We have shown, then, that exists a positive vector $h\in{R^{3}}$ with $h_{1}$
being a large enough positive number, $h_{2}$ and $h_{3}$ being small enough
positive numbers as required so that $(R,\theta)$ under the static-priority
policy 1-5-6 is Chen-$\mathcal{S}$. ∎

∎

###### Proof.

Proof for priority 4-5-6 (Chen-S).

In the balanced DHV network under priority 4-5-6, we have
$J=3,K=6,\rho=\left[\alpha_{1}\ \alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}1&0&-m_{4}\mcr-\displaystyle\frac{m_{2}}{m_{1}}&1&\displaystyle\frac{m_{2}}{m_{1}}-1\mcr
0&-\displaystyle\frac{m_{3}}{m_{2}}&\displaystyle\frac{m_{3}}{m_{2}}\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)$
$\displaystyle=(\alpha_{1}-1)\left[\begin{matrix}1-m_{4}\mcr
1-\displaystyle\frac{m_{2}}{m_{1}}+\displaystyle\frac{m_{2}}{m_{1}}-1\mcr\displaystyle\frac{m_{3}}{m_{2}}-\displaystyle\frac{m_{3}}{m_{2}}\end{matrix}\right]\mcr$
$\displaystyle=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\mcr 0\mcr
0\end{matrix}\right].$

Similar to the priority 2-4-6 case, it can be proved that the reflection
matrix $R$ is completely-$\mathcal{S}$, we omit the details here.

Partition 1: $a=1;b=2,3$

$\theta_{b}=\theta_{23}=\left[\begin{matrix}0\mcr
0\end{matrix}\right],\;R_{b}^{-1}=R_{23}^{-1}=\left[\begin{matrix}\displaystyle\frac{m_{1}}{m_{2}}&\displaystyle\frac{m_{1}m_{5}-m_{2}m_{4}}{m_{3}}\mcr\displaystyle\frac{m_{1}}{m_{2}}&\displaystyle\frac{m_{1}}{m_{3}}\end{matrix}\right]\Rightarrow
u=-R_{b}^{-1}\theta_{b}=\left[\begin{matrix}0\mcr 0\end{matrix}\right];$

$R_{ab}=\left[\begin{matrix}0&-m_{4}\end{matrix}\right]\Rightarrow R_{ab}u=0.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)m_{1}$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1,3$

$\theta_{b}=\theta_{13}=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\mcr
0\end{matrix}\right],\;R_{b}^{-1}=R_{13}^{-1}=\left[\begin{matrix}1&\displaystyle\frac{m_{2}m_{4}}{m_{3}}\mcr
0&\displaystyle\frac{m_{2}}{m_{3}}\end{matrix}\right]\Rightarrow
u=-R_{b}^{-1}\theta_{b}=\left[\begin{matrix}(1-\alpha_{1})m_{1}\mcr
0\end{matrix}\right];$
$R_{ab}=\left[\begin{matrix}-\displaystyle\frac{m_{2}}{m_{1}}&\displaystyle\frac{m_{2}}{m_{1}}-1\end{matrix}\right]\Rightarrow
R_{ab}u=(\alpha_{1}-1)m_{2}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=(\alpha_{1}-1)m_{2}h_{2}$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=3;b=1,2$

$\theta_{b}=\theta_{12}=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\\\
0\end{matrix}\right],\;R_{b}^{-1}=R_{12}^{-1}=\left[\begin{matrix}1&0\mcr\displaystyle\frac{m_{2}}{m_{1}}&1\end{matrix}\right]\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})\left[\begin{matrix}m_{1}\mcr
m_{2}\end{matrix}\right];$
$R_{ab}=\left[\begin{matrix}0&-\displaystyle\frac{m_{3}}{m_{2}}\end{matrix}\right]\Rightarrow
R_{ab}u=(\alpha_{1}-1)m_{3}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{3}(\alpha_{1}-1)m_{3}$.
Since $\alpha_{1}<1$, taking $h_{3}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 4: $a=1,2;b=3$

$\theta_{b}=\theta_{3}=0\;,R_{b}^{-1}=R_{3}^{-1}=\frac{m_{2}}{m_{3}}\;\Rightarrow
u=-R_{b}^{-1}\theta_{b}=0;$
$R_{ab}=\left[\begin{matrix}-m_{4}\mcr\displaystyle\frac{m_{2}}{m_{1}}-1\end{matrix}\right]\Rightarrow
R_{ab}u=\left[\begin{matrix}0\mcr
0\end{matrix}\right];\theta_{a}=\theta_{12}=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\mcr
0\end{matrix}\right]$

Thus,

$\displaystyle h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]$
$\displaystyle=(\alpha_{1}-1)\left[\begin{matrix}h_{1}&h_{2}\end{matrix}\right]\left[\begin{matrix}m_{1}\mcr
0\end{matrix}\right]$ $\displaystyle=(\alpha_{1}-1)m_{1}h_{1}$

Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 5: $a=1,3;b=2$

$\theta_{b}=\theta_{2}=0\;,R_{b}^{-1}=R_{2}^{-1}=1\;\Rightarrow
u=-R_{b}^{-1}\theta_{b}=0;$
$R_{ab}=\left[\begin{matrix}0\mcr-\displaystyle\frac{m_{3}}{m_{2}}\end{matrix}\right]\Rightarrow
R_{ab}u=\left[\begin{matrix}0\mcr
0\end{matrix}\right];\theta_{a}=\theta_{13}=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\mcr
0\end{matrix}\right]$

Thus,

$\displaystyle h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]$
$\displaystyle=(\alpha_{1}-1)\left[\begin{matrix}h_{1}&h_{3}\end{matrix}\right]\left[\begin{matrix}m_{1}\mcr
0\end{matrix}\right]$ $\displaystyle=(\alpha_{1}-1)m_{1}h_{1}$

Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 6: $a=2,3;b=1$

$\theta_{b}=\theta_{1}=(\alpha_{1}-1)m_{1}\;,R_{b}^{-1}=R_{1}^{-1}=1\;\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})m_{1};$
$R_{ab}=\left[\begin{matrix}-\displaystyle\frac{m_{2}}{m_{1}}\mcr
0\end{matrix}\right]\Rightarrow
R_{ab}u=(\alpha_{1}-1)\left[\begin{matrix}m_{2}\mcr
0\end{matrix}\right];\theta_{a}=\theta_{23}=\left[\begin{matrix}0\mcr
0\end{matrix}\right]$

Thus,

$\displaystyle h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]$
$\displaystyle=(\alpha_{1}-1)\left[\begin{matrix}h_{2}&h_{3}\end{matrix}\right]\left[\begin{matrix}m_{2}\mcr
0\end{matrix}\right]$ $\displaystyle=(\alpha_{1}-1)m_{2}h_{2}$

Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 7: $a=1,2,3;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=(\alpha_{1}-1)m_{1}h_{1}$

Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{3}}$ as required
so that $(R,\theta)$ under the static-priority policy 4-5-6 is
Chen-$\mathcal{S}$. ∎
∎

###### Proof.

Proof for priority 1-2-6 (Chen-S).

In the balanced DHV network under priority 1-2-6, we have
$J=3,K=6,\rho=\left[\alpha_{1}\ \alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}1&\displaystyle\frac{m_{4}m_{6}}{m_{3}m_{5}}&-\displaystyle\frac{m_{4}}{m_{3}}\mcr-\displaystyle\frac{m_{5}}{m_{4}}&1&0\mcr
0&-\displaystyle\frac{m_{6}}{m_{5}}&1\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)$
$\displaystyle=(\alpha_{1}-1)\left[\begin{matrix}1+\displaystyle\frac{m_{4}m_{6}}{m_{3}m_{5}}-\displaystyle\frac{m_{4}}{m_{3}}\mcr
1-\displaystyle\frac{m_{5}}{m_{4}}\mcr
1-\displaystyle\frac{m_{6}}{m_{5}}\end{matrix}\right]\mcr$
$\displaystyle=\frac{\alpha_{1}-1}{m_{3}m_{4}m_{5}}\left[\begin{matrix}m_{3}m_{4}m_{5}+m_{4}^{2}m_{6}-m_{4}^{2}m_{5}\mcr
m_{3}m_{5}(m_{4}-m_{5})\mcr m_{3}m_{4}(m_{5}-m_{6})\end{matrix}\right].$

Similar to the priority 2-4-6 case, it can be proved that the reflection
matrix $R$ is completely-$\mathcal{S}$, we omit the details here.

Partition 1: $a=1;b=2,3$

$\theta_{b}=\theta_{23}=\frac{\alpha_{1}-1}{m_{4}m_{5}}\left[\begin{matrix}m_{5}(m_{4}-m_{5})\mcr
m_{4}(m_{5}-m_{6})\end{matrix}\right],\;R_{b}^{-1}=R_{23}^{-1}=\left[\begin{matrix}1&0\mcr\displaystyle\frac{m_{6}}{m_{5}}&1\end{matrix}\right]\Rightarrow
u=-R_{b}^{-1}\theta_{b}=\frac{1-\alpha_{1}}{m_{4}}\left[\begin{matrix}m_{4}-m_{5}\mcr
m_{4}-m_{6}\end{matrix}\right];$

$R_{ab}=\left[\begin{matrix}\displaystyle\frac{m_{4}m_{6}}{m_{3}m_{5}}&-\displaystyle\frac{m_{4}}{m_{3}}\end{matrix}\right]\Rightarrow
R_{ab}u=\displaystyle\frac{1-\alpha_{1}}{m_{3}m_{5}}m_{4}(m_{6}-m_{5}).$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1,3$

$\theta_{b}=\theta_{13}=\frac{\alpha_{1}-1}{m_{3}m_{5}}\left[\begin{matrix}m_{3}m_{5}+m_{4}m_{6}-m_{4}m_{5}\mcr
m_{3}(m_{5}-m_{6})\end{matrix}\right],\;R_{b}^{-1}=R_{13}^{-1}=\left[\begin{matrix}1&\displaystyle\frac{m_{4}}{m_{3}}\mcr
0&1\end{matrix}\right]\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})\left[\begin{matrix}1\mcr\displaystyle\frac{m_{5}-m_{6}}{m_{5}}\end{matrix}\right];$
$R_{ab}=\left[\begin{matrix}-\displaystyle\frac{m_{5}}{m_{4}}&0\end{matrix}\right]\Rightarrow
R_{ab}u=(\alpha_{1}-1)\frac{m_{5}}{m_{4}}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=(\alpha_{1}-1)h_{2}$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=3;b=1,2$

$\theta_{b}=\theta_{12}=\frac{\alpha_{1}-1}{m_{3}m_{4}m_{5}}\left[\begin{matrix}m_{3}m_{4}m_{5}+m_{4}^{2}m_{6}-m_{4}^{2}m_{5}\mcr
m_{3}m_{4}m_{5}-m_{3}m_{5}^{2}\end{matrix}\right],\;R_{b}^{-1}=R_{12}^{-1}=\left[\begin{matrix}m_{3}&-\displaystyle\frac{m_{4}m_{6}}{m_{5}}\\\
\displaystyle\frac{m_{3}m_{5}}{m_{4}}&m_{3}\end{matrix}\right]$ $\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})\left[\begin{matrix}m_{3}-m_{4}+m_{6}\mcr
m_{6}-m_{5}+m_{3}\end{matrix}\right];$
$R_{ab}=\left[\begin{matrix}0&-\displaystyle\frac{m_{6}}{m_{5}}\end{matrix}\right]\Rightarrow
R_{ab}u=(\alpha_{1}-1)\frac{m_{6}}{m_{5}}(1-m_{5}).$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{3}(\alpha_{1}-1)m_{3}$.
Since $\alpha_{1}<1$, taking $h_{3}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 4: $a=1,2;b=3$

$\theta_{b}=\theta_{3}=(\alpha_{1}-1)(1-\frac{m_{6}}{m_{5}})\;,R_{b}^{-1}=R_{3}^{-1}=1\;\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})(1-\frac{m_{6}}{m_{5}});$
$R_{ab}=\left[\begin{matrix}-\displaystyle\frac{m_{4}}{m_{3}}\mcr
0\end{matrix}\right]\Rightarrow
R_{ab}u=\left[\begin{matrix}\displaystyle\frac{\alpha_{1}-1}{m_{3}m_{5}}(m_{4}m_{5}-m_{4}m_{6})\mcr
0\end{matrix}\right];\theta_{a}=\theta_{12}=\frac{\alpha_{1}-1}{m_{3}m_{4}m_{5}}\left[\begin{matrix}m_{3}m_{4}m_{5}+m_{4}^{2}m_{6}-m_{4}^{2}m_{5}\mcr
m_{3}m_{5}(m_{4}-m_{5})\end{matrix}\right]$

Thus,

$\displaystyle h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]$
$\displaystyle=(\alpha_{1}-1)\left[\begin{matrix}h_{1}&h_{2}\end{matrix}\right]\left[\begin{matrix}1\mcr
1-\displaystyle\frac{m_{5}}{m_{4}}\end{matrix}\right]$
$\displaystyle=(\alpha_{1}-1)\left[h_{1}+(1-\frac{m_{5}}{m_{4}})h_{2}\right]$

Since $\alpha_{1}<1$, fixing $h_{1}>0$ large enough and $h_{2}>0$ small
enough, we ensure $h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 5: $a=1,3;b=2$

$\theta_{b}=\theta_{2}=(\alpha_{1}-1)(1-\frac{m_{5}}{m_{4}})\;,R_{b}^{-1}=R_{2}^{-1}=1\;\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})(1-\frac{m_{5}}{m_{4}});$
$R_{ab}=\left[\begin{matrix}\displaystyle\frac{m_{4}m_{6}}{m_{3}m_{5}}\mcr-\displaystyle\frac{m_{6}}{m_{5}}\end{matrix}\right]\Rightarrow
R_{ab}u=(\alpha_{1}-1)\left[\begin{matrix}\displaystyle\frac{m_{6}(m_{5}-m_{4})}{m_{3}m_{5}}\mcr\displaystyle\frac{m_{6}(m_{4}-m_{5})}{m_{4}m_{5}}\end{matrix}\right];\theta_{a}=\theta_{13}=\frac{\alpha_{1}-1}{m_{3}m_{5}}\left[\begin{matrix}m_{3}m_{5}+m_{4}m_{6}-m_{4}m_{5}\mcr
m_{3}m_{5}-m_{3}m_{6}\end{matrix}\right]$

Thus,

$\displaystyle h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]$
$\displaystyle=\frac{\alpha_{1}-1}{m_{3}m_{4}}\left[\begin{matrix}h_{1}&h_{3}\end{matrix}\right]\left[\begin{matrix}m_{1}m_{4}\mcr
m_{3}(m_{4}-m_{6})\end{matrix}\right]$
$\displaystyle=(\alpha_{1}-1)\left[\frac{m_{1}}{m_{3}}h_{1}+\frac{m_{4}-m_{6}}{m_{4}}h_{3}\right]$

Since $\alpha_{1}<1$, fixing $h_{1}>0$ large enough and $h_{3}>0$ small
enough, we ensure $h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 6: $a=2,3;b=1$

$\theta_{b}=\theta_{1}=\frac{\alpha_{1}-1}{m_{3}m_{5}}(m_{3}m_{5}+m_{4}m_{6}-m_{4}m_{5})\;,R_{b}^{-1}=R_{1}^{-1}=1\;\Rightarrow
u=-R_{b}^{-1}\theta_{b}=\frac{1-\alpha_{1}}{m_{3}m_{5}}(m_{3}m_{5}+m_{4}m_{6}-m_{4}m_{5});$
$R_{ab}=\left[\begin{matrix}-\displaystyle\frac{m_{5}}{m_{4}}\mcr
0\end{matrix}\right]\Rightarrow
R_{ab}u=\frac{\alpha_{1}-1}{m_{3}m_{4}}\left[\begin{matrix}m_{3}m_{5}+m_{4}m_{6}-m_{4}m_{5}\mcr
0\end{matrix}\right];\theta_{a}=\theta_{23}=\frac{\alpha_{1}-1}{m_{3}m_{4}m_{5}}\left[\begin{matrix}m_{3}m_{4}m_{5}-m_{3}m_{5}^{2}\mcr
m_{3}m_{4}m_{5}-m_{3}m_{4}m_{6}\end{matrix}\right]$

Thus,

$\displaystyle h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]$
$\displaystyle=\frac{\alpha_{1}-1}{m_{3}m_{5}}\left[\begin{matrix}h_{2}&h_{3}\end{matrix}\right]\left[\begin{matrix}m_{5}m_{2}\mcr
m_{3}(m_{5}-m_{6})\end{matrix}\right]$
$\displaystyle=(\alpha_{1}-1)(\frac{m_{2}}{m_{3}}h_{2}+\frac{m_{5}-m_{6}}{m_{5}}h_{3})$

Since $\alpha_{1}<1$, fixing $h_{2}>0$ large enough and $h_{3}>0$ small
enough, we ensure $h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 7: $a=1,2,3;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=(\alpha_{1}-1)\left[(1+\frac{m_{4}m_{6}}{m_{3}m_{5}}-\frac{m_{4}}{m_{3}})h_{1}+(1-\frac{m_{5}}{m_{4}})h_{2}+(1-\frac{m_{6}}{m_{5}})h_{3}\right]$

Let
$K=1+\displaystyle\frac{m_{4}m_{6}}{m_{3}m_{5}}-\frac{m_{4}}{m_{3}}=\frac{1}{m_{3}m_{5}}(m_{3}m_{5}+m_{4}m_{6}-m_{4}m_{5})$.

If $m_{6}-m_{5}\geq 0$, then

$K=\frac{1}{m_{3}m_{5}}(m_{3}m_{5}+m_{4}m_{6}-m_{4}m_{5})=\frac{1}{m_{3}m_{5}}(m_{3}m_{5}+m_{4}(m_{6}-m_{5}))>0;$

If $m_{6}-m_{5}\leq 0$, then

$\begin{split}K=\;&\frac{1}{m_{3}m_{5}}(m_{3}m_{5}+m_{4}m_{6}-m_{4}m_{5})=\;\frac{1}{m_{3}m_{5}}((1-m_{6})m_{5}+(1-m_{1})m_{6}-m_{4}m_{5})\mcr=\;&\frac{1}{m_{3}m_{5}}((m_{5}-m_{5}m_{6}+m_{6}-m_{1}m_{6}-m_{4}m_{5})=\;\frac{1}{m_{3}m_{5}}((1-m_{4})m_{5}+(1-m_{5})m_{6}-m_{1}m_{6})\mcr=\;&\frac{1}{m_{3}m_{5}}(m_{1}m_{5}+m_{2}m_{6}-m_{1}m_{6})=\;\frac{1}{m_{3}m_{5}}(m_{1}(m_{5}-m_{6})+m_{2}m_{6})>0.\end{split}$

Thus, we have $K>0$. Since $\alpha_{1}<1$, fixing $h_{1}>0$ large enough,
$h_{2}>0$ and $h_{3}>0$ small enough, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{3}}$ with $h_{1}$
being a relatively large enough positive number comparing to $h_{2}$, and
$h_{2}$ being a relatively large enough positive number comparing to $h_{3}$
as required so that $(R,\theta)$ under the static-priority policy 1-2-6 is
Chen-$\mathcal{S}$. ∎
∎

###### Lemma 7

([5] Lemma 3.2) A lower triangular matrix whose inverse exists and is non-
negative is both completely-$\mathcal{S}$ and Schur-S.

###### Lemma 8

([5] Lemma 3.1) A nonsingular M-matrix is both completely-$\mathcal{S}$ and
Schur-S.

###### Proof.

Proof for priority 1-2-3/4-2-3/4-5-3 (Chen-S). In the balanced DHV network
under priority 1,2,3, we have

$R_{1,2,3}^{-1}(\Delta)=\left[\begin{matrix}1&0&0\mcr\displaystyle\frac{m_{5}}{m_{4}}&1&0\mcr\displaystyle\frac{m_{6}}{m_{4}}&\displaystyle\frac{m_{6}}{m_{5}}&1\end{matrix}\right],\;\;R_{1,2,3}(\Delta)=\left[\begin{matrix}1&0&0\mcr-\displaystyle\frac{m_{5}}{m_{4}}&1&0\mcr
0&-\displaystyle\frac{m_{6}}{m_{5}}&1\end{matrix}\right],\\\ $

where $R_{1,2,3}(\Delta)$ is a lower triangular matrix whose inverse exists
and is non-negative. Thus, with Lemma 7, $R_{1,2,3}(\Delta)$ is both
completely-$\mathcal{S}$ and Schur-S, and with $\alpha_{1}<1$, it satisfies
Chen-$\mathcal{S}$ according to Lemma 2.

Similarly, with

$R_{4,2,3}(\Delta)^{-1}=\left[\begin{matrix}\displaystyle\frac{1}{m_{1}}&0&0\mcr\displaystyle\frac{1}{m_{1}}&1&0\mcr\displaystyle\frac{1}{m_{1}}&\displaystyle\frac{m_{6}}{m_{5}}&1\end{matrix}\right],\;\;R_{4,2,3}(\Delta)=\left[\begin{matrix}m_{1}&0&0\mcr-1&1&0\mcr\displaystyle\frac{m_{2}-m_{3}}{m_{5}}&-\displaystyle\frac{m_{6}}{m_{5}}&1\end{matrix}\right];\\\
$

$R_{4,5,3}(\Delta)^{-1}=\left[\begin{matrix}\displaystyle\frac{1}{m_{1}}&\displaystyle\frac{m_{4}}{m_{2}}&0\mcr\displaystyle\frac{1}{m_{1}}&\displaystyle\frac{1}{m_{2}}&0\mcr\displaystyle\frac{1}{m_{1}}&\displaystyle\frac{1}{m_{2}}&1\end{matrix}\right],\;\;R_{4,5,3}(\Delta)=\left[\begin{matrix}1&m_{1}-1&0\mcr-\displaystyle\frac{m_{2}}{m_{1}}&\displaystyle\frac{m_{2}}{m_{1}}&0\mcr
0&-1&1\end{matrix}\right];\\\ $

we know that $R_{4,2,3}(\Delta)$ is lower triangular matrix whose inverse
exists and is non-negative; $R_{4,5,3}(\Delta)$ is a non-singular M-matrix.
Thus, with Lemma 7 and Lemma 8, we can conclude that $R_{4,2,3}(\Delta)$ and
$R_{4,5,3}(\Delta)$ satisfies Chen-$\mathcal{S}$ with $\alpha_{1}<1$. ∎∎

This completes the proof that the $(R,\theta)$ for the 8 static-priority cases
are all Chen-$\mathcal{S}$.

