流体模型是一个输入-输出系统, 其中流体连续地流入和流出一个缓冲器, 其流入率和流出率由一个外部随机过程(或称驱动系统)控制.在这个信息技术高速发展的时代, 传统的离散时间排队模型有很大的局限性, 流体模型已经成为通信网络和计算机技术性能评价的重要指标, 并受到众多学者的关注.

Virtamo和Norro^[1]对M/M/1队列驱动的流体模型进行了研究, 推广了有限状态环境使用的谱分析方法, 并且使用第二类切比雪夫多项式给出库存量平稳分布. Parthasarathy, Vijayashree和Lenin^[2]运用了连分数法给出了由M/M/1队列驱动的流体队列缓冲器占用率和缓冲器容量的平稳分布的完整解. Adan和Resing^[3]运用嵌入时间点的方法分析了有M/M/1驱动的流体模型, 并根据第一类修正的Bessel函数表示出了其库存量的平稳分布. Sericoal, Parthasarathy和Vijayashree^[4]使用连续分数法明确地获得了由M/M/1排队驱动的流体模型的瞬态解, 并说明了缓冲器在指定时间为空的概率.

近年来, 一些学者将休假策略运用到流体模型的外界环境中, Mao等^[5]分析了多重休假的M/G/1排队系统驱动的流体模型, 通过Laplace变换的方法得到稳态库存量的均值. Xu等^[6]研究了工作休假的M/PH/1排队系统驱动的流体模型, 得到平均吞吐量, 服务器利用率及平稳库存量的均值.随着研究的进一步深入, 众多学者将驱动系统由单服务台扩展为多服务台, 以便能更好地应用到实际生产生活中.徐秀丽等^[7]分析了M/M/c排队驱动的流体模型, 得到流体模型稳态库存量的LST, 并求得空库概率及均值的表达式. Xu等^[8]探讨了多重工作休假排队系统驱动的流体模型, 给出其稳态下系统性能指标.

无线Mesh网络具有易布置、可治愈、高传输率的特点, 被认为是下一代无线移动终端网络接入的关键技术, 近几年来受到广泛关注.对于多信道无线Mesh网络, 主要是通过合理分配信道来提高系统性能.魏振春^[10]等分析了一种用于多信道无线Mesh网络的信道分配方法, 谢鲲^[11]等研究了多接口多信道无线Mesh网络中面向协作通信的接口分配算法, 谢桂芳^[12]分析了多接口多信道无线Mesh网络动态信道分配中的路由协议, 都很有效地提高了网络性能.

如果把网络中多个信息节点看作一个多服务台的排队系统, 节点中的信息流看作流体, 信道相当于流体模型的缓冲器.当信道没有收到干扰, 信息节点正常传输, 系统处于忙期; 当信道收到干扰时, 信息节点传输速率降低, 相当于系统处于工作休假期; 当信道未受干扰时, 但信息节点中无信息传输时, 系统处于闲期.在此基础上, 本文构建了单重工作休假M/M/c排队系统驱动的流体模型, 研究系统的性能指标, 并将其应用到无线Mesh网络中, 分析参数变化对系统性能的影响.

在驱动系统中, 假设顾客的到达时间间隔和正规忙期的服务时间分别服从参数为$ \lambda $和$ \mu_b $的指数分布.当系统中的所有顾客都完成服务后, 服务员进入一次随机长度的工作休假, 休假时间服从参数为$ \theta $的指数分布.在工作休假期间, 服务员进入一种低速服务状态, 服务时间服从参数为$ \mu_v $ ($ \mu_v $<$ \mu_b $)的指数分布.一次休假完成时, 若系统中有顾客正在接受服务, 则从低速率$ \mu_v $切换到正常服务率$ \mu_b $, 重新开始服务被中断的顾客, 系统进入一个正规忙期; 若休假完成时, 系统中无顾客, 则服务员进入闲期, 处于待机状态, 直到有顾客进入系统, 立即为顾客服务, 系统进入一个正规忙期.为了方便, 把上述单重工作休假模型记为M/M/c/SWV排队.

假设到达时间间隔、服务时间和工作休假时间之间相互独立.另外, 采用先到先服务(FCFS)的排队规则.

设$ L(t) $表示时刻$ t $系统中的顾客数, 记

$ J(t) = \left\{ \begin{array}{ll} 0, & \mbox{ 时刻$t$系统处于工作休假, }\\ 1, &\mbox{ 时刻$t$系统处于闲期, }\\ 2, &\mbox{ 时刻$t$系统处于忙期.} \end{array} \right. $

则$ \{L(t), J(t), t\geq0\} $形成一个状态空间为$ \Omega = \{(0, 0)\cup(0, 1)\cup(k, j), k\geq1, j = 0, 2\} $的拟生灭过程, 把状态空间按照字典的顺序排列, 便得到二维随机过程的最小生成元矩阵Q, 即

$ \mathit{\boldsymbol{Q}} = \left( \begin{array}{ccccccccc} \mathit{\boldsymbol{A}}_{00}&\mathit{\boldsymbol{C}}_{01}\\ \mathit{\boldsymbol{B}}_{10}&\mathit{\boldsymbol{A}}_1&\mathit{\boldsymbol{C}}\\ &\mathit{\boldsymbol{B}}_2&\mathit{\boldsymbol{A}}_2&\mathit{\boldsymbol{C}}\\ & &\ddots&\ddots&\ddots\\ & & &\mathit{\boldsymbol{B}}_{c-1}&\mathit{\boldsymbol{A}}_{c-1}&\mathit{\boldsymbol{C}}\\ & & & &\mathit{\boldsymbol{B}}&\mathit{\boldsymbol{A}}&\mathit{\boldsymbol{C}}\\ & & & & &\ddots&\ddots&\ddots\\ \end{array} \right), $

其中

$ \mathit{\boldsymbol{A}}_{00} = \left(\begin{array}{cc}-(\lambda+\theta)&\theta\\0&-\lambda \end{array}\right), \; \mathit{\boldsymbol{C}}_{01} = \left(\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right), \; \mathit{\boldsymbol{B}}_{10} = (\mu_v, \mu_b)^T, $

$ \mathit{\boldsymbol{A}}_k = \left(\begin{array}{cc}-(\lambda+k\mu_v+\theta)&\theta\\0&-(\lambda+k\mu_b) \end{array}\right), 1\leq k\leq c-1, \; \mathit{\boldsymbol{B}}_k = \left(\begin{array}{cc}k\mu_v&0\\0&k\mu_b\end{array}\right), 2\leq k \leq c-1, \\ $

$ \mathit{\boldsymbol{B}} = \left(\begin{array}{cc}c\mu_v&0\\0&c\mu_b\end{array}\right), \; \mathit{\boldsymbol{A}} = \left(\begin{array}{cc}-(\lambda+c\mu_v+\theta)&\theta\\0&-(\lambda+c\mu_b) \end{array}\right), \; \mathit{\boldsymbol{C}} = \left(\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right). $

引理2.1^[8]  如果系统负载$ \rho = \lambda/c\mu_b<1 $, 则二次矩阵方程R²B+RA+C = 0 存在最小非负解

(2.1) $ \begin{equation} \mathit{\boldsymbol{R}} = \left(\begin{array}{cc}r{\quad}& { } \frac{\theta r}{c\mu_b(1-r)}\\ 0{\quad} &\rho\end{array}\right), \end{equation} $

其中r = $ \frac{1}{2c\mu_v}(\lambda+c\mu_v+\theta-\sqrt{(\lambda+c\mu_v+\theta)^2-4c\lambda\mu_v}), 0<r<1 $.

