SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} C _ {\theta , t} ^ {p} = \left\{\left(x _ {0}, x _ {1}, 0\right): x _ {0} <   t \right\} \cup \left\{\left(x _ {0}, x _ {1}, x _ {2}\right): x _ {0} <   \frac {2 \tilde {\lambda}}{1 - 2 \tilde {\lambda}}, x _ {1} + x _ {2} \geq 1 \right\}, \\ \beta_ {\theta , t} ^ {p} = 1 - 2 \bar {\lambda}, \\ \alpha_ {\theta , t} ^ {p} = \frac {1}{2}, \\ V _ {\theta , t} ^ {p} (\boldsymbol {x}) = \| \boldsymbol {x} \| _ {1} ^ {2}, \\ b _ {\theta , t} ^ {p} = \max  _ {\boldsymbol {x} \in C _ {\theta , t} ^ {p}} \left(\| \boldsymbol {x} \| _ {1} + 1\right) ^ {2}), \\ \end{array}
$$
where the last line holds because  $PV_{\theta ,t}^{p}(\boldsymbol {x})\leq V_{\theta ,t}^{p}(\boldsymbol {y})$  for  $\mathbf{y}$  such that  $\| \pmb {y}\| _1 = \| \pmb {x}\| _1 + 1$ .
F.3 Proof of Proposition 3
Proof. To show geometric ergodicity of the chain that follows  $\pi_{\omega}$ , we verify (70). Take  $a_{\theta, \omega}^{g} > 0$  and
$$
V _ {\theta , \omega} ^ {g} (\boldsymbol {x}) = \frac {\omega}{\omega + 1} \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 1}{\omega}\right) + \frac {1}{\omega + 1} \exp \left(a _ {\theta , \omega} ^ {g} \left(x _ {2} + 1\right)\right). \tag {86}
$$
First, we find  $PV_{\theta, \omega}^{g}(\pmb{x})$  for the function defined above. We have
$$
P V _ {\theta , \omega} ^ {g} (\boldsymbol {x}) = \mathbb {E} _ {\boldsymbol {x}} ^ {\pi_ {\omega}} \left[ \frac {\omega}{\omega + 1} \exp \left(a _ {\theta , \omega} ^ {g} \frac {X _ {1} (2) + 1}{\omega}\right) \right] + \mathbb {E} _ {\boldsymbol {x}} ^ {\pi_ {\omega}} \left[ \frac {1}{\omega + 1} \exp \left(a _ {\theta , \omega} ^ {g} \left(X _ {2} (2) + 1\right)\right) \right], \tag {87}
$$
where  $X(2) = (X_{1}(2), X_{2}(2))$  is the state of the system at the second arrival, starting from state  $x$ . To find the above expectations, we first find the corresponding transition probabilities. If the number of departures from server  $i$  during a fixed interval with length  $t$  is less than the total number of jobs in the queue of that server, the number of departures follows a Poisson distribution with parameter  $\theta_{i}t$ . Let  $\mathbb{P}\left((x_1, x_2) \to (x_1', \mathcal{X})\right)$  be the probability of transitioning from a system with  $x_i$  jobs in server-queue pair  $i$  (just after the assignment of the arrival) to a queueing system with  $x_1'$  jobs in the first server-queue pair (just before the upcoming arrival). For  $1 \leq x_1' \leq x_1$ , we have
$$
\mathbb {P} \left(\left(x _ {1}, x _ {2}\right)\rightarrow \left(x _ {1} ^ {\prime}, \mathcal {X}\right)\right) = \int_ {0} ^ {\infty} \lambda \exp (- \lambda t) \frac {\left(\theta_ {1} t\right) ^ {x _ {1} - x _ {1} ^ {\prime}}}{\left(x _ {1} - x _ {1} ^ {\prime}\right) !} \exp (- \theta_ {1} t) d t = \frac {\lambda}{\theta_ {1} + \lambda} \left(\frac {\theta_ {1}}{\theta_ {1} + \lambda}\right) ^ {x _ {1} - x _ {1} ^ {\prime}}, \tag {88}
$$
$$
\mathbb {P} \left(\left(x _ {1}, x _ {2}\right)\rightarrow (0, \mathcal {X})\right) = 1 - \sum_ {i = 1} ^ {x _ {1}} \frac {\lambda}{\theta_ {1} + \lambda} \left(\frac {\theta_ {1}}{\theta_ {1} + \lambda}\right) ^ {x _ {1} - i} = \left(\frac {\theta_ {1}}{\theta_ {1} + \lambda}\right) ^ {x _ {1}}. \tag {89}
