这是一份 Imperial帝国理工大学 MATH97119作业代写的成功案例
where $P_{Q}$ denotes the probability that all $m$ servers are busy and $p(0)$ denotes the probability that the queue length is 0 . We have
$$
P_{Q}=\left(m \cdot \rho^{\prime}\right)^{m} \frac{p(0)}{m !} \cdot \frac{1}{1-\rho^{\prime}}
$$
$$
p(0)=\left[\frac{\left(m \cdot \rho^{\prime}\right)^{m}}{m !\left(1-\rho^{\prime}\right)}+\left(\sum_{i=0}^{m-1} \frac{\left(m \cdot \rho^{\prime}\right)^{\prime}}{i !}\right)\right]^{-1}
$$
Average population: $n^{\prime}=P_{Q} \cdot \frac{\rho^{\prime}}{\left(1-\rho^{\prime}\right)}$
MATH97119 COURSE NOTES ：
$$
V_{i, j}{ }^{m}\left(t, X_{i, j}{ }^{w}\right)=1-\frac{1}{\mu_{i, j}\left(X_{i, j}{ }^{w x}\right)} \times e^{\frac{1}{\mu_{v}\left(X_{i, j}{ }^{n \prime}\right)}} .
$$
Note that $\mu_{i j}\left(X_{i, j}{ }^{w s}\right.$ denotes the service rate of $w s_{i, f}$
$$
\mu_{i, j}\left(X_{i, j}{ }^{k x}\right)=\frac{1}{t_{i, j}\left(\mu_{i, j}\left(X_{i, j}{ }^{* x}\right)\right)}
$$