系统负载$ \rho<1 $时, 随机过程$ \{L(t), J(t), t\geq0\} $存在稳态分布, 记为$ \pi_{kj} = \lim\limits_{t\to\infty}P\{L(t) = k, J(t) = j\}, (k, j)\in\Omega $.为了得到稳态分布, 引入函数序列

$ \left\{ \begin{array}{ll} \tau_0 = 1, \\ { } \tau_k = \frac{\theta r}{\lambda(1-r)}+\frac{(c-k)\mu_v}{\lambda}\tau_{k-1}+\frac{\theta}{\lambda}\sum\limits_{v = 0}^{k-1}\tau_v, 1\leq k\leq c-1. \end{array} \right. $

可以得到如下定理.

定理2.1  若$ \rho<1 $, 拟生灭过程$ \{L(t), J(t), t\geq0\} $的平稳分布为

$ \pi_{01} = \frac{\theta}{\lambda}\pi_{00}, $

(2.2) $ \begin{equation} \pi_{k0} = \left\{ \begin{array}{ll} K\tau_{c-1-k}, &0\leq k\leq c-1, \\ Kr, &k = c, \\ \pi_{c0}r^{k-c}, &k>c. \end{array} \right. \end{equation} $

(2.3) $ \begin{equation} \pi_{k2} = \left\{ \begin{array}{ll} { } K\left(\frac{\lambda+\theta}{\mu_b}-\frac{\mu_v}{\mu_b}\right), k = 1, \\ { } K\frac{1}{k!}\left\{\left(\frac{\lambda}{\mu_b}\right)^{k-1}\left(\frac{\lambda+\theta-\mu_v}{\mu_b}\right)+\frac{\theta r}{\lambda(1-r)}\sum\limits_{i = 1}^{k-1}\left(\frac{\lambda}{\mu_b}\right)^i(k-i)!\right.\\ { } \qquad\left.+\frac{\theta}{\lambda}\sum\limits_{j = 1}^{k-1}\left(\frac{\lambda}{\mu_b}\right)^i(k-i)!\sum\limits_{m = k+1-i}^{c-1}\tau_{c-1-m}\right\}, 2\leq k \leq c, \\ { } \pi_{c2}\rho^{k-c}+\frac{\theta r}{c\mu_b(1-r)}\pi_{c0}\sum\limits_{m = 0}^{k-c-1}r^m\rho^{k-c-1-m}, k>c, \end{array} \right. \end{equation} $

其中$ K = \pi_{c-1, 0} $可由临界条件$ \pi_{01}+\sum\limits_{k = 0}^\infty\pi_{k0}+\sum\limits_{k = 0}^\infty\pi_{k2} = 1 $得到.

证  记π₀$ = (\pi_{00}, \pi_{01}), $π_k $= (\pi_{k0}, \pi_{k2}), k\geq1, $x=(π₀,π₁,···,π_c).众所周知, 随机过程$ \{L(t), J(t), t\geq0\} $的稳态分布存在当且仅当二次矩阵方程R²B+RA+C = 0的最小非负解的谱半径SPR<1, 并且xB[R] = 0 有正解, 其中

$ \mathit{\boldsymbol{B}}\left[ R \right] = \left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{A}}_{00}}}&{{\mathit{\boldsymbol{C}}_{01}}}&{}&{}&{}&{}\\ {{\mathit{\boldsymbol{B}}_{10}}}&{{\mathit{\boldsymbol{A}}_1}}&\mathit{\boldsymbol{C}}&{}&{}&{}\\ {}&{{\mathit{\boldsymbol{B}}_2}}&{{\mathit{\boldsymbol{A}}_2}}&\mathit{\boldsymbol{C}}&{}&{}\\ {}&{}& \ddots & \ddots & \ddots &{}\\ {}&{}&{}&{{\mathit{\boldsymbol{B}}_{c - 1}}}&{{\mathit{\boldsymbol{A}}_{c - 1}}}&\mathit{\boldsymbol{C}}\\ {}&{}&{}&{}&\mathit{\boldsymbol{B}}&{{\mathit{\boldsymbol{RB}} + \mathit{\boldsymbol{A}}}} \end{array}} \right). $

由xB[R] = 0, 可以得到方程组

(2.4) $ \begin{equation} \left\{ \begin{array}{ll} { } \pi_{00} = \frac{\lambda}{\theta}\pi_{01}, \\ -(\lambda+\theta)\pi_{00}+\mu_v\pi_{10}+\mu_b\pi_{12} = 0, \\ \lambda\pi_{01}+\theta\pi_{10}-(\lambda+\mu_b)\pi_{12}+2\mu_b\pi_{22} = 0, \\ \lambda\pi_{k-1, 0}-(\lambda+\theta+k\mu_v)\pi_{k0}+(k+1)\mu_v\pi_{k+1, 0} = 0, 1\leq k\leq c-1, \\ \lambda\pi_{k-1, 2}+\theta\pi_{k0}-(\lambda+k\mu_b)\pi_{k2}+(k+1)\mu_b\pi_{k+1, 2} = 0, 2\leq k\leq c-1, \\ { } \lambda\pi_{c-1, 0}-\frac{\lambda}{r}\pi_{c0} = 0, \\ { } \lambda\pi_{c-1, 2}+\frac{\theta}{1-r}\pi_{c0}-c\mu_b\pi_{c2} = 0. \end{array} \right. \end{equation} $

方程组各式依次定义为(2.4.1)–(2.4.7).

从(2.4.1)式可以得到$ \pi_{01} = \frac{\theta}{\lambda}\pi_{00} $.记$ \pi_{c-1, 0} = K $.由(2.4.6)式可知

(2.5) $ \begin{equation} \pi_{c0} = r\pi_{c-1, 0} = Kr. \end{equation} $

把(2.5)式代入(2.4.4)式中递推得到$ \pi_{k0} = \pi_{c-1, 0}\tau_{c-1-k} = K\tau_{c-1-k}, 0\leq k\leq c-1 $.

根据(2.4.7)式, 可以得到

(2.6) $ \begin{equation} c\mu_b\pi_{c2}-\lambda\pi_{c-1, 2} = \frac{\theta}{1-r}\pi_{c0}. \end{equation} $

结合(2.6)式和(2.4.5)式, 得$ k\mu_b\pi_{k2} = \lambda\pi_{k-1, 2}+\frac{\theta}{1-r}\pi_{c0}+\theta\sum\limits_{m = k}^{c-1}\pi_{m0}, 2\leq k\leq c-1 $.递推计算可得

(2.7) $ \begin{eqnarray} \pi_{k2}& = &\frac{1}{k!}\left\{\left(\frac{\lambda}{\mu_b}\right)^{k-1}\pi_{12}+\frac{\theta}{\lambda(1-r)} \pi_{c0}\sum\limits_{i = 1}^{k-1}\left(\frac{\lambda}{\mu_b}\right)^i(k-i)!\right.{}\\ &&\left.+\frac{\theta}{\lambda}\sum\limits_{i = 1}^{k-1}\left(\frac{\lambda}{\mu_b}\right)^i(k-i)!\sum\limits_{m = k+1-i}^{c-1}\pi_{c0}\right\}. \end{eqnarray} $

根据(2.4.2)式和(2.2)式, 可以得到

(2.8) $ \begin{equation} \pi_{12} = K\left(\frac{\lambda+\theta}{\mu_b}\tau_{c-1}-\frac{\mu_v}{\mu_b}\tau_{c-2}\right). \end{equation} $