$$
Assume  $1 + x_{1} \leq \omega(1 + x_{2})$ , which results in the new arrival being assigned to the first server. For the first term in (87), we have
$$
\begin{array}{l} \mathbb {E} _ {\boldsymbol {x}} ^ {\pi_ {\omega}} \left[ \exp \left(a _ {\theta , \omega} ^ {g} \frac {X _ {1} (2)}{\omega}\right) \right] \\ = \sum_ {i = 0} ^ {x _ {1} + 1} \mathbb {P} \left(\left(x _ {1} + 1, x _ {2}\right)\rightarrow (i, \mathcal {X})\right) \exp \left(a _ {\theta , \omega} ^ {g} \frac {i}{\omega}\right) \\ = \left(\frac {\theta_ {1}}{\theta_ {1} + \lambda}\right) ^ {x _ {1} + 1} + \sum_ {i = 1} ^ {x _ {1} + 1} \exp \left(a _ {\theta , \omega} ^ {g} \frac {i}{\omega}\right) \frac {\lambda}{\theta_ {1} + \lambda} \left(\frac {\theta_ {1}}{\theta_ {1} + \lambda}\right) ^ {x _ {1} + 1 - i} \\ = \left(\frac {\theta_ {1}}{\theta_ {1} + \lambda}\right) ^ {x _ {1} + 1} + \frac {\lambda}{\theta_ {1} + \lambda} \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 1}{\omega}\right) \frac {1 - \exp \left(- a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 1}{\omega}\right) \left(\frac {\theta_ {1}}{\theta_ {1} + \lambda}\right) ^ {x _ {1} + 1}}{1 - \exp \left(- \frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) \frac {\theta_ {1}}{\theta_ {1} + \lambda}}, \\ <   \left(\frac {\theta_ {1}}{\theta_ {1} + \lambda}\right) ^ {x _ {1} + 1} + \frac {\lambda}{\theta_ {1} + \lambda} \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 1}{\omega}\right) \frac {1}{1 - \exp \left(- \frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) \frac {\theta_ {1}}{\theta_ {1} + \lambda}}. \tag {90} \\ \end{array}
$$
Similarly, for the second term in (87), we have
$$
\mathbb {E} _ {\boldsymbol {x}} ^ {\pi_ {\omega}} \left[ \exp \left(a _ {\theta , \omega} ^ {g} X _ {2} (2)\right) \right] \leq \left(\frac {\theta_ {2}}{\theta_ {2} + \lambda}\right) ^ {x _ {2}} + \frac {\lambda}{\theta_ {2} + \lambda} \exp \left(a _ {\theta , \omega} ^ {g} x _ {2}\right) \frac {1}{1 - \exp \left(- a _ {\theta , \omega} ^ {g}\right) \frac {\theta_ {2}}{\theta_ {2} + \lambda}}. \tag {91}
$$
To satisfy (70), for some  $0 < \gamma_{\theta, \omega}^{g} < 1$  and all but finitely many  $x$ , the following should hold,
$$
P V _ {\theta , \omega} ^ {g} (\boldsymbol {x}) \leq \gamma_ {\theta , \omega} ^ {g} V _ {\theta , \omega} ^ {g} (\boldsymbol {x}),
$$
or from (86) and (87),
$$
\begin{array}{l} \mathbb {E} _ {\boldsymbol {x}} ^ {\pi_ {\omega}} \left[ \omega \exp \left(a _ {\theta , \omega} ^ {g} \frac {X _ {1} (2) + 1}{\omega}\right) \right] + \mathbb {E} _ {\boldsymbol {x}} ^ {\pi_ {\omega}} \left[ \exp \left(a _ {\theta , \omega} ^ {g} \left(X _ {2} (2) + 1\right)\right) \right] \\ \leq \gamma_ {\theta , \omega} ^ {g} \left(\omega \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 1}{\omega}\right) + \exp \left(a _ {\theta , \omega} ^ {g} (x _ {2} + 1)\right)\right). \\ \end{array}
$$
Notice that
$$
\omega \left(\frac {\theta_ {1}}{\theta_ {1} + \lambda}\right) ^ {x _ {1} + 1} + \left(\frac {\theta_ {2}}{\theta_ {2} + \lambda}\right) ^ {x _ {2}} \leq c _ {R} R + 1.
