# 佇列系統期末整理

::: spoiler 目錄
[TOC]
:::

## Birth-Death Process(BDP)

* 在 CTMC 的基礎上定義出來的一種 process

* 基礎定義：
1. $P_j(t) = P(X(t) = j), t \geq 0, j \geq 0, X(0) = r, r \geq 0$，其中 $X(0) = r$ 代表 initial state 在 $r$。
2. $P_r(0) = P(X(0) = r) = 1$ (initial state probability)
3. $P_{i,k}(s,t) = P(X(t) = k | P(s) = i), \ 0 \leq s \leq t < \infty$ (同前面 DTMC 和 CTMC 定義)
4. $P_{i,k}(t) = P_{i,k}(0, t) = P(X(t) = k | P(0) = i)$ (time homogeneous)

* 討論間隔時間為極短時 ($h \rightarrow 0^+$，可假設只會發生一種事件)：
1. if $| k - i | \geq 2, P_{i,k}(t,t+h) = o(h)$，也就是發生改變兩個狀態值的機率極低。
2. if $| k - i | \leq 1, P_{i,k}(t,t+h) =$
\begin{cases}
[a_i(t) + \kappa(t)]*h + o(h), k = i + 1, i \geq 0\\
1 - [a_i(t) + b_i(t) + \kappa(t)]*h + o(h), k = i, i > 0\\
1 - [a_0(t) + \kappa(t)]*h + o(h), k = i = 0\\
[b_i(t)]*h + o(h), k = i - 1, i \geq 1
\end{cases}
* $a(t)$ 代表 birth rate
* $b(t)$ 代表 death rate
* $\kappa(t)$ 代表 immigration rate (immegrate in)，底下的部分假設這個數值為 $0$。

### time-homogeneous BDP

* $a_i(t) \equiv a_i,\ b_i(t) \equiv b_i$

* transitional functions:
\begin{cases}
P^\prime_{i, 0}(t) = -a_0P_{i,0}(t) + b_1P_{i,1}(t) \\
P^\prime_{i, k}(t) = -(a_k + b_k)P_{i,k}(t) + a_{k-1}P_{i,k-1}(t) + b_{k+1}P_{i,k+1}(t), k > 1
\end{cases}

* **sojourn time**(waiting time $d_k$ for change of state from state $k$) is **exponentially distributed**, $d_k = a_k + b_k$
* 並且這個數值與到達 state $k$ 前的路徑無關
* 在 state $k$ 時，前往 $k + 1$ 的機率 $p_k = \frac{a_k}{d_k}$，前往 $k - 1$ 的機率 $q_k = \frac{b_k}{d_k}$
* 對於 state $0$ 來說，一開始的 $p_0 = 1$；再次抵達 state $0$ 後，$q_0 = 1$，也就是不再離開 state $0$。

* If $p_0 = 1, 0 < p_k < 1, k \geq 1$
1. $P_k(t)$ 滿足下列 ODE:
\begin{cases}
P^\prime_0(t) = -a_0P_0(t) + b_1P_1(t) \\
P^\prime_{i, k}(t) = -(a_k + b_k)P_k(t) + a_{k-1}P_{k-1}(t) + b_{k+1}P_{k+1}(t)
\end{cases}
2. 對於所有 $k \geq 0$，極限 $\lim_{t \rightarrow \infty} P_k(t) = \pi_k$ 存在，而且與 initial distribution 無關。(limiting distribution)
* 如果 $\lim^\infty_{k = 0} \rho_k < \infty$，其中 $\rho_0 = 1, \rho_k = \frac{a_0 a_1 \cdots a_{k-1}}{b_1 b_2 \cdots b_k}, k \geq 1$，那 $\pi_k > 0, k \geq 0$ 而且符合下式
\begin{cases}
\pi_0 = [\sum^\infty_{j=0} \rho_j]^{-1} \\
\pi_k = \rho_k \pi_0
\end{cases}
* 如果 $\lim^\infty_{k = 0} \rho_k = \infty$，那 $\pi_k = 0, k \geq 0$。 (limiting distribution 不存在)

* balance equation
* local balance equation(只討論單邊的鄰居)
\begin{cases}
-a_kP_k + b_{k+1}P_{k+1} = 0 \\
-b_kP_k + a_{k-1}P_{k-1} = 0
\end{cases}
* global balance equation(所有鄰居都要討論)
$-(a_k + b_k)P_k + a_{k-1}P_{k-1} + b_{k+1}P_{k+1} = 0$