把(2.8)式和(2.2)式代入(2.7)式中有

$ \begin{eqnarray*} \pi_{k2}& = &K\frac{1}{k!}\left\{\left(\frac{\lambda}{\mu_b}\right)^{k-1}\left(\frac{\lambda+\theta}{\mu_b}-\frac{\mu_v}{\mu_b}\right)+\frac{\theta r}{\lambda(1-r)}\sum\limits_{i = 1}^{k-1}\left(\frac{\lambda}{\mu_b}\right)^i(k-i)!\right.\\ && +\left.\frac{\theta}{\lambda}\sum\limits_{i = 1}^{k-1}\left(\frac{\lambda}{\mu_b}\right)^i(k-i)!\sum\limits_{m = k+1-i}^{c-1}\tau_{c-1-m}\right\}, 2\leq k\leq c. \end{eqnarray*} $

另一方面, 由矩阵几何解方法知π_k = π_cR^k-c = (π_c0, π_c2)R^k-c, k>c .由(2.1)式可得

$ \mathit{\boldsymbol{R}}^k = \left(\begin{array}{cc} r^k{\quad}&{ }\frac{\theta r}{c\mu_b(1-r)}\sum\limits_{m = 0}^{k-1}r^m\rho^{k-1-m}\\ 0{\quad}&\rho^k \end{array}\right) , k\geq1, $

进而可得(2.2)式和(2.3)式中最后一式, 其中$ K $由临界条件$ \pi_{01}+\sum\limits_{k = 0}^\infty\pi_{k0}+\sum\limits_{k = 0}^\infty\pi_{k2} = 1 $得到.

用$ X(t) $来表示缓冲器在$ t $时刻的库存量, 则$ X(t) $就是一个非负的随机变量, 设缓冲器的净流入率(流入率-流出率)为过程$ \{X(t), L(t), J(t), t\geq0\} $的函数

$ \omega[X(t), L(t), J(t)] = \frac{{\rm d}X(t)}{{\rm d}t} = \left\{ \begin{array}{ll} \sigma_1, (L(t), J(t)) = (k, 0), k\geq1, \\ \sigma_2, (L(t), J(t)) = (k, 2), k\geq1, \\ \sigma_3, (L(t), J(t)) = (0, 1), X(t)>0, \\ \sigma, (L(t), J(t)) = (0, 0), X(t)>0, \\ 0, (L(t), J(t)) = (0, 0)\cap(0, 1), X(t) = 0, \\ \end{array} \right. $

其中$ \sigma<0, \sigma_2>\sigma_1>\sigma_3>0 $.当驱动系统处于工作休假时期, 并且驱动系统中有顾客时, 缓冲器库存量以速率$ \sigma_1 $增加; 当驱动系统处于忙期时, 缓冲器库存量以速率$ \sigma_2 $增加; 当驱动系统处于闲期时, 缓冲器库存量以速率$ \sigma_3 $增加; 当驱动系统处于工作休假时期, 并且系统中没有顾客时, 缓冲器库存量以速率$ -\sigma $减少, 直至缓冲器库存量为空.一直保持这种状态直至驱动系统中再次有顾客进入则开始增加, 一直按照这种状态循环下去.

流体模型的平均漂移率为

$ d = \sigma\pi_{00}+\sigma_1\sum\limits_{k = 1}^{+\infty}\pi_{k0}+\sigma_2 \sum\limits_{k = 1}^{+\infty}\pi_{k2}+\sigma_3\pi_{01}, $

其中$ \pi_{kj}, (k, j)\in\Omega $, 见定理2.1.易知, 当$ d<0 $, $ \rho<1 $时, 流体模型是稳态的$ ^{[9]} $.记其稳态随机向量为$ (X, L, J) $, 这里的$ X $为稳态下的缓冲器库存量.

设流体模型的稳态联合分布为

$ F_{kj}(x) = P\{X\leq x, L = k, J = j\} = \lim\limits_{t\to\infty}P\{X(t)\leq x, L(t) = k, J(t) = j\}, x\geq0, (k, j)\in\Omega. $

则由全概率公式知, 流体模型库存量的稳态概率分布为

$ F(x) = P\{X\leq x\} = F_{00}(x)+F_{01}(x)+\sum\limits_{k = 1}^\infty F_{k0}(x)+\sum\limits_{k = 1}^\infty F_{k2}(x). $

定理3.1  $ \sum\limits_{k = 1}^\infty F_{k0}(x) $及$ \sum\limits_{k = 1}^\infty F_{k2}(x) $在$ (0, \infty) $上一致收敛.

证  由$ F_{k0}(x) $及$ F_{k2}(x) $关于$ x $递增, 可得

$ \left|F_{k0}(x)\right| = F_{k0}(x)\leq \lim\limits_{x\to\infty} = \pi_{k0}, k = 1, 2, \cdots, $

$ \left|F_{k2}(x)\right| = F_{k2}(x)\leq \lim\limits_{x\to\infty} = \pi_{k2}, k = 1, 2, \cdots. $

由(2.2)和(2.3)式可知

$ \begin{eqnarray*} \sum\limits_{k = 1}^\infty\pi_{k0}& = &K\sum\limits_{k = 1}^{c-1}\tau_{c-1-k}+Kr+\pi_c0\sum\limits_{k = c+1}^\infty r^{k-c}\\ & = &K\sum\limits_{k = 1}^{c-1}\tau_{c-1-k}+Kr+\pi_{c0}\frac{r}{1-r}<\infty, \\ \sum\limits_{k = 1}^\infty\pi_{k2}& = &\sum\limits_{k = 1}^c\pi_{k2}+\pi_{c2}\sum\limits_{k = c+1}^\infty\rho^{k-c}+\frac{\theta r}{c\mu_b(1-r)}\pi_{c0}\sum\limits_{k = c+1}^\infty\sum\limits_{m = 0}^{k-c-1}r^m\rho^{k-c-1-m}\\ & = &\sum\limits_{k = 1}^c\pi_{k2}+\pi_{c2}\frac{\rho}{1-\rho}+\frac{\theta r}{c\mu_b(1-r)}\pi_{c0}\frac{1}{\rho-r}\left(\frac{\rho}{1-\rho}-\frac{r}{1-r}\right)<\infty. \end{eqnarray*} $

由Weierstrass判别法, $ \sum\limits_{k = 1}^\infty F_{k0}(x) $及$ \sum\limits_{k = 1}^\infty F_{k2}(x) $在$ (0, \infty) $上一致收敛.证毕.

运用标准方法容易得到以下稳态联合分布$ F_{kj}(x) $满足的矩阵微分方程组

(3.1) $ \begin{equation} \left\{ \begin{array}{ll} { }\sigma\frac{{\rm d}F_{00}(x)}{{\rm d}x} = -(\lambda+\theta)F_{00}(x)+\mu_vF_{10}(x)+\mu_bF_{12}(x), \\ { }\sigma_3\frac{{\rm d}F_{01}(x)}{{\rm d}x} = \theta F_{00}(x)-\lambda F_{01}(x), \\ { }\sigma_1\frac{{\rm d}F_{k0}(x)}{{\rm d}x} = -(\lambda+\theta+k\mu_v)F_{k0}(x)+\lambda F_{k-1, 0}(x)+(k+1)\mu_vF_{k+1, 0}(x), 1\leq k\leq c-1, \\ { }\sigma_2\frac{{\rm d}F_{12}(x)}{{\rm d}x} = \lambda F_{01}(x)+\theta F_{10}(x)-(\lambda+\mu_b)F_{12}(x)+2\mu_bF_{22}(x), \\ { }\sigma_2\frac{{\rm d}F_{k2}(x)}{{\rm d}x} = -(\lambda+k\mu_b)F_{k2}(x)+\lambda F_{k-1, 2}(x)+(k+1)\mu_bF_{k+1, 2}(x), 2\leq k\leq c-1, \\ { }\sigma_1\frac{{\rm d}F_{k0}(x)}{{\rm d}x} = -(\lambda+\theta+c\mu_v)F_{k0}(x)+\lambda F_{k-1, 0}(x)+c\mu_vF_{k+1, 0}(x), k\geq c, \\ { }\sigma_2\frac{{\rm d}F_{k2}(x)}{{\rm d}x} = -(\lambda+c\mu_b)F_{k2}(x)+\lambda F_{k-1, 2}(x)+c\mu_bF_{k+1, 2}(x), k\geq c, \end{array} \right. \end{equation} $