$$
From (90) and (91), it suffices to have
$$
\begin{array}{l} \left(c _ {R} R + 1\right) \exp \left(c _ {R} R a _ {\theta , \omega} ^ {g}\right) + \frac {\omega \frac {\lambda}{\theta_ {1} + \lambda} \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 2}{\omega}\right)}{1 - \exp \left(- \frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) \frac {\theta_ {1}}{\theta_ {1} + \lambda}} + \frac {\frac {\lambda}{\theta_ {2} + \lambda} \exp \left(a _ {\theta , \omega} ^ {g} \left(x _ {2} + 1\right)\right)}{1 - \exp \left(- a _ {\theta , \omega} ^ {g}\right) \frac {\theta_ {2}}{\theta_ {2} + \lambda}} \\ \leq \gamma_ {\theta , \omega} ^ {g} \left(\omega \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 1}{\omega}\right) + \exp \left(a _ {\theta , \omega} ^ {g} \left(x _ {2} + 1\right)\right)\right). \tag {92} \\ \end{array}
$$
$$
\zeta_ {1, \theta , \omega} = \frac {\frac {\lambda}{\theta_ {1} + \lambda}}{1 - \exp \left(- \frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) \frac {\theta_ {1}}{\theta_ {1} + \lambda}}, \quad \zeta_ {2, \theta , \omega} = \frac {\frac {\lambda}{\theta_ {2} + \lambda}}{1 - \exp \left(- a _ {\theta , \omega} ^ {g}\right) \frac {\theta_ {2}}{\theta_ {2} + \lambda}}.
$$
Simplifying (92), we need the following to hold
$$
\begin{array}{l} (c _ {R} R + 1) \exp (c _ {R} R a _ {\theta , \omega} ^ {g}) + \omega \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 1}{\omega}\right) \left(\zeta_ {1, \theta , \omega} \exp \left(\frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) - \gamma_ {\theta , \omega} ^ {g}\right) \\ + \exp \left(a _ {\theta , \omega} ^ {g} \left(x _ {2} + 1\right)\right) \left(\zeta_ {2, \theta , \omega} - \gamma_ {\theta , \omega} ^ {g}\right) \leq 0. \tag {93} \\ \end{array}
$$
As  $\zeta_{i,\theta ,\omega} < 1$ , there exists  $\gamma_{\theta ,\omega}^{g}$  such that
$$
\zeta_ {2, \theta , \omega} <   \gamma_ {\theta , \omega} ^ {g} <   1.
$$
From the assumption  $1 + x_{1} \leq \omega(1 + x_{2})$  and the above equation, (93) can be further simplified as
$$
\left(c _ {R} R + 1\right) \exp \left(c _ {R} R a _ {\theta , \omega} ^ {g}\right) + \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 1}{\omega}\right) \left(\omega \zeta_ {1, \theta , \omega} \exp \left(\frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) + \zeta_ {2, \theta , \omega} - (\omega + 1) \gamma_ {\theta , \omega} ^ {g}\right) \leq 0. \tag {94}
$$
For the above to hold outside a finite set, we need to have
$$
\frac {\zeta_ {1 , \theta , \omega} \omega}{1 + \omega} \exp \left(\frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) + \frac {\zeta_ {2 , \theta , \omega}}{1 + \omega} <   \gamma_ {\theta , \omega} ^ {g}. \tag {95}
$$
$$
\zeta_ {3} = \frac {1}{1 + \delta}, \quad \zeta_ {4} = \frac {1 - 0 . 5 \delta}{1 - \delta}. \tag {96}
$$
Note that  $\zeta_3 < 1$  and  $\zeta_4 > 1$ . Defining function  $f(y) \coloneqq 1 + \zeta_4y - \exp(y)$ , we note that for  $y \leq \log \zeta_4$ ,  $f(y) > 0$ , where  $\log \zeta_4$  is the maximizer of  $f(y)$ . Similarly, taking  $g(y) \coloneqq 1 - \zeta_3y - \exp(-y)$ , for  $y \leq -\log \zeta_3$ ,  $g(y) > 0$ , where  $-\log \zeta_3$  is the maximizer of  $g(y)$ . Thus, we conclude that for  $a_{\theta,\omega}^g \leq \min(-\omega \log \zeta_3, -\log \zeta_3, \omega \log \zeta_4)$ ,
$$