As stated above, the reflection matrices for the 8 balanced DHV networks
following static-priority policies are all invertible. Now we check the sign
of their determinants. We can calculate that $det(R_{4,2,3})=m_{1}$,
$det(R_{4,5,3})=m_{2}$, $det(R_{4,5,6})=m_{3}$,
$det(R_{4,2,6})=\displaystyle\frac{m_{1}m_{3}m_{5}}{m_{1}m_{3}m_{5}+m_{2}m_{4}m_{6}}$,
$det(R_{1,2,3})=det(R_{1,5,3})=det(R_{1,2,6})=det(R_{1,5,6})=1$, where the
subscript label indicates the classes with high priority in each station. The
sign of the determinants are all positive. Thus, we can conclude that the
reflection matrix for a balanced DHV network is invertible for any ratio
matrix $\Delta$.

According to Theorem 4, in a balanced DHV network with $\alpha_{1}<1$, the
reflection matrix is invertible and $(R,\theta)$ is Chen-$\mathcal{S}$ for any
ratio matrix $\Delta$.

###  B.2 The SSC Inequalities for Balanced DHV Networks

The requirements in Theorem 7 are satisfied if for all static priority
policies, there exists a vector $h\in\mathbb{R}^{K}_{++}$ and a constant
$\hbar>0$ common to all static priority policies such that

$\displaystyle\sum_{k\in\mathcal{H}_{\pi}:z_{k}>0}h_{k}\dot{\overline{Z}}_{k}(t)\leq-\hbar\text{
for any regular time }t\text{ with
}\lVert\overline{Z}_{\mathcal{H}_{\pi}}(t)\rVert>0.$ (B.1)

We check the 8 static priority policies for (B.1) of balanced DHV networks in
this appendix.

Given a policy $\pi$, let $\overline{Z}^{\pi}_{k}(t)$ be the fluid level of
class $k$ at time $t$ under the policy $\pi$, and
$\dot{Y}^{\pi}(t)=\sum_{k\in\mathcal{H}_{\pi}:z_{k}>0}h_{k}\dot{\overline{Z}^{\pi}_{k}}(t)$.
Then, we need $\dot{Y}^{\pi}(t)<0$ with a common vector
$h\in\mathbb{R}^{K}_{++}$ for any regular time $t$ with
$\lVert\overline{Z}_{\mathcal{H}_{\pi}}(t)\rVert>0$, i.e., at least one of the
high-priority queues is not empty, under all the 8 static priority policies.
We omit $\pi$ when fixed. Note that in balanced DHV networks, $m_{k}<1$ and
thus $\mu_{k}>1$ for all $k\in\mathcal{K}$. Denote $I_{k}(t)$ as the input
rate of class $k$ at time $t$; $O_{k}(t)$ as the output rate of class $k$ at
time $t$. Then, $\dot{\overline{Z}}_{k}(t)=I_{k}(t)-O_{k}(t)\text{ for
}k\in\mathcal{K};I_{k}(t)=O_{k-1}(t)\text{ for
}k\in\mathcal{K}/\\{1\\};I_{1}(t)=\alpha_{1}$. In this section, we use a set
of high-priority classes to represent static priority policies, for example,
(4, 2, 6) stands for a static priority policy that class 4, 2 and 6 have high
priorities in each station.

Case 1: (4, 2, 6)

  1. 1.

$\overline{Z}_{2}(t)>0,\overline{Z}_{4}(t)>0,\overline{Z}_{6}(t)>0.$

Since class 4,2,6 are all not empty and have priorities in their stations,
servers focus on class 4,2,6 jobs, and $O_{1}(t)=O_{3}(t)=O_{5}(t)=0$, which
means that $I_{4}(t)=I_{2}(t)=I_{6}(t)=0$. Thus,
$\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)+h_{4}\dot{\overline{Z}}_{4}(t)+h_{6}\dot{\overline{Z}}_{6}(t)<0$,
since $O_{2}(t),O_{4}(t),O_{6}(t)>0$.

  2. 2.

$\overline{Z}_{2}(t)>0,\overline{Z}_{6}(t)>0,\overline{Z}_{4}(t)=0.$

Since class 2 and 6 are not empty and have priorities in their stations,
$O_{3}(t)=O_{5}(t)=0$. Thus, $\dot{\overline{Z}}_{6}(t)=0-\mu_{6}=-\mu_{6}$
and $O_{2}(t)=\mu_{2}$. Since class 4 have high priority in station 1 but
queue 4 is empty and $I_{4}(t)=O_{3}(t)=0$, station 1 focuses on class 1 jobs.
If queue 1 is not empty, $O_{1}(t)=\mu_{1}$, and
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{2}$; if queue 1 is empty,
$O_{1}(t)=\alpha_{1}$ (since $\alpha_{1}<1<\mu_{1}$), and
$\dot{\overline{Z}}_{2}(t)=\alpha_{1}-\mu_{2}$. Thus,
$\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)+h_{6}\dot{\overline{Z}}_{6}(t)=h_{2}(\mu_{1}-\mu_{2})-h_{6}\mu_{6}$
or $h_{2}(\alpha_{1}-\mu_{2})-h_{6}\mu_{6}$. Since
$h_{2}>0,\alpha_{1}<\mu_{1}$, we have
$h_{2}(\mu_{1}-\mu_{2})-h_{6}\mu_{6}>h_{2}(\alpha_{1}-\mu_{2})-h_{6}\mu_{6}$.
Thus, we need $h_{2}(\mu_{1}-\mu_{2})-h_{6}\mu_{6}<0$.

  3. 3.

$\overline{Z}_{2}(t)>0,\overline{Z}_{4}(t)>0,\overline{Z}_{6}(t)=0.$

Similarly, since class 2 and 4 are not empty and have priorities in their
stations, $O_{1}(t)=O_{5}(t)=0$. Thus,
$\dot{\overline{Z}}_{2}(t)=0-\mu_{2}=-\mu_{2}$ and $O_{4}(t)=\mu_{4}$. Since
class 6 have high priority in station 3 but queue 6 is empty and $I_{6}(t)=0$,
station 3 focuses on class 3 jobs. If queue 3 is not empty, the
$O_{3}(t)=\mu_{3}$, and $\dot{\overline{Z}}_{4}(t)=\mu_{3}-\mu_{4}$; if queue
3 is empty, $O_{3}(t)=min\\{\mu_{2},\mu_{3}\\}$, and
$\dot{\overline{Z}}_{4}(t)=min\\{\mu_{2},\mu_{3}\\}-\mu_{4}<\mu_{3}-\mu_{4}$.
Thus, we need
$\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)+h_{4}\dot{\overline{Z}}_{4}(t)\leq
h_{4}(\mu_{3}-\mu_{4})-h_{2}\mu_{2}<0$.

  4. 4.

$\overline{Z}_{4}(t)>0,\overline{Z}_{6}(t)>0,\overline{Z}_{2}(t)=0.$

Since class 4 and 6 are not empty and have priorities in their stations,
$O_{1}(t)=O_{3}(t)=0$. Thus, $\dot{\overline{Z}}_{4}(t)=0-\mu_{4}=-\mu_{4}$
and $O_{6}(t)=\mu_{6}$. Since class 2 have high priority in station 2 but
queue 2 is empty and $I_{2}(t)=0$, station 2 focuses on class 5 jobs. If queue
5 is not empty, $O_{5}(t)=\mu_{5}$, and
$\dot{\overline{Z}}_{6}(t)=\mu_{5}-\mu_{6}$; if queue 5 is empty,
$O_{5}(t)=min\\{\mu_{4},\mu_{5}\\}$, and
$\dot{\overline{Z}}_{6}(t)=min\\{\mu_{4},\mu_{5}\\}-\mu_{6}<\mu_{5}-\mu_{6}$.
Thus, we need
$\dot{Y}(t)=h_{4}\dot{\overline{Z}}_{4}(t)+h_{6}\dot{\overline{Z}}_{6}(t)\leq
h_{6}(\mu_{5}-\mu_{6})-h_{4}\mu_{4}<0$.

  5. 5.

$\overline{Z}_{2}(t)>0,\overline{Z}_{4}(t)=\overline{Z}_{6}(t)=0.$

We need $\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{2}(t)<0$. Since class 2 have priority in station 2 and
queue 2 is not empty, $O_{5}(t)=0$ and $O_{2}(t)=\mu_{2}$. As queue 6 is
empty, $I_{6}(t)=O_{5}(t)=0$, we know that $O_{6}(t)=0$ and station 3 focus on
class 3 jobs. If queue 3 is not empty, $O_{3}(t)=\mu_{3}$; if queue 3 is
empty, $O_{3}(t)=min\\{\mu_{2},\mu_{3}\\}\leq\mu_{3}$. Thus,
$\dot{\overline{T}}_{4}(t)=\frac{O_{3}(t)}{\mu_{4}}\geq\frac{min\\{\mu_{2},\mu_{3}\\}}{\mu_{4}}$.

Since class 4 have priority in station 1 and queue 4 is empty, we have
$\mu_{4}\dot{\overline{T}}_{4}(t)=O_{3}(t)$. $\dot{\overline{Z}}_{2}(t)<0$ is
equivalent to $O_{1}(t)=\mu_{1}\dot{\overline{T}}_{1}(t)<O_{2}(t)=\mu_{2}$.
Since $0\leq\dot{\overline{T}}_{1}(t)\leq 1-\dot{\overline{T}}_{4}(t)$, it is
sufficient to consider
$\mu_{1}(1-\frac{min\\{\mu_{2},\mu_{3}\\}}{\mu_{4}})<\mu_{2}$, which requires
$\mu_{1}(1-\frac{\mu_{2}}{\mu_{4}})<\mu_{2}$ and
$\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})<\mu_{2}$.

$\mu_{1}(1-\frac{\mu_{2}}{\mu_{4}})<\mu_{2}\Leftrightarrow
1-\frac{\mu_{2}}{\mu_{4}}<\frac{\mu_{2}}{\mu_{1}}\Leftrightarrow\frac{\mu_{2}}{\mu_{4}}+\frac{\mu_{2}}{\mu_{1}}>1\Leftrightarrow\frac{m_{4}}{m_{2}}+\frac{m_{1}}{m_{2}}>1\Leftrightarrow\frac{1}{m_{2}}>1$
is true since $m_{2}<1$.

$\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})<\mu_{2}\Leftrightarrow
1-\frac{\mu_{3}}{\mu_{4}}<\frac{\mu_{2}}{\mu_{1}}\Leftrightarrow\frac{\mu_{2}}{\mu_{1}}+\frac{\mu_{3}}{\mu_{4}}>1\Leftrightarrow\frac{m_{1}}{m_{2}}+\frac{m_{4}}{m_{3}}>1$
is true since $\frac{m_{1}}{m_{2}}+\frac{m_{4}}{m_{3}}>m_{1}+m_{4}=1$.

Thus, $\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)<0$ is satisfied.

  6. 6.

$\overline{Z}_{4}(t)>0,\overline{Z}_{2}(t)=\overline{Z}_{6}(t)=0.$

We need $\dot{Y}(t)=h_{4}\dot{\overline{Z}}_{4}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{4}(t)<0$. Since class 4 have priority in station 1 and
queue 4 is not empty, $O_{1}(t)=0$ and $O_{4}(t)=\mu_{4}$. As queue 2 is
empty, $I_{2}(t)=O_{1}(t)=0$, we know that $O_{2}(t)=0$ and station 2 focus on
class 5 jobs. If queue 5 is not empty, $O_{5}(t)=\mu_{5}$; if queue 5 is
empty, $O_{5}(t)=min\\{\mu_{4},\mu_{5}\\}\leq\mu_{5}$. Thus,
$\dot{\overline{T}}_{6}(t)=\frac{O_{5}(t)}{\mu_{6}}\geq\frac{min\\{\mu_{4},\mu_{5}\\}}{\mu_{6}}$.

Since class 6 have priority in station 3 and queue 6 is empty, we have
$\mu_{6}\dot{\overline{T}}_{6}(t)=O_{5}(t)$. $\dot{\overline{Z}}_{4}(t)<0$ is
equivalent to $O_{3}(t)=\mu_{3}\dot{\overline{T}}_{3}(t)<O_{4}(t)=\mu_{4}$.
Since $0\leq\dot{\overline{T}}_{3}(t)\leq 1-\dot{\overline{T}}_{6}(t)$, it is
sufficient to consider
$\mu_{3}(1-\frac{min\\{\mu_{4},\mu_{5}\\}}{\mu_{6}})<\mu_{4}$, which requires
$\mu_{3}(1-\frac{\mu_{4}}{\mu_{6}})<\mu_{4}$ and
$\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})<\mu_{4}$.

$\mu_{3}(1-\frac{\mu_{4}}{\mu_{6}})<\mu_{4}\Leftrightarrow
1-\frac{\mu_{4}}{\mu_{6}}<\frac{\mu_{4}}{\mu_{3}}\Leftrightarrow\frac{\mu_{4}}{\mu_{3}}+\frac{\mu_{4}}{\mu_{6}}>1\Leftrightarrow\frac{m_{3}}{m_{4}}+\frac{m_{6}}{m_{4}}>1\Leftrightarrow\frac{1}{m_{4}}>1$
is true since $m_{4}<1$.

$\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})<\mu_{4}\Leftrightarrow
1-\frac{\mu_{5}}{\mu_{6}}<\frac{\mu_{4}}{\mu_{3}}\Leftrightarrow\frac{\mu_{4}}{\mu_{3}}+\frac{\mu_{5}}{\mu_{6}}>1\Leftrightarrow\frac{m_{3}}{m_{4}}+\frac{m_{6}}{m_{5}}>1$
is true since $\frac{m_{3}}{m_{4}}+\frac{m_{6}}{m_{5}}>m_{3}+m_{6}=1$.

Thus, $\dot{Y}(t)=h_{4}\dot{\overline{Z}}_{4}(t)<0$ is satisfied.

  7. 7.

$\overline{Z}_{6}(t)>0,\overline{Z}_{2}(t)=\overline{Z}_{4}(t)=0.$

We need $\dot{Y}(t)=h_{6}\dot{\overline{Z}}_{6}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{6}(t)<0$. Since class 6 have priority in station 3 and
queue 6 is not empty, $O_{3}(t)=0$ and $O_{6}(t)=\mu_{6}$. As queue 4 is
empty, $I_{4}(t)=O_{3}(t)=0$, we know that $O_{4}(t)=0$ and station 1 focus on
class 1 jobs. If queue 1 is not empty, $O_{1}(t)=\mu_{1}$; if queue 1 is
empty, $O_{1}(t)=min\\{\alpha_{1},\mu_{1}\\}=\alpha_{1}$. Thus,
$\dot{\overline{T}}_{2}(t)=\frac{O_{1}(t)}{\mu_{2}}\geq\frac{\alpha_{1}}{\mu_{2}}$.

Since class 2 have priority in station 5 and queue 2 is empty, we have
$\mu_{2}\dot{\overline{T}}_{2}(t)=O_{1}(t)$. $\dot{\overline{Z}}_{6}(t)<0$ is
equivalent to $O_{5}(t)=\mu_{5}\dot{\overline{T}}_{5}(t)<O_{6}(t)=\mu_{6}$.
Since $0\leq\dot{\overline{T}}_{5}(t)\leq 1-\dot{\overline{T}}_{6}(t)$, it is
sufficient to consider $\mu_{5}(1-\frac{\alpha_{1}}{\mu_{2}})<\mu_{6}$.
$\mu_{5}(1-\frac{\alpha_{1}}{\mu_{2}})<\mu_{6}\Leftrightarrow
1-\frac{\alpha_{1}}{\mu_{2}}<\frac{\mu_{6}}{\mu_{5}}\Leftrightarrow\frac{\mu_{6}}{\mu_{5}}+\frac{\alpha_{1}}{\mu_{2}}>1\Leftrightarrow\frac{m_{5}}{m_{6}}+\alpha_{1}m_{2}>1\Leftrightarrow\alpha_{1}>\frac{m_{6}-m_{5}}{m_{2}m_{6}}$.

Thus, $\dot{Y}(t)=h_{6}\dot{\overline{Z}}_{6}(t)<0$ is satisfied when
$\alpha_{1}>\frac{m_{6}-m_{5}}{m_{2}m_{6}}$.

Note that $\frac{m_{6}-m_{5}}{m_{2}m_{6}}<1$, thus, the condition
$\alpha_{1}>\frac{m_{6}-m_{5}}{m_{2}m_{6}}$ does not conflict with the nominal
condition($\alpha_{1}<1$). It is clear that $\frac{m_{6}-m_{5}}{m_{2}m_{6}}<1$
since $\frac{m_{6}-m_{5}}{m_{2}m_{6}}<1\Leftrightarrow
m_{6}-m_{5}<m_{2}m_{6}\Leftrightarrow m_{5}>m_{6}(1-m_{2})\Leftrightarrow
m_{5}>m_{6}m_{5}\Leftrightarrow m_{5}(1-m_{6})>0\Leftrightarrow m_{5}m_{3}>0$,
which is true obviously.

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 1:

$\displaystyle h_{2}(\mu_{1}-\mu_{2})-h_{6}\mu_{6}<0,$ $\displaystyle
h_{4}(\mu_{3}-\mu_{4})-h_{2}\mu_{2}<0,$ $\displaystyle
h_{6}(\mu_{5}-\mu_{6})-h_{4}\mu_{4}<0,$
$\displaystyle\alpha_{1}>\frac{m_{6}-m_{5}}{m_{2}m_{6}}.$

Case 2: (1, 2, 3)

  1. 1.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)>0.$

It is clear that $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{2}$ and
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-\mu_{3}$. Thus, we need
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{2}\dot{\overline{Z}}_{2}(t)+h_{3}\dot{\overline{Z}}_{3}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})+h_{3}(\mu_{2}-\mu_{3})<0$.

  2. 2.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)=0.$

It is clear that $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$ and
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{2}$. Thus, we need
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{2}\dot{\overline{Z}}_{2}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})<0$.

  3. 3.

$\overline{Z}_{1}(t)>0,\overline{Z}_{3}(t)>0,\overline{Z}_{2}(t)=0.$

It is clear that $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$. Since class 2
has priority in station 2 and queue 2 is empty, we have
$O_{2}(t)=min\\{\mu_{1},\mu_{2}\\}$. If $O_{2}(t)=\mu_{2}$, $t$ is not
differentiable, since it will turn to the situation that
$\overline{Z}_{2}(t)>0$ immediately. Thus, $O_{2}(t)=\mu_{1}$. Then we can get
$\dot{\overline{Z}}_{3}(t)=\mu_{1}-\mu_{3}$ and we need
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{3}\dot{\overline{Z}}_{3}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{3}(\mu_{1}-\mu_{3})<0$.

  4. 4.

$\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)>0,\overline{Z}_{1}(t)=0.$

Since class 1 has priority in station 1 and queue 1 is empty, we have
$O_{1}(t)=min\\{\mu_{1},\alpha_{1}\\}=\alpha_{1}$. Then we can get
$\dot{\overline{Z}}_{2}(t)=\alpha_{1}-\mu_{2}$,
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-\mu_{3}$ and we need
$\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)+h_{3}\dot{\overline{Z}}_{3}(t)=h_{2}(\alpha_{1}-\mu_{2})+h_{3}(\mu_{2}-\mu_{3})<0$.

  5. 5.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)=\overline{Z}_{3}(t)=0.$

It is clear that $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}<0$ and
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)<0$.

  6. 6.

$\overline{Z}_{2}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{3}(t)=0.$

Since class 1 has priority in station 1 and queue 1 is empty, we have
$O_{1}(t)=min\\{\mu_{1},\alpha_{1}\\}=\alpha_{1}$. Thus, it is clear that
$\dot{\overline{Z}}_{2}(t)=\alpha_{1}-\mu_{2}<0$ and
$\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)<0$.

  7. 7.

$\overline{Z}_{3}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{2}(t)=0.$

Similarly, it is clear that $\dot{\overline{Z}}_{3}(t)=\alpha_{1}-\mu_{3}<0$
and $\dot{Y}(t)=h_{3}\dot{\overline{Z}}_{3}(t)<0$.

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 2:

$\displaystyle
h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})+h_{3}(\mu_{2}-\mu_{3})<0,$
$\displaystyle h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})<0,$
$\displaystyle h_{1}(\alpha_{1}-\mu_{1})+h_{3}(\mu_{1}-\mu_{3})<0,$
$\displaystyle h_{2}(\alpha_{1}-\mu_{2})+h_{3}(\mu_{2}-\mu_{3})<0.$

Case 3: (4, 2, 3)

  1. 1.

$\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)>0,\overline{Z}_{4}(t)>0.$

Since class 4 is not empty and has high priority in station 1, thus
$O_{1}(t)=0$ and $\dot{\overline{Z}}_{2}(t)=0-\mu_{2}=-\mu_{2}$. It is easy to
know that $\dot{\overline{Z}}_{3}(t)=\mu_{2}-\mu_{3}$ and
$\dot{\overline{Z}}_{4}(t)=\mu_{3}-\mu_{4}$. Thus, we need
$\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)+h_{3}\dot{\overline{Z}}_{3}(t)+h_{4}\dot{\overline{Z}}_{4}(t)=h_{3}(\mu_{2}-\mu_{3})+h_{4}(\mu_{3}-\mu_{4})-h_{2}\mu_{2}<0$.

  2. 2.

$\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)>0,\overline{Z}_{4}(t)=0.$

Since class 2 has priority in station 2 and queue 2 is not empty, we have
$O_{2}(t)=\mu_{2}$. Similarly, $O_{3}(t)=\mu_{3}$. Then,
$\dot{\overline{Z}}_{3}(t)=\mu_{3}-\mu_{2}$. To make $t$ differentiable, we
have
$\dot{\overline{Z}}_{4}(t)=\mu_{3}-\mu_{4}\dot{\overline{T}}_{4}(t)=0\Rightarrow\dot{\overline{T}}_{4}(t)=\frac{\mu_{3}}{\mu_{4}}\Rightarrow\dot{\overline{T}}_{1}(t)_{max}=1-\frac{\mu_{3}}{\mu_{4}}\Rightarrow
O_{1}(t)_{max}=\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})\Rightarrow\dot{\overline{Z}}_{2}(t)_{max}=\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{2}$.
Thus, we need
$\dot{Y}(t)_{max}=h_{2}\dot{\overline{Z}}_{2}(t)_{max}+h_{3}\dot{\overline{Z}}_{3}(t)=h_{2}\left[\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{2}\right]+h_{3}(\mu_{2}-\mu_{3})<0$.