且满足边界条件为

$ F_{00}(0) = a, F_{01}(0) = 0, F_{kj}(0) = 0, k\geq1, j = 0, 1, \quad F_{kj}(\infty) = \pi_{kj}, (k, j)\in\Omega. $

其中概率$ a = F_{00}(0) = P\{X = 0, L = 0, J = 0\} $称为缓冲器的空库概率.

为了方便计算, 记向量

$ \mathit{\boldsymbol{F}}_k(x) = (F_{k0}(x), F_{k2}(x)), k\geq1, \quad \mathit{\boldsymbol{{\cal F}}}(x) = (F_{00}(x), F_{01}(x), \mathit{\boldsymbol{F}}_1(x), \mathit{\boldsymbol{F}}_2(x), \cdots), $

则微分方程组(3.1)可以写成下面的矩阵形式

(3.2) $ \begin{equation} \frac{\rm d}{{\rm d}x}\mathit{\boldsymbol{{\cal F}}}(x)\mathit{\boldsymbol{\Lambda}} = \mathit{\boldsymbol{{\cal F}}}(x)\mathit{\boldsymbol{Q}}, \end{equation} $

其中$ \mathit{\boldsymbol{\Lambda}} = {\rm diag}((\sigma, \sigma_3), (\sigma_1, \sigma_2), (\sigma_1, \sigma_2), \cdots) $, $ \mathit{\boldsymbol{Q}} $是拟生灭过程$ \{L(t), J(t), t\geq0\} $的无穷小生成元.

直接用边界条件来解矩阵方程(3.2)是非常困难的, 从而引入稳态联合分布$ F_{kj}(x) $和库存量稳态概率分布$ F(x) $的LT变换如下

$ \hat{F}_{kj}(s) = \int_{0}^{+\infty}e^{-sx}F_{kj}(x)\, {\rm d}x, s>0, (k, j)\in\Omega, \quad\hat{F}(s) = \int_{0}^{+\infty}e^{-sx}F(x)\, {\rm d}x, s>0. $

记

$ \mathit{\boldsymbol{\hat{F}}}_k(s) = (\hat{F}_{k0}(s), \hat{F}_{k2}(s)), k\geq1, \quad\mathit{\boldsymbol{{\hat{\cal F}}}}(s) = ((\hat{F}_{00}(s), \hat{F}_{01}(s)), \mathit{\boldsymbol{\hat{F}}}_1(s), \mathit{\boldsymbol{\hat{F}}}_2(s), \cdots). $

对矩阵方程(3.2)两边进行Laplace变换, 可得

$ \left(-\mathit{\boldsymbol{{\cal F}}}(0)+s\mathit{\boldsymbol{{\hat{\cal F}}}}(s)\right)\mathit{\boldsymbol{{\Lambda}}} = \mathit{\boldsymbol{{\hat{\cal F}}}}(s)\mathit{\boldsymbol{Q}}. $

利用边界条件, 整理可得

(3.3) $ \begin{equation} \mathit{\boldsymbol{{\hat{\cal F}}}}(s)(\mathit{\boldsymbol{Q}}-s\mathit{\boldsymbol{{\Lambda}}}) = -\mathit{\boldsymbol{{\cal F}}}(0)\mathit{\boldsymbol{{\Lambda}}} = (-a\sigma, {\bf 0}, {\bf 0}, \cdots). \end{equation} $

为了导出流体模型的稳态分布及性能指标, 引入下面两个二次方程.

引理3.1  若驱动系统的负载$ \rho<1 $, 对任意$ s\geq0 $, 二次方程$ c\mu_vz^2-(\lambda+\theta+c\mu_v+s\sigma_1)z+\lambda = 0 $存在两个实根$ \gamma_0(s) $和$ \gamma_1(s) $, 且

(3.4) $ \begin{equation} \gamma_0(s)(\gamma_1(s)) = \frac{(\lambda+\theta+c\mu_v+s\sigma_1)\mp\sqrt{(\lambda+\theta+c\mu_v+s\sigma_1) ^2-4\lambda c\mu_v}}{2c\mu_v}, \end{equation} $

易得$ 0<\gamma_0(s)<1 $, $ \gamma_1(s)>1 $, $ \gamma_0(0) = r $, 这里的$ r $如$ (2.1) $式中所示.

引理3.2  若驱动系统的负载$ \rho<1 $, 对任意$ s\geq0 $, 二次方程$ c\mu_bz^2-(\lambda+c\mu_b+s\sigma_2)z+\lambda = 0 $存在两个实根$ \beta_0(s) $和$ \beta_1(s) $, 且

(3.5) $ \begin{equation} \beta_0(s)(\beta_1(s)) = \frac{(\lambda+c\mu_b+s\sigma_2)\mp\sqrt{(\lambda+c\mu_b+s\sigma_2)^2-4\lambda c\mu_b}}{2c\mu_b} \end{equation} $

易得$ 0<\beta_0(s)<1 $, $ \beta_1(s)\geq1 $, $ \beta_0(0) = \frac{\lambda}{c\mu_b} = \rho $, $ \beta_1(0) = 1 $.

对于任意$ s\geq0 $, 引入矩阵函数

$ \mathit{\boldsymbol{A}}(s) = \left(\begin{array}{cc} -(\lambda+\theta+c\mu_v+s\sigma_1){\quad} &\theta\\0{\quad} &-(\lambda+c\mu_b+s\sigma_2) \end{array}\right), $

可以得到如下引理.

引理3.3  对于任意$ s>0 $, 二次矩阵方程R²B+RA(s)+C = 0有最小非负解

(3.6) $ \begin{equation} \mathit{\boldsymbol{R}}(s) = \left(\begin{array}{cc} \gamma_0(s){\quad}&{ }\frac{\theta\gamma_0(s)}{c\mu_b(\beta_1(s)-\gamma_0(s))}\\0{\quad}&\beta_0(s) \end{array}\right), \end{equation} $

这里$ \gamma_0(s) $, $ \beta_0(s) $和$ \beta_1(s) $见$ (3.4) $和$ (3.5) $, 从方程$ (2.1) $可以知道R(0)=R.

引理3.3的证明见文献[8].