\exp (- y) \leq 1 - \zeta_ {3} y \quad \text {h o l d s f o r} \quad y \leq \max  \left(\frac {a _ {\theta , \omega} ^ {g}}{\omega}, a _ {\theta , \omega} ^ {g}\right), \tag {97}
$$
$$
\exp (y) \leq 1 + \zeta_ {4} y \quad \text {h o l d s f o r} \quad y \leq \frac {a _ {\theta , \omega} ^ {g}}{\omega}. \tag {98}
$$
To guarantee the existence of  $0 < \gamma_{\theta, \omega}^{g} < 1$  that satisfies (95), we need to ensure the left-hand side of (95) is strictly less than 1. Using the bounds found in (97) and (98) and the definition of  $\zeta_{1, \theta, \omega}$  and  $\zeta_{2, \theta, \omega}$ , we simplify (95) to get
$$
\frac {\frac {\lambda}{1 + \omega} \left(\omega + a _ {\theta , \omega} ^ {g} \zeta_ {4}\right)}{\lambda + \frac {\theta_ {1} a _ {\theta , \omega} ^ {g} \zeta_ {3}}{\omega}} + \frac {\frac {\lambda}{1 + \omega}}{\lambda + \theta_ {2} a _ {\theta , \omega} ^ {g} \zeta_ {3}} <   1,
$$
$$
a _ {\theta , \omega} ^ {g} \zeta_ {3} \theta_ {2} \left(\lambda \zeta_ {4} - \frac {\zeta_ {3} \theta_ {1} (1 + \omega)}{\omega}\right) <   \lambda \zeta_ {3} \left(\theta_ {1} + \theta_ {2}\right) - \lambda^ {2} \zeta_ {4}. \tag {99}
$$
To make sure there exists  $a_{\theta, \omega}^{g} > 0$  that satisfies (99), the right-hand side of (99) needs to be positive, which follows as below:
$$
\begin{array}{l} \lambda \zeta_ {3} (\theta_ {1} + \theta_ {2}) - \lambda^ {2} \zeta_ {4} = \lambda \left(\frac {\theta_ {1} + \theta_ {2}}{1 + \delta} - \lambda \frac {1 - 0 . 5 \delta}{1 - \delta}\right) \\ = \lambda \left(\theta_ {1} + \theta_ {2} + \lambda\right) \left(\frac {1 - \tilde {\lambda}}{1 + \delta} - \tilde {\lambda} \frac {1 - 0 . 5 \delta}{1 - \delta}\right) \\ = \lambda \left(\theta_ {1} + \theta_ {2} + \lambda\right) \left(\frac {1}{1 + \delta} - \tilde {\lambda} \left(\frac {1}{1 + \delta} + \frac {1 - 0 . 5 \delta}{1 - \delta}\right)\right) \\ \geq \lambda \left(\theta_ {1} + \theta_ {2} + \lambda\right) \left(\frac {1}{1 + \delta} - \frac {1 - \delta}{2} \left(\frac {1}{1 + \delta} + \frac {1 - 0 . 5 \delta}{1 - \delta}\right)\right) \\ = \frac {\delta}{4} \lambda \left(\theta_ {1} + \theta_ {2} + \lambda\right) \tag {100} \\ \end{array}
$$
where  $\tilde{\lambda},\tilde{\theta}_1$  , and  $\tilde{\theta}_{2}$  are the normalized rates defined in (83) and we have used the stability condition  $\tilde{\lambda}\leq 0.5 - 0.5\delta$  . We further simplify the left-hand side of (99) as
$$
\zeta_ {3} \theta_ {2} \left(\lambda \zeta_ {4} - \frac {\zeta_ {3} \theta_ {1} (1 + \omega)}{\omega}\right) <   \theta_ {2} \lambda \zeta_ {3} \zeta_ {4} <   \frac {1 - 0 . 5 \delta}{1 - \delta^ {2}} (\theta_ {1} + \theta_ {2} + \lambda) \lambda .
$$
From the above equation and (100),  $a_{\theta ,\omega}^{g}$  needs to satisfy
$$
a _ {\theta , \omega} ^ {g} \leq \frac {\delta (1 - \delta^ {2})}{8 (1 - 0 . 5 \delta)}.
$$
Finally, we take  $a_{\theta ,\omega}^{g}$  as
$$
a _ {\theta , \omega} ^ {g} = \min  \left(- \omega \log \zeta_ {3}, - \log \zeta_ {3}, \omega \log \zeta_ {4}, \frac {\delta (1 - \delta^ {2})}{8 (1 - 0 . 5 \delta)}\right).
$$
After finding an appropriate  $a_{\theta, \omega}^{g}$ , we can choose  $0 < \gamma_{\theta, \omega}^{g} < 1$  such that (95) holds or
$$
\gamma_ {\theta , \omega} ^ {g} \geq \frac {1}{2} \left(1 + \frac {\zeta_ {1 , \theta , \omega} \omega}{1 + \omega} \exp \left(\frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) + \frac {\zeta_ {2 , \theta , \omega}}{1 + \omega}\right).
