# Stalling in Queuing Systems with Heterogeneous Channels
## Appendix A  
Let for \(0\leq j\leq n\), denote \(A_{-1,j}=1\). At first, note the easily verifiable identity:  
\[\sum_{s=j}^{n}\frac{(s-j)\cdot s!}{j!\cdot(s-j)!}P_{min(i-1,m),s,max(i-m,0)}= \sum_{s=j}^{n-1}\frac{\cdot(s+1)\cdot s!}{j!\cdot(s-j)!}P_{min(i-1,m),s+1,max  
Established relationship and definition (3) imply the Lemma.\(\square\)  
**Proof of Lemma A2.** Indeed, \(h_{n}=1\). Hence, according to (5),  
\[h_{j}=\frac{R_{m+K,j}}{R_{m+K,n}}=\frac{\sum_{s=j}^{n}\frac{s!}{j!(s-j)!}P_{m,s, K}}{P_{m,n,K}},\ 0\leq j\leq n.\]  
After elementary steps by virtue of latter relationship and Lemma 1, one can make sure that  
\[ \begin{array}{c}h_{j}\cdot\Big{(}\frac{A_{m+K+1,j}}{A_{m+K,j}}-1\Big{)}+\frac {n!}{(j-1)!(n-j+1)!}\\ =\Big{(}\frac{j}{r\cdot m\cdot q}+\frac{1}{q}\Big{)}\cdot\sum_{s=j}^{n}\frac{s!} {j!(s-j)!}\frac{P_{m,s,K}}{P_{m,n,K}}-\sum_{s=j}^{n-1}\frac{s!}{j!(s-j)!}\frac{ P_{m,s,K-1}}{P_{m,n,K}}\\ \qquad\qquad-\frac{n!}{j!(n-j)!}\cdot\frac{P_{m,s,K-1}}{P_{m,n,K}}+\frac{n!}{ j!(n-j)!}(n-j+1)!\\ =\Big{(}\frac{j}{r\cdot m\cdot q}+\frac{1}{q}\Big{)}\cdot\sum_{s=j}^{n}\frac{ s!}{j!(s-j)!}\frac{P_{m,s,K}}{P_{m,n,K}}-\sum_{s=j}^{n-1}\frac{s!}{j!(s-j)!}\frac{ P_{m,s,K-1}}{P_{m,n,K}}\\ \qquad\qquad-\frac{n!}{j!(n-j)!}\cdot\frac{P_{m,n,K-1}}{P_{m,n,K}}+\frac{n!}{ (j-1)!(n-j+1)!}\\ \qquad\qquad=\Big{(}\frac{j}{r\cdot m\cdot q}+\frac{1}{q}\Big{)}\cdot\sum_{s=j} ^{n}\frac{s!}{j!(s-j)!}\cdot\frac{P_{m,s,K}}{P_{m,n,K}}\\ \qquad\qquad\qquad-\sum_{s=j}^{n-1}\frac{s!}{j!(s-j)!}\\ \qquad\qquad\qquad\cdot\frac{P_{m,s,K}\cdot\Big{(}1+\frac{1}{q}+\frac{r}{r \cdot m\cdot q}\Big{)}-P_{m,n-1,K}-P_{m,n+1,K}\cdot\frac{s+1}{r\cdot m\cdot q} }{P_{m,n,K}}\\ \qquad\qquad\qquad-\frac{n!}{j!(n-j)!}\cdot\frac{P_{m,n,K}\left(\frac{1}{q}+ \frac{n}{r\cdot m\cdot q}\right)-P_{m,n-1,K}}{P_{m,n,K}}\\ \qquad\qquad\qquad+\frac{n!}{(j-1)!(n-j+1)!}\\ =\sum_{s=j}^{n-1}\frac{s!}{j!(s-j)!}\cdot\frac{P_{m,s,s-1}K-P_{m,n,K}}{P_{m,n,n,K}}-\frac{n!}{j!(n-j)!}\cdot\frac{P_{m,n,K-1}}{P_{m,n,n,K}}\\ \qquad\qquad\qquad\qquad+\frac{n!}{(j-1)!(n-j+1)!}\\ =\sum_{s=j}^{n-1}\frac{s!}{(j-1)!(s-j-1)!}\cdot\frac{P_{m,n,s,K}}{P_{m,n,n,K} }+\frac{n!}{(j-1)!(n-j+1)!}=h_{j-1}\square\end{array} \]  
**Proof of Theorem A1. ** Let us consider Equation (1), for \(0\leq i<m+\text{K,1}\leq j\leq n\), or for \(i=m+\text{K,1}\leq\text{j}<n\), and assume that the waiting line is empty. Respectively, define, \(P_{-1,j,0}=0\), and \(P_{i,-1,k}=0\), \(P_{i,n+1,k}=0\). Then, according to (6), we have that  
\[ \begin{array}{c}\Big{(}q+min\Big{(}\frac{i}{m},1\Big{)}+\frac{j}{r\cdot m} \Big{)}\cdot P_{min\big{(}i,m\big{)},j,max(i-m,0)}\\ =\Big{(}q+min\Big{(}\frac{i}{m},1\Big{)}+\frac{j}{r\cdot m}\Big{)}\cdot P_{m,n,K }\ \sum_{s=j}^{n}h_{s}\cdot\frac{A_{s}}{A_{m+K,s}}\frac{(-1)^{s-j},s!}{j!(s-j)!} \cdot\\ \qquad\qquad=P_{m,n,K}\cdot\sum_{s=j}^{n}h_{s}\cdot\frac{A_{s,s}}{A_{m+K,s}} \frac{(j-s)}{r\cdot m}\frac{(-1)^{s-j},s!}{j!(s-j)!}+P_{m,n,K}\\ \qquad\qquad\cdot\sum_{s=j}^{n}h_{s}\cdot\Big{(}q+min\Big{(}\frac{i}{m},1 \Big{)}+\frac{s}{r\cdot m}\Big{)}\cdot\frac{A_{s,s}}{A_{m+K,s}}\frac{(-1)^{s-j},s!}{j!(s-j)!}\\ \end{array}  \tag{14}\]  
In addition, in virtue of Lemma 1,  
\[ \begin{array}{c}\sum_{s=j}^{n}h_{s}\cdot\Big{(}q+min\Big{(}\frac{i}{m},1 \Big{)}+\frac{s}{r\cdot m}\Big{)}\cdot\frac{A_{s,s}}{A_{m+K,s}}\frac{(-1)^{s-j},s!}{j!(s-j)!}\\ =q\cdot\sum_{s=j}^{n}h_{s}\cdot\frac{A_{i-1,s}}{A_{m+K,s}}\frac{(-1)^{s-j},s!}{ j!(s-j)!}+\sum_{s=j}^{n}h_{s}\cdot\frac{A_{i-1,s}}{A_{m+K,s}}\frac{(-1)^{s-j},s!}{j!(s-j)!}\\ =\Big{(}P_{min\big{(}i-1,m\big{)},j,max(i-1-m,0\big{)}\cdot q+P_{min\big{(}i+1, m\big{)},j,max(i+1-m,0\big{)}}}\Big{)}/P_{m,n,K}\end{array}  \tag{15}\]Thus, (A2) and (A3) imply that relations among steady-state probabilities in (6) satisfy Equation (1), if \(0\leq i<m+\text{K,1}\leq j\leq n\), or \(i=m+\text{K, 1}\leq\text{j}<n\), and the waiting line is empty. Next, it is easy to see that Equation (7) is equivalent to recursive relation \(P_{m,n,K+k}=\rho\cdot P_{m,n,K+k-1}\), which easily follow from respective Equation (1).