## Definition of Queueing system

* Notation
* $(A/B/c/d/e/x)$

* Arrival process($A$)
* 使用 **intearrival time** 的 distribution 描述，並且假設各段 interarrival time 互相獨立。
* 可能受到顧客數量影響
* Service process($B$)
* 用來描述 server 服務 customer 的 process
* 在基礎的 queueing model 內假設是 i.i.d random variables
* 可能受到顧客數量影響

* 對於 arrival process 和 service process 可能有以下幾種參數：
* $M$: Memoryless，也就是 exponential distribution
* $E_r$: order-r Erlang distribution
* $H_r$: order-r hyperexponential distribution
* $D$: deterministic (固定不變的常數)
* $G$ or $GI$: 某個 i.i.d 的 distribution

* System structure
* $c$: server 的數量有幾個
* $d$: System capacity，也就是整個 queueing system 可以容納多少顧客(包含 queue size 和 server count)。(default value: $\infty$)
* 如果 $c = d$，代表這個 queueing system 內沒有任何 waiting room(buffer)。
* 如果 $c < d$，顧客從開始進入 queueing system 到開始被服務的時間稱為 waiting time。
* $e$: 顧客的總數量 (default value: $\infty$)

* Service displine($x$)
* server 是如何服務 customer 的，預設是 FIFO(FCFS)。


### Work-conserving property

* 如果顧客進入 queueing system 時有空閒的 server，這個顧客會立即被服務。

### Customer loss probability

* $q = \lim_{n \rightarrow \infty} \frac{m_n}{n}$，$m_n$ 代表這 $n$ 個顧客中沒成功進入 system 的顧客數量。

### Waiting time distribution

* $F(x)$: waiting time distribution 的 CDF
\begin{cases}
F(x) = \lim_{n \rightarrow \infty} F_n(x), \forall x > 0 \\
F_n(x) = \frac{1}{n} \sum^n_{n = 1} \mathcal{J}_{w_i \leq x}\ , x > 0
\end{cases}
其中 $w_i$ 是第 $i$ 位顧客的 waiting time。

* Mean waiting time
* $W^{(1)} = \lim_{n \rightarrow \infty} \frac{1}{n} (W_1 + \cdots + W_n)$，此時的 $W^{(1)}$ 就是 mean waiting time。
* $W^{(k)} = \lim_{n \rightarrow \infty} \frac{1}{n} (W_1^k + \cdots + W_n^k)$，這個算法是計算更高維度的 mean waiting time。

* Busy period
* $[a_n, b_n), a_n < b_n < a_{n+1}, n \geq 1$
* idle period: $[b_n, a_n+1)$
* 這寫法代表 第 $n$ 次的 busy period

* Queue Length distribution
* $L(t), t \geq 0$: 在時間 $t$ 時 system 內的顧客數量。
* $\bar{L}_k(t)$: 在時間 $(0, t)$ 內，有 $k$ 個顧客的時間比例。
* $p_k = \lim_{t \rightarrow \infty} \bar{L}_k(t), k \geq 0$

### Little's law(formula)

* 系統中平均顧客數 = arrival rate * 平均顧客在系統內的時間
* $L = \lambda * T$

* 這個算法與以下條件無關：
* 系統的定義
* Arrival(Service) time distribution
* Server 數量
* Waiting room(buffer) 大小
* Service discipline


* 其他變種
* 只針對 Queue 來看
* 只針對 Server 來看

### Poisson Arrivals See time Average(PASTA)

* 在穩定態時，在 state $k (k > 0)$ 的機率相等於在一個 customer arrives 時看到 system 在 state $k$ 的機率。
* 用數學式來表示為 $\lim_{t \rightarrow \infty}P(L(t) = k) = \lim_{t \rightarrow \infty}P(L(t) = k | E_k(t)), k \geq 0$
* 其中 $L(t)$ 為在 $t$ 時間時 system 內的顧客數，$E_k(t)$ 則是在時間 t 時發生了 arrival。

## M/M/1 queue
* 1個 server，system size 為無限
* arrival process: poisson process with rate $\lambda$
* service process: exponentially distributed with rate $\mu$
* service disciplne: FCFS
* **work-conserving** service process