$$
Moreover, from (94) a lower bound  $x_{1,\theta,\omega}^{g_1}$  for  $x_1$  is derived; In other words, (94) holds for  $x_1 > x_{1,\theta,\omega}^{g_1}$ . From (93), we can find the corresponding  $x_{2,\theta,\omega}^{g_1}$  and take  $\pmb{x}_{\theta,\omega}^{g_1} = (x_{1,\theta,\omega}^{g_1}, x_{2,\theta,\omega}^{g_1})$ . By repeating the same arguments when  $1 + x_1 < \omega(1 + x_2)$ , we finally conclude that
$$
\Delta V _ {\theta , \omega} ^ {g} (\boldsymbol {x}) \leq - \left(1 - \gamma_ {\theta , \omega} ^ {g}\right) V _ {\theta , \omega} ^ {g} (\boldsymbol {x}) + b _ {\theta , \omega} ^ {g} \mathbb {I} _ {C _ {\theta , \omega} ^ {g}} (\boldsymbol {x}), \quad \boldsymbol {x} \in \mathcal {X},
$$
$$
a _ {\theta , \omega} ^ {g} = \min \left(\omega \log (1 + \delta), \log (1 + \delta), \omega \log \frac {1 - 0 . 5 \delta}{1 - \delta}, \log \frac {1 - 0 . 5 \delta}{1 - \delta}, \frac {\delta (1 - \delta^ {2})}{4 c _ {R} R (1 - 0 . 5 \delta)}\right),
$$
$$
C _ {\theta , \omega} ^ {g} = \left\{\left(x _ {1}, x _ {2}\right) \in \mathcal {X}: x _ {i} \leq \max  \left(x _ {i, \theta , \omega} ^ {g _ {j}}, 0\right), i, j = 1, 2 \right\},
$$
$$
\gamma_ {\theta , \omega} ^ {g} = \frac {1}{2} + \frac {1}{2} \max  \left(\zeta_ {1, \theta , \omega}, \zeta_ {2, \theta , \omega}, \frac {\zeta_ {1 , \theta , \omega} \omega}{1 + \omega} \exp \left(\frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) + \frac {\zeta_ {2 , \theta , \omega}}{1 + \omega}, \frac {\zeta_ {1 , \theta , \omega} \omega}{1 + \omega} + \frac {\zeta_ {2 , \theta , \omega}}{1 + \omega} \exp \left(a _ {\theta , \omega} ^ {g}\right)\right),
$$
$$
b _ {\theta , \omega} ^ {g} = \max _ {\boldsymbol {x} \in C _ {\theta , \omega} ^ {g}} \left(\frac {2 \omega}{\omega + 1} \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1} + 2}{\omega}\right) + \frac {2}{\omega + 1} \exp \left(a _ {\theta , \omega} ^ {g} (x _ {2} + 2)\right)\right),
$$
$$
\zeta_ {1, \theta , \omega} = \frac {\frac {\lambda}{\theta_ {1} + \lambda}}{1 - \exp \left(- \frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) \frac {\theta_ {1}}{\theta_ {1} + \lambda}},
$$
$$
\zeta_ {2, \theta , \omega} = \frac {\frac {\lambda}{\theta_ {2} + \lambda}}{1 - \exp \left(- a _ {\theta , \omega} ^ {g}\right) \frac {\theta_ {2}}{\theta_ {2} + \lambda}},
$$
$$
x _ {1, \theta , \omega} ^ {g _ {1}} = \frac {\omega}{a _ {\theta , \omega} ^ {g}} \log \frac {(c _ {R} R + 1) \exp (c _ {R} R a _ {\theta , \omega} ^ {g})}{(\omega + 1) \gamma_ {\theta , \omega} ^ {g} - \omega \zeta_ {1 , \theta , \omega} \exp \left(\frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) - \zeta_ {2 , \theta , \omega}},
$$
$$
x _ {2, \theta , \omega} ^ {g _ {1}} = \frac {1}{a _ {\theta , \omega} ^ {g}} \log \frac {(c _ {R} R + 1) \exp (c _ {R} R a _ {\theta , \omega} ^ {g}) + \omega \exp \left(a _ {\theta , \omega} ^ {g} \frac {x _ {1 , \theta , \omega} ^ {g _ {1}} + 1}{\omega}\right) (\zeta_ {1 , \theta , \omega} \exp \left(\frac {a _ {\theta , \omega} ^ {g}}{\omega}\right) - \gamma_ {\theta , \omega} ^ {g})}{\gamma_ {\theta , \omega} ^ {g} - \zeta_ {2 , \theta , \omega}},
