Queuing theory-Multiple server (reducing simple recurrence formulas)

Question

The equations given in 6.3 have been reduced which really eases the computation in further studies. But I tried to find the method of reducing these but I could not find a way at all. Any hints will be very helpful. Thanks

Answer 1

This follows fairly directly by induction. What is probably confusing you is that the value of $n$ in the equilibrium equations is different from the $n$ in the reduced equations. Let us start with the no queue situation. (I'll rename $n$ to $m$ in the equilibrium equations to reduce confusion.)

We have \begin{align*} 0 &= -\lambda P_0 + \mu P_1&\qquad &\text{for $m=0$}\\ 0 &= -(\lambda + m \mu) P_m + \lambda P_{m-1} + \mu(m+1) P_{m+1} &\qquad &\text{for $1\le m \le k-1$} \end{align*} and we want to prove that $$0 = -\lambda P_{n-1} + n\mu P_n \qquad \text{for $1\le n\le k$}.$$

It is easy to see that the equation holds for $n=1$ as that is just the $m=0$ equilibrium equation. Let us proceed by induction and assume that the reduced equation holds for $n-1$ (assume $n>1$ as we have already dealt with that.) From the equilibrium equation for $m=n-1$ we have $$0 = -(\lambda + (n-1) \mu) P_{n-1} + \lambda P_{n-2} + \mu n P_{n}$$ and from the last equilibrium equation we have $$ \lambda P_{n-2} = (n-1) \mu P_{n-1}. $$ Combining the equations we get $0 = -\lambda P_{n-1} + n \mu P_{n}$ the $n$th reduced equation (the $-(n-1)\mu P_{n-1}$ and $(n-1)\mu P_{n-1} terms cancel.)

The other reduced equation is proved in exactly the same way. We note that it holds for $n=k$ (by the above induction) and then substitute the last reduced equation into the appropriate equilibrium equation. Lather, rinse, repeat.