### state transition probability
* $P_{i, j}(t) = P(N(t) = j | N(0) = i)$, $\Delta$ 值極小

\begin{cases}
p_{i, i + 1}(\Delta) = \lambda * \Delta + o(\Delta), i = 0. 1, \cdots \\
p_{i, i - 1}(\Delta) = \mu * \Delta + o(\Delta), i = 1, 2, \cdots \\
p_{i, i}(\Delta) = 1 - (\lambda + \mu) * \Delta + o(\Delta), i = 1, 2, \cdots \\
p_{0, 0}(\Delta) = 1 - \lambda * \Delta + o(\Delta)
\end{cases}

### global balance equation

\begin{cases}
\lambda p_{n-1} + \mu p_{n+1} - (\lambda + \mu) p_n = 0, n \geq 1 \\
\mu p_1 - \lambda p_0 = 0, n = 0
\end{cases}

### 迭代法解

* 目標是求得 $p_0$ 和 $p_n$ 的 close form

$\mu p_1 - \lambda p_0 = 0 \Rightarrow p_1 = \frac{\lambda}{\mu}p_0$
$\lambda p_0 + \mu p_2 - (\lambda + \mu) p_1 = 0$

\begin{aligned}
p_2 &= \frac{1}{\mu}((\lambda + \mu) p_1 - \lambda p_0) \\
&= \frac{1}{\mu}((\lambda + \mu) \frac{\lambda}{\mu}p_0 - \lambda p_0) \\
&= \frac{1}{\mu}(\frac{\lambda^2}{\mu}p_0) \\
&= \frac{\lambda^2}{\mu^2}p_0
\end{aligned}

$p_n = (\dfrac{\lambda}{\mu})^np_0$

* 定義 $\rho \equiv \dfrac{\lambda}{\mu}$ 為 ultilization factor(rate)

$\displaystyle \sum^\infty_{n = 0} \rho^n p_0 = 1 \\
\Rightarrow \dfrac{1}{1-\rho}p_0 = 1,\ \text{if}\ \rho < 1 \\
\Rightarrow p_0 = 1 - \rho \\
\Rightarrow p_n = \rho^n (1 - \rho)$

### 透過 generating function 解

* 定義 $P(z) = \displaystyle \sum^\infty_{n = 0} z^n p_n ,\ |z| \leq 1$ ($z$ 是複數平面上一點，在單位圓內)
\begin{cases}
(\lambda+\mu)p_n = \lambda p_{n-1} + \mu p_{n+1},\ n \geq 1 \\
\lambda p_0 = \mu p_1,\ n = 0
\end{cases}
$/ \mu$ 後寫成
\begin{cases}
\rho \cdot p_n = \rho \cdot p_{n-1} + p_{n+1},\ n \geq 1 \\
\rho \cdot p_0 = p_1,\ n = 0
\end{cases}
同乘上 $z^n$ 並將所有式子相加起來
$(\rho + 1) \displaystyle \sum^\infty_{n=1} z^n p_n + \rho \cdot p_0 = \rho \sum^\infty_{n = 1} z^n p_{n - 1} + \sum^\infty_{n = 0} z^n p_{n + 1}$
將 $\sum$ 的係數盡可能調成與 $P(z)$ 一致
$(\rho + 1) \displaystyle \sum^\infty_{n=0} z^n p_n - p_0 = \rho z\sum^\infty_{n = 1} z^{n-1} p_{n - 1} + \dfrac{1}{z} \sum^\infty_{n = 1} z^{n+1} p_{n + 1} \\
\Rightarrow (\rho + 1) P(z) - p_0 = \rho z P(z) + \dfrac{1}{z}(P(z) - p_0) \\
\Rightarrow (\rho z^2 - \rho z - z + 1)P(z) = (1-z) p_0 \\
\Rightarrow P(z) = \dfrac{p_0}{1 - \rho z}$
boundary condition: z = 1
$P(z = 1) = \displaystyle \sum^\infty_{n = 0} 1^n p_n = \sum^\infty_{n = 0} p_n = 1 = \dfrac{p_0}{1 - \rho} \\
\Rightarrow p_0 = 1 - \rho, P(z) = \dfrac{1 - \rho}{1 - \rho z},\ \rho < 1,\ |z| \leq 1$
$P(z) = \dfrac{1 - \rho}{1 - \rho z} = \displaystyle \sum^\infty_{n = 0} (1 - \rho) \rho^n z^n = \sum^\infty_{n = 0} z^n p^n \\
\Rightarrow p^n = (1 - \rho) \rho^n$