为了得到库存量及状态的稳态联合分布函数的LT, 构造三个函数序列

(3.7) $ \begin{equation} \left\{ \begin{array}{ll} \psi_0(s) = 1, \\ { }\psi_k(s) = 1-\frac{\gamma_0(s)}{\rho}+\frac{(c-k)\mu_v}{\lambda}\psi_{k-1}(s) +\frac{\theta+s\sigma_1}{\lambda}\sum\limits_{m = 0}^{k-1}\psi_m(s), 1\leq k\leq c-1. \end{array} \right. \end{equation} $

(3.8) $ \begin{equation} \left\{ \begin{array}{ll} \eta_0(s) = 1, \\ { } \eta_1(s) = \frac{\lambda+(c-1)\mu_b+s\sigma_2-c\mu_b\beta_0(s)}{\lambda}, \\ { } \eta_k(s) = \frac{\lambda+(c-k)\mu_b+s\sigma_2}{\lambda}\eta_{k-1}(s)-\frac{c-k-1}{\lambda}\eta_{k-2}(s), 2\leq k\leq c-1. \end{array} \right. \end{equation} $

(3.9) $ \begin{equation} \left\{ \begin{array}{ll} { }\varphi_0(s) = 0, \\ { }\varphi_1(s) = \frac{\theta\beta_1(s)}{\lambda(\beta_1(s)-\gamma_0(s))}, \\ { }\varphi_k(s) = \frac{\lambda+(c-k)\mu_b+s\sigma_2}{\lambda}\varphi_{k-1}(s)-\frac{(c-k-1)\mu_b}{\lambda}\varphi_{k-2}(s)+\frac{\theta}{\lambda}\psi_{k-1}(s), 2\leq k\leq c-1. \end{array} \right. \end{equation} $

可以得到如下定理.

定理3.2  若$ \rho<1 $和$ d<0 $, 库存量及状态的稳态联合分布函数的LT为

(3.10) $ \begin{equation} \left\{ \begin{array}{ll} \hat{F}_{k0}(s) = K_0(s)\psi_{c-1-k}(s), 0\leq k\leq c-1, \\ { }\hat{F}_{01}(s) = K_0(s)\frac{\theta}{\lambda+s\sigma_3}\psi_{c-1}(s), \\ \hat{F}_{k2}(s) = K_2(s)\eta_{c-1-k}(s)-K_0(s)\varphi_{c-1-k}(s), 1\leq k\leq c-1, \\ \hat{F}_{k}(s) = (K_0(s), K_2(s))\mathit{\boldsymbol{R}}^{k-c+1}(s), k\geq c, \end{array} \right. \end{equation} $

其中

$ \begin{eqnarray*} &&K_0(s) = \frac{-a\sigma[(\lambda+\mu_b+s\sigma_2)\eta_{c-2}(s)-2\mu_b\eta_{c-3}(s)]}{H(s)}, \\ &&K_2(s) = \frac{a\sigma[\frac{\lambda\theta}{\lambda+s\sigma_3}\psi_{c-1}(s)+\theta\psi_{c-2}(s)-(\lambda+\mu_b+s\sigma_2)\varphi_{c-2}(s)+2\mu_b\varphi_{c-3}(s)]}{H(s)}, \\ &&H(s) = [(\lambda+\mu_b+s\sigma_2)\eta_{c-2}(s)-2\mu_b\eta_{c-3}(s)][\mu_v\psi_{c-2}(s)-(\lambda+\theta+s\sigma)\psi_{c-1}(s)+\mu_b\varphi_{c-2}(s)]\\ &&{\qquad}{\qquad}+\mu_b\eta_{c-2}(s)\left[\frac{\lambda\theta}{\lambda+s\sigma_3}\psi_{c-1}(s)+\theta\psi_{c-2}(s)-(\lambda+\mu_b+s\sigma_2)\varphi_{c-2}(s)+2\mu_b\varphi_{c-3}(s)\right]. \end{eqnarray*} $

证  矩阵方程(3.3)可以写成下面的形式

(3.11) $ \begin{equation} \left\{ \begin{array}{ll} -(\lambda+\theta+s\sigma)\hat{F}_{00}(s)+\mu_v\hat{F}_{10}(s)+\mu_b\hat{F}_{12}(s) = -a\sigma, \\ \theta\hat{F}_{00}(s)-(\lambda+s\sigma_3)\hat{F}_{01}(s) = 0, \\ \lambda\hat{F}_{01}(s)+\theta\hat{F}_{10}(s)-(\lambda+\mu_b+s\sigma_2)\hat{F}_{12}(s)+2\mu_b\hat{F}_{22}(s) = 0, \\ \lambda\hat{F}_{k-1, 0}(s)-(\lambda+k\mu_v+\theta+s\sigma_1)\hat{F}_{k0}(s)+(k+1)\mu_v\hat{F}_{k+1, 0}(s) = 0, 1\leq k\leq c-1, \\ \lambda\hat{F}_{k-1, 2}(s)+\theta\hat{F}_{k0}(s)-(\lambda+k\mu_b+s\sigma_2)\hat{F}_{k2}(s)+(k+1)\mu_b\hat{F}_{k+1, 2}(s) = 0, 2\leq k\leq c-1, \\ \mathit{\boldsymbol{\hat{F}}}_{k-1}(s)\mathit{\boldsymbol{C}}+\mathit{\boldsymbol{\hat{F}}}_k(s)\mathit{\boldsymbol{A}}(s)+\mathit{\boldsymbol{\hat{F}}}_{k+1}(s)\mathit{\boldsymbol{B}} = 0, k\geq c. \end{array} \right. \end{equation} $

方程组各式依次记为(3.11.1)$ - $(3.11.6)式.

如果$ \rho<1 $和$ d<0 $成立, 三维马尔可夫过程$ \{X(t), L(t), J(t), t\geq0\} $有唯一的平稳概率分布$ \{F_{kj}(x), (k, j)\in\Omega\} $$ ^{[9]} $.因此, 上述方程组(3.11)存在唯一解.下面只需验证(3.10)式满足(3.11)式.

首先, 对于$ k\geq c $, 把$ \mathit{\boldsymbol{{\hat{\cal F}}}}_k(s) = \mathit{\boldsymbol{{\hat{\cal F}}}}_{c-1}(s)\mathit{\boldsymbol{R}}^{k-c+1}(s) $代入(3.11.6)式中, 得到

$ \mathit{\boldsymbol{\hat{F}}}_{k-1}(s)\mathit{\boldsymbol{C}}+\mathit{\boldsymbol{\hat{F}}}_k(s)\mathit{\boldsymbol{A}}(s)+\mathit{\boldsymbol{\hat{F}}}_{k+1}(s)\mathit{\boldsymbol{B}} = \mathit{\boldsymbol{\hat{F}}}_{c-1}(s)\mathit{\boldsymbol{R}}^{k-c}(s)[\mathit{\boldsymbol{C}}+\mathit{\boldsymbol{R}}(s)\mathit{\boldsymbol{A}}(s)+\mathit{\boldsymbol{R}}^2(s)\mathit{\boldsymbol{B}}] = 0. $

记$ \mathit{\boldsymbol{\hat{F}}}_{c-1}(s) = (\hat{F}_{c-1, 0}(s), \hat{F}_{c-1, 2}(s)) = (K_0(s), K_2(s)) $, 得

$ \mathit{\boldsymbol{\hat{F}}}_k(s) = (K_0(s), K_2(s))\mathit{\boldsymbol{R}}^{k-c+1}(s), k\geq c. $

特殊地, 当$ k = c $时, 有

$ \hat{F}_{c0}(s) = K_0(s)\gamma_0(s), \quad\hat{F}_{c2}(s) = K_0(s)\frac{\theta\gamma_0(s)}{c\mu_b(\beta_1(s)-\gamma_0(s))}+K_2(s)\beta_0(s). $

把$ \hat{F}_{c0}(s) = K_0(s)\gamma_0(s) $代入(3.11.4)式中, 结合(3.7)式, 可得

$ \hat{F}_{k0}(s) = K_0(s)\psi_{c-1-k}(s), \quad 0\leq k\leq c-1. $

然后把$ \hat{F}_{00}(s) = K_0(s)\psi_{c-1}(s) $代入(3.11.2)式中, 得到$ \hat{F}_{01}(s) = K_0(s)\frac{\theta}{\lambda+s\sigma_3}\psi_{c-1}(s) $.