$$
$$
x _ {2, \theta , \omega} ^ {g _ {2}} = \frac {1}{a _ {\theta , \omega} ^ {g}} \log \frac {(c _ {R} R + 1) \exp (c _ {R} R a _ {\theta , \omega} ^ {g})}{(\omega + 1) \gamma_ {\theta , \omega} ^ {g} - \omega \zeta_ {1 , \theta , \omega} - \zeta_ {2 , \theta , \omega} \exp (a _ {\theta , \omega} ^ {g})},
$$
$$
x _ {1, \theta , \omega} ^ {g _ {2}} = \frac {\omega}{a _ {\theta , \omega} ^ {g}} \log \frac {(c _ {R} R + 1) \exp (c _ {R} R a _ {\theta , \omega} ^ {g}) + \exp (a _ {\theta , \omega} ^ {g} (x _ {2 , \theta , \omega} ^ {g _ {2}} + 1)) (\zeta_ {2 , \theta , \omega} \exp (a _ {\theta , \omega} ^ {g}) - \gamma_ {\theta , \omega} ^ {g})}{\omega (\gamma_ {\theta , \omega} ^ {g} - \zeta_ {1 , \theta , \omega})}.
$$
F.4 Proof of Proposition 4
Proof. Define  $V_{\theta, \omega}^p(\boldsymbol{x}) = \frac{x_1^2}{\omega} + x_2^2$ , and  $\alpha_{\theta, \omega}^p = 1/2$ . Assume that  $x_1 = 0$  and  $x_2 > (1 - \omega)/\omega$ ; which means the new job will be assigned to the first server. The transition probabilities of the discrete-time chain sampled at Poisson arrivals is given in (88) and (89), and we calculate  $PV_{\theta, \omega}^p(\boldsymbol{x})$  as
$$
P V _ {\theta , \omega} ^ {p} (\boldsymbol {x}) = \frac {\lambda}{\omega (\lambda + \theta_ {1})} + \sum_ {i = 1} ^ {x _ {2}} i ^ {2} \frac {\lambda}{\lambda + \theta_ {2}} \left(\frac {\theta_ {2}}{\theta_ {2} + \lambda}\right) ^ {x _ {2} - i} <   c _ {R} R + \sum_ {i = 1} ^ {x _ {2}} i ^ {2} \frac {\lambda}{\lambda + \theta_ {2}} \left(\frac {\theta_ {2}}{\theta_ {2} + \lambda}\right) ^ {x _ {2} - i}. \tag {101}
$$
We define  $d_{i}\coloneqq \theta_{i} / (\theta_{i} + \lambda)$  for  $i = 1,2$  and
$$
\begin{array}{l} \sum_ {i = 1} ^ {x _ {2}} i ^ {2} \frac {\lambda}{\lambda + \theta_ {2}} \left(\frac {\theta_ {2}}{\theta_ {2} + \lambda}\right) ^ {x _ {2} - i} \\ = \frac {1}{(1 - d _ {2}) ^ {2}} \left(- d _ {2} ^ {x _ {2}} \left(d _ {2} + d _ {2} ^ {2}\right) + d _ {2} ^ {2} \left(x _ {2} ^ {2} + 2 x _ {2} + 1\right) + d _ {2} \left(- 2 x _ {2} ^ {2} - 2 x _ {2} + 1\right) + x _ {2} ^ {2}\right) \\ = \frac {1}{(1 - d _ {2}) ^ {2}} \left((1 - d _ {2} ^ {x _ {2}}) (d _ {2} + d _ {2} ^ {2}) + x _ {2} ^ {2} (d _ {2} ^ {2} - 2 d _ {2} + 1) + x _ {2} (2 d _ {2} ^ {2} - 2 d _ {2})\right) \\ = x _ {2} ^ {2} - \frac {2 d _ {2}}{1 - d _ {2}} x _ {2} + \frac {\left(1 - d _ {2} ^ {x _ {2}}\right) \left(d _ {2} + d _ {2} ^ {2}\right)}{\left(1 - d _ {2}\right) ^ {2}}. \tag {102} \\ \end{array}
$$
From (101),
$$
P V _ {\theta , \omega} ^ {p} (\boldsymbol {x}) - V _ {\theta , \omega} ^ {p} (\boldsymbol {x}) + \beta_ {\theta , \omega} ^ {p} x _ {2} <   \left(- \frac {2 d _ {2}}{1 - d _ {2}} + \beta_ {\theta , \omega} ^ {p}\right) x _ {2} + \frac {\left(1 - d _ {2} ^ {x _ {2}}\right) \left(d _ {2} + d _ {2} ^ {2}\right)}{\left(1 - d _ {2}\right) ^ {2}} + c _ {R} R.