### long-term system

* mean system size $\bar{N} = \displaystyle \sum^\infty_{i=0} iP_i = \dfrac{\rho}{1 - \rho} = \dfrac{\lambda}{\mu - \lambda}$

* 因為 service time 是 exponential distirbution，剩餘的 service time (waiting)也是 exponential distribution

* mean system time $D(t) = \displaystyle \sum^{N(t)}_{i = 1}Y_i + Y$
* $Y_i$: service time of customer $i$
* $Y$: remaining service time

* $E(D(t)) = \bar{D} = E(Y_1) * \bar{N} + E(Y) = \dfrac{1}{\mu} \cdot \dfrac{\lambda}{\mu - \lambda} + \dfrac{1}{\mu} = \dfrac{1}{\mu - \lambda}$

### waiting time distribution

* $T_q$: random variable representing time spent waiting in queue
* $D_q(t)$: CDF of $T_q$

1. t = 0
$q(n) \equiv Pr\{\text{n in system at arrival}\}$
* $q(n) = p(n)$ in M/M/1

$D_q(0) = Pr\{T_q \leq 0\} = Pr\{T_q = 0\} \text{(waiting time can not be negative)} \\
= Pr\{\text{system is empty at arrival}\} = q_0 = p_0 = 1 - \rho$

2. t > 0
$D_q(t) = Pr\{T_q \leq t\} \\
= \displaystyle \sum^\infty_{n = 1} Pr\{T_q \leq t | \text{n in system at arrival}\} \times Pr\{\text{n in system at arrival}\} + Pr\{T_q = 0\} \\
= \displaystyle \sum^\infty_{n = 1} Pr\{\text{time for n service completion} | \text{n in system at arrival}\} \times p_n + p_0 \\
= \displaystyle \sum^\infty_{n = 1} (1 - \rho) \rho^n \int^t_0 \dfrac{\mu(\mu x)^{n - 1}}{(n - 1)!} e^{-\mu x} dx + (1 - \rho) \\
= (1 - \rho) \rho \int^t_0 \mu e^{-\mu x} \displaystyle \sum^\infty_{n = 1} \dfrac{(\rho \mu x)^{n - 1}}{(n - 1)!} + (1 - \rho) \\
= (1 - \rho) \rho \int^t_0 \mu e^{-\mu x(1 - \rho)} + (1 - \rho) \\
= \rho \int^t_0 \mu (1 - \rho) e^{-\mu (1 - \rho)x} + (1 - \rho) \\
= \rho[1 - e^{-\mu(1 - \rho)t}] + (1 - \rho) \\
= 1 - \rho e^{-\mu(1 - \rho)t}$

$D_q = E[T_q] = \displaystyle \int^\infty_0 t \cdot d D_q(t) \\
= \displaystyle \int^\infty_0 t [\rho \mu (1 - \rho) e^{-\mu(1 - \rho)t}] dt
= \rho \dfrac{1}{\mu(1 - \rho)}$

### system time distribution
* $T$: random variable representing time a customer stays in the system
* $D(t)$: CDF of $T$

$D(t) = Pr\{T \leq t\} \\
= \displaystyle \sum^\infty_{n = 0} Pr\{T \leq t | \text{n in system at arrival}\} \times q_n \\
= \displaystyle \sum^\infty_{n = 0} (1 - \rho) \rho^n \int^t_0 \mu e^{-\mu x} \dfrac{(\mu x)^{n}}{(n)!} dx \\
= (1 - \rho) \displaystyle \int^t_0 \mu e^{-\mu x} \sum^\infty_{n = 0} \dfrac{(\rho \mu x)^{n}}{(n)!} dx \\
= (1 - \rho) \displaystyle \int^t_0 \mu e^{-(1 - \rho)\mu x} dx \\
= 1 - e^{-(1 - \rho)\mu t} \\
\Rightarrow \bar{D} = \dfrac{1}{\mu(1 - \rho)} = \dfrac{1}{\mu - \lambda}$