  3. 3.

$\overline{Z}_{2}(t)>0,\overline{Z}_{4}(t)>0,\overline{Z}_{3}(t)=0.$

Since class 4 has priority in station 1 and queue 4 is not empty, we have
$O_{1}(t)=0$ and $O_{4}(t)=\mu_{4}$. It is clear that $O_{2}(t)=\mu_{2}$.
Then, $\dot{\overline{Z}}_{2}(t)=0-\mu_{2}=-\mu_{2}$. To make $t$
differentiable, we have
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-O_{3}(t)=0\Rightarrow O_{3}(t)=\mu_{2}$.
Then $\dot{\overline{Z}}_{4}(t)=\mu_{2}-\mu_{4}$. Thus, we need
$\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)+h_{4}\dot{\overline{Z}}_{4}(t)=h_{4}(\mu_{2}-\mu_{4})-h_{2}\mu_{2}<0$.

  4. 4.

$\overline{Z}_{3}(t)>0,\overline{Z}_{4}(t)>0,\overline{Z}_{2}(t)=0.$

It is clear that $O_{3}(t)=\mu_{3}$, $O_{4}(t)=\mu_{4}$, then
$\dot{\overline{Z}}_{4}(t)=\mu_{3}-\mu_{4}$. Since class 4 has priority in
station 1 and queue 4 is not empty, $O_{1}(t)=0$. Since queue 2 is empty, we
have $O_{2}(t)=0\Rightarrow dot{\overline{Z}}_{3}(t)=0-\mu_{3}=-\mu_{3}$.
Thus, we need
$\dot{Y}(t)=h_{3}\dot{\overline{Z}}_{3}(t)+h_{4}\dot{\overline{Z}}_{4}(t)=h_{4}(\mu_{3}-\mu_{4})-h_{3}\mu_{3}<0$.

  5. 5.

$\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)=\overline{Z}_{4}(t)=0.$

We need $\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{2}(t)<0$. It is clear that $O_{2}(t)=\mu_{2}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-O_{3}(t)=0\Rightarrow O_{3}(t)=\mu_{2}$ and
$\dot{\overline{Z}}_{4}(t)=\mu_{2}-\mu_{4}\dot{\overline{T}}_{4}(t)=0\Rightarrow\dot{\overline{T}}_{4}(t)=\frac{\mu_{2}}{\mu_{4}}\Rightarrow\dot{\overline{T}}_{1}(t)_{max}=1-\frac{\mu_{2}}{\mu_{4}}\Rightarrow
O_{1}(t)_{max}=\mu_{1}(1-\frac{\mu_{2}}{\mu_{4}})\Rightarrow\dot{\overline{Z}}_{2}(t)_{max}=O_{1}(t)_{max}-\mu_{2}=\mu_{1}(1-\frac{\mu_{2}}{\mu_{4}})-\mu_{2}$,
which is needed to be negative. $\mu_{1}(1-\frac{\mu_{2}}{\mu_{4}})-\mu_{2}<0$
is true since
$\mu_{1}(1-\frac{\mu_{2}}{\mu_{4}})-\mu_{2}<0\Leftrightarrow\frac{\mu_{2}}{\mu_{4}}+\frac{\mu_{2}}{\mu_{1}}>1\Leftrightarrow\frac{m_{4}}{m_{2}}+\frac{m_{1}}{m_{2}}>1\Leftrightarrow\frac{1}{m_{2}}>1$,
which is satisfied for sure.

  6. 6.

$\overline{Z}_{3}(t)>0,\overline{Z}_{4}(t)=\overline{Z}_{2}(t)=0.$

We need $\dot{Y}(t)=h_{3}\dot{\overline{Z}}_{3}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{3}(t)<0$. It is clear that $O_{3}(t)=\mu_{3}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{4}(t)=\mu_{3}-O_{4}(t)=0\Rightarrow
O_{4}(t)=\mu_{4}\dot{\overline{T}}_{4}(t)=\mu_{3}\Rightarrow\dot{\overline{T}}_{4}(t)=\frac{\mu_{3}}{\mu_{4}}\Rightarrow\dot{\overline{T}}_{1}(t)_{max}=1-\frac{\mu_{3}}{\mu_{4}}\Rightarrow
O_{1}(t)_{max}=\mu_{1}\dot{\overline{T}}_{1}(t)_{max}=\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})$
and $\dot{\overline{Z}}_{2}(t)=O_{1}(t)-O_{2}(t)=0\Rightarrow
O_{2}(t)_{max}=O_{1}(t)_{max}=\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})\Rightarrow\dot{\overline{Z}}_{3}(t)_{max}=O_{2}(t)_{max}-\mu_{3}=\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{3}$,
which is needed to be negative. $\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{3}<0$
is true since
$\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{3}<0\Leftrightarrow\frac{\mu_{3}}{\mu_{1}}+\frac{\mu_{3}}{\mu_{4}}>1\Leftrightarrow\frac{m_{4}}{m_{3}}+\frac{m_{1}}{m_{3}}>1\Leftrightarrow\frac{1}{m_{3}}>1$,
which is satisfied for sure.

  7. 7.

$\overline{Z}_{4}(t)>0,\overline{Z}_{2}(t)=\overline{Z}_{3}(t)=0.$

We need $\dot{Y}(t)=h_{4}\dot{\overline{Z}}_{4}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{4}(t)<0$. Since class 4 has priority in station 1 and
queue 4 is not empty, we have $O_{4}(t)=\mu_{4}$ and $O_{1}(t)=0$. Since queue
2 is empty, we have $O_{2}(t)=0$. Again, since queue 3 is empty, we have
$O_{3}(t)=0$. Thus, $\dot{\overline{Z}}_{4}(t)=0-\mu_{4}<0$ for sure.

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 3:

$\displaystyle h_{3}(\mu_{2}-\mu_{3})+h_{4}(\mu_{3}-\mu_{4})-h_{2}\mu_{2}<0,$
$\displaystyle
h_{2}\left[\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{2}\right]+h_{3}(\mu_{2}-\mu_{3})<0,$
$\displaystyle h_{4}(\mu_{2}-\mu_{4})-h_{2}\mu_{2}<0,$ $\displaystyle
h_{4}(\mu_{3}-\mu_{4})-h_{3}\mu_{3}<0.$

Case 4: (4, 5, 3)

  1. 1.

$\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)>0,\overline{Z}_{3}(t)>0.$

It is clear that $O_{3}(t)=\mu_{3}$, $O_{4}(t)=\mu_{4}$, $O_{5}(t)=\mu_{5}$,
$O_{2}(t)=0$, then $\dot{\overline{Z}}_{3}(t)=0-\mu_{3}=-\mu_{3}$,
$\dot{\overline{Z}}_{4}(t)=\mu_{3}-\mu_{4}$ and
$\dot{\overline{Z}}_{5}(t)=\mu_{4}-\mu_{5}$. Thus, we need
$\dot{Y}(t)=h_{3}\dot{\overline{Z}}_{3}(t)+h_{4}\dot{\overline{Z}}_{4}(t)+h_{5}\dot{\overline{Z}}_{5}(t)=h_{4}(\mu_{3}-\mu_{4})+h_{5}(\mu_{4}-\mu_{5})-h_{3}\mu_{3}<0$.

  2. 2.

$\overline{Z}_{3}(t)>0,\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)=0.$

It is clear that $O_{3}(t)=\mu_{3}$, $O_{4}(t)=\mu_{4}$, $O_{1}(t)=0$, then
$\dot{\overline{Z}}_{4}(t)=\mu_{3}-\mu_{4}$. To make $t$ differentiable, we
have $\dot{\overline{Z}}_{5}(t)=\mu_{4}-O_{5}(t)=0\Rightarrow
O_{5}(t)=\mu_{5}\dot{\overline{T}}_{5}(t)=\mu_{4}\Rightarrow\dot{\overline{T}}_{5}(t)=\frac{\mu_{4}}{\mu_{5}}\Rightarrow\dot{\overline{T}}_{2}(t)_{max}=1-\frac{\mu_{4}}{\mu_{5}}\Rightarrow
O_{2}(t)_{max}=\mu_{2}\dot{\overline{T}}_{2}(t)_{max}=\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})\Rightarrow\dot{\overline{Z}}_{3}(t)_{max}=O_{2}(t)_{max}-\mu_{3}=\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{3}$.
Thus, we need
$\dot{Y}(t)_{max}=h_{3}\dot{\overline{Z}}_{3}(t)_{max}+h_{4}\dot{\overline{Z}}_{4}(t)=h_{3}\left[\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{3}\right]+h_{4}(\mu_{3}-\mu_{4})<0$.

  3. 3.

$\overline{Z}_{3}(t)>0,\overline{Z}_{5}(t)>0,\overline{Z}_{4}(t)=0.$

It is clear that $O_{3}(t)=\mu_{3}$, $O_{5}(t)=\mu_{5}$, $O_{2}(t)=0$, then
$\dot{\overline{Z}}_{3}(t)=0-\mu_{3}=-\mu_{3}$. To make $t$ differentiable, we
have $\dot{\overline{Z}}_{4}(t)=\mu_{3}-O_{4}(t)=0\Rightarrow
O_{4}(t)==\mu_{3}\Rightarrow\dot{\overline{Z}}_{5}(t)=\mu_{3}-\mu_{5}$. Thus,
we need
$\dot{Y}(t)=h_{3}\dot{\overline{Z}}_{3}(t)+h_{5}\dot{\overline{Z}}_{5}(t)=h_{5}(\mu_{3}-\mu_{5})-h_{3}\mu_{3}<0$.

  4. 4.

$\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)>0,\overline{Z}_{3}(t)=0.$

It is clear that $O_{4}(t)=\mu_{4}$, $O_{5}(t)=\mu_{5}$, $O_{2}(t)=0$, then
$\dot{\overline{Z}}_{5}(t)=\mu_{4}-\mu_{5}$. To make $t$ differentiable, we
have $\dot{\overline{Z}}_{3}(t)=0-O_{3}(t)=0\Rightarrow
O_{3}(t)=0\Rightarrow\dot{\overline{Z}}_{4}(t)=0-\mu_{4}=-\mu_{4}$. Thus, we
need
$\dot{Y}(t)=h_{4}\dot{\overline{Z}}_{4}(t)+h_{5}\dot{\overline{Z}}_{5}(t)=h_{5}(\mu_{4}-\mu_{5})-h_{4}\mu_{4}<0$.

  5. 5.

$\overline{Z}_{3}(t)>0,\overline{Z}_{4}(t)=\overline{Z}_{5}(t)=0.$

We need $\dot{Y}(t)=h_{3}\dot{\overline{Z}}_{3}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{3}(t)<0$. It is clear that $O_{3}(t)=\mu_{3}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{4}(t)=O_{4}(t)-\mu_{3}=0\Rightarrow O_{4}(t)=\mu_{3}$ and
$\dot{\overline{Z}}_{5}(t)=O_{4}(t)-O_{5}(t)=0\Rightarrow
O_{5}(t)=\mu_{5}\dot{\overline{T}}_{5}(t)=\mu_{3}\Rightarrow\dot{\overline{T}}_{5}(t)=\frac{\mu_{3}}{\mu_{5}}\Rightarrow\dot{\overline{T}}_{2}(t)_{max}=1-\frac{\mu_{3}}{\mu_{5}}\Rightarrow
O_{2}(t)_{max}=\mu_{2}\dot{\overline{T}}_{2}(t)_{max}=\mu_{2}(1-\frac{\mu_{3}}{\mu_{5}})\Rightarrow\dot{\overline{Z}}_{3}(t)_{max}=O_{2}(t)_{max}-\mu_{3}=\mu_{2}(1-\frac{\mu_{3}}{\mu_{5}})-\mu_{3}$,
which is needed to be negative. $\mu_{2}(1-\frac{\mu_{3}}{\mu_{5}})-\mu_{3}<0$
is true since
$\mu_{2}(1-\frac{\mu_{3}}{\mu_{5}})-\mu_{3}<0\Leftrightarrow\frac{\mu_{3}}{\mu_{2}}+\frac{\mu_{3}}{\mu_{5}}>1\Leftrightarrow\frac{m_{2}}{m_{3}}+\frac{m_{5}}{m_{3}}>1\Leftrightarrow\frac{1}{m_{3}}>1$,
which is satisfied for sure.

  6. 6.

$\overline{Z}_{4}(t)>0,\overline{Z}_{3}(t)=\overline{Z}_{5}(t)=0.$

We need $\dot{Y}(t)=h_{4}\dot{\overline{Z}}_{4}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{4}(t)<0$. It is clear that $O_{4}(t)=\mu_{4}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{5}(t)=\mu_{4}-O_{5}(t)=0\Rightarrow
O_{5}(t)=\mu_{5}\dot{\overline{T}}_{5}(t)=\mu_{4}\Rightarrow\dot{\overline{T}}_{5}(t)=\frac{\mu_{4}}{\mu_{5}}\Rightarrow\dot{\overline{T}}_{2}(t)_{max}=1-\frac{\mu_{4}}{\mu_{5}}\Rightarrow
O_{2}(t)_{max}=\mu_{2}\dot{\overline{T}}_{2}(t)_{max}=\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})$
and $\dot{\overline{Z}}_{3}(t)=O_{2}(t)-O_{3}(t)=0\Rightarrow
O_{2}(t)_{max}=O_{3}(t)_{max}=\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})\Rightarrow\dot{\overline{Z}}_{4}(t)_{max}=O_{3}(t)_{max}-\mu_{4}=\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{4}$,
which is needed to be negative. $\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{4}<0$
is true since
$\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{4}<0\Leftrightarrow\frac{\mu_{4}}{\mu_{2}}+\frac{\mu_{4}}{\mu_{5}}>1\Leftrightarrow\frac{m_{2}}{m_{4}}+\frac{m_{5}}{m_{4}}>1\Leftrightarrow\frac{1}{m_{4}}>1$,
which is satisfied for sure.

  7. 7.

$\overline{Z}_{5}(t)>0,\overline{Z}_{3}(t)=\overline{Z}_{4}(t)=0.$

We need $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{5}(t)<0$. Since class 5 has priority in station 2 and
queue 5 is not empty, we have $O_{5}(t)=\mu_{5}$ and $O_{2}(t)=0$. Since queue
3 is empty, we have $O_{3}(t)=0$. Again, since queue 4 is empty, we have
$O_{4}(t)=0$. Thus, $\dot{\overline{Z}}_{5}(t)=0-\mu_{5}<0$ for sure.

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 4:

$\displaystyle h_{4}(\mu_{3}-\mu_{4})+h_{5}(\mu_{4}-\mu_{5})-h_{3}\mu_{3}<0,$
$\displaystyle
h_{3}\left[\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{3}\right]+h_{4}(\mu_{3}-\mu_{4})<0,$
$\displaystyle h_{5}(\mu_{3}-\mu_{5})-h_{3}\mu_{3}<0,$ $\displaystyle
h_{5}(\mu_{4}-\mu_{5})-h_{4}\mu_{4}<0.$

Case 5: (4, 5, 6)

  1. 1.

$\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)>0,\overline{Z}_{6}(t)>0.$

It is clear that $O_{4}(t)=\mu_{4}$, $O_{5}(t)=\mu_{5}$, $O_{6}(t)=\mu_{6}$,
$O_{3}(t)=0$, then $\dot{\overline{Z}}_{4}(t)=0-\mu_{4}=-\mu_{4}$,
$\dot{\overline{Z}}_{5}(t)=\mu_{4}-\mu_{5}$ and
$\dot{\overline{Z}}_{6}(t)=\mu_{5}-\mu_{6}$. Thus, we need
$\dot{Y}(t)=h_{4}\dot{\overline{Z}}_{4}(t)+h_{5}\dot{\overline{Z}}_{5}(t)+h_{6}\dot{\overline{Z}}_{6}(t)=h_{5}(\mu_{4}-\mu_{5})+h_{6}(\mu_{5}-\mu_{6})-h_{4}\mu_{4}<0$.

  2. 2.

$\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)>0,\overline{Z}_{6}(t)=0.$

Since class 4 has priority in station 1 and queue 4 is not empty, we have
$O_{4}(t)=\mu_{4}$. Similarly, $O_{5}(t)=\mu_{5}$. Then,
$\dot{\overline{Z}}_{5}(t)=\mu_{4}-\mu_{5}$. To make $t$ differentiable, we
have
$\dot{\overline{Z}}_{6}(t)=\mu_{5}-\mu_{6}\dot{\overline{T}}_{6}(t)=0\Rightarrow\dot{\overline{T}}_{6}(t)=\frac{\mu_{5}}{\mu_{6}}\Rightarrow\dot{\overline{T}}_{3}(t)_{max}=1-\frac{\mu_{5}}{\mu_{6}}\Rightarrow
O_{3}(t)_{max}=\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})\Rightarrow\dot{\overline{Z}}_{4}(t)_{max}=\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{4}$.
Thus, we need
$\dot{Y}(t)_{max}=h_{4}\dot{\overline{Z}}_{4}(t)_{max}+h_{5}\dot{\overline{Z}}_{5}(t)=h_{4}\left[\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{4}\right]+h_{5}(\mu_{4}-\mu_{5})<0$.

  3. 3.

$\overline{Z}_{4}(t)>0,\overline{Z}_{6}(t)>0,\overline{Z}_{5}(t)=0.$

Since class 6 has priority in station 3 and queue 6 is not empty, we have
$O_{3}(t)=0$ and $O_{6}(t)=\mu_{6}$. It is clear that $O_{4}(t)=\mu_{4}$.
Then, $\dot{\overline{Z}}_{4}(t)=0-\mu_{4}=-\mu_{4}$. To make $t$
differentiable, we have
$\dot{\overline{Z}}_{5}(t)=\mu_{4}-O_{5}(t)=0\Rightarrow O_{5}(t)=\mu_{4}$.
Then $\dot{\overline{Z}}_{6}(t)=\mu_{4}-\mu_{6}$. Thus, we need
$\dot{Y}(t)=h_{4}\dot{\overline{Z}}_{4}(t)+h_{6}\dot{\overline{Z}}_{6}(t)=h_{6}(\mu_{4}-\mu_{6})-h_{4}\mu_{4}<0$.

  4. 4.

$\overline{Z}_{5}(t)>0,\overline{Z}_{6}(t)>0,\overline{Z}_{4}(t)=0.$

It is clear that $O_{5}(t)=\mu_{5}$, $O_{6}(t)=\mu_{6}$, $O_{3}(t)=0$, then
$\dot{\overline{Z}}_{6}(t)=\mu_{5}-\mu_{6}$. To make $t$ differentiable, we
have $\dot{\overline{Z}}_{4}(t)=0-O_{4}(t)=0\Rightarrow
O_{4}(t)=0\Rightarrow\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$. Thus, we
need
$\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)+h_{6}\dot{\overline{Z}}_{6}(t)=h_{6}(\mu_{5}-\mu_{6})-h_{5}\mu_{5}<0$.

  5. 5.

$\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)=\overline{Z}_{6}(t)=0.$

We need $\dot{Y}(t)=h_{4}\dot{\overline{Z}}_{4}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{4}(t)<0$. It is clear that $O_{4}(t)=\mu_{4}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{5}(t)=\mu_{4}-O_{5}(t)=0\Rightarrow O_{5}(t)=\mu_{4}$ and
$\dot{\overline{Z}}_{6}(t)=O_{5}(t)-O_{6}(t)=0\Rightarrow
O_{6}(t)=\mu_{6}\dot{\overline{T}}_{6}(t)=\mu_{4}\Rightarrow\dot{\overline{T}}_{6}(t)=\frac{\mu_{4}}{\mu_{6}}\Rightarrow\dot{\overline{T}}_{3}(t)_{max}=1-\frac{\mu_{4}}{\mu_{6}}\Rightarrow
O_{3}(t)_{max}=\mu_{3}\dot{\overline{T}}_{3}(t)_{max}=\mu_{3}(1-\frac{\mu_{4}}{\mu_{6}})\Rightarrow\dot{\overline{Z}}_{4}(t)_{max}=O_{3}(t)_{max}-\mu_{4}=\mu_{3}(1-\frac{\mu_{4}}{\mu_{6}})-\mu_{4}$,
which is needed to be negative. $\mu_{3}(1-\frac{\mu_{4}}{\mu_{6}})-\mu_{4}<0$
is true since
$\mu_{3}(1-\frac{\mu_{4}}{\mu_{6}})-\mu_{4}<0\Leftrightarrow\frac{\mu_{4}}{\mu_{3}}+\frac{\mu_{4}}{\mu_{6}}>1\Leftrightarrow\frac{m_{3}}{m_{4}}+\frac{m_{6}}{m_{4}}>1\Leftrightarrow\frac{1}{m_{4}}>1$,
which is satisfied for sure.

  6. 6.

$\overline{Z}_{5}(t)>0,\overline{Z}_{4}(t)=\overline{Z}_{6}(t)=0.$

We need $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{5}(t)<0$. It is clear that $O_{5}(t)=\mu_{5}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{6}(t)=\mu_{5}-O_{6}(t)=0\Rightarrow
O_{6}(t)=\mu_{6}\dot{\overline{T}}_{6}(t)=\mu_{5}\Rightarrow\dot{\overline{T}}_{6}(t)=\frac{\mu_{5}}{\mu_{6}}\Rightarrow\dot{\overline{T}}_{3}(t)_{max}=1-\frac{\mu_{5}}{\mu_{6}}\Rightarrow
O_{3}(t)_{max}=\mu_{3}\dot{\overline{T}}_{3}(t)_{max}=\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})$
and $\dot{\overline{Z}}_{4}(t)=O_{3}(t)-O_{4}(t)=0\Rightarrow
O_{4}(t)_{max}=O_{3}(t)_{max}=\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})\Rightarrow\dot{\overline{Z}}_{5}(t)_{max}=O_{4}(t)_{max}-\mu_{5}=\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{5}$,
which is needed to be negative. $\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{5}<0$
is true since
$\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{5}<0\Leftrightarrow\frac{\mu_{5}}{\mu_{6}}+\frac{\mu_{5}}{\mu_{3}}>1\Leftrightarrow\frac{m_{3}}{m_{5}}+\frac{m_{6}}{m_{5}}>1\Leftrightarrow\frac{1}{m_{5}}>1$,
which is satisfied for sure.