结合$ \hat{F}_{c2}(s) $的表达式和(3.11.5)式, 可求得

$ \hat{F}_{k2}(s) = K_2(s)\eta_{c-1-k}(s)-K_0(s)\varphi_{c-1-k}(s), 1\leq k\leq c-1. $

最后通过(3.11.1)和(3.11.3)式可以得到$ K_0(s) $和$ K_2(s) $的表达式.定理3.2证毕.

定理3.3  $ \sum\limits_{k = 1}^\infty F_{k0}(x) $和$ \sum\limits_{k = 1}^\infty F_{k2}(x) $可以逐项求Laplace变换, 即

$ \int_{0}^{+\infty}e^{-sx}\sum\limits_{k = 1}^\infty F_{k0}(x){\rm d}x = \sum\limits_{k = 1}^\infty\int_{0}^{+\infty}e^{-sx}F_{k0}(x){\rm d}x, $

$ \int_{0}^{+\infty}e^{-sx}\sum\limits_{k = 1}^\infty F_{k2}(x){\rm d}x = \sum\limits_{k = 1}^\infty\int_{0}^{+\infty}e^{-sx}F_{k2}(x){\rm d}x. $

证  设$ u_k(x) = e^{-sx}F_{k0}(x) $, 由$ \sum\limits_{k = 1}^\infty F_{k0}(x) $在$ (0, \infty) $上一致收敛, 可知$ \sum\limits_{k = 1}^\infty u_k(x) $在$ (0, \infty) $上一致收敛.

再设$ S(x) = \sum\limits_{k = 1}^\infty u_k(x), S_n(x) = \sum\limits_{k = 1}^n, r_n(x) = S(x)-S_n(x) $, 由一致收敛级数的性质1可知, $ S(x), r_n(x) $在$ (0, \infty) $上连续, 所以积分存在.

又$ S(x) = \sum\limits_{k = 1}^\infty u_k(x)<e^{-sx} $, 所以

$ \int_{0}^xS(t){\rm d}t<\int_{0}^xe^{-st}{\rm d}t = \frac{1}{s}(1-e^{-sx})<\frac{1}{s}, $

则$ S(x) $在$ (0, \infty) $有界.由无穷限广义积分审敛法, $ \int_{0}^{\infty}S(x){\rm d}x $收敛, 而$ r_n(x)<S(x) $, 所以有

$ \int_{0}^xr_n(t){\rm d}t<\int_{0}^xS(t){\rm d}t<\frac{1}{s}, $

从而$ r_n(x) $在$ (0, \infty) $上有界且收敛.

由级数的一致收敛性, 对于任给的$ \varepsilon $, 存在$ N = N(\varepsilon) $, 当$ n>N $时, 对于一切$ x\in(0, a) $, 都有

$ \left|r_n(x)\right|<\frac{\varepsilon}{a}. $

从而

$ \int_{0}^a\left|r_n(x)\right|{\rm d}x<\varepsilon. $

于是, 当$ n>N $时, 有

$ \left|\int_{0}^{\infty}S(x){\rm d}x-\int_{0}^{\infty}S_n(x){\rm d}x\right| = \left|\int_{0}^{\infty}r_n(x){\rm d}x\right|\leq\int_{0}^{\infty}\left|r_n(x)\right|{\rm d}x = \lim\limits_{a\to\infty}\int_{0}^a\left|r_n(x)\right|{\rm d}x<\varepsilon. $

根据极限的定义, 有

$ \int_{0}^{\infty}S(x){\rm d}x = \lim\limits_{n\to\infty}\int_{0}^{\infty}S_n(x){\rm d}x = \lim\limits_{n\to\infty}\sum\limits_{k = 1}^n\int_{0}^{\infty}u_k(x){\rm d}x. $

同理可知$ \sum\limits_{k = 1}^\infty F_{k2}(x) $也可以逐项求Laplace变换.证毕.

定理3.4  当$ d<0 $和$ \rho<1 $, 有平稳库存量的LST和空库概率及均值的表达式

$ \begin{eqnarray*} \tilde{F}(s)& = &sK_0(s)\sum\limits_{k = 0}^{c-2}\psi_{c-1-k}(s)+\frac{K_0(s)}{\sigma_2}\left[\theta\sum\limits_{k = 1}^{c-1} \psi_{c-1-k}(s)+\frac{\theta\gamma_0(s)}{\beta_1(s)-\gamma_0(s)} \right.\\ &&\left.-(\lambda+\theta+s\sigma)\psi_{c-1}(s)+\frac{\lambda\theta}{\lambda+s\sigma_3}\psi_{c-1}(s) +\mu_v\psi_{c-2}(s)\right] \\ &&+\frac{sK_0(s)}{1-\gamma_0(s)}+\frac{sK_0(s)\theta\gamma_0(s)}{(1-\gamma_0(s))(1-\beta_0(s))c\mu_b(\beta_1(s)-\gamma_0(s))} \\ && +K_2(s)\left(\frac{c\mu_b\beta_0(s)}{\sigma_2}+\frac{s\beta_0(s)}{1-\beta_0(s)}\right)+\frac{a\sigma}{\sigma_2}, \\ a& = &\frac{\sigma_2}{\sigma}\left[\frac{\theta(\psi_{c-1}+\psi_{c-2})-(\lambda+\mu_b)\eta_{c-2}+2\mu_b\eta_{c-3}}{H(0)} (c\mu_b-\lambda) \right.\\ &&\left.+\frac{(\lambda+\mu_b)\varphi_{c-2}-2\mu_b\varphi_{c-3}}{H(0)} \left(\theta\sum\limits_{k = 0}^{c-2}\psi_k-\lambda\psi_{c-1}+\mu_v\psi_{c-2}+\frac{\theta r}{1-r}\right)+1\right]^{-1}, \\ E(X)& = &K_0\left\{\frac{\lambda+\theta}{\sigma_2}\psi_{c-1}'(0)+\frac{\lambda+\theta+\sigma}{\sigma_2}\psi_{c-1}-\frac{\theta}{\sigma_2}\left[\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}'(0)-\frac{r\sigma_1}{c\mu_v(r_1-r)(1-r)} \right.\right.\\ &&\left.\left.+\theta\psi_{c-1}'(0)-\frac{\sigma_3}{\lambda}\psi_{c-1}-\frac{r}{(1-r)^2}\left(\frac{\sigma_2}{c\mu_b(1-\rho)} +\frac{r\sigma_1}{c\mu_v(r_1-r)}\right)\right] \right.\\ &&\left.-\mu_v\psi_{c-2}'(0)-\frac{1}{1-r}-\frac{\theta r}{(1-r)^2(1-\rho)c\mu_b}\right\} \\ &&+K_0'(0)\left[\frac{\theta}{\sigma_2}\left(\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}+\frac{r}{1-r}\right)+\frac{\mu_v}{\sigma_2}\psi_{c-2}-\frac{\lambda}{\sigma_2}\psi_c-1\right], \end{eqnarray*} $

其中$ K_0 = K_0(0) $, $ K_2 = K_2(0) $和$ \psi_k = \psi_k(0) $, $ \eta_k = \eta_k(0) $, $ \varphi_k = \varphi_k(0) $, $ 1\leq k\leq c-1 $.

