For investigating a queueing model in my research, I need to obtain the probability generating function (PGF), $G(z)$, by solving the following functional equation $$\lambda(z(1-p)+p)G((1-p)z)-(\mu+\lambda p)(1-p)G(z)=(\lambda p-(\mu+\lambda p)(1-p))\pi_0$$ where $\lambda, \mu, p, \pi_0$ are parameters, and $0<p<1$.
Since $G(z)$ is a PGF, it is known that $G(0)=\pi_0$ and $G(1)=1$. Also, by substituting $z=1$, I get $$G(1-p)=\frac{(\lambda p-(\mu+\lambda p)(1-p))\pi_0+(\mu+\lambda p)(1-p)}{\lambda}$$ In addition, by substituting $z=1/(1-p)$, $z=1-p$ and $z=(1-p)^2$, I have $$G(\frac{1}{1-p})=\frac{\lambda(1+p)-(\lambda p-(\mu+\lambda p)(1-p))\pi_0}{(\mu+\lambda p)(1-p)})$$ $$G((1-p)^2)=\frac{(\lambda p-(\mu+\lambda p)(1-p))\pi_0+(\mu+\lambda p)(1-p)G(1-p)}{\lambda}$$ $$G((1-p)^3)=\frac{(\lambda p-(\mu+\lambda p)(1-p))\pi_0+(\mu+\lambda p)(1-p)G(1-p)^2}{\lambda}$$ Since $G(z)$ is a PGF it cannot be a polynomial function. So, I assumed that the function is of form $$G(z)=\frac{az^2+bz+c}{dz^2+ez+f}$$ and tried to calculate the coefficients $a, b, c, d, e, f$ by solving the set of six equations using the six expressions above, i.e., $$\frac{c}{f}=\pi_0$$ $$\frac{a+b+c}{d+e+f}=1$$ $$\frac{a(1-p)^2+b(1-p)+c}{d(1-p)^2+e(1-p)+f}=G(1-p)$$ $$\frac{a(1/(1-p))^2+b(1/(1-p))+c}{d(1/(1-p))^2+e(1/(1-p))+f}=G(1/(1-p))$$ $$\frac{a(1-p)^4+b(1-p)^2+c}{d(1-p)^4+e(1-p)^2+f}=G((1-p)^2)$$ $$\frac{a(1-p)^6+b(1-p)^3+c}{d(1-p)^6+e(1-p)^3+f}=G((1-p)^3)$$
However, there is no possible solution for this set. I will appreciate if some one could help me.