  7. 7.

$\overline{Z}_{6}(t)>0,\overline{Z}_{4}(t)=\overline{Z}_{5}(t)=0.$

We need $\dot{Y}(t)=h_{6}\dot{\overline{Z}}_{6}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{6}(t)<0$. Since class 6 has priority in station 3 and
queue 6 is not empty, we have $O_{6}(t)=\mu_{6}$ and $O_{3}(t)=0$. Since queue
4 is empty, we have $O_{4}(t)=0$. Again, since queue 5 is empty, we have
$O_{5}(t)=0$. Thus, $\dot{\overline{Z}}_{6}(t)=0-\mu_{6}<0$ for sure.

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 5:

$\displaystyle h_{5}(\mu_{4}-\mu_{5})+h_{6}(\mu_{5}-\mu_{6})-h_{4}\mu_{4}<0,$
$\displaystyle
h_{4}\left[\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{4}\right]+h_{5}(\mu_{4}-\mu_{5})<0,$
$\displaystyle h_{6}(\mu_{4}-\mu_{6})-h_{4}\mu_{4}<0,$ $\displaystyle
h_{6}(\mu_{5}-\mu_{6})-h_{5}\mu_{5}<0.$

Case 6: (1, 5, 6)

  1. 1.

$\overline{Z}_{1}(t)>0,\overline{Z}_{5}(t)>0,\overline{Z}_{6}(t)>0.$

It is clear that $O_{1}(t)=\mu_{1}$, $O_{5}(t)=\mu_{5}$, $O_{6}(t)=\mu_{6}$,
$O_{4}(t)=0$, then $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$ and
$\dot{\overline{Z}}_{6}(t)=\mu_{5}-\mu_{6}$. Thus, we need
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{5}\dot{\overline{Z}}_{5}(t)+h_{6}\dot{\overline{Z}}_{6}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{6}(\mu_{5}-\mu_{6})-h_{5}\mu_{5}<0$.

  2. 2.

$\overline{Z}_{1}(t)>0,\overline{Z}_{5}(t)>0,\overline{Z}_{6}(t)=0.$

It is clear that $O_{1}(t)=\mu_{1}$, $O_{5}(t)=\mu_{5}$, $O_{4}(t)=0$, then
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$. Thus, we need
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{5}\dot{\overline{Z}}_{5}(t)=h_{1}(\alpha_{1}-\mu_{1})-h_{5}\mu_{5}<0$,
which is obviously satisfied since $\alpha_{1}<1<\mu_{1}$.

  3. 3.

$\overline{Z}_{1}(t)>0,\overline{Z}_{6}(t)>0,\overline{Z}_{5}(t)=0.$

It is clear that $O_{1}(t)=\mu_{1}$, $O_{6}(t)=\mu_{6}$, $O_{4}(t)=0$, then
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$. Since $O_{4}(t)=0$ and queue 5
is empty, we have
$O_{5}(t)=0\Rightarrow\dot{\overline{Z}}_{6}(t)=0-\mu_{6}=-\mu_{6}$. Thus, we
need
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{6}\dot{\overline{Z}}_{6}(t)=h_{1}(\alpha_{1}-\mu_{1})-h_{6}\mu_{6}<0$,
which is obviously satisfied since $\alpha_{1}<1<\mu_{1}$.

  4. 4.

$\overline{Z}_{5}(t)>0,\overline{Z}_{6}(t)>0,\overline{Z}_{1}(t)=0.$

Since class 6 has priority in station 3 and queue 6 is not empty, we have
$O_{6}(t)=\mu_{6}$ and $O_{3}(t)=0$. Similarly, $O_{5}(t)=\mu_{5}$. Then,
$\dot{\overline{Z}}_{6}(t)=\mu_{5}-\mu_{6}$. To make $t$ differentiable, we
have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}\dot{\overline{T}}_{1}(t)=0\Rightarrow\dot{\overline{T}}_{1}(t)=\frac{\alpha_{1}}{\mu_{1}}\Rightarrow\dot{\overline{T}}_{4}(t)_{max}=1-\frac{\alpha_{1}}{\mu_{1}}\Rightarrow
O_{4}(t)_{max}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})\Rightarrow\dot{\overline{Z}}_{5}(t)_{max}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}$.
Thus, we need
$\dot{Y}(t)_{max}=h_{5}\dot{\overline{Z}}_{5}(t)_{max}+h_{6}\dot{\overline{Z}}_{6}(t)=h_{5}\left[\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}\right]+h_{6}(\mu_{5}-\mu_{6})<0$.

  5. 5.

$\overline{Z}_{1}(t)>0,\overline{Z}_{5}(t)=\overline{Z}_{6}(t)=0.$

We need $\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{1}(t)<0$. It is clear that
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}<0$.

  6. 6.

$\overline{Z}_{5}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{6}(t)=0.$

We need $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{5}(t)<0$. It is clear that $O_{5}(t)=\mu_{5}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}\dot{\overline{T}}_{1}(t)=0\Rightarrow\dot{\overline{T}}_{1}(t)=\frac{\alpha_{1}}{\mu_{1}}\Rightarrow\dot{\overline{T}}_{4}(t)_{max}=1-\frac{\alpha_{1}}{\mu_{1}}\Rightarrow
O_{4}(t)_{max}=\mu_{4}\dot{\overline{T}}_{4}(t)_{max}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})\Rightarrow\dot{\overline{Z}}_{5}(t)_{max}=O_{4}(t)_{max}-\mu_{5}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}$,
which is needed to be negative.
$\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}<0\Leftrightarrow
1-m_{1}\alpha_{1}<\frac{m_{4}}{m_{5}}\Leftrightarrow\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$.
Thus, $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$ is satisfied when
$\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$.

Note that $\frac{m_{5}-m_{4}}{m_{1}m_{5}}<1$, thus, the condition
$\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$ does not conflict with the nominal
condition($\alpha_{1}<1$). It is clear that $\frac{m_{5}-m_{4}}{m_{1}m_{5}}<1$
since $\frac{m_{5}-m_{4}}{m_{1}m_{5}}<1\Leftrightarrow
m_{5}-m_{4}<m_{1}m_{5}\Leftrightarrow m_{4}>m_{5}(1-m_{1})\Leftrightarrow
m_{4}>m_{5}m_{4}\Leftrightarrow m_{4}(1-m_{5})>0\Leftrightarrow m_{4}m_{2}>0$,
which is true obviously.

  7. 7.

$\overline{Z}_{6}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{5}(t)=0.$

We need $\dot{Y}(t)=h_{6}\dot{\overline{Z}}_{6}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{6}(t)<0$. It is clear that $O_{6}(t)=\mu_{6}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{5}(t)=O_{4}(t)-O_{5}(t)=0\Rightarrow O_{5}(t)=O_{4}(t)$
and
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}\dot{\overline{T}}_{1}(t)=0\Rightarrow\dot{\overline{T}}_{1}(t)=\frac{\alpha_{1}}{\mu_{1}}\Rightarrow\dot{\overline{T}}_{4}(t)_{max}=1-\frac{\alpha_{1}}{\mu_{1}}\Rightarrow
O_{4}(t)_{max}=\mu_{4}\dot{\overline{T}}_{4}(t)_{max}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})\Rightarrow
O_{5}(t)_{max}=O_{4}(t)_{max}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})\Rightarrow\dot{\overline{Z}}_{6}(t)_{max}=O_{5}(t)_{max}-\mu_{6}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{6}$,
which is needed to be negative.
$\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{6}<0\Leftrightarrow
1-m_{1}\alpha_{1}<\frac{m_{4}}{m_{6}}\Leftrightarrow\alpha_{1}>\frac{m_{6}-m_{4}}{m_{1}m_{6}}$.
Thus, $\dot{Y}(t)=h_{6}\dot{\overline{Z}}_{6}(t)<0$ is satisfied when
$\alpha_{1}>\frac{m_{6}-m_{4}}{m_{1}m_{6}}$.

Note that $\frac{m_{6}-m_{4}}{m_{1}m_{6}}<1$, thus, the condition
$\alpha_{1}>\frac{m_{6}-m_{4}}{m_{1}m_{6}}$ does not conflict with the nominal
condition($\alpha_{1}<1$). It is clear that $\frac{m_{6}-m_{4}}{m_{1}m_{6}}<1$
since $\frac{m_{6}-m_{4}}{m_{1}m_{6}}<1\Leftrightarrow
m_{6}-m_{4}<m_{1}m_{6}\Leftrightarrow m_{4}>m_{6}(1-m_{1})\Leftrightarrow
m_{4}>m_{6}m_{4}\Leftrightarrow m_{4}(1-m_{6})>0\Leftrightarrow m_{4}m_{3}>0$,
which is true obviously.

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 6:

$\displaystyle
h_{1}(\alpha_{1}-\mu_{1})+h_{6}(\mu_{5}-\mu_{6})-h_{5}\mu_{5}<0,$
$\displaystyle
h_{5}\left[\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}\right]+h_{6}(\mu_{5}-\mu_{6})<0,$
$\displaystyle\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}},$
$\displaystyle\alpha_{1}>\frac{m_{6}-m_{4}}{m_{1}m_{6}}.$

Case 7: (1, 2, 6)

  1. 1.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)>0,\overline{Z}_{6}(t)>0.$

It is clear that $O_{1}(t)=\mu_{1}$, $O_{2}(t)=\mu_{2}$, $O_{6}(t)=\mu_{6}$,
$O_{5}(t)=0$, then $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{2}$ and
$\dot{\overline{Z}}_{6}(t)=0-\mu_{6}=-\mu_{6}$. Thus, we need
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{2}\dot{\overline{Z}}_{2}(t)+h_{6}\dot{\overline{Z}}_{6}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})-h_{6}\mu_{6}<0$.

  2. 2.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)>0,\overline{Z}_{6}(t)=0.$

It is clear that $O_{1}(t)=\mu_{1}$, $O_{2}(t)=\mu_{2}$, then
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{2}$. Thus, we need
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{2}\dot{\overline{Z}}_{2}(t)=h_{1}(\alpha_{1}-\mu_{1})-h_{2}(\mu_{1}-\mu_{2})<0$.

  3. 3.

$\overline{Z}_{1}(t)>0,\overline{Z}_{6}(t)>0,\overline{Z}_{2}(t)=0.$

It is clear that $O_{1}(t)=\mu_{1}$, $O_{6}(t)=\mu_{6}$, $O_{4}(t)=0$, then
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$. To make $t$ differentiable, we
have
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-O_{2}(t)=\mu_{1}-\mu_{2}\dot{\overline{T}}_{2}(t)=0\Rightarrow\dot{\overline{T}}_{2}(t)=\frac{\mu_{1}}{\mu_{2}}\Rightarrow\dot{\overline{T}}_{5}(t)_{max}=1-\frac{\mu_{1}}{\mu_{2}}\Rightarrow
O_{5}(t)_{max}=\mu_{5}\dot{\overline{T}}_{5}(t)_{max}=\mu_{5}(1-\frac{\mu_{1}}{\mu_{2}})\Rightarrow\dot{\overline{Z}}_{6}(t)_{max}=O_{5}(t)_{max}-\mu_{6}=\mu_{5}(1-\frac{\mu_{1}}{\mu_{2}})-\mu_{6}$.
Thus, we need
$\dot{Y}(t)_{max}=h_{1}\dot{\overline{Z}}_{1}(t)+h_{6}\dot{\overline{Z}}_{6}(t)_{max}=h_{1}(\alpha_{1}-\mu_{1})+h_{6}\left[\mu_{5}(1-\frac{\mu_{1}}{\mu_{2}})-\mu_{6}\right]<0$,
which is true since $\alpha_{1}-\mu_{1}$ is obviously negative, and we can
prove that $\mu_{5}(1-\frac{\mu_{1}}{\mu_{2}})-\mu_{6}<0$ as the following. If
$1-\frac{\mu_{1}}{\mu_{2}}\leq 0$, then
$\mu_{5}(1-\frac{\mu_{1}}{\mu_{2}})-\mu_{6}<0$ for sure; if
$1-\frac{\mu_{1}}{\mu_{2}}>0$, then
$\mu_{5}(1-\frac{\mu_{1}}{\mu_{2}})-\mu_{6}<0\Leftrightarrow\frac{\mu_{1}}{\mu_{2}}+\frac{\mu_{6}}{\mu_{5}}>1\Leftrightarrow\frac{m_{2}}{m_{1}}+\frac{m_{5}}{m_{6}}>1$,
which is satisfied since
$\frac{m_{2}}{m_{1}}+\frac{m_{5}}{m_{6}}>m_{5}+m_{2}=1$.

  4. 4.

$\overline{Z}_{2}(t)>0,\overline{Z}_{6}(t)>0,\overline{Z}_{1}(t)=0.$

It is clear that $O_{2}(t)=\mu_{2}$, $O_{6}(t)=\mu_{6}$, $O_{5}(t)=0$, then
$\dot{\overline{Z}}_{6}(t)=0-\mu_{6}=-\mu_{6}$. To make $t$ differentiable, we
have $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=0\Rightarrow
O_{1}(t)=\alpha_{1}\Rightarrow\dot{\overline{Z}}_{2}(t)=\alpha_{1}-\mu_{2}$.
It obviously holds that
$\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)+h_{6}\dot{\overline{Z}}_{6}(t)=h_{2}(\alpha_{1}-\mu_{2})-h_{6}\mu_{6}<0$
as required.

  5. 5.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)=\overline{Z}_{6}(t)=0.$

We need $\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{1}(t)<0$. It is clear that
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}<0$.

  6. 6.

$\overline{Z}_{2}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{6}(t)=0.$

We need $\dot{Y}(t)=h_{2}\dot{\overline{Z}}_{2}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{2}(t)<0$. It is clear that $O_{2}(t)=\mu_{2}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=0\Rightarrow
O_{1}(t)=\alpha_{1}$. Then, $\dot{\overline{Z}}_{2}(t)=\alpha_{1}-\mu_{2}<0$
for sure.

  7. 7.

$\overline{Z}_{6}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{2}(t)=0.$

We need $\dot{Y}(t)=h_{6}\dot{\overline{Z}}_{6}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{6}(t)<0$. It is clear that $O_{6}(t)=\mu_{6}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=0\Rightarrow
O_{1}(t)=\alpha_{1}$ and
$\dot{\overline{Z}}_{2}(t)=\alpha_{1}-O_{2}(t)=0\Rightarrow
O_{2}(t)=\mu_{2}\dot{\overline{T}}_{2}(t)=\alpha_{1}\Rightarrow\dot{\overline{T}}_{2}(t)=\frac{\alpha_{1}}{\mu_{2}}\Rightarrow\dot{\overline{T}}_{5}(t)_{max}=1-\frac{\alpha_{1}}{\mu_{2}}\Rightarrow
O_{5}(t)_{max}=\mu_{5}\dot{\overline{T}}_{5}(t)_{max}=\mu_{5}(1-\frac{\alpha_{1}}{\mu_{2}})\Rightarrow\dot{\overline{Z}}_{6}(t)_{max}=O_{5}(t)_{max}-\mu_{6}=\mu_{5}(1-\frac{\alpha_{1}}{\mu_{2}})-\mu_{6}$,
which is needed to be negative.
$\mu_{5}(1-\frac{\alpha_{1}}{\mu_{2}})-\mu_{6}<0\Leftrightarrow
1-m_{2}\alpha_{1}<\frac{m_{5}}{m_{6}}\Leftrightarrow\alpha_{1}>\frac{m_{6}-m_{5}}{m_{2}m_{6}}$.
Thus, $\dot{Y}(t)=h_{6}\dot{\overline{Z}}_{6}(t)<0$ is satisfied when
$\alpha_{1}>\frac{m_{6}-m_{5}}{m_{2}m_{6}}$.

We state in the analysis of case 1
($\overline{Z}_{6}(t)>0,\overline{Z}_{2}(t)=\overline{Z}_{4}(t)=0$) that
$\frac{m_{6}-m_{5}}{m_{2}m_{6}}<1$, thus, the condition
$\alpha_{1}>\frac{m_{6}-m_{5}}{m_{2}m_{6}}$ does not conflict with the nominal
condition($\alpha_{1}<1$).

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 7:

$\displaystyle
h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})-h_{6}\mu_{6}<0,$
$\displaystyle h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})<0,$
$\displaystyle\alpha_{1}>\frac{m_{6}-m_{5}}{m_{2}m_{6}}.$

Case 8: (1, 5, 3)

  1. 1.

$\overline{Z}_{1}(t)>0,\overline{Z}_{3}(t)>0,\overline{Z}_{5}(t)>0.$

It is clear that $O_{1}(t)=\mu_{1}$, $O_{3}(t)=\mu_{3}$, $O_{5}(t)=\mu_{5}$,
$O_{4}(t)=0$, $O_{2}(t)=0$, then
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{3}(t)=0-\mu_{3}=-\mu_{3}$ and
$\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$. It obviously holds that
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{3}\dot{\overline{Z}}_{3}(t)+h_{5}\dot{\overline{Z}}_{5}(t)=h_{1}(\alpha_{1}-\mu_{1})-h_{3}\mu_{3}-h_{5}\mu_{5}<0$
as required.

  2. 2.

$\overline{Z}_{1}(t)>0,\overline{Z}_{3}(t)>0,\overline{Z}_{5}(t)=0.$

It is clear that $O_{1}(t)=\mu_{1}$, $O_{3}(t)=\mu_{3}$, $O_{4}(t)=0$, then
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$. Since queue 5 is empty and
$O_{4}(t)=0$, we have $O_{5}(t)=0\Rightarrow
O_{2}(t)_{max}=\mu_{2}\Rightarrow\dot{\overline{Z}}_{3}(t)_{max}=O_{2}(t)_{max}-\mu_{3}=\mu_{2}-\mu_{3}$.
Thus, we need
$\dot{Y}(t)_{max}=h_{1}\dot{\overline{Z}}_{1}(t)+h_{3}\dot{\overline{Z}}_{3}(t)_{max}=h_{1}(\alpha_{1}-\mu_{1})+h_{3}(\mu_{2}-\mu_{3})<0$.

  3. 3.

$\overline{Z}_{1}(t)>0,\overline{Z}_{5}(t)>0,\overline{Z}_{3}(t)=0.$

It is clear that $O_{1}(t)=\mu_{1}$, $O_{5}(t)=\mu_{5}$, $O_{4}(t)=0$, then
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$. It obviously holds that
$\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)+h_{5}\dot{\overline{Z}}_{5}(t)=h_{1}(\alpha_{1}-\mu_{1})-h_{5}\mu_{5}<0$
as required.

  4. 4.

$\overline{Z}_{3}(t)>0,\overline{Z}_{5}(t)>0,\overline{Z}_{1}(t)=0.$

It is clear that $O_{3}(t)=\mu_{3}$, $O_{5}(t)=\mu_{5}$, $O_{2}(t)=0$, then
$\dot{\overline{Z}}_{3}(t)=0-\mu_{3}=-\mu_{3}$. To make $t$ differentiable, we
have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=\alpha_{1}-\mu_{1}\dot{\overline{T}}_{1}(t)=0\Rightarrow\dot{\overline{T}}_{1}(t)=\frac{\alpha_{1}}{\mu_{1}}\Rightarrow\dot{\overline{T}}_{4}(t)_{max}=1-\frac{\alpha_{1}}{\mu_{1}}\Rightarrow
O_{4}(t)_{max}=\mu_{4}\dot{\overline{T}}_{4}(t)_{max}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})\Rightarrow\dot{\overline{Z}}_{5}(t)_{max}=O_{4}(t)_{max}-\mu_{5}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}$.
Thus, we need
$\dot{Y}(t)_{max}=h_{3}\dot{\overline{Z}}_{3}(t)+h_{5}\dot{\overline{Z}}_{5}(t)_{max}=h_{5}\left[\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}\right]-h_{3}\mu_{3}<0$.

In the analysis of case 6
($\overline{Z}_{5}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{6}(t)=0$), we need
the condition $\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$, with which
$\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}<0$. Thus, without adding more
conditions, we have
$\dot{Y}(t)_{max}=h_{5}\left[\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}\right]-h_{3}\mu_{3}<0$.

  5. 5.

$\overline{Z}_{1}(t)>0,\overline{Z}_{3}(t)=\overline{Z}_{5}(t)=0.$

We need $\dot{Y}(t)=h_{1}\dot{\overline{Z}}_{1}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{1}(t)<0$. It is clear that
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}<0$.

  6. 6.

$\overline{Z}_{3}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{5}(t)=0.$

We need $\dot{Y}(t)=h_{3}\dot{\overline{Z}}_{3}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{3}(t)<0$. It is clear that $O_{3}(t)=\mu_{3}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=0\Rightarrow
O_{1}(t)=\mu_{1}\dot{\overline{T}}_{1}(t)=\alpha_{1}\Rightarrow\dot{\overline{T}}_{1}(t)=\frac{\alpha_{1}}{\mu_{1}}$
and $\dot{\overline{Z}}_{5}(t)=O_{4}(t)-O_{5}(t)=0\Rightarrow
O_{5}(t)=O_{4}(t)\Rightarrow\dot{\overline{T}}_{5}(t)=\frac{O_{5}(t)}{\mu_{5}}=\frac{O_{4}(t)}{\mu_{5}}\Rightarrow\dot{\overline{T}}_{5}(t)_{min}=\frac{O_{4}(t)_{min}}{\mu_{5}}\Rightarrow\dot{\overline{T}}_{2}(t)_{max}=1-\dot{\overline{T}}_{5}(t)_{min}=1-\frac{O_{4}(t)_{min}}{\mu_{5}}$.
Then
$O_{2}(t)_{max}=\mu_{2}\dot{\overline{T}}_{2}(t)_{max}=\mu_{2}(1-\frac{O_{4}(t)_{min}}{\mu_{5}})$.
If queue 4 is not empty,
$O_{4}(t)=\mu_{4}(1-\dot{\overline{T}}_{1}(t))=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})$;
if queue 4 is empty,
$O_{4}(t)=min\\{\mu_{3},\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})\\}$. Thus,
$O_{4}(t)_{min}=\mu_{3}$ or $\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})$, and
$O_{2}(t)_{max}=\mu_{2}(1-\frac{\mu_{3}}{\mu_{5}})$ or
$\mu_{2}(1-\frac{\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})}{\mu_{5}})$. Then,
$\dot{\overline{Z}}_{3}(t)<0$ is equivalent to
$\mu_{2}(1-\frac{\mu_{3}}{\mu_{5}})-\mu_{3}<0$ and
$\mu_{2}\left[1-\frac{\mu_{4}}{\mu_{5}}(1-\frac{\alpha_{1}}{\mu_{1}})\right]-\mu_{3}<0$.