证  将(3.11.5)式从$ k = 2 $到$ k = c-1 $相加, 得到

$ \lambda\sum\limits_{k = 2}^{c-1}\hat{F}_{k-1, 2}(s)+\theta\sum\limits_{k = 2}^{c-1}\hat{F}_{k0}(s)-\sum\limits_{k = 2}^{c-1}(\lambda+k\mu_b+s\sigma_2)\hat{F}_{k2}(s)+\sum\limits_{k = 2}^{c-1}(k+1)\mu_b\hat{F}_{k+1, 2}(s) = 0. $

经过移项合并之后, 得到

(3.12) $ \begin{equation} \theta\sum\limits_{k = 2}^{c-1}\hat{F}_{k0}(s)-s\sigma_2\sum\limits_{k = 2}^{c-1}\hat{F}_{k2}(s)+\lambda(\hat{F}_{12}(s)-\hat{F}_{c-1, 2}(s))-2\mu_b\hat{F}_{22}(s)+c\mu_b\hat{F}_{c2}(s) = 0. \end{equation} $

把(3.11.3)式代入到(3.12)式中, 可得

(3.13) $ \begin{equation} s\sigma_2\sum\limits_{k = 1}^{c-1}\hat{F}_{k2}(s) = \theta\sum\limits_{k = 1}^{c-1}\hat{F}_{k0}(s)+\lambda\hat{F}_{01}(s)-\mu_b\hat{F}_{12}(s)-\lambda\hat{F}_{c-1, 2}(s)+c\mu_b\hat{F}_{c2}(s). \end{equation} $

把(3.11.1)式和$ \hat{F}_{c2}(s) $的表达式代入(3.13)式, 可得

$ \begin{eqnarray*} \sum\limits_{k = 1}^{c-2}\hat{F}_{k2}(s) & = &\frac{K_0(s)}{s\sigma_2}\left[\theta\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}(s)+\frac{\lambda\theta}{\lambda+s\sigma_3}\psi_{c-1}(s)-(\lambda+\theta+s\sigma)\psi_{c-1}(s) \right.\\ &&\left.+\mu_v\psi_{c-2}(s)+\frac{\theta\gamma_0(s)}{\beta_1(s)-\gamma_0(s)}\right]+\frac{K_2(s)}{s\sigma_2}(c\mu_b\beta_0(s)-\lambda-s\sigma_2)+\frac{a\sigma}{s\sigma_2}. \end{eqnarray*} $

结合定理3.3, 进而平稳分布$ F(x) $缓冲器容量的LT为

$ \begin{eqnarray*} \hat{F}(s)& = &\hat{F}_{00}(s)+\hat{F}_{01}(s)+\sum\limits_{k = 1}^{\infty}\mathit{\boldsymbol{\hat{F}}}_k(s){\mathit{\boldsymbol{e}}}\\ & = &\sum\limits_{k = 0}^{c-2}\hat{F}_{k0}(s)+\sum\limits_{k = 1}^{c-2}\hat{F}_{k2}(s)+\sum\limits_{k = c-1}^{\infty}\mathit{\boldsymbol{\hat{F}}}_k(s){\mathit{\boldsymbol{e}}}+\hat{F}_{01}(s)\\ & = &K_0(s)\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}(s)+(K_0(s), K_2(s))(\mathit{\boldsymbol{I}}-\mathit{\boldsymbol{R}}(s))^{-1}{\mathit{\boldsymbol{e}}}+K_0(s)\frac{\theta}{\lambda+s\sigma_3}\psi_{c-1}(s)\\ &&+\frac{K_0(s)}{s\sigma_2}\left[\theta\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}(s)+\frac{\lambda\theta}{\lambda+s\sigma_3}\psi_{c-1}(s)-(\lambda+\theta+s\sigma)\psi_{c-1}(s)+\mu_v\psi_{c-2}(s)\right.\\ &&\left.+\frac{\theta\gamma_0(s)}{\beta_1(s)-\gamma_0(s)}\right]+\frac{K_2(s)}{s\sigma_2}(c\mu_b\beta_0(s)-\lambda-s\sigma_2)+\frac{a\sigma}{s\sigma_2}, \end{eqnarray*} $

其中e是二维单位列向量.

由(3.4)和(3.5)式知, R(s)特征值$ \gamma_0(s) $和$ \beta_0(s) $的绝对值小于1, 可得其谱半径SP[R(s)] <1, I-R(s)) 是可逆的, 并且

$ (\mathit{\boldsymbol{I}}-\mathit{\boldsymbol{R}}(s))^{-1} = \left(\begin{array}{cc} { } \frac{1}{1-\gamma_0(s)}{\quad} & { }\frac{\theta\gamma_0(s)}{(1-\gamma_0(s))(1-\beta_0(s))c\mu_b(\beta_1(s)-\gamma_0(s))}\\ 0{\quad}&{ }\frac{1}{1-\beta_0(s)} \end{array}\right). $

代入$ \hat{F}(s) $, 得到

$ \begin{eqnarray*} \hat{F}(s)& = &K_0(s)\sum\limits_{k = 0}^{c-2}\psi_{c-1-k}(s)+\frac{K_0(s)}{s\sigma_2}\left[\theta\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}(s)+\frac{\theta\gamma_0(s)}{\beta_1(s)-\gamma_0(s)} \right.\\ &&\left.-(\lambda+\theta+s\sigma)\psi_{c-1}(s)+\frac{\lambda\theta}{\lambda+s\sigma_3}\psi_{c-1}(s)+\mu_v\psi_{c-2}(s)\right] +\frac{K_0(s)}{1-\gamma_0(s)}\\ &&+\frac{K_0(s)\theta\gamma_0(s)}{(1-\gamma_0(s))(1-\beta_0(s))c\mu_b(\beta_1(s)-\gamma_0(s))} +K_2(s)\left(\frac{c\mu_b\beta_0(s)}{s\sigma_2}+\frac{\beta_0(s)}{1-\beta_0(s)}\right)+\frac{a\sigma}{s\sigma_2}. \end{eqnarray*} $

为了得到平稳库存量的均值, 引入$ F_{kj}(s), (k, j)\in\Omega $和$ F(x) $的LST, 记

$ \tilde{F}_{kj}(s) = \int_{0}^{+\infty}e^{-sx}\, {\rm d}F_{kj}(x), (k, j)\in\Omega, \quad\tilde{F}_(s) = \int_{0}^{+\infty}e^{-sx}\, {\rm d}F(x). $

易得$ \tilde{F}_(s) = s\hat{F}(s) $.