$$
Outside a finite set, we need the above equation to be non-positive; which is equivalent to
$$
\left(- 2 + \beta_ {\theta , \omega} ^ {p} \frac {1 - d _ {2}}{d _ {2}}\right) x _ {2} + \frac {\left(1 - d _ {2} ^ {x _ {2}}\right) (1 + d _ {2})}{1 - d _ {2}} + c _ {R} R \frac {1 - d _ {2}}{d _ {2}} \leq 0.
$$
As  $d_2 < 1$
$$
\frac {1 - d _ {2} ^ {y}}{1 - d _ {2}} = 1 + d _ {2} + \dots + d _ {2} ^ {y - 1} \leq y \quad \text {f o r} y \geq 1. \tag {103}
$$
$$
\begin{array}{l} \left(- 2 + \beta_ {\theta , \omega} ^ {p} \frac {1 - d _ {2}}{d _ {2}}\right) x _ {2} + \frac {\left(1 - d _ {2} ^ {x _ {2}}\right) (1 + d _ {2})}{1 - d _ {2}} + c _ {R} R \frac {1 - d _ {2}}{d _ {2}} \\ \leq \left(d _ {2} - 1 + \beta_ {\theta , \omega} ^ {p} \frac {1 - d _ {2}}{d _ {2}}\right) x _ {2} + c _ {R} R \frac {1 - d _ {2}}{d _ {2}}. \\ \end{array}
$$
By taking  $\beta_{\theta ,\omega}^p\leq d_2 / 2$  , it suffices for the following to be non-positive,
$$
- \frac {1 - d _ {2}}{2} x _ {2} + c _ {R} R \frac {1 - d _ {2}}{d _ {2}} \leq 0,
$$
which holds for  $x_{2} \geq 2c_{R}R / d_{2}$ . Thus, for  $x_{1} = 0$  and  $x_{2} \geq \max(2c_{R}R(\lambda + \theta_{2}) / \theta_{2}, (1 - \omega) / \omega) = 2c_{R}R(\lambda + \theta_{2}) / \theta_{2}$ , (76) holds. The case of  $x_{2} = 0$  and non-zero  $x_{1}$  follows same arguments and (76) holds for  $\beta_{\theta, \omega}^{p} \leq d_{1} / 2\sqrt{\omega}$ ,  $x_{2} = 0$ , and  $x_{1} \geq \max(2c_{R}R(\lambda + \theta_{1}) / \theta_{1}, \omega - 1) = 2c_{R}R(\lambda + \theta_{1}) / \theta_{1}$ . We now consider the case of  $x_{1}, x_{2} > 0$  and  $x_{1} + 1 \leq \omega(x_{2} + 1)$ , and note that
$$
\sqrt {V _ {\theta , \omega} ^ {p} (\boldsymbol {x})} = \sqrt {\frac {x _ {1} ^ {2}}{\omega} + x _ {2} ^ {2}} \leq \sqrt {\frac {(x _ {1} + 1) ^ {2}}{\omega} + (x _ {2} + 1) ^ {2}} \leq \sqrt {\omega + 1} (x _ {2} + 1).
$$
Hence, it suffices to find finite set  $C_{\theta, \omega}^p$ , constants  $b_{\theta, \omega}^p$  and  $\beta_{\theta, \omega}^p > 0$ , such that the following holds for  $V_{\theta, \omega}^p(\boldsymbol{x}) = \frac{x_1^2}{\omega} + x_2^2$ ,
$$
\Delta V _ {\theta , \omega} ^ {p} (\boldsymbol {x}) \leq - \sqrt {\omega + 1} \beta_ {\theta , \omega} ^ {p} (x _ {2} + 1) + b _ {\theta , \omega} ^ {p} \mathbb {I} _ {C _ {\theta , \omega} ^ {p}} (\boldsymbol {x}).
$$
As  $x_{1} + 1 \leq \omega (x_{2} + 1)$ , the new arrival is assigned to the first queue and we find  $\Delta V_{\theta ,\omega}^{p}(\boldsymbol {x}) + \sqrt{\omega + 1}\beta_{\theta ,\omega}^{p}(x_{2} + 1)$  using the same calculations as (102).