* $X_s$: number of customers being served
* $E(X_s) = 0 \cdot q_0 + 1 \cdot (1 - q_0) = (1 - p_0) = \rho$
* with Little's Law: $E(X_s) = \bar{\lambda} \cdot \bar{x} = \lambda \dfrac{1}{\mu} = \dfrac{\lambda}{\mu} = \rho$
* $X$: number of customers in the system
* $E(X) = \displaystyle \sum^\infty_{k = 0} k \cdot p_k = \dfrac{\rho}{1 - \rho}$
* $X_w$: number of customers waiting in the system
* $E(X_w) = \displaystyle \sum^\infty_{k = 0} (k-1) \cdot p_k = \sum^\infty_{k = 0} k \cdot p_k - \sum^\infty_{k = 0} p_k = \dfrac{\rho^2}{1 - \rho}$

### System time, waiting time, service time 的關係

* $T$: system time
* $W$: waiting time
* $x$: service time

$T = W + x \Rightarrow \bar{T} = \bar{W} + \bar{x}$
用 Little's Law 可以得到
* $\bar{T} = \dfrac{E(X)}{\lambda} = \dfrac{1}{\mu(1-\rho)}$
* $\bar{W} = \dfrac{E(X_w)}{\lambda} = \dfrac{\rho}{\mu(1-\rho)}$
* $\bar{x} = \dfrac{1}{\mu}$

$\bar{T}$ 也可用另一種方式計算得到
$\bar{T} = \bar{x} + \displaystyle \sum^\infty_{k = 0} k \cdot \bar{x} \cdot p_k$

## M/M/c queue

### state transition probability

\begin{cases}
P_{i, i + 1}(\Delta) = \lambda \Delta + o(\Delta), i = 0, 1, \cdots \\
P_{i, i - 1}(\Delta) = i \mu \Delta + o(\Delta), 0 \leq i \leq c \\
P_{i, i}(\Delta) = 1 - (\lambda + i \mu) \Delta + o(\Delta), 0 \leq i \leq c \\
P_{i, i}(\Delta) = 1 - (\lambda + c \mu) \Delta + o(\Delta), i \geq c
\end{cases}

### global balance equation

\begin{cases}
0 = -(\lambda + n \mu)p_n + \lambda p_{n - 1} + (n + 1)\mu p_{n + 1}, 0 < n < c \\
0 = -(\lambda + c \mu)p_n + \lambda p_{n - 1} + (n + 1)\mu p_{n + 1}, n \geq c \\
0 = -\lambda p_0 + \mu p_1
\end{cases}

$p_n = \dfrac{\lambda}{n \mu} p_{n - 1} = (\dfrac{\lambda}{\mu})^n \dfrac{1}{n!} p_0, n = 1, 2, \cdots, c-1$
$p_n = \dfrac{\lambda}{c \mu} p_{n - 1} = (\dfrac{\lambda}{\mu})^n \dfrac{1}{c!} \dfrac{1}{c^{n - c}} p_0, n \geq c$

$p_0 = [\displaystyle \dfrac{1}{c!} (\dfrac{\lambda}{\mu})^c \dfrac{1}{(1 - \frac{\lambda}{c \mu})} + \sum^{c-1}_{j=0} (\dfrac{\lambda}{\mu})^j \dfrac{1}{j!}]^{-1}$

* $\rho = \dfrac{\lambda}{c \mu}, r = \dfrac{\lambda}{\mu}$
* $p_n =$
\begin{cases}
\dfrac{r^n}{n!} p_0, 0 < n < c \\
\dfrac{r^n}{c^{n - c} c!} p_0, n \geq c
\end{cases}

### L, Lq, W(T), Wq(Tq)


* $L$: mean number of customers in the system
* $L_q$: mean number of customers in the queue
* $W$: mean system time
* $W_q$: mean waiting time in the queue

$L_q = \displaystyle \sum^\infty_{n = c + 1} (n - c) p_n \\
= \displaystyle \sum^\infty_{n = c + 1} (n - c) \dfrac{r^n}{c^{n - c} c!} p_0 \\
= \dfrac{r^c}{c!} p_0 \displaystyle \sum^\infty_{n = c + 1} (n - c) (\dfrac{r}{c})^{n - c} \\
= \dfrac{r^c}{c!} p_0 \displaystyle \sum^\infty_{n = c + 1} (n - c) (\rho)^{n - c} \\
= \dfrac{r^c}{c!} p_0 \dfrac{\rho}{(1 - \rho)^2} \\
= \dfrac{r^c}{c! (1 - \rho)^2} p_0$