$\mu_{2}(1-\frac{\mu_{3}}{\mu_{5}})-\mu_{3}<0$ is satisfied since
$\mu_{2}(1-\frac{\mu_{3}}{\mu_{5}})-\mu_{3}<0\Leftrightarrow\frac{\mu_{3}}{\mu_{2}}+\frac{\mu_{3}}{\mu_{5}}>1\Leftrightarrow\frac{m_{2}}{m_{3}}+\frac{m_{5}}{m_{3}}>1\Leftrightarrow\frac{1}{m_{3}}>1$,
which is true obviously.

Let
$f(\alpha_{1})=\mu_{2}\left[1-\frac{\mu_{4}}{\mu_{5}}(1-\frac{\alpha_{1}}{\mu_{1}})\right]-\mu_{3}$.
Then $\dot{f}(\alpha_{1})=\frac{\mu_{2}\mu_{4}}{\mu_{5}\mu_{1}}>0$. We know
that
$f(1)=\mu_{2}\left[1-\frac{\mu_{4}}{\mu_{5}}(1-\frac{1}{\mu_{1}})\right]-\mu_{3}=\mu_{2}\left[1-\frac{\mu_{4}}{\mu_{5}}(1-m_{1})\right]-\mu_{3}=\mu_{2}(1-\frac{m_{5}}{m_{4}}\times
m_{4})-\mu_{3}=\mu_{2}(1-m_{5})-\mu_{3}=1-\mu_{3}<0$. So, for any
$\alpha_{1}<1$, we have $f(\alpha_{1})<f(1)<0$, which means
$\mu_{2}\left[1-\frac{\mu_{4}}{\mu_{5}}(1-\frac{\alpha_{1}}{\mu_{1}})\right]-\mu_{3}<0$.

  7. 7.

$\overline{Z}_{5}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{3}(t)=0.$

We need $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$, which is equivalent to
$\dot{\overline{Z}}_{5}(t)<0$. It is clear that $O_{5}(t)=\mu_{5}$. To make
$t$ differentiable, we have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=0\Rightarrow
O_{1}(t)=\mu_{1}\dot{\overline{T}}_{1}(t)=\alpha_{1}\Rightarrow\dot{\overline{T}}_{1}(t)=\frac{\alpha_{1}}{\mu_{1}}\Rightarrow\dot{\overline{T}}_{4}(t)_{max}=1-\frac{\alpha_{1}}{\mu_{1}}\Rightarrow
O_{4}(t)_{max}=\mu_{4}\dot{\overline{T}}_{4}(t)_{max}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})\Rightarrow\dot{\overline{Z}}_{5}(t)_{max}=O_{4}(t)_{max}-\mu_{5}=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}$,
which is needed to be negative.
$\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}<0\Leftrightarrow
1-m_{1}\alpha_{1}<\frac{m_{4}}{m_{5}}\Leftrightarrow\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$.
Thus, $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$ is satisfied when
$\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$.

We state in the analysis of case 6
($\overline{Z}_{5}(t)>0,\overline{Z}_{1}(t)=\overline{Z}_{6}(t)=0$) that
$\frac{m_{5}-m_{4}}{m_{1}m_{5}}<1$, thus, the condition
$\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$ does not conflict with the nominal
condition($\alpha_{1}<1$).

To sum up, except for the positive condition for common vector $h$, we need
the following condition from case 8:

$\displaystyle h_{1}(\alpha_{1}-\mu_{1})+h_{3}(\mu_{2}-\mu_{3})<0,$
$\displaystyle\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}.$

###  B.3 Sufficient Conditions for SSC of Balanced DHV Networks

The question of whether there exists such a vector $h$ satisfying the 23
inequalities that we mentioned in §4.1 can be turned into a LP problem
$(P_{h})$. If $(P_{h})$ has a strictly positive solution, then a vector $h$
that satisfies all 23 inequalities exists. The inequalities below are
reordered by their algebraic form.

$\displaystyle(P_{h})=\;$ max $\displaystyle p$ s.t. $\displaystyle
h_{1},h_{2},h_{3},h_{4},h_{5},h_{6}\geq p$ $\displaystyle
h_{2}(\mu_{1}-\mu_{2})-h_{6}\mu_{6}\leq-p$ $\displaystyle
h_{4}(\mu_{3}-\mu_{4})-h_{2}\mu_{2}\leq-p$ $\displaystyle
h_{6}(\mu_{5}-\mu_{6})-h_{4}\mu_{4}\leq-p$ $\displaystyle
h_{1}(\alpha_{1}-\mu_{1})+h_{3}(\mu_{2}-\mu_{3})\leq-p$ $\displaystyle
h_{3}(\mu_{1}-\mu_{3})+h_{1}(\alpha_{1}-\mu_{1})\leq-p$ $\displaystyle
h_{4}(\mu_{2}-\mu_{4})-h_{2}\mu_{2}\leq-p$ $\displaystyle
h_{5}(\mu_{3}-\mu_{5})-h_{3}\mu_{3}\leq-p$ $\displaystyle
h_{6}(\mu_{4}-\mu_{6})-h_{4}\mu_{4}\leq-p$ $\displaystyle
h_{3}(\mu_{2}-\mu_{3})+h_{2}(\alpha_{1}-\mu_{2})\leq-p$ $\displaystyle
h_{4}(\mu_{3}-\mu_{4})-h_{3}\mu_{3}\leq-p$ $\displaystyle
h_{5}(\mu_{4}-\mu_{5})-h_{4}\mu_{4}\leq-p$ $\displaystyle
h_{6}(\mu_{5}-\mu_{6})-h_{5}\mu_{5}\leq-p$ $\displaystyle
h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})\leq-p$ $\displaystyle
h_{2}\left[\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{2}\right]+h_{3}(\mu_{2}-\mu_{3})\leq-p$
$\displaystyle
h_{3}\left[\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{3}\right]+h_{4}(\mu_{3}-\mu_{4})\leq-p$
$\displaystyle
h_{4}\left[\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{4}\right]+h_{5}(\mu_{4}-\mu_{5})\leq-p$
$\displaystyle
h_{5}\left[\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}\right]+h_{6}(\mu_{5}-\mu_{6})\leq-p$
$\displaystyle
h_{2}(\mu_{1}-\mu_{2})+h_{3}(\mu_{2}-\mu_{3})+h_{1}(\alpha_{1}-\mu_{1})\leq-p$
$\displaystyle
h_{3}(\mu_{2}-\mu_{3})+h_{4}(\mu_{3}-\mu_{4})-h_{2}\mu_{2}\leq-p$
$\displaystyle
h_{4}(\mu_{3}-\mu_{4})+h_{5}(\mu_{4}-\mu_{5})-h_{3}\mu_{3}\leq-p$
$\displaystyle
h_{5}(\mu_{4}-\mu_{5})+h_{6}(\mu_{5}-\mu_{6})-h_{4}\mu_{4}\leq-p$
$\displaystyle
h_{6}(\mu_{5}-\mu_{6})+h_{1}(\alpha_{1}-\mu_{1})-h_{5}\mu_{5}\leq-p$
$\displaystyle p\leq 1$

The constraint $p\leq 1$ guarantees that the objective value is bounded from
above, without which, if ($P_{h}$) has a positive solution, then by scaling
both $h$ and $p$, this solution is infinite. We care only about whether a
positive solution to ($P_{h}$) exists.

One obvious feasible solution to $(P_{h})$ is
$h_{1}=h_{2}=h_{3}=h_{4}=h_{5}=h_{6}=p=0$, with a corresponding objective
value of 0. Since $(P_{h})$ is a maximization problem, the optimal value of
$(P_{h})$ is non-negative. If the optimal value is strictly positive, then we
can scale $h$ and $p$ to make it larger. Thus, the optimal value of $(P_{h})$
is always $0$ or $1$. According to strong duality, the dual optimal of
$(P_{h})$ is also $0$ or $1$; call the dual problem as $(D_{h})$.

$\displaystyle(D_{h})=\max\quad$ $\displaystyle x_{29}$ s.t. $\displaystyle-
x_{1}+(\alpha_{1}-\mu_{1})(x_{10}+x_{11}+x_{19}+x_{24}+x_{28})=0$ (B.2)
$\displaystyle-
x_{2}+(\mu_{1}-\mu_{2})(x_{7}+x_{19}+x_{24})-\mu_{2}(x_{8}+x_{12}+x_{25})+(\alpha_{1}-\mu_{2})x_{15}+\left[\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{2}\right]x_{20}=0$
(B.3) $\displaystyle-
x_{3}+(\mu_{1}-\mu_{3})x_{11}+(\mu_{2}-\mu_{3})(x_{10}+x_{15}+x_{20}+x_{24}+x_{25})-\mu_{3}(x_{13}+x_{16}+x_{26})$
(B.4)
$\displaystyle+\left[\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{3}\right]x_{21}=0$
(B.5) $\displaystyle-
x_{4}+(\mu_{3}-\mu_{4})(x_{8}+x_{16}+x_{21}+x_{25}+x_{26})-\mu_{4}(x_{9}+x_{14}+x_{17}+x_{27})+(\mu_{2}-\mu_{4})x_{12}$
(B.6)
$\displaystyle+\left[\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{4}\right]x_{22}=0$
(B.7) $\displaystyle-
x_{5}+(\mu_{3}-\mu_{5})x_{13}+(\mu_{4}-\mu_{5})(x_{17}+x_{22}+x_{26}+x_{27})-\mu_{5}(x_{18}+x_{28})+\left[\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}\right]x_{23}=0$
(B.8) $\displaystyle-
x_{6}+(\mu_{4}-\mu_{6})x_{14}+(\mu_{5}-\mu_{6})(x_{9}+x_{18}+x_{23}+x_{27}+x_{28})-\mu_{6}x_{7}=0$
(B.9) $\displaystyle\sum_{i=1}^{29}x_{i}=1$ (B.10) $\displaystyle
x_{1},x_{2},...,x_{29}\geq 0$ (B.11)

If the optimal value of of $(D_{h})$ is $0$, it is clear that $x_{29}=0$. If
the optimal value is $1$, with constraints (B.10) and (B.11), $x_{29}=1$,
$x_{1}=x_{2}=...=x_{28}=0$. This means that if $(D_{h})$ has $x_{29}=1$,
$x_{1}=x_{2}=...=x_{28}=0$ as its optimal solution, then the vector $h$ that
satisfies all the 23 inequalities exists.

Let us now consider the signs of coefficients of the variables in $(D_{h})$.
We have

$\displaystyle\frac{m_{1}}{m_{2}}+\frac{m_{4}}{m_{3}}>m_{1}+m_{4}=1$
$\displaystyle\Rightarrow\;$
$\displaystyle\frac{\mu_{2}}{\mu_{1}}+\frac{\mu_{3}}{\mu_{4}}>1$
$\displaystyle\Rightarrow\;$ $\displaystyle
1-\frac{\mu_{3}}{\mu_{4}}<\frac{\mu_{2}}{\mu_{1}}$
$\displaystyle\Rightarrow\;$
$\displaystyle\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{2}<0.$

Similarly, we have

$\displaystyle\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{3}<0,$
$\displaystyle\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{4}<0,$
$\displaystyle\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}<0.$

Form constraints (B.2) and (B.11), since $\alpha_{1}-\mu_{1}<0$,
$x_{1},x_{10},x_{11},x_{19},x_{24}$ and $x_{28}$ should always be $0$.
$(D_{h})$ can be simplified as:

$\displaystyle(D_{h1})=\max\quad$ $\displaystyle x_{29}$ s.t. $\displaystyle-
x_{2}+(\mu_{1}-\mu_{2})x_{7}-\mu_{2}(x_{8}+x_{12}+x_{25})+(\alpha_{1}-\mu_{2})x_{15}+\left[\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{2}\right])x_{20}=0$
(B.12) $\displaystyle-
x_{3}+(\mu_{2}-\mu_{3})(x_{15}+x_{20}+x_{25})-\mu_{3}(x_{13}+x_{16}+x_{26})+\left[\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{3}\right]x_{21}=0$
(B.13) $\displaystyle-
x_{4}+(\mu_{3}-\mu_{4})(x_{8}+x_{16}+x_{21}+x_{25}+x_{26})-\mu_{4}(x_{9}+x_{14}+x_{17}+x_{27})+(\mu_{2}-\mu_{4})x_{12}$
(B.14)
$\displaystyle+\left[\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{4}\right]x_{22}=0$
(B.15) $\displaystyle-
x_{5}+(\mu_{3}-\mu_{5})x_{13}+(\mu_{4}-\mu_{5})(x_{17}+x_{22}+x_{26}+x_{27})-\mu_{5}x_{18}+\left[\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}\right]x_{23}=0$
(B.16) $\displaystyle-
x_{6}+(\mu_{4}-\mu_{6})x_{14}+(\mu_{5}-\mu_{6})(x_{9}+x_{18}+x_{23}+x_{27})-\mu_{6}x_{7}=0$
(B.17) $\displaystyle\sum_{i=1}^{29}x_{i}=1$ (B.18) $\displaystyle
x_{1},x_{2},...,x_{29}\geq 0$ (B.19)

We divide our analysis into 4 cases. The universe

$\displaystyle U$
$\displaystyle=\\{\mu_{1}\leq\mu_{2}\\}\cup\\{(\mu_{4}\leq\mu_{6})\cap(\mu_{5}\leq\mu_{6})\\}\cup\\{(\mu_{1}>\mu_{2})\cap((\mu_{4}>\mu_{6})\cup(\mu_{5}>\mu_{6}))\\}$
$\displaystyle=\\{\mu_{1}\leq\mu_{2}\\}\cup\\{(\mu_{4}\leq\mu_{6})\cap(\mu_{5}\leq\mu_{6})\\}\cup\\{(\mu_{1}>\mu_{2})\cap((\mu_{1}<\mu_{3})\cup(\mu_{2}<\mu_{3}))\\}$
$\displaystyle=\\{\mu_{1}\leq\mu_{2}\\}\cup\\{(\mu_{4}\leq\mu_{6})\cap(\mu_{5}\leq\mu_{6})\\}\cup\\{(\mu_{1}>\mu_{2})\cap(\mu_{3}>\mu_{2})\\}$
$\displaystyle=\\{\mu_{1}\leq\mu_{2}\\}\cup\\{(\mu_{4}\leq\mu_{6})\cap(\mu_{5}\leq\mu_{6})\\}\cup\\{(\mu_{1}>\mu_{2})\cap(\mu_{3}>\mu_{2})\cap(\mu_{3}\leq\mu_{4})\cap(\mu_{2}\leq\mu_{4})\\}$
$\displaystyle\;\;\;\;\;\cup\\{(\mu_{1}>\mu_{2})\cap(\mu_{3}>\mu_{2})\cap((\mu_{2}>\mu_{4})\cup(\mu_{3}>\mu
4))\\}$
$\displaystyle=\\{\mu_{1}\leq\mu_{2}\\}\cup\\{(\mu_{4}\leq\mu_{6})\cap(\mu_{5}\leq\mu_{6})\\}\cup\\{(\mu_{1}>\mu_{2})\cap(\mu_{4}\geq\mu_{3}>\mu_{2})\\}\cup\\{(\mu_{1}>\mu_{2})\cap(\mu_{3}>\mu_{2})\cap(\mu_{3}>\mu_{4})\\}.$

Case 1: $\mu_{1}-\mu_{2}\leq 0$.

From constraint (B.12), since $\mu_{1}-\mu_{2}\leq 0$, $\alpha_{1}-\mu_{2}<0$
and $\mu_{1}(1-\frac{\mu_{3}}{\mu_{4}})-\mu_{2}<0$, we have
$x_{2}=x_{7}=x_{8}=x_{12}=x_{25}=x_{15}=x_{20}=0$. $(D_{h1})$ can be further
simplified as:

$\displaystyle(D_{h2})=\max\quad$ $\displaystyle x_{29}$ s.t. $\displaystyle-
x_{3}-\mu_{3}(x_{13}+x_{16}+x_{26})+\left[\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-\mu_{3}\right]x_{21}=0$
(B.20) $\displaystyle-
x_{4}+(\mu_{3}-\mu_{4})(x_{16}+x_{21}+x_{26})-\mu_{4}(x_{9}+x_{14}+x_{17}+x_{27})+\left[\mu_{3}(1-\frac{\mu_{5}}{\mu_{6}})-\mu_{4}\right]x_{22}=0$
(B.21) $\displaystyle-
x_{5}+(\mu_{3}-\mu_{5})x_{13}+(\mu_{4}-\mu_{5})(x_{17}+x_{22}+x_{26}+x_{27})-\mu_{5}x_{18}+\left[\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}\right]x_{23}=0$
(B.22) $\displaystyle-
x_{6}+(\mu_{4}-\mu_{6})x_{14}+(\mu_{5}-\mu_{6})(x_{9}+x_{18}+x_{23}+x_{27})=0$
(B.23) $\displaystyle\sum_{i=1}^{29}x_{i}=1$ (B.24) $\displaystyle
x_{1},x_{2},...,x_{29}\geq 0$ (B.25)

Similarly, from constraint (B.20), since
$\mu_{2}(1-\displaystyle\frac{\mu_{4}}{\mu_{5}})-\mu_{3}<0$, we have
$x_{3}=x_{13}=x_{16}=x_{26}=x_{21}=0$. Repeat the above steps, from constraint
(B.21), we have $x_{4}=x_{9}=x_{14}=x_{17}=x_{27}=x_{22}=0$; from constraint
(B.22), we have $x_{5}=x_{18}=x_{23}=0$; from constraint (B.23), we have
$x_{6}=0$. So far, we have claimed that $x_{1}=x_{2}=...=x_{28}=0$, thus,
$x_{29}=1$, which is the optimal solution of $(D_{h})$.

Case 2: $(\mu_{4}\leq\mu_{6})\cap(\mu_{5}\leq\mu_{6})$.

From constraint (B.17) of $(D_{h1})$, since $\mu_{4}-\mu_{6}\leq 0$ and
$\mu_{5}-\mu_{6}\leq 0$, we have $x_{7}=0$. Once we have $x_{7}=0$, we follow
the same argument as in Case 1. Thus, $x_{1}=x_{2}=...=x_{28}=0$, $x_{29}=1$
is the optimal solution of $(D_{h})$ under Case 2.

Before we analyze Case 3 and Case 4 separately, we find that Case 3 and Case 4
both satisfy $\mu_{2}-\mu_{3}<0$ and $\mu_{1}-\mu_{2}>0$. Thus, from
constraint (B.13), since $\mu_{2}-\mu_{3}<0$ and
$\mu_{2}(1-\displaystyle\frac{\mu_{4}}{\mu_{5}})-\mu_{3}<0$, we have
$x_{3}=x_{15}=x_{20}=x_{25}=x_{13}=x_{16}=x_{26}=x_{21}=0$. Also, from
constraint (B.16) in $(D_{h1})$, since $\mu_{1}-\mu_{2}>0$ implies
$\mu_{4}-\mu_{5}<0$, along with
$\mu_{4}(1-\displaystyle\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}<0$, the only
coefficient that could be positive is $(\mu_{3}-\mu_{5})$. However,
$(\mu_{3}-\mu_{5})$ is the coefficient of $x_{13}$ and we already know that
$x_{13}=0$. Thus, we have $x_{5}=x_{17}=x_{22}=x_{26}=x_{27}=x_{18}=x_{23}=0$.
$(D_{h1})$ can be simplified as:

$\displaystyle(D_{h3})=\max\quad$ $\displaystyle x_{29}$ s.t. $\displaystyle-
x_{2}+(\mu_{1}-\mu_{2})x_{7}-\mu_{2}(x_{8}+x_{12})=0$ (B.26) $\displaystyle-
x_{4}+(\mu_{3}-\mu_{4})x_{8}-\mu_{4}(x_{9}+x_{14})+(\mu_{2}-\mu_{4})x_{12}=0$
(B.27) $\displaystyle-
x_{6}+(\mu_{4}-\mu_{6})x_{14}+(\mu_{5}-\mu_{6})x_{9}-\mu_{6}x_{7}=0$ (B.28)
$\displaystyle x_{2}+x_{4}+x_{6}+x_{7}+x_{8}+x_{9}+x_{12}+x_{14}=1$ (B.29)
$\displaystyle x_{2},x_{4}...,x_{29}\geq 0$ (B.30)

Case 3: $(\mu_{1}>\mu_{2})\cap(\mu_{4}\geq\mu_{3}>\mu_{2})$.

From constraint (B.27) in $(D_{h3})$, since $\mu_{3}-\mu_{4}\leq 0$ and
$\mu_{2}-\mu_{4}<0$, we have $x_{4}=x_{9}=x_{14}=0$. Note that we cannot
decide if $x_{8}=x_{12}=0$ here since $\mu_{3}-\mu_{4}$ could be $0$. Then,
from constraint (B.28) in $(D_{h3})$, since $x_{14}=x_{9}=0$, we have
$x_{6}=x_{7}=0$. Consequently, from constraint (B.26) in $(D_{h3})$, since
$x_{7}=0$, we have $x_{2}=x_{8}=x_{12}=0$. Thus, $x_{1}=x_{2}=...=x_{28}=0$,
$x_{29}=1$ is the optimal solution of $(D_{h})$ under Case 3.

Case 4: $(\mu_{1}>\mu_{2})\cap(\mu_{3}>\mu_{2})\cap(\mu_{3}>\mu_{4})$.

