How does the rewriting of the following two equations work?

Question

I am failing to understand the proof of coming to the steady-state formula in queueing theory. This is probably due to the fact that I may have forgotten (and cannot find it back) some of the algebra from my school days. The step I am missing is the transition from the first to the second statement below:

$$(\lambda + \mu)\frac{\lambda}{\mu}P_0 = \lambda P_0 + \mu P2$$

$$\frac{\lambda^2}{\mu}P0 + \lambda P_0 = \lambda P_0 + \mu P2$$

My main question is how does $$(\lambda + \mu)\frac{\lambda}{\mu}P_0$$ become $$\frac{\lambda^2}{\mu}P_0 + \lambda P_0$$ and what did happen to: $$\mu$$

Can anybody help me out?

Answer 1

Recall the distributive law: If $a,b,c$ are terms, we have $(a+b)c=ac+bc$. In this case,

$$(\lambda+\mu)\frac{\lambda}{\mu}P_0 = \lambda\frac{\lambda}{\mu}P_0 + \mu\frac{\lambda}{\mu}P_0= \frac{\lambda \cdot \lambda}{\mu}P_0 + \frac{\mu \cdot \lambda}{\mu}P_0 = \frac{\lambda^2}{\mu}P_0 + \lambda P_0.$$

I hope this helps. Let me know if you need more detail.

Answer 2

This is rather simple actually, all you need to do is expand (distribute the elements).

Expanding this: $$ (\lambda + \mu)\frac{\lambda}{\mu}P0 $$

We have this: $$ \frac{\lambda^2}{\mu}P0+\frac{\mu\lambda}{\mu}P0 $$

And the final result after simplification is : $$ \frac{\lambda^2}{\mu}P0 + \lambda P0 $$

To answer your question about what happened to $\mu$. When you simplify: $\frac{\mu\lambda}{\mu}$ it becomes $1\lambda$, which is the same as just $\lambda$.