$L = L_q + r = \dfrac{r^c}{c! (1 - \rho)^2} p_0 + r$

---

$\bar{T} = \bar{x} + \bar{W} = \frac{1}{\mu} + \displaystyle \sum^\infty_{k = c} E(W | k) p^{(a)}_k$
$E(T) = \dfrac{1}{\mu} + \displaystyle \sum^\infty_{k = c} \dfrac{k - c + 1}{c \mu} p_k$
$\bar{T} = \frac{1}{\mu} + \displaystyle \sum^\infty_{k = c} \dfrac{k - c + 1}{c \mu} p_c (\dfrac{\lambda}{c \mu})^{k - c} \\
= \frac{1}{\mu} + \dfrac{p_c}{c \mu}\displaystyle \sum^\infty_{k = c} (k - c + 1) (\dfrac{\lambda}{c \mu})^{k - c} \\
= \frac{1}{\mu} + \dfrac{p_c}{c \mu}\displaystyle \sum^\infty_{i = 1} i (\dfrac{\lambda}{c \mu})^{i - 1} \\
= \frac{1}{\mu} + \dfrac{c \mu p_c}{(c \mu - \lambda)^2}$

---

$D_q(0) = Pr\{T_q = 0\} = Pr\{ \leq c - 1 \text{ in system at arrival}\} \\
= \displaystyle \sum^{c - 1}_{n = 0} q_n = p_0 \sum^{c - 1}_{n = 0} \dfrac{r^n}{n!}, p_0 = (\dfrac{r^c}{c!(1 - \rho)} + \sum^{c - 1}_{n = 0} \dfrac{r^n}{n!})^{-1} \\
= 1 - \dfrac{r^c}{c! (1 - \rho)} p_0 \Rightarrow Pr\{T_q > 0\} = \frac{r^c}{c!(1 - \rho)} p_0$

---

$D_q(t) = Pr\{T_q \leq t\} \\
= \displaystyle \sum^\infty_{n = c} Pr\{n - c + 1 \text{ completions} \leq t | n \text{ in system at arrival}\} q_n + D_q(0) \\
= \displaystyle \sum^\infty_{n = 0} \dfrac{r^c}{c^{n-c} c!} p_0 \int^t_0 \dfrac{c \mu (c \mu x)^{n - c}}{(n - c)!}e^{-c \mu x} dx + D_q(0) \\
= \dfrac{r^c}{(c - 1)!} p_0 \displaystyle \int^t_0 \mu e^{-c \mu x} \sum^\infty_{n = 0} \dfrac{(r \mu x)^{n - c}}{(n - c)!} dx + D_q(0) \\
= \dfrac{r^c}{(c - 1)!} p_0 \displaystyle \int^t_0 \mu e^{-\mu (c - r)x} dx + D_q(0) \\
= \dfrac{r^c}{(c - 1)! (c - r)} p_0 \displaystyle \int^t_0 \mu(c - r) e^{-\mu (c - r)x} dx + D_q(0) \\
= \dfrac{r^c}{c! (1 - \rho)} (1 - e^{-(c \mu - \lambda)t} ) + D_q(0), D_q(0) = \frac{r^c}{c!(1 - \rho)} p_0 \\
= 1 - \dfrac{r^c}{c! (1 - \rho)} (1 - e^{-(c \mu - \lambda)t} ) = 1 - Pr\{T_q > t\} \\
\Rightarrow Pr\{T_q > t\} = \dfrac{r^c}{c! (1 - \rho)} (1 - e^{-(c \mu - \lambda)t} )$

---

$Pr\{T_q > 0\} = \dfrac{r^c p_0}{c! (1 - \rho)}, Pr\{T_q > t\} = \dfrac{r^c p_0}{c! (1 - \rho)} \cdot e^{-(c \mu - \lambda)t}$
$Pr\{T_q > t | T_q > 0\} = \dfrac{Pr\{T_q > t\}}{Pr\{T_q > 0\}} = 1 - e^{-(c \mu - \lambda)t}$

## Finite-Source Queue


* $\lambda_n =$
\begin{cases}
(M-n)\lambda,\ 0 \leq n < M \\
0,\ n \geq M
\end{cases}
* $\mu_n =$
\begin{cases}
n\mu,\ 0 \leq n < c \\
c\mu,\ n \geq c
\end{cases}

###### tags: `1111_courses` `queueing system`