In $(D_{h3})$, we see $x_{2},x_{4}$ and $x_{6}$ as slackness variables, and
constraints (B.26) - (B.28) can be rewritten as:

$\begin{split}&\left\\{\begin{array}[]{lr}(\mu_{1}-\mu_{2})x_{7}\geq\mu_{2}(x_{8}+x_{12})\\\
(\mu_{3}-\mu_{4})x_{8}\geq\mu_{4}(x_{9}+x_{14})+(\mu_{4}-\mu_{2})x_{12}\\\
(\mu_{5}-\mu_{6})x_{9}\geq\mu_{6}x_{7}+(\mu_{6}-\mu_{4})x_{14}\end{array}\right.\\\
\Leftrightarrow&\left\\{\begin{array}[]{lr}(\mu_{1}-\mu_{2})x_{7}\geq\mu_{2}(x_{8}+x_{12})\\\
(\mu_{3}-\mu_{4})(x_{8}+x_{12})\geq\mu_{4}(x_{9}+x_{14})+(\mu_{3}-\mu_{2})x_{12}\\\
(\mu_{5}-\mu_{6})(x_{9}+x_{14})\geq\mu_{6}x_{7}+(\mu_{5}-\mu_{4})x_{14}\end{array}\right.\\\
\Rightarrow&\;(\mu_{1}-\mu_{2})x_{7}\geq\mu_{2}(x_{8}+x_{12})\\\
\Rightarrow&\;(\mu_{1}-\mu_{2})x_{7}\geq\frac{\mu_{2}\mu_{4}}{\mu_{3}-\mu_{4}}(x_{9}+x_{14})+\frac{\mu_{2}(\mu_{3}-\mu_{2})}{\mu_{3}-\mu_{4}}x_{12}\\\
\Rightarrow&\;(\mu_{1}-\mu_{2})x_{7}\geq\frac{\mu_{2}\mu_{4}}{\mu_{3}-\mu_{4}}\left[\frac{\mu_{6}}{\mu_{5}-\mu_{6}}x_{7}+\frac{\mu_{5}-\mu_{4}}{\mu_{5}-\mu_{6}}x_{14}\right]+\frac{\mu_{2}(\mu_{3}-\mu_{2})}{\mu_{3}-\mu_{4}}x_{12}\\\
\Leftrightarrow&\;(\mu_{1}-\mu_{2})(\mu_{3}-\mu_{4})(\mu_{5}-\mu_{6})x_{7}\geq\mu_{2}\mu_{4}\mu_{6}x_{7}+\mu_{2}\mu_{4}(\mu_{5}-\mu_{4})x_{14}+\mu_{2}(\mu_{3}-\mu_{2})(\mu_{5}-\mu_{6})x_{12}\\\
\Leftrightarrow&\;\left[(\mu_{1}-\mu_{2})(\mu_{3}-\mu_{4})(\mu_{5}-\mu_{6})-\mu_{2}\mu_{4}\mu_{6}\right]x_{7}\geq\mu_{2}\mu_{4}(\mu_{5}-\mu_{4})x_{14}+\mu_{2}(\mu_{3}-\mu_{2})(\mu_{5}-\mu_{6})x_{12}\end{split}$

Note that $(\mu_{1}>\mu_{2})\cap(\mu_{3}>\mu_{2})\cap(\mu_{3}>\mu_{4})$
implies $(\mu_{5}>\mu_{4})\cap(\mu_{5}>\mu_{6})\cap(\mu_{1}>\mu_{6})$. So we
have $\mu_{5}-\mu_{4}>0$, $\mu_{3}-\mu_{2}>0$ and $\mu_{5}-\mu_{6}>0$. Thus,
the right hand side of the above inequality is larger than or equal to $0$. As
Chen and Ye proved in [6], under $m_{2}+m_{4}+m_{6}<2$, we have
$\left[(\mu_{1}-\mu_{2})(\mu_{3}-\mu_{4})(\mu_{5}-\mu_{6})-\mu_{2}\mu_{4}\mu_{6}\right]<0$
since this is equivalent to
$m_{1}m_{3}m_{5}-(m_{2}-m_{1})(m_{4}-m_{3})(m_{6}-m_{5})>0$ and we can verify
that
$m_{1}m_{3}m_{5}-(m_{2}-m_{1})(m_{4}-m_{3})(m_{6}-m_{5})=(m_{1}+m_{3}-m_{2})\left[(m_{2}-m_{1})m_{6}+m_{1}m_{5}\right]=(2-m_{4}-m_{6}-m_{2})\left[(m_{2}-m_{1})m_{6}+m_{1}m_{5}\right]>0$.

Consequently, under $m_{2}+m_{4}+m_{6}<2$, we have $x_{7}=x_{14}=x_{12}=0$.
From constraint (B.26), since $x_{7}=0$,we have $x_{2}=x_{8}=0$. From
constraint (B.27), since $x_{8}=x_{12}=0$, we have $x_{4}=x_{9}=0$. From
constraint (B.28), since $x_{14}=x_{9}=0$, we have $x_{6}=0$. So far, we have
argued that under $m_{2}+m_{4}+m_{6}<2$, $x_{1}=x_{2}=...=x_{28}=0$, thus
$x_{29}=1$, and this is the optimal solution of $(D_{h})$ under Case 4 when
$m_{2}+m_{4}+m_{6}<2$.

To sum up, if we have $m_{2}+m_{4}+m_{6}<2$, then the optimal value of
$(D_{h})$ is $1$, which means that there exists a vector $h$ that satisfies
all the 23 inequalities we derived for a balanced DHV network. According to
Theorem 7, we can conclude that under $m_{2}+m_{4}+m_{6}<2$,
$\alpha_{1}\in(\max\\{\frac{m_{6}-m_{5}}{m_{2}m_{6}},\frac{m_{5}-m_{4}}{m_{1}m_{5}},\frac{m_{6}-m_{4}}{m_{1}m_{6}},0\\},1)$,
a balanced DHV network satisfies SSC under any queue-ratio policies.

##  Appendix C Detailed Proofs for Balanced Push Started Lu-Kumar Networks

###  C.1 Chen-$\mathcal{S}$ in Balanced push started Lu-Kumar Networks

In this appendix, we prove that the reflection matrices for all the static-
priority cases of a balanced push started Lu-Kumar network
($m_{1}+m_{3}+m_{4}=1$; $m_{2}+m_{5}=1$) with $\alpha_{1}<1$ are
Chen-$\mathcal{S}$, and invertible and have same-sign determinants if
$m_{2}m_{4}>m_{3}m_{5}$. In turn, Theorem 4 allows us to say that the
refection matrix is invertible and Chen-$\mathcal{S}$ for any ratio matrix
$\Delta$ of a balanced push started Lu-Kumar network if
$m_{2}m_{4}>m_{3}m_{5}$.

For each of the 6 static priority policies, to prove that the corresponding
$(R,\theta)$ is Chen-$\mathcal{S}$, we must show that $R$ is
completely-$\mathcal{S}$, and consider all the 3 partitions of $\mathcal{J}$
and show that there exists a positive vector $h\in\mathbb{R}^{2}$ such that
(2.8) satisfies for all the 3 partitions.

###### Proof.

Proof for lowest-priority 1, 2 (Chen-S).

In the balanced push started Lu-Kumar network with class 1 having the lowest
priority in station 1 and class 2 having the lowest priority in station 2, we
have $J=2$, $K=5$, $\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}1&m_{1}-1\mcr-\displaystyle\frac{m_{2}}{m_{1}}&\displaystyle\frac{m_{2}}{m_{1}}\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\mcr
0\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$, since
$R$ is $2\times 2$, $R^{-1}=CMQ\Delta>0$ and all its diagonal elements are
positive.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=0,\;R_{b}^{-1}=R_{2}^{-1}=\frac{m_{1}}{m_{2}}\Rightarrow
u=-R_{b}^{-1}\theta_{b}=0\Rightarrow R_{ab}u=0.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)m_{1}$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=(\alpha_{1}-1)m_{1},\;R_{b}^{-1}=R_{1}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})m_{1};$

$R_{ab}=-\frac{m_{2}}{m_{1}}\Rightarrow R_{ab}u=(\alpha_{1}-1)m_{2}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{2}(\alpha_{1}-1)m_{2}$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=(\alpha_{1}-1)h_{1}m_{1}$

Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{2}}$ as required
so that $(R,\theta)$ under the static-priority policy 1, 2 is
Chen-$\mathcal{S}$. ∎
∎

###### Proof.

Proof for lowest-priority 1, 5 (Chen-S).

In the balanced push started Lu-Kumar network with class 1 having the lowest
priority in station 1 and class 5 having the lowest priority in station 2, we
have $J=2,K=5,\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}m_{1}&0\mcr-1&1\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\mcr
0\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$, since
$R$ is $2\times 2$, $R^{-1}=CMQ\Delta>0$ and all its diagonal elements are
positive.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=0,\;R_{b}^{-1}=R_{2}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=0\Rightarrow R_{ab}u=0.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)m_{1}$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=(\alpha_{1}-1)m_{1},\;R_{b}^{-1}=R_{1}^{-1}=\frac{1}{m_{1}}\Rightarrow
u=-R_{b}^{-1}\theta_{b}=1-\alpha_{1};$

$R_{ab}=-1\Rightarrow R_{ab}u=\alpha_{1}-1.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{2}(\alpha_{1}-1)$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=(\alpha_{1}-1)h_{1}m_{1}$

Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{2}}$ as required
so that $(R,\theta)$ under the static-priority policy 1, 5 is
Chen-$\mathcal{S}$. ∎
∎

###### Proof.

Proof for lowest-priority 3,2 (Chen-S).

In the balanced push started Lu-Kumar network with class 3 having the lowest
priority in station 1 and class 2 having the lowest priority in station 2, we
have $J=2,K=5,\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}\displaystyle\frac{m_{3}}{m_{2}(1-m_{1})}&-\displaystyle\frac{m_{3}}{m_{2}}\mcr-\displaystyle\frac{m_{5}}{1-m_{1}}&1\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=(\alpha_{1}-1)\left[\begin{matrix}\displaystyle\frac{m_{3}m_{1}}{m_{2}(1-m_{1})}\mcr-\displaystyle\frac{m_{2}-m_{1}}{1-m_{1}}\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$, since
$R$ is $2\times 2$, $R^{-1}=CMQ\Delta>0$ and all its diagonal elements are
positive.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=(\alpha_{1}-1)\frac{m_{2}-m_{1}}{1-m_{1}},\;R_{b}^{-1}=R_{2}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})\frac{m_{2}-m_{1}}{1-m_{1}};$

$R_{ab}=-\frac{m_{3}}{m_{2}}\Rightarrow
R_{ab}u=(\alpha_{1}-1)\frac{m_{3}(m_{2}-m_{1})}{m_{2}(1-m_{1})}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)\displaystyle\frac{m_{1}m_{3}+m_{3}(m_{2}-m_{1})}{m_{2}(1-m_{1})}=h_{1}(\alpha_{1}-1)\frac{m_{3}}{1-m_{1}}$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=(\alpha_{1}-1)\frac{m_{3}m_{1}}{m_{2}(1-m_{1})},\;R_{b}^{-1}=R_{1}^{-1}=\frac{m_{2}(1-m_{1})}{m_{3}}\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})m_{1};$

$R_{ab}=-\frac{m_{5}}{1-m_{1}}\Rightarrow
R_{ab}u=(\alpha_{1}-1)\frac{m_{1}m_{5}}{1-m_{1}}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{2}(\alpha_{1}-1)\displaystyle\frac{m_{2}-m_{1}+m_{1}m_{5}}{1-m_{1}}=h_{2}(\alpha_{1}-1)\frac{m_{2}-m_{1}(1-m_{5})}{1-m_{1}}=h_{2}(\alpha_{1}-1)\frac{m_{2}-m_{1}m_{2}}{1-m_{1}}=(\alpha_{1}-1)h_{2}m_{2}$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=(\alpha_{1}-1)\left[\frac{m_{3}m_{1}}{m_{2}(1-m_{1})}h_{1}+\frac{m_{2}-m_{1}}{1-m_{1}}h_{2}\right]$

Since $\alpha_{1}<1$, $\displaystyle\frac{m_{3}m_{1}}{m_{2}(1-m_{1})}>0$,
fixing $h_{1}>0$ large enough, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{2}}$ with $h_{1}$
being large enough as required so that $(R,\theta)$ under the static-priority
policy 3, 2 is Chen-$\mathcal{S}$. ∎
∎

###### Proof.

Proof for lowest-priority 3, 5 (Chen-S).

In the balanced push started Lu-Kumar network with class 3 having lowest
priority in station 1 and class 5 having lowest priority in station 2, we have
$J=2$, $K=5$, $\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}\displaystyle\frac{m_{3}}{1-m_{1}}&0\mcr-\displaystyle\frac{m_{5}}{1-m_{1}}&1\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=\frac{\alpha_{1}-1}{1-m_{1}}\left[\begin{matrix}m_{3}\mcr
m_{2}-m_{1}\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$, since
$R$ is $2\times 2$, $R^{-1}=CMQ\Delta>0$ and all its diagonal elements are
positive.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=(\alpha_{1}-1)\frac{m_{2}-m_{1}}{1-m_{1}},\;R_{b}^{-1}=R_{2}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(\alpha_{1}-1)\frac{m_{1}-m_{2}}{1-m_{1}};$

$R_{ab}=0\Rightarrow R_{ab}u=0.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)\displaystyle\frac{m_{3}}{1-m_{1}}$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=(\alpha_{1}-1)\frac{m_{3}}{1-m_{1}},\;R_{b}^{-1}=R_{1}^{-1}=\frac{1-m_{1}}{m_{3}}\Rightarrow
u=-R_{b}^{-1}\theta_{b}=1-\alpha_{1};$

$R_{ab}=-\frac{m_{5}}{1-m_{1}}\Rightarrow
R_{ab}u=(\alpha_{1}-1)\frac{m_{5}}{1-m_{1}}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{2}(\alpha_{1}-1)\displaystyle\frac{m_{2}-m_{1}+m_{5}}{1-m_{1}}=(\alpha_{1}-1)h_{2}$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=\frac{\alpha_{1}-1}{1-m_{1}}\left[m_{3}h_{1}+(m_{2}-m_{1})h_{2}\right]$

Since $\alpha_{1}<1$, $1-m_{1}>0$, fixing $h_{1}>0$ large enough, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{2}}$ with $h_{1}$
being large enough as required so that $(R,\theta)$ under the static-priority
policy 3, 5 is Chen-$\mathcal{S}$. ∎
∎

###### Proof.

Proof for lowest-priority 4, 2 (Chen-S).

In the balanced push started Lu-Kumar network with class 4 having lowest
priority in station 1 and class 2 having lowest priority in station 2, we have
$J=2$, $K=5$, $\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}\displaystyle\frac{m_{4}}{m_{2}m_{4}-m_{3}m_{5}}&-\displaystyle\frac{m_{4}(1-m_{1})}{m_{2}m_{4}-m_{3}m_{5}}\mcr-\displaystyle\frac{m_{2}m_{5}}{m_{2}m_{4}-m_{3}m_{5}}&\displaystyle\frac{m_{2}m_{4}}{m_{2}m_{4}-m_{3}m_{5}}\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=\frac{\alpha_{1}-1}{m_{2}m_{4}-m_{3}m_{5}}\left[\begin{matrix}m_{1}m_{4}\mcr
m_{2}(m_{4}-m_{5})\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$ if
$m_{2}m_{4}>m_{3}m_{5}$, since $R$ is $2\times 2$, $R^{-1}=CMQ\Delta>0$ and
all its diagonal elements are positive when $m_{2}m_{4}>m_{3}m_{5}$.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=\frac{\alpha_{1}-1}{m_{2}m_{4}-m_{3}m_{5}}m_{2}(m_{4}-m_{5}),\;R_{b}^{-1}=R_{2}^{-1}=\frac{m_{2}m_{4}-m_{3}m_{5}}{m_{2}m_{4}}\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})\frac{m_{4}-m_{5}}{m_{4}};$

$R_{ab}=-\frac{m_{4}(1-m_{1})}{m_{2}m_{4}-m_{3}m_{5}}\Rightarrow
R_{ab}u=\frac{\alpha_{1}-1}{m_{2}m_{4}-m_{3}m_{5}}(m_{4}-m_{5})(1-m_{1}).$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=\displaystyle\frac{\alpha_{1}-1}{m_{2}m_{4}-m_{3}m_{5}}h_{1}(m_{4}-m_{5})(1-m_{1})(m_{1}m_{4}+m_{4}-m_{1}m_{4}-m_{5}+m_{1}m_{5})=\frac{\alpha_{1}-1}{m_{2}m_{4}-m_{3}m_{5}}h_{1}(m_{4}-m_{5}+m_{1}m_{5})$.
Since
$m_{4}-m_{5}+m_{1}m_{5}=m_{4}-m_{5}(1-m_{1})=m_{4}-m_{5}(m_{3}+m_{4})=m_{4}(1-m_{5})-m_{3}m_{5}=m_{4}m_{2}-m_{3}m_{5}$,
$\left[\theta_{a}+R_{ab}u\right]=(\alpha_{1}-1)h_{1}$. As $\alpha_{1}<1$,
taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=\frac{\alpha_{1}-1}{m_{2}m_{4}-m_{3}m_{5}}m_{1}m_{4},\;R_{b}^{-1}=R_{1}^{-1}=\frac{m_{2}m_{4}-m_{3}m_{5}}{m_{4}}\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})m_{1};$

$R_{ab}=-\frac{m_{2}m_{5}}{m_{2}m_{4}-m_{3}m_{5}}\Rightarrow
R_{ab}u=(\alpha_{1}-1)\frac{m_{1}m_{2}m_{5}}{m_{2}m_{4}-m_{3}m_{5}}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=\displaystyle\frac{\alpha_{1}-1}{m_{2}m_{4}-m_{3}m_{5}}h_{2}\left[m_{2}(m_{4}-m_{5})+m_{1}m_{2}m_{5}\right]$.
Since
$m_{2}(m_{4}-m_{5})+m_{1}m_{2}m_{5}=m_{2}m_{4}+m_{2}m_{5}(m_{1}-1)=m_{2}m_{4}+m_{2}m_{5}(-m_{3}-m_{4})=m_{2}m_{5}(1-m_{5})-m_{2}m_{3}m_{5}=m_{2}^{2}m_{4}-m_{2}m_{3}m_{5}=m_{2}(m_{2}m_{4}-m_{3}m_{5})$,
we have $\left[\theta_{a}+R_{ab}u\right]=(\alpha_{1}-1)h_{2}m_{2}$. As
$\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=\frac{\alpha_{1}-1}{m_{2}m_{4}-m_{3}m_{5}}\left[m_{1}m_{4}h_{1}+m_{2}(m_{4}-m_{5})h_{2}\right]$

Since $\alpha_{1}<1$, $m_{1}m_{4}>0$, and we are not sure the sigh of
$m_{2}(m_{4}-m_{5})$, if $m_{2}m_{4}>m_{3}m_{5}$ is satisfied, fixing
$h_{1}>0$ large enough, and $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, if $m_{2}m_{4}>m_{3}m_{5}$, then, that exists a positive vector
$h\in{R^{2}}$ with $h_{1}$ being large enough as required so that $(R,\theta)$
under the static-priority policy 4, 2 is Chen-$\mathcal{S}$. ∎
∎

###### Proof.

Proof for lowest-priority 4, 5 (Chen-S).

In the balanced push started Lu-Kumar network with class 4 having lowest
priority in station 1 and class 5 having lowest priority in station 2, we have
$J=2$, $K=5$, $\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}1&0\mcr-\displaystyle\frac{m_{5}}{m_{4}}&1\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=(\alpha_{1}-1)\left[\begin{matrix}1\mcr\displaystyle\frac{m_{4}-m_{5}}{m_{4}}\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$, since
$R$ is $2\times 2$, $R^{-1}=CMQ\Delta>0$ and all its diagonal elements are
positive.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=(\alpha_{1}-1)\frac{m_{4}-m_{5}}{m_{4}},\;R_{b}^{-1}=R_{2}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})\frac{m_{4}-m_{5}}{m_{4}};$

$R_{ab}=0\Rightarrow R_{ab}u=0.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=\alpha_{1}-1,\;R_{b}^{-1}=R_{1}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=1-\alpha_{1};$

$R_{ab}=-\frac{m_{5}}{m_{4}}\Rightarrow
R_{ab}u=(\alpha_{1}-1)\frac{m_{5}}{m_{4}}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=(\alpha_{1}-1)h_{2}\displaystyle\frac{m_{4}-m_{5}+m_{5}}{m_{4}}=(\alpha_{1}-1)h_{2}$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=(\alpha_{1}-1)(h_{1}+\frac{m_{4}-m_{5}}{m_{4}}h_{2})$

Since $\alpha_{1}<1$, fixing $h_{1}>0$ large enough, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{2}}$ with $h_{1}$
being large enough as required so that $(R,\theta)$ under the static-priority
policy 4, 5 is Chen-$\mathcal{S}$. ∎∎

Now, we complete the proof that the 6 static-priority cases are all
Chen-$\mathcal{S}$ if $m_{2}m_{4}>m_{3}m_{5}$.

The reflection matrices for the 6 balanced push started Lu-Kumar networks
following static-priority policies are all invertible. Now we check the sign
of determinants of them. We can calculate that $det(R_{1,2})=m_{2}>0$,
$det(R_{1,5})=m_{1}>0$, $det(R_{3,2})=\displaystyle\frac{m_{3}}{1-m_{1}}>0$,
$det(R_{3,5})=\displaystyle\frac{m_{3}}{1-m_{1}}>0$,
$det(R_{4,2})=\displaystyle\frac{m_{2}m_{4}}{m_{2}m_{4}-m_{3}m_{5}}$,
$det(R_{4,5})=1>0$, where the subscript label indicates the classes with the
lowest priority in each station. If we want the sign of the determinants to be
the same, we need $m_{2}m_{4}>m_{3}m_{5}$. Thus, we can conclude that the
reflection matrix for a balanced push started Lu-Kumar network is invertible
for any ratio matrix $\Delta$ if $m_{2}m_{4}>m_{3}m_{5}$.

According to Theorem 4, in a balanced push started Lu-Kumar network with
$\alpha_{1}<1$, the reflection matrix is invertible and $(R,\theta)$ is
Chen-$\mathcal{S}$ for any ratio matrix $\Delta$ if $m_{2}m_{4}>m_{3}m_{5}$.

