sharing it in the sense that both customer classes use the same queue. It's still possible that this is Jacksonian, but I have to give it some more thought. – Brian Tung Jun 1 '15 at 5:01
• Still looking for a solution of this problem. – marcella Jul 4 '15 at 5:38