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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?

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2 Answers 2

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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.

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I recall that, but then you would expect that somewhere you get a value along the lines this would result in: $$\frac{\lambda^2}{\mu}P_0 + \lambda P_0 + \mu P_0$$ –  Jochem May 12 '13 at 8:22
    
I'm sorry, I don't follow. Could you expand on your question a little bit? –  Potato May 12 '13 at 8:24
    
@Jochem It might help to note that $\frac{\lambda}{\mu}P_0$ is one term and plays the role of $c$ in the distributive law. –  Potato May 12 '13 at 8:26
    
This part I understand: $$(\lambda+\mu)\frac{\lambda}{\mu}P_0 = \lambda\frac{\lambda}{\mu}P_0 + \mu\frac{\lambda}{\mu}P_0$$ But when you add them up you get: $$\frac{\lambda * \lambda}{\mu}P_0 + \frac{\mu * \lambda}{\mu}P_0$$ Is it here that multiplying of the $$\mu$$ in the second part results in $$P_0$$? –  Jochem May 12 '13 at 8:28
    
@Jochem That is correct. The first term you wrote at the bottom becomes $\frac{\lambda^2}{\mu}P_0$, and the second is $\lambda P_0$. Does this make sense? –  Potato May 12 '13 at 8:32
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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 $\lambda$.

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Thank you. Just went through the same line of thought with @Potato. –  Jochem May 12 '13 at 8:39
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