###  C.2 The SSC Inequalities for Balanced Push Started Lu-Kumar Networks

Similarly as the analysis in Appendix B.2, we get the SSC conditions for
balanced push started Lu-Kumar networks in this appendix. Note that in
balanced push started Lu-Kumar networks, we have $m_{1}+m_{3}+m_{4}=1$ and
$m_{2}+m_{5}=1$. We check the 6 static priority policies of balanced push
started Lu-Kumar networks for (B.1) in the following analysis. As mentioned in
§4.2, we use a fixed priority order as $1>3>4$ (in station 1) within the set
$\\{\ell\in\mathcal{C}(j):Z_{l}(t)-\delta_{l}W_{j}(t)>0\\}$ as the “tie-
breaking” rule.

Same as Appendix B.2, given a policy $\pi$, let
$\dot{Y}^{\pi}(t)=\sum_{k\in\mathcal{H}_{\pi}:z_{k}>0}h_{k}\dot{\overline{Z}_{k}^{\pi}}(t)$.
Then, we want $\dot{Y}^{\pi}(t)<0$ with a common vector
$h\in\mathbb{R}^{K}_{++}$ for any regular time $t$ with
$\lVert\overline{Z}^{\pi}_{\mathcal{H}_{\pi}}(t)\rVert>0$, i.e., at least one
of the high-priority queues is not empty, under all the 6 static priority
policies. We omit $\pi$ when it is fixed. In balanced push started Lu-Kumar
networks, $m_{k}<1$ and thus $\mu_{k}>1$ for all $k\in\mathcal{K}$. Denote
$I_{k}(t)$ as the input rate of class $k$ at time $t$; $O_{k}(t)$ as the
output rate of class $k$ at time $t$. Then,
$\dot{\overline{Z}}_{k}(t)=I_{k}(t)-O_{k}(t)\text{ for
}k\in\mathcal{K};I_{k}(t)=O_{k-1}(t)\text{ for
}k\in\mathcal{K}/\\{1\\};I_{1}(t)=\alpha_{1}$. In this section, we use a set
of lowest-priority classes to represent static priority policies, for example,
(1, 2) stands for a static priority policy that class 1 and 2 have the lowest
priorities in each station.

Case 1: (1, 2)

  1. 1.

$\overline{Z}_{3}(t)>0,\overline{Z}_{5}(t)>0.$

It is clear that server 1 serves queue 3, server 2 serves queue 5, then
$O_{3}(t)=\mu_{3}$, $O_{5}(t)=\mu_{5}$, $O_{1}(t)=O_{2}(t)=O_{4}(t)=0$. So
$\dot{\overline{Z}}_{3}(t)=0-\mu_{3}=-\mu_{3}$,
$\dot{\overline{Z}}_{4}(t)=\mu_{3}-0=\mu_{3}$ and
$\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$. Thus, we need
$\dot{Y}(t)=-h_{3}\mu_{3}+h_{4}\mu_{3}-h_{5}\mu_{5}<0$.

  2. 2.

$\overline{Z}_{3}(t)=0,\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)>0.$

It is clear that $O_{1}(t)=0$, server 2 serves queue 5, then
$O_{5}(t)=\mu_{5}$, $O_{2}(t)=0$. Since $I_{3}(t)=O_{2}(t)=0$ and
$\overline{Z}_{3}(t)=0$, we have $O_{3}(t)=0$ and serve 1 serves queue 4,
which means $O_{4}(t)=\mu_{4}$. So $\dot{\overline{Z}}_{3}(t)=0-0=0$,
$\dot{\overline{Z}}_{4}(t)=0-\mu_{4}=-\mu_{4}$ and
$\dot{\overline{Z}}_{5}(t)=\mu_{4}-\mu_{5}$. Thus, we need
$\dot{Y}(t)=-h_{4}\mu_{4}+h_{5}(\mu_{4}-\mu_{5})<0$.

  3. 3.

$\overline{Z}_{3}(t)>0,\overline{Z}_{5}(t)=0.$

It is clear that server 1 serves queue 3, then $O_{3}(t)=\mu_{3}$,
$O_{1}(t)=O_{4}(t)=0$. Since $O_{4}(t)=0$ and $\overline{Z}_{5}(t)=0$, we have
$O_{5}(t)=0$. So $\dot{\overline{Z}}_{4}(t)=\mu_{3}-0=\mu_{3}$ and
$\dot{\overline{Z}}_{5}(t)=0-0=0$.

If $\overline{Z}_{2}(t)=0$, since $O_{1}(t)=0$, we have $O_{2}(t)=0$, then
$\dot{\overline{Z}}_{3}(t)=0-\mu_{3}=-\mu_{3}$. Thus, we need
$\dot{Y}(t)=-h_{3}\mu_{3}+h_{4}\mu_{3}<0$.

If $\overline{Z}_{2}(t)>0$, since $O_{5}(t)=0$, server 2 serves queue 2, we
have $O_{2}(t)=\mu_{2}$, then $\dot{\overline{Z}}_{3}(t)=\mu_{2}-\mu_{3}$.
Thus, we need $\dot{Y}(t)=h_{3}(\mu_{2}-\mu_{3})+h_{4}\mu_{3}<0$.

Note that $h_{3}(\mu_{2}-\mu_{3})+h_{4}\mu_{3}\geq-h_{3}\mu_{3}+h_{4}\mu_{3}$,
then it is enough to only consider
$\dot{Y}(t)=h_{3}(\mu_{2}-\mu_{3})+h_{4}\mu_{3}<0$ in this situation.

  4. 4.

$\overline{Z}_{3}(t)=0,\overline{Z}_{4}(t)=0,\overline{Z}_{5}(t)>0.$

It is clear that server 2 serves queue 5, then $O_{5}(t)=\mu_{5}$,
$O_{2}(t)=0$. Since $I_{3}(t)=O_{2}(t)=0$ and $\overline{Z}_{3}(t)=0$, we have
$O_{3}(t)=0$. Since $I_{4}(t)=O_{3}(t)=0$ and $\overline{Z}_{4}(t)=0$, we have
$O_{4}(t)=0$. So $\dot{\overline{Z}}_{3}(t)=0-0=0$,
$\dot{\overline{Z}}_{4}(t)=0-0=0$ and
$\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$. It obviously holds that
$\dot{Y}(t)=-h_{5}\mu_{5}<0$ as required.

  5. 5.

$\overline{Z}_{3}(t)=0,\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)=0.$

It is clear that server 1 serves queue 4, then $O_{4}(t)=\mu_{4}$,
$O_{1}(t)=O_{3}(t)=0$. So $\dot{\overline{Z}}_{4}(t)=0-\mu_{4}=-\mu_{4}$. If
$\mu_{4}>\mu_{5}$, $\dot{\overline{Z}}_{5}(t)=\mu_{4}-O_{5}(t)>0$, then $t$ is
not differentiable; if $\mu_{4}\leq\mu_{5}$, $O_{5}(t)=\mu_{5}$,
$\dot{\overline{Z}}_{5}(t)=\mu_{4}-\mu_{4}=0$,
$\dot{\overline{T}}_{5}(t)=\frac{\mu_{4}}{\mu_{5}}$. If
$\overline{Z}_{2}(t)>0$, we have
$O_{2}(t)=\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})$, and
$\dot{\overline{Z}}_{3}(t)=\mu_{2}(1-\frac{\mu_{4}}{\mu_{5}})-0>0$, then $t$
is not differentiable; if $\overline{Z}_{2}(t)=0$, since
$I_{2}(t)=O_{1}(t)=0$, we have $O_{2}(t)=0$, so
$\dot{\overline{Z}}_{3}(t)=0-0=0$. It obviously holds that
$\dot{Y}(t)=-h_{4}\mu_{4}<0$ as required.

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 1:

$\displaystyle-h_{3}\mu_{3}+h_{4}\mu_{3}-h_{5}\mu_{5}<0,$ $\displaystyle-
h_{4}\mu_{4}+h_{5}(\mu_{4}-\mu_{5})<0,$ $\displaystyle
h_{3}(\mu_{2}-\mu_{3})+h_{4}\mu_{3}<0.$

Note that $-h_{3}\mu_{3}+h_{4}\mu_{3}-h_{5}\mu_{5}\leq
h_{3}(\mu_{2}-\mu_{3})+h_{4}\mu_{3}$, thus, it is enough to consider only the
following two conditions from case 1:

$\displaystyle-h_{4}\mu_{4}+h_{5}(\mu_{4}-\mu_{5})<0,$ $\displaystyle
h_{3}(\mu_{2}-\mu_{3})+h_{4}\mu_{3}<0.$

Case 2: (1, 5)

  1. 1.

$\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)>0.$

It is clear that server 1 serves queue 3, server 2 serves queue 2, then
$O_{3}(t)=\mu_{3}$, $O_{2}(t)=\mu_{2}$, $O_{1}(t)=O_{4}(t)=O_{5}(t)=0$. So
$\dot{\overline{Z}}_{2}(t)=0-\mu_{2}=-\mu_{2}$,
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-\mu_{3}$ and
$\dot{\overline{Z}}_{4}(t)=\mu_{3}-0=\mu_{3}$. Thus, we need
$\dot{Y}(t)=-h_{2}\mu_{2}+h_{3}(\mu_{2}-\mu_{3})+h_{4}\mu_{3}<0$.

  2. 2.

$\overline{Z}_{2}(t)=0,\overline{Z}_{3}(t)>0$

It is clear that server 1 serves queue 3, then $O_{3}(t)=\mu_{3}$,
$O_{1}(t)=O_{4}(t)=0$. Since $I_{2}(t)=O_{1}(t)=0$ and
$\overline{Z}_{2}(t)=0$, we have $O_{2}(t)=0$. So
$\dot{\overline{Z}}_{2}(t)=0-0=0$,
$\dot{\overline{Z}}_{3}(t)=0-\mu_{3}=-\mu_{3}$ and
$\dot{\overline{Z}}_{4}(t)=\mu_{3}-0=\mu_{3}$. Thus, we need
$\dot{Y}(t)=-h_{3}\mu_{3}+h_{4}\mu_{3}<0$.

  3. 3.

$\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)=0,\overline{Z}_{4}(t)>0.$

It is clear that server 1 serves queue 4, and server 2 serves queue 2, then
$O_{4}(t)=\mu_{4}$, $O_{2}(t)=\mu_{2}$, $O_{1}(t)=O_{3}(t)=O_{5}(t)=0$. Thus,
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-0=\mu_{2}>0$, which means that $t$ is not
differentiable.

  4. 4.

$\overline{Z}_{2}(t)=0,\overline{Z}_{3}(t)=0,\overline{Z}_{4}(t)>0.$

It is clear that server 1 serves queue 4, then $O_{4}(t)=\mu_{4}$,
$O_{1}(t)=O_{3}(t)=0$. Since $I_{2}(t)=O_{1}(t)=0$ and
$\overline{Z}_{2}(t)=0$, we have $O_{2}(t)=0$. Since $I_{3}(t)=O_{2}(t)=0$ and
$\overline{Z}_{3}(t)=0$, we have $O_{3}(t)=0$. So
$\dot{\overline{Z}}_{2}(t)=0-0=0$, $\dot{\overline{Z}}_{3}(t)=0-0=0$ and
$\dot{\overline{Z}}_{4}(t)=0-\mu_{4}=-\mu_{4}$. It obviously holds
that$\dot{Y}(t)=-h_{4}\mu_{4}<0$ as required.

  5. 5.

$\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)=0,\overline{Z}_{4}(t)=0.$

It is clear that server 2 serves queue 2, then $O_{2}(t)=\mu_{2}$,
$O_{5}(t)=0$. If $\mu_{2}>\mu_{3}$, we have $O_{3}(t)=\mu_{3}$, and then
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-\mu_{3}>0$, then $t$ is not differentiable;
if $\mu_{2}\leq\mu_{3}$, $O_{3}(t)=\mu_{2}$,
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-\mu_{2}=0$,
$\dot{\overline{T}}_{3}(t)=\frac{\mu_{2}}{\mu_{3}}$. If
$\mu_{2}>\mu_{4}(1-\frac{\mu_{2}}{\mu_{3}})$, we have
$O_{4}(t)=\mu_{4}(1-\frac{\mu_{2}}{\mu_{3}})$, and
$\dot{\overline{Z}}_{4}(t)=\mu_{2}-\mu_{4}(1-\frac{\mu_{2}}{\mu_{3}})>0$, then
$t$ is not differentiable; if $\mu_{2}\leq\mu_{4}(1-\frac{\mu_{2}}{\mu_{3}})$,
then $O_{4}(t)=\mu_{2}$, $\dot{\overline{Z}}_{4}(t)=\mu_{2}-\mu_{2}=0$ and
$\dot{\overline{T}}_{1}(t)=1-\frac{\mu_{2}}{\mu_{3}}-\frac{\mu_{2}}{\mu_{4}}$.
So, $O_{1}(t)=\alpha_{1}$ or
$O_{1}(t)=\mu_{1}(1-\frac{\mu_{2}}{\mu_{3}}-\frac{\mu_{2}}{\mu_{4}})$. Thus,
$\dot{\overline{Z}}_{2}(t)=\alpha_{1}-\mu_{2}<0$ or
$\dot{\overline{Z}}_{2}(t)=\mu_{1}(1-\frac{\mu_{2}}{\mu_{3}}-\frac{\mu_{2}}{\mu_{4}})-\mu_{2}$.
We have $\mu_{1}(1-\frac{\mu_{2}}{\mu_{3}}-\frac{\mu_{2}}{\mu_{4}})-\mu_{2}<0$
since
$\mu_{1}(1-\frac{\mu_{2}}{\mu_{3}}-\frac{\mu_{2}}{\mu_{4}})-\mu_{2}<0\Leftrightarrow\frac{\mu_{2}}{\mu_{1}}+\frac{\mu_{2}}{\mu_{3}}+\frac{\mu_{2}}{\mu_{4}}>1\Leftrightarrow\frac{m_{1}+m_{3}+m_{4}}{m_{2}}>1\Leftrightarrow\frac{1}{m_{2}}>1$,
which is true obviously. So, we have $\dot{\overline{Z}}_{2}(t)<0$,
$\dot{\overline{Z}}_{3}(t)=0$ and $\dot{\overline{Z}}_{4}(t)=0$. Thus,
$\dot{Y}(t)<0$ is satisfied obviously, as required.

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 2:

$\displaystyle-h_{2}\mu_{2}+h_{3}(\mu_{2}-\mu_{3})+h_{4}\mu_{3}<0$
$\displaystyle-h_{3}\mu_{3}+h_{4}\mu_{3}<0.$

Case 3: (3, 2)

  1. 1.

$\overline{Z}_{1}(t)>0,\overline{Z}_{5}(t)>0.$

It is clear that $I_{1}(t)=\alpha_{1}$, server 1 serves queue 1, server 2
serves queue 5, then $O_{1}(t)=\mu_{1}$, $O_{5}(t)=\mu_{5}$,
$O_{2}(t)=O_{3}(t)=O_{4}(t)=0$. So
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{4}(t)=0-0=0$ and
$\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$. Thus, we need
$\dot{Y}(t)=h_{1}(\alpha_{1}-\mu_{1})-h_{5}\mu_{5}<0$, which is satisfied
obviously since $\alpha_{1}-\mu_{1}<0$.

  2. 2.

$\overline{Z}_{1}(t)=0,\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)>0$

It is clear that server 1 serves queue 4, then $O_{4}(t)=\mu_{4}$,
$O_{1}(t)=O_{3}(t)=0$. Thus, $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-0>0$, which
means that $t$ is not differentiable.

  3. 3.

$\overline{Z}_{1}(t)>0,\overline{Z}_{5}(t)=0.$

It is clear that server 1 serves queue 1, then $O_{1}(t)=\mu_{1}$,
$O_{3}(t)=O_{4}(t)=0$, so $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{4}(t)=0-0=0$. Since $I_{5}(t)=O_{4}(t)=0$ and
$\overline{Z}_{5}(t)=0$, we have $O_{5}(t)=0$. So
$\dot{\overline{Z}}_{5}(t)=0-0=0$. It obviously holds that
$\dot{Y}(t)=h_{1}(\alpha_{1}-\mu_{1})<0$ as required since
$\alpha_{1}-\mu_{1}<0$.

  4. 4.

$\overline{Z}_{1}(t)=0,\overline{Z}_{4}(t)=0,\overline{Z}_{5}(t)>0.$

It is clear that server 2 serves queue 5, then $O_{5}(t)=\mu_{5}$,
$O_{2}(t)=0$. To make $t$ differentiable, we have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=0$, then $O_{1}(t)=\alpha_{1}$
and $\dot{\overline{T}}_{1}(t)=\frac{\alpha_{1}}{\mu_{1}}$.

If $\overline{Z}_{3}(t)=0$, since $I_{3}(t)=O_{2}(t)=0$, we have $O_{3}(t)=0$.
Since $I_{4}(t)=O_{3}(t)=0$, $\overline{Z}_{4}(t)=0$, we have $O_{4}(t)=0$.
Then $\dot{\overline{Z}}_{4}(t)=0-0=0$,
$\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$. It obviously holds that
$\dot{Y}(t)=-h_{5}\mu_{5}<0$ as required.

If $\overline{Z}_{3}(t)>0$, to keep $t$ differentiable, we have
$\dot{\overline{Z}}_{4}(t)=O_{3}(t)-O_{4}(t)=0$, which means
$\mu_{3}(1-\frac{\alpha_{1}}{\mu_{1}}-\dot{\overline{T}}_{4}(t))=\mu_{4}\dot{\overline{T}}_{4}(t)\Leftrightarrow(\mu_{3}+\mu_{4})\dot{\overline{T}}_{4}(t)=\mu_{3}(1-\frac{\alpha_{1}}{\mu_{1}})\Leftrightarrow\dot{\overline{T}}_{4}(t)=\frac{\mu_{3}(1-\frac{\alpha_{1}}{\mu_{1}})}{\mu_{3}+\mu_{4}}$.
Thus, if we want $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$, we need
$\dot{\overline{Z}}_{5}(t)=O_{4}(t)-\mu_{5}=\mu_{4}\dot{\overline{T}}_{4}(t)-\mu_{5}=\frac{\mu_{3}\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})}{\mu_{3}+\mu_{4}}-\mu_{5}<0$.
$\frac{\mu_{3}\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})}{\mu_{3}+\mu_{4}}-\mu_{5}<0\Leftrightarrow
1-\frac{\alpha_{1}}{\mu_{1}}<\frac{\mu_{5}(\mu_{3}+\mu_{4})}{\mu_{3}\mu_{4}}\Leftrightarrow
1-m_{1}\alpha_{1}<\frac{m_{3}+m_{4}}{m_{5}}\Leftrightarrow
m_{1}\alpha_{1}>\frac{m_{5}-m_{3}-m_{4}}{m_{5}}\Leftrightarrow\alpha_{1}>\frac{m_{5}-(1-m_{1})}{m_{1}m_{5}}\Leftrightarrow\alpha_{1}>\frac{m_{1}-m_{2}}{m_{1}m_{5}}$.
Thus, $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$ is satisfied if
$\alpha_{1}>\frac{m_{1}-m_{2}}{m_{1}m_{5}}$.

Note that $\frac{m_{1}-m_{2}}{m_{1}m_{5}}<1$, thus, the condition
$\alpha_{1}>\frac{m_{1}-m_{2}}{m_{1}m_{5}}$ does not conflict with the nominal
condition($\alpha_{1}<1$). It is clear that $\frac{m_{1}-m_{2}}{m_{1}m_{5}}<1$
since $\frac{m_{1}-m_{2}}{m_{1}m_{5}}<1\Leftrightarrow
m_{1}-m_{2}<m_{1}m_{5}\Leftrightarrow m_{2}>m_{1}(1-m_{5})\Leftrightarrow
m_{2}>m_{1}m_{2}\Leftrightarrow m_{2}(1-m_{1})>0\Leftrightarrow
m_{2}(m_{3}+m_{4})>0$, which is true obviously.

  5. 5.

$\overline{Z}_{1}(t)=0,\overline{Z}_{4}(t)>0,\overline{Z}_{5}(t)=0.$

It is clear that server 1 serves queue 4, then $O_{4}(t)=\mu_{4}$,
$O_{1}(t)=O_{3}(t)=0$. Thus, $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-0>0$, which
means that $t$ is not differentiable.

To sum up, except for the positive condition for common vector $h$, we need
the following conditions from case 3:

$\displaystyle\alpha_{1}>\frac{m_{1}-m_{2}}{m_{1}m_{5}}.$

Case 4: (3, 5)

  1. 1.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)>0.$

It is clear that server 1 serves queue 1, server 2 serves queue 2, then
$O_{1}(t)=\mu_{1}$, $O_{2}(t)=\mu_{2}$, $O_{3}(t)=O_{4}(t)=O_{5}(t)=0$. So
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{2}$ and
$\dot{\overline{Z}}_{4}(t)=0-0=0$. Thus, we need
$\dot{Y}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})<0$.

  2. 2.

$\overline{Z}_{1}(t)=0,\overline{Z}_{2}(t)>0,\overline{Z}_{4}(t)>0$

It is clear that server 1 serves queue 4, then $O_{4}(t)=\mu_{4}$,
$O_{1}(t)=O_{3}(t)=0$. Thus, $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-0>0$, which
means that $t$ is not differentiable.

  3. 3.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)=0.$

It is clear that server 1 serves queue 1, then $O_{1}(t)=\mu_{1}$,
$O_{4}(t)=O_{3}(t)=0$. Thus, $\dot{\overline{Z}}_{4}(t)=0-0=0$,
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}=0$. To make $t$ differentiable,
we have $\dot{\overline{Z}}_{2}(t)=\mu_{1}-O_{2}(t)=0$. It obviously holds
that $\dot{Y}(t)=h_{1}(\alpha_{1}-\mu_{1})<0$ as required since
$\alpha_{1}<\mu_{1}$.

  4. 4.

$\overline{Z}_{1}(t)=0,\overline{Z}_{2}(t)=0,\overline{Z}_{4}(t)>0.$

It is clear that server 1 serves queue 4, then $O_{4}(t)=\mu_{4}$,
$O_{1}(t)=O_{3}(t)=0$. Thus, $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-0>0$, which
means that $t$ is not differentiable.

  5. 5.