进而得到缓冲器库存量稳态分布的LST

(3.14) $ \begin{eqnarray} \tilde{F}(s)& = &sK_0(s)\sum\limits_{k = 0}^{c-2}\psi_{c-1-k}(s)+\frac{K_0(s)}{\sigma_2}\left[\theta\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}(s)+\frac{\theta\gamma_0(s)}{\beta_1(s)-\gamma_0(s)} \right.{}\\ &&\left.-(\lambda+\theta+s\sigma)\psi_{c-1}(s)+\frac{\lambda\theta}{\lambda+s\sigma_3}\psi_{c-1}(s)+\mu_v\psi_{c-2}(s)\right]+\frac{sK_0(s)}{1-\gamma_0(s)} {}\\ &&+\frac{sK_0(s)\theta\gamma_0(s)}{(1-\gamma_0(s))(1-\beta_0(s))c\mu_b(\beta_1(s)-\gamma_0(s))}+K_2(s)\left(\frac{c\mu_b\beta_0(s)}{\sigma_2}+\frac{s\beta_0(s)}{1-\beta_0(s)}\right)+\frac{a\sigma}{\sigma_2}. {}\\ \end{eqnarray} $

根据正规化条件$ \lim\limits_{s\to0}\tilde{F}(s) = 1 $, 可以得到空库概率$ a $的表达式

$ a = \frac{K_0}{\sigma}\left[\lambda\psi_{c-1}-\mu_v\psi_{c-2}-\theta\sum\limits_{k = 0}^{c-2}\psi_k+\frac{K_2}{\sigma}(c\mu_b\rho-\lambda)\right], $

进一步整理得

$ \begin{eqnarray*} a& = &\frac{\sigma_2}{\sigma}\left[\frac{\theta(\psi_{c-1}+\psi_{c-2})-(\lambda+\mu_b)\eta_{c-2}+2\mu_b\eta_{c-3}}{H(0)} (c\mu_b-\lambda) \right.\\ &&\left.+\frac{(\lambda+\mu_b)\varphi_{c-2}-2\mu_b\varphi_{c-3}}{H(0)}\left(\theta\sum\limits_{k = 0}^{c-2}\psi_k-\lambda\psi_{c-1}+\mu_v\psi_{c-2}+\frac{\theta r}{1-r}\right)+1\right]^{-1}. \end{eqnarray*} $

另一方面, 根据(3.5)式, 可以得到关系式

$ \left\{ \begin{array}{ll} c\mu_b[\beta_0(s)+\beta_1(s)] = \lambda+c\mu_b+s\sigma_2, \\ c\mu_b\beta_0(s)\beta_1(s) = \lambda. \end{array} \right. $

由$ \beta_0(0) = \rho $和$ \beta_1(0) = 1 $, 可得

(3.15) $ \begin{equation} \beta_0'(0) = \frac{-\rho\sigma_2}{c\mu_b(1-\rho)}, {\quad} \beta_1'(0) = \frac{\sigma_2}{c\mu_b(1-\rho)}. \end{equation} $

同样地, 根据(3.4)式可以得到

$ \left\{ \begin{array}{ll} c\mu_v[\gamma_0(s)+\gamma_1(s)] = \lambda+c\mu_v+\theta+s\sigma_1, \\ c\mu_v\gamma_0(s)\gamma_1(s) = \lambda. \end{array} \right. $

由$ \gamma_0(0) = r $, 可得

(3.16) $ \begin{equation} \gamma_0'(0) = \frac{-r\sigma_1}{c\mu_v(r_1-r)}, {\quad} \gamma_1'(0) = \frac{r_1\sigma_1}{c\mu_v(r^1-r)}. \end{equation} $

其中$ r_1 = \gamma_1(0) $由方程(3.4)来决定.

众所周知

$ E(X) = -\lim\limits_{s\to0}\frac{\rm d}{{\rm d}s}\tilde{F}(s). $

由此可得

$ \begin{eqnarray*} E(X)& = &K_0\left\{-\frac{\theta}{\sigma_2}\left[\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}'(0)+\frac{\gamma_0'(0)(1-r)-r(\beta_1'(0)-\gamma_0'(0))}{(1-r)^2}+\psi_{c-1}'(0)-\frac{\sigma_3}{\lambda}\psi_{c-1}\right]\right.\\ &&\left.+\frac{\lambda+\theta}{\sigma_2}\psi_{c-1}'(0)+\frac{\lambda+\theta+\sigma}{\sigma_2}\psi_{c-1}-\mu_v\psi_{c-2}'(0)-\frac{1}{1-r}-\frac{\theta r}{(1-r)^2(1-\rho)c\mu_b}\right\}\\ &&+K_0'(0)\left[\frac{\theta}{\sigma_2}\left(\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}+\frac{r}{1-r}\right)+\frac{\mu_v}{\sigma_2}-\frac{\lambda}{\sigma_2}\psi_{c-1}\right]\\ &&-K_2\left[\frac{\rho}{1-\rho}+\frac{c\mu_b\beta_0'(0)}{\sigma_2}\right]-K_2'(0)\frac{c\mu_b\beta_0(s)-\lambda}{\sigma_2}, \end{eqnarray*} $

其中$ K_0'(0) = K_0'(s)|_{s = 0} $, $ K_2'(0) = K_2'(s)|_{s = 0} $, $ \psi_k'(0) = \psi_k'(s)|_{s = 0}, 1\leq k\leq c-1 $.

最后, 结合(3.15)和(3.16)式, 得到流体模型平稳库存量的均值表达式

$ \begin{eqnarray*} E(X)& = &K_0\left\{\frac{\lambda+\theta}{\sigma_2}\psi_{c-1}'(0) +\frac{\lambda+\theta+\sigma}{\sigma_2}\psi_{c-1}-\frac{\theta}{\sigma_2}\left[\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}'(0) -\frac{r\sigma_1}{c\mu_v(r_1-r)(1-r)}\right.\right.\\ &&\left.+\theta\psi_{c-1}'(0)-\frac{\sigma_3}{\lambda}\psi_{c-1}-\frac{r}{(1-r)^2}\left(\frac{\sigma_2}{c\mu_b(1-\rho)} +\frac{r\sigma_1}{c\mu_v(r_1-r)}\right)\right]\\ &&\left.-\mu_v\psi_{c-2}'(0)-\frac{1}{1-r}-\frac{\theta r}{(1-r)^2(1-\rho)c\mu_b}\right\}\\ &&+K_0'(0)\left[\frac{\theta}{\sigma_2}\left(\sum\limits_{k = 1}^{c-1}\psi_{c-1-k}+\frac{r}{1-r}\right)+\frac{\mu_v}{\sigma_2}\psi_{c-2}-\frac{\lambda}{\sigma_2}\psi_c-1\right], \end{eqnarray*} $

其中$ K_0 = K_0(0) $, $ K_2 = K_2(0) $和$ \psi_k = \psi_k(0) $, $ \eta_k = \eta_k(0) $, $ \varphi_k = \varphi_k(0) $, $ 1\leq k\leq c-1 $.定理3.4证毕.

在无线Mesh网络中, 假设信息节点数为3时, 也就是$ c = 3 $时, 当$ \theta = 0.2, \mu_b = 10, $$ \sigma_1 = 10, \sigma_2 = 20, \sigma_3 = 5 $时, 系统的平均有效带宽随传输速率$ \mu_v $和到达率$ \lambda $的变化如图 1和图 2所示.

观察图 1, 设节点到达率$ \lambda = 5 $, 由图像可以看出, 当$ \mu_v $一定时, $ E(X) $随$ \sigma $的增大而减小; 当$ \sigma $一定时, $ E(X) $随传输率$ \mu_v $的增大而增大.观察图 2, 设收到干扰时的传输率$ \mu_v = 5 $, 从图像中可以看出, 当$ \lambda $一定时, $ E(X) $随$ \sigma $的增大而减小; 当$ \sigma $一定时, $ E(X) $随到达率$ \lambda $增大而减小.由此可见, 若想提高无线Mesh的网络性能, 必须设置好合适的信道传输率和节点到达率, 从而有效提高平均有效带宽以及网络利用率, 降低能耗.

本文讨论了一种由单重工作休假排队系统驱动的流体模型, 构建了净输入率结构, 运用LT的方法进行矩阵分析, 得到了稳态库存量的空库概率以及均值表达式, 进一步完善流体模型理论.最后将分析结果运用到多信道无线Mesh网中, 运用Matlab软件进行数值分析, 通过调节参数提高网络性能.本文所得结果为今后多服务台驱动系统的流体模型研究提供了一定的理论基础, 更好地应用到通信、交通、能源等各个领域中.