How were the right hand sides of equations 6 and 7 in the paper in the question details derived?

Question

I read a paper on Queuing theory and the equations have been a bit of a problem. I have been able to solve the left hand sides somehow but for the right hand sides, I've hit a little bit of a snag. Could you please walk me through the derivation of the right hand sides of equations 6 and 7? Here is the link of the paper: https://drive.google.com/open?id=0Bzd_DcKn0DRqLXJ6X0J4ZFR3LXc Any help will be really appreciated.

Answer 1

What do you mean solve the left hand side but not the right hand side? It leads to one equation that relies on terms from both sides.

Anyway, for equation (6), multiplying by $z^n$ and summing all the terms on the lhs of (1) and (2) yields: \begin{align} (*) &= (\lambda+\gamma)p_{0,0}z^0+\sum_{n=1}^\infty(\lambda+\gamma+n\xi)p_{0,n}z^n \\ &=(\lambda+\gamma)\sum_{n=0}^\infty p_{0,n}z^n+\xi z\sum_{n=1}^\infty np_{0,n}z^{n-1} \\ &= (\lambda+\gamma)P_0(z)+\xi zP_0^{'}(z), \end{align} and doing the same for the rhs yields: \begin{align} (**) &= (\xi p_{0,1}+\mu p_{1,1})z^0+\sum_{n=1}^\infty\lambda p_{0,n-1}z^n+\sum_{n=1}^\infty(n+1)\xi p_{0,n+1}z^n \\ &= \mu p_{1,1}+\lambda z\sum_{n=0}^\infty p_{0,n}z^n+\xi \sum_{n=1}^\infty p_{0,n}nz^{n-1} \\ &= \mu p_{1,1}+\lambda zP_0(z)+\xi P_0^{'}(z). \end{align} Now equating $(*)=(**)$ and rearranging terms leads to $$ (\lambda(1-z)+\gamma)P_0(z)-\xi(1-z)P_0^{'}(z)=\mu p_{1,1}, $$ which is equivalent to (6). Equation (7) can be obtained in the same manner.