$$
\begin{array}{l} \Delta V _ {\theta , \omega} ^ {p} (\boldsymbol {x}) + \sqrt {\omega + 1} \beta_ {\theta , \omega} ^ {p} \left(x _ {2} + 1\right) \\ = \frac {1}{\omega} \left((x _ {1} + 1) ^ {2} - \frac {2 d _ {1}}{1 - d _ {1}} (x _ {1} + 1) + \frac {\left(1 - d _ {1} ^ {x _ {1} + 1}\right) \left(d _ {1} + d _ {1} ^ {2}\right)}{\left(1 - d _ {1}\right) ^ {2}} - x _ {1} ^ {2}\right) \\ - \frac {2 d _ {2}}{1 - d _ {2}} x _ {2} + \frac {\left(1 - d _ {2} ^ {x _ {2}}\right) \left(d _ {2} + d _ {2} ^ {2}\right)}{\left(1 - d _ {2}\right) ^ {2}} + \sqrt {\omega + 1} \beta_ {\theta , \omega} ^ {p} \left(x _ {2} + 1\right) \\ = \frac {x _ {1}}{\omega} \left(2 - \frac {2 d _ {1}}{1 - d _ {1}}\right) + \frac {1 - 3 d _ {1}}{\omega \left(1 - d _ {1}\right)} + \frac {\left(1 - d _ {1} ^ {x _ {1} + 1}\right) \left(d _ {1} + d _ {1} ^ {2}\right)}{\omega \left(1 - d _ {1}\right) ^ {2}} (104) \\ + \left(x _ {2} + 1\right) \left(- \frac {2 d _ {2}}{1 - d _ {2}} + \sqrt {\omega + 1} \beta_ {\theta , \omega} ^ {p}\right) + \frac {2 d _ {2}}{1 - d _ {2}} + \frac {\left(1 - d _ {2} ^ {x _ {2}}\right) \left(d _ {2} + d _ {2} ^ {2}\right)}{\left(1 - d _ {2}\right) ^ {2}}. (105) \\ \end{array}
$$
We next consider two different cases based on the value of  $d_{1}$  and analyze them separately.
One.  $0.8 \leq d_{1} < 1$ : We first notice that the coefficient of  $x_{1}$  in (104) is negative, as  $d_{1} > 1/2$ . For  $x_{1} \geq 1$ , (104) is equal to
$$
\begin{array}{l} \frac {1}{\omega \left(1 - d _ {1}\right)} \left((2 - 4 d _ {1}) x _ {1} + 1 - 3 d _ {1} + \left(d _ {1} + d _ {1} ^ {2}\right) \sum_ {i = 0} ^ {x _ {1}} d _ {1} ^ {i}\right) \\ = \frac {1}{\omega \left(1 - d _ {1}\right)} \left((2 - 4 d _ {1}) \left(x _ {1} - 1\right) + d _ {1} ^ {3} \left(1 + d _ {1}\right) \sum_ {i = 0} ^ {x _ {1} - 2} d _ {1} ^ {i} + d _ {1} \left(1 + d _ {1}\right) ^ {2} + 3 - 7 d _ {1}\right) \\ \leq \frac {1}{\omega (1 - d _ {1})} \left((2 - 4 d _ {1}) (x _ {1} - 1) + d _ {1} ^ {3} (1 + d _ {1}) (x _ {1} - 1) + d _ {1} (1 + d _ {1}) ^ {2} + 3 - 7 d _ {1}\right) \\ = \frac {1}{\omega \left(1 - d _ {1}\right)} \left(\left(d _ {1} ^ {4} + d _ {1} ^ {3} - 4 d _ {1} + 2\right) \left(x _ {1} - 1\right) + d _ {1} \left(1 + d _ {1}\right) ^ {2} + 3 - 7 d _ {1}\right) \\ = \frac {- d _ {1} ^ {3} - 2 d _ {1} ^ {2} - 2 d _ {1} + 2}{\omega} (x _ {1} - 1) + \frac {- d _ {1} ^ {2} - 3 d _ {1} + 3}{\omega} \\ <   0, \\ \end{array}
$$
where the third line follows from (103), and the last line from the fact that when  $0.8 \leq d_{1} < 1$ , both terms  $-d_{1}^{3} - 2d_{1}^{2} - 2d_{1} + 2$  and  $-d_{1}^{2} - 3d_{1} + 3$  are negative. Next, we notice that (105) is equal to