$\overline{Z}_{1}(t)=0,\overline{Z}_{2}(t)>0,\overline{Z}_{4}(t)=0.$

It is clear that server 2 serves queue 2, then $O_{2}(t)=\mu_{2}$,
$O_{5}(t)=0$. To make $t$ differentiable, we have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=0$,
$\dot{\overline{Z}}_{4}(t)=O_{3}(t)-O_{4}(t)=0$, then $O_{1}(t)=\alpha_{1}$
$\dot{\overline{T}}_{1}(t)=\frac{\alpha_{1}}{\mu_{1}}$,
$\dot{\overline{Z}}_{2}(t)=\alpha_{1}-\mu_{2}<0$, $O_{3}(t)=O_{4}(t)$. To have
$O_{3}(t)=O_{4}(t)$, we need
$\mu_{3}(1-\frac{\alpha_{1}}{\mu_{1}}-\dot{\overline{T}}_{4}(t))=\mu_{4}\dot{\overline{T}}_{4}(t)\Leftrightarrow\dot{\overline{T}}_{4}(t)=\frac{1-\frac{\alpha_{1}}{\mu_{1}}}{\mu_{3}+\mu_{4}}$,
which is possible since $\dot{\overline{T}}_{4}(t)>0$. Thus, we have
$\dot{Y}(t)<0$ for sure.

To sum up, except for the positive condition for common vector $h$, we need
the following condition from case 4:

$\displaystyle h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})<0.$

Case 5: (4, 2)

  1. 1.

$\overline{Z}_{1}(t)>0,\overline{Z}_{5}(t)>0.$

It is clear that server 1 serves queue 1, server 2 serves queue 5, then
$O_{1}(t)=\mu_{1}$, $O_{5}(t)=\mu_{5}$, $O_{2}(t)=O_{3}(t)=O_{4}(t)=0$. So
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{3}(t)=0-0=0$ and
$\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$. It obviously holds that
$\dot{Y}(t)=h_{1}(\alpha_{1}-\mu_{1})-h_{5}\mu_{5}<0$ as required since
$\alpha_{1}<\mu_{1}$.

  2. 2.

$\overline{Z}_{1}(t)=0,\overline{Z}_{3}(t)>0,\overline{Z}_{5}(t)>0$

It is clear that server 2 serves queue 5, then $O_{5}(t)=\mu_{5}$,
$O_{2}(t)=0$. Thus, $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-0>0$, which means
that $t$ is not differentiable.

  3. 3.

$\overline{Z}_{1}(t)>0,\overline{Z}_{5}(t)=0.$

It is clear that server 1 serves queue 1, then $O_{1}(t)=\mu_{1}$,
$O_{3}(t)=O_{4}(t)=0$. So $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$.
Since $I_{5}(t)=O_{4}(t)=0$, $\overline{Z}_{5}(t)=0$, we have $O_{5}(t)=0$ and
then $\dot{\overline{Z}}_{5}(t)=0-0=0$.

If $\overline{Z}_{2}(t)>0$, then server 2 serves queue 2, we have
$O_{2}(t)=\mu_{2}$ and $\dot{\overline{Z}}_{3}(t)=\mu_{2}-0=\mu_{2}$. Thus, we
need $\dot{Y}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{3}\mu_{2}<0$.

If $\overline{Z}_{2}(t)=0$, when $\mu_{1}>\mu_{2}$,
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{2}>0$ and $t$ is not differentiable;
when $\mu_{1}\leq\mu_{2}$, $O_{2}(t)=\mu_{1}$,
$\dot{\overline{Z}}_{3}(t)=\mu_{1}-0=\mu_{1}$, and we need
$\dot{Y}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{3}\mu_{1}<0$.

It is clear that when $\mu_{1}\leq\mu_{2}$,
$h_{1}(\alpha_{1}-\mu_{1})+h_{3}\mu_{2}\geq
h_{1}(\alpha_{1}-\mu_{1})+h_{3}\mu_{1}$. So it is sufficient to consider only
$h_{1}(\alpha_{1}-\mu_{1})+h_{3}\mu_{2}<0$ in this situation.

  4. 4.

$\overline{Z}_{1}(t)=0,\overline{Z}_{3}(t)=0,\overline{Z}_{5}(t)>0.$

It is clear that server 2 serves queue 5, then $O_{5}(t)=\mu_{5}$,
$O_{2}(t)=0$. Since $I_{3}(t)=O_{2}(t)=0$, $\overline{Z}_{3}(t)=0$, we have
$O_{3}(t)=0$ and $\dot{\overline{Z}}_{3}(t)=0-0=0$. To keep $t$
differentiable, we have $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=0$,
then $O_{1}(t)=\alpha_{1}$ and
$\dot{\overline{T}}_{1}=\frac{\alpha_{1}}{\mu_{1}}$.

If $\overline{Z}_{4}(t)=0$, since $I_{4}(t)=O_{3}(t)=0$, we have $O_{4}(t)=0$.
Then $\dot{\overline{Z}}_{5}(t)=0-\mu_{5}=-\mu_{5}$, It obviously holds that
$\dot{Y}(t)=-h_{5}\mu_{5}<0$ as required.

If $\overline{Z}_{4}(t)>0$, then
$O_{4}(t)=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})$, so
$\dot{\overline{Z}}_{5}(t)=\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}$.
Thus, if we want $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$, we need
$\mu_{4}(1-\frac{\alpha_{1}}{\mu_{1}})-\mu_{5}<0\Leftrightarrow
1-m_{1}\alpha_{1}<\frac{m_{4}}{m_{5}}\Leftrightarrow\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$.
Thus, $\dot{Y}(t)=h_{5}\dot{\overline{Z}}_{5}(t)<0$ is satisfied if
$\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$.

Note that to make the condition $\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}$
does not conflict with the nominal condition($\alpha_{1}<1$), we need
$\frac{m_{5}-m_{4}}{m_{1}m_{5}}<1\Leftrightarrow
m_{5}-m_{4}<m_{1}m_{5}\Leftrightarrow m_{4}>m_{5}(1-m_{1})\Leftrightarrow
m_{4}>m_{5}(m_{3}+m_{4})\Leftrightarrow
m_{4}(1-m_{5})>m_{3}m_{5}\Leftrightarrow m_{2}m_{4}>m_{3}m_{5}$.

  5. 5.

$\overline{Z}_{1}(t)=0,\overline{Z}_{3}(t)>0,\overline{Z}_{5}(t)=0.$

It is clear that server 1 serves queue 3, then $O_{3}(t)=\mu_{3}$,
$O_{1}(t)=O_{4}(t)=0$. Thus, $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-0>0$, which
means that $t$ is not differentiable.

To sum up, except for the positive condition for common vector $h$, we need
the following condition from case 5:

$\displaystyle h_{1}(\alpha_{1}-\mu_{1})+h_{3}\mu_{2}<0$ $\displaystyle
m_{2}m_{4}>m_{4}m_{5},$
$\displaystyle\alpha_{1}>\frac{m_{5}-m_{4}}{m_{1}m_{5}}.$

Case 6: (4, 5)

  1. 1.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)>0.$

It is clear that server 1 serves queue 1, server 2 serves queue 2, then
$O_{1}(t)=\mu_{1}$, $O_{2}(t)=\mu_{2}$, $O_{3}(t)=O_{4}(t)=O_{5}(t)=0$. So
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$,
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{2}$ and
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-0=\mu_{2}$. Thus, we need
$\dot{Y}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})+h_{3}\mu_{2}<0$.

  2. 2.

$\overline{Z}_{1}(t)>0,\overline{Z}_{2}(t)=0.$

It is clear that server 1 serves queue 1, then $O_{1}(t)=\mu_{1}$,
$O_{3}(t)=O_{4}(t)=0$. So $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-\mu_{1}$.
Since $O_{1}(t)=\mu_{1}$, $\overline{Z}_{2}(t)=0$, when $\mu_{1}>\mu_{2}$, we
have $O_{2}(t)=\mu_{2}$, and then
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{2}>0$, so $t$ is not differentiable;
when $\mu_{1}\leq\mu_{2}$, we have $O_{2}(t)=\mu_{1}$, then
$\dot{\overline{Z}}_{2}(t)=\mu_{1}-\mu_{1}=0$,
$\dot{\overline{Z}}_{3}(t)=\mu_{1}-0=\mu_{1}$, and we need
$\dot{Y}(t)=h_{1}(\alpha_{1}-\mu_{1})+h_{3}\mu_{1}<0$ (if
$\mu_{1}\leq\mu_{2}$).

  3. 3.

$\overline{Z}_{1}(t)=0,\overline{Z}_{3}(t)>0$

It is clear that server 1 serves queue 3, then $O_{3}(t)=\mu_{3}$,
$O_{1}(t)=O_{4}(t)=0$. Thus, $\dot{\overline{Z}}_{1}(t)=\alpha_{1}-0>0$, which
means that $t$ is not differentiable.

  4. 4.

$\overline{Z}_{1}(t)=0,\overline{Z}_{2}(t)>0,\overline{Z}_{3}(t)=0.$

It is clear that server 2 serves queue 2, then $O_{2}(t)=\mu_{2}$,
$O_{5}(t)=0$. To make $t$ differentiable, we have
$\dot{\overline{Z}}_{1}(t)=\alpha_{1}-O_{1}(t)=0$, then $O_{1}(t)=\alpha_{1}$,
$\dot{\overline{T}}_{1}(t)=\frac{\alpha_{1}}{\mu_{1}}$ and
$\dot{\overline{Z}}_{2}(t)=\alpha_{1}-\mu_{2}$. Thus, the maximum of
$O_{3}(t)=\mu_{3}(1-\frac{\alpha_{1}}{\mu_{1}})$. When
$\mu_{2}>\mu_{3}(1-\frac{\alpha_{1}}{\mu_{1}})$, we have
$O_{3}(t)=\mu_{3}(1-\frac{\alpha_{1}}{\mu_{1}})$, and then
$\dot{\overline{Z}}_{3}(t)=\mu_{2}-\mu_{3}(1-\frac{\alpha_{1}}{\mu_{1}})>0$,
so $t$ is not differentiable; when
$\mu_{2}\leq\mu_{3}(1-\frac{\alpha_{1}}{\mu_{1}})$, we have
$O_{3}(t)=\mu_{2}$, then $\dot{\overline{Z}}_{3}(t)=\mu_{2}-\mu_{2}=0$, and it
obviously holds that $\dot{Y}(t)=h_{2}(\alpha_{1}-\mu_{2})<0$ as required
since $\alpha_{1}<\mu_{2}$.

To sum up, except for the positive condition for common vector $h$, we need
the following condition from case 6:

$\displaystyle
h_{1}(\alpha_{1}-\mu_{1})+h_{2}(\mu_{1}-\mu_{2})+h_{3}\mu_{2}<0$
$\displaystyle
h_{1}(\alpha_{1}-\mu_{1})+h_{3}\mu_{1}<0\;\;\text{if}\;\;\mu_{1}\leq\mu_{2}.$

##  Appendix D Detailed Proofs for Balanced Lu-Kumar Networks

###  D.1 Chen-$\mathcal{S}$ in Balanced Lu-Kumar Networks

In this appendix, we prove that the reflection matrices for the four static-
priority polices of a balanced Lu-Kumar network ($m_{1}+m_{4}=1$;
$m_{2}+m_{3}=1$) with $\alpha_{1}<1$ are Chen-$\mathcal{S}$, invertible and
have same-sign determinants if $m_{2}+m_{4}<1$. Theorem 4 then guarantees that
the refection matrix is invertible and Chen-$\mathcal{S}$ for any ratio matrix
$\Delta$ of a balanced Lu-Kumar network if $m_{2}+m_{4}<1$.

Given a static priority ratio matrix $\Delta$, to prove that the corresponding
$(R_{\Delta},\theta_{\Delta})$ is Chen-$\mathcal{S}$, we must show that (i)
$R$ is completely-$\mathcal{S}$, and (ii) that there exists a positive vector
$h\in\mathbb{R}^{2}$ such that (2.8) satisfies for each of the 3 partitions of
$\mathcal{J}$.

###### Proof.

Proof for priority 1, 2 (Chen-S).

In the balanced Lu-Kumar network with class 1 having the priority in station 1
and class 2 having the priority in station 2, we have $J=2$, $K=4$,
$\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}1&-\displaystyle\frac{m_{4}}{m_{3}}\mcr
0&1\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=(\alpha_{1}-1)\left[\begin{matrix}1-\displaystyle\frac{m_{4}}{m_{3}}\mcr
1\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$, since
$R$ is $2\times 2$, $R^{-1}=CMQ\Delta>0$ and all its diagonal elements are
positive.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=\alpha_{1}-1,\;R_{b}^{-1}=R_{2}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=1-\alpha_{1}\Rightarrow
R_{ab}u=(\alpha_{1}-1)\displaystyle\frac{m_{4}}{m_{3}}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=(\alpha_{1}-1)(1-\displaystyle\frac{m_{4}}{m_{3}}),\;R_{b}^{-1}=R_{1}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})\left(1-\displaystyle\frac{m_{4}}{m_{3}}\right);$
$R_{ab}=0\Rightarrow R_{ab}u=0.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{2}(\alpha_{1}-1)$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=(\alpha_{1}-1)\left[\left(1-\displaystyle\frac{m_{4}}{m_{3}}\right)h_{1}+h_{2}\right].$

Since $\alpha_{1}<1$, fixing $h_{2}>0$ large enough, we guarantee that
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{2}}$ as required
so that $(R,\theta)$ under the static-priority policy 1, 2 is
Chen-$\mathcal{S}$. ∎
∎

###### Proof.

Proof for priority 1, 3 (Chen-S).

In the balanced Lu-Kumar network with class 1 having the priority in station 1
and class 3 having the priority in station 2, we have
$J=2,K=4,\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}1&-m_{4}\mcr 0&m_{2}\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\mcr
m_{2}\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$, since
$R$ is $2\times 2$, $R^{-1}=CMQ\Delta>0$ and all its diagonal elements are
positive.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=(\alpha_{1}-1)m_{2},\;R_{b}^{-1}=R_{2}^{-1}=\displaystyle\frac{1}{m_{2}}\Rightarrow
u=-R_{b}^{-1}\theta_{b}=1-\alpha_{1}\Rightarrow R_{ab}u=(\alpha_{1}-1)m_{4}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=(\alpha_{1}-1)m_{1},\;R_{b}^{-1}=R_{1}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})m_{1};$ $R_{ab}=0\Rightarrow R_{ab}u=0.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{2}(\alpha_{1}-1)m_{2}$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=(\alpha_{1}-1)[m_{1}h_{1}+m_{2}h_{2}]$

Since $\alpha_{1}<1$, taking $h_{1}>0$, $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{2}}$ as required
so that $(R,\theta)$ under the static-priority policy 1, 3 is
Chen-$\mathcal{S}$. ∎
∎

###### Proof.

Proof for priority 4,2 (Chen-S).

In the balanced Lu-Kumar network with class 4 having the priority in station 1
and class 2 having the priority in station 2, we have
$J=2,K=4,\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\displaystyle\frac{1}{m_{1}-m_{2}}\left[\begin{matrix}m_{1}m_{3}&-m_{1}m_{4}\mcr-
m_{3}&m_{3}\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\mcr
0\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$ if
$m_{1}-m_{2}>0$, equivalently $m_{2}+m_{4}<1$, since $R$ is $2\times 2$,
$R^{-1}=CMQ\Delta>0$ and all its diagonal elements are positive when
$m_{1}-m_{2}>0$.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=0,\;R_{b}^{-1}=R_{2}^{-1}=\displaystyle\frac{m_{1}-m_{2}}{m_{3}}\Rightarrow
u=-R_{b}^{-1}\theta_{b}=0;$

$R_{ab}=\frac{m_{1}m_{4}}{m_{2}-m_{1}}\Rightarrow R_{ab}u=0.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)m_{1}$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=(\alpha_{1}-1)m_{1},\;R_{b}^{-1}=R_{1}^{-1}=\frac{m_{1}-m_{2}}{m_{1}m_{3}}\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(\alpha_{1}-1)\frac{m_{2}-m_{1}}{m_{3}};$

$R_{ab}=\frac{m_{3}}{m_{2}-m_{1}}\Rightarrow R_{ab}u=\alpha_{1}-1.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{2}(\alpha_{1}-1)$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=(\alpha_{1}-1)m_{1}h_{1}$

Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{2}}$ as required
so that $(R,\theta)$ under the static-priority policy 4, 2 is
Chen-$\mathcal{S}$. ∎
∎

###### Proof.

Proof for priority 4, 3 (Chen-S).

In the balanced Lu-Kumar network with class 4 having priority in station 1 and
class 3 having priority in station 2, we have $J=2$, $K=4$,
$\rho=\left[\alpha_{1}\ \alpha_{1}\right]^{\prime},$

$R=\left[\begin{matrix}1&-m_{4}\mcr-\displaystyle\frac{m_{2}}{m_{1}}&\displaystyle\frac{m_{2}}{m_{1}}\end{matrix}\right],$
$\displaystyle\theta=R(\rho-e)=(\alpha_{1}-1)\left[\begin{matrix}m_{1}\mcr
0\end{matrix}\right].$

It is clear that the reflection matrix $R$ is completely-$\mathcal{S}$, since
$R$ is $2\times 2$, $R^{-1}=CMQ\Delta>0$ and all its diagonal elements are
positive.

Partition 1: $a=1;b=2$

$\theta_{b}=\theta_{2}=0,\;R_{b}^{-1}=R_{2}^{-1}=\displaystyle\frac{m_{1}}{m_{2}}\Rightarrow
u=0;$ $R_{ab}=-m_{4}\Rightarrow R_{ab}u=0.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{1}(\alpha_{1}-1)m_{1}$.
Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 2: $a=2;b=1$

$\theta_{b}=\theta_{1}=(\alpha_{1}-1)m_{1},\;R_{b}^{-1}=R_{1}^{-1}=1\Rightarrow
u=-R_{b}^{-1}\theta_{b}=(1-\alpha_{1})m_{1};$

$R_{ab}=-\frac{m_{2}}{m_{1}}\Rightarrow R_{ab}u=(\alpha_{1}-1)m_{2}.$

Thus,
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h_{2}(\alpha_{1}-1)m_{2}$.
Since $\alpha_{1}<1$, taking $h_{2}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

Partition 3: $a=1,2;b=\varnothing$

$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]=h^{\prime}\theta=h_{1}(\alpha_{1}-1)m_{1}$

Since $\alpha_{1}<1$, taking $h_{1}>0$, we ensure
$h_{a}^{{}^{\prime}}\left[\theta_{a}+R_{ab}u\right]<0$.

We have shown, then, that exists a positive vector $h\in{R^{2}}$ as required
so that $(R,\theta)$ under the static-priority policy 4, 3 is
Chen-$\mathcal{S}$. ∎
∎

We can now complete the proof that the 4 static-priority cases are all
Chen-$\mathcal{S}$ if $m_{2}+m_{4}<1$.

The reflection matrices for the 4 static-priorty policies are invertible. We
can calculate that $det(R_{1,2})=1>0$, $det(R_{1,3})=m_{2}>0$,
$det(R_{4,2})=\displaystyle\frac{m_{1}m_{3}}{1-m_{4}-m_{2}}$,
$det(R_{4,3})=m_{2}>0$, where the subscript label indicates the classes with
the priority in each station. These determinants have the same sign if
$m_{2}+m_{4}<1$. We can conclude that the reflection matrix for a balanced
push started Lu-Kumar network is invertible for any ratio matrix $\Delta$
provided that $m_{2}+m_{4}<1$.

According to Theorem 4, in a balanced Lu-Kumar network with $\alpha_{1}<1$,
the reflection matrix is invertible and $(R,\theta)$ is Chen-$\mathcal{S}$ for
any ratio matrix $\Delta$ if $m_{2}+m_{4}<1$.

###  D.2 The SSC Inequalities for Balanced Lu-Kumar Networks

In balanced Lu-Kumar networks, we have $m_{1}+m_{4}=m_{2}+m_{3}=1$. We
consider the 4 static priority policies of balanced Lu-Kumar networks for
(B.1) in the following analysis.

Given a policy $\pi$, we let
$\dot{Y}^{\pi}(t)=\sum_{k\in\mathcal{H}_{\pi}:z_{k}>0}h_{k}\dot{\overline{Z}^{\pi}_{k}}(t)$.
We need $\dot{Y}^{\pi}(t)<0$ with a common vector $h\in\mathbb{R}^{K}_{++}$
for any regular time $t$ with
$\lVert\overline{Z}^{\pi}_{\mathcal{H}_{\pi}}(t)\rVert>0$, i.e., with at least
one non-empty high-priority queue. We omit $\pi$ when it is fixed. In the
balanced Lu-Kumar network, $m_{k}<1$ and thus $\mu_{k}>1$ for all
$k\in\mathcal{K}$. Denote $I_{k}(t)$ as the input rate of class $k$ at time
$t$; $O_{k}(t)$ as the output rate of class $k$ at time $t$. Then,
$\dot{\overline{Z}}_{k}(t)=I_{k}(t)-O_{k}(t)\text{ for
}k\in\mathcal{K};I_{k}(t)=O_{k-1}(t)\text{ for
}k\in\mathcal{K}/\\{1\\};I_{1}(t)=\alpha_{1}$.

We use the set of high priority classes to represent static priority policies;
(1,2) for example stands for a static priority policy that class 1 and 2 have
the priorities in their respective stations.

Case 1: (4, 2)

  1. 1.

$\overline{Z}_{2}(t)>0,\overline{Z}_{4}(t)>0.$

In this case, by the priority policy, at time $t$ server 1 serves queue 4 and
server 2 serves queue 2 so that $O_{4}(t)=\mu_{4}$, $O_{2}(t)=\mu_{2}$,
$O_{1}(t)=O_{3}(t)=0$. Then, $\dot{\overline{Z}}_{2}(t)=0-\mu_{2}=-\mu_{2}$
and $\dot{\overline{Z}}_{4}(t)=0-\mu_{4}=-\mu_{4}$. It obviously holds that
$\dot{Y}(t)=-h_{2}\mu_{2}-h_{4}\mu_{4}<0$ as required.

  2. 2.

$\overline{Z}_{2}(t)>0,\overline{Z}_{4}(t)=0$

In this scenario, server 2 serves queue 2 so that $O_{2}(t)=\mu_{2}$,
$O_{3}(t)=0$. Since $I_{4}(t)=O_{3}(t)=0$ and $\overline{Z}_{4}(t)=0$, we have
$O_{4}(t)=0$. We have then that $\dot{\overline{Z}}_{4}(t)=0-0=0$. If