## C.4 Summary of Queueing Formulas

This section provides the formulas for basic single queue stations and is meant simply as a resource where the formulas are readily available for potential applicaiton. The following notation is used within this section.

Let $$N$$ represent the steady state number of customers in the system, where $$N \in \{0,1,2,...,k\}$$ where $$k$$ is the maximum number of customers in the system and may be infinite ($$\infty$$).

Let $$\lambda_{n}$$ be the arrival rate when there are $$N=n$$ customers in the system.

Let $$\mu_{n}$$ be the service rate when there are $$N=n$$ customers in the system.

Let $$P_n = P[N=n]$$ be the probability that there are $$n$$ customers in the system in steady state.

When $$\lambda_{n}$$ is constant for all $$n$$, we write $$\lambda_{n} = \lambda$$.

When $$\mu_{n}$$ is constant for all $$n$$, we write $$\mu_{n} = \mu$$.

Let $$\lambda_e$$ be the effective arrival rate for the system, where

$\lambda_e = \sum_{n=0}^{\infty} \lambda_n P_n$ Since $$\lambda_n = 0$$ for $$n \geq k$$ for a finite system size, $$k$$, we have:

$\lambda_e = \sum_{n=0}^{k-1} \lambda_n P_n$

Let $$\rho = \frac{\lambda}{c\mu}$$ be the utilization.

Let $$r = \frac{\lambda}{\mu}$$ be the offered load.

### C.4.1 M/M/1 Queue

\begin{aligned} \lambda_n & =\lambda \\ \mu_n & = \mu \\ r & = \lambda/\mu \end{aligned}

$P_0 = 1 - r$ $P_n = P_0 r^n$ $L_q = \dfrac{r^2}{1 - r}$

Results M/G/1 and M/D/1
Model Parameters $$L_q$$
M/G/1 $$E[ST] = \dfrac{1}{\mu}$$; $$Var[ST] = \sigma^2$$; $$r = \lambda/\mu$$ $$L_q = \dfrac{\lambda^2 \sigma^2 + r^2}{2(1 - r)}$$
M/D/1 $$E[ST] = \dfrac{1}{\mu}$$; $$Var[ST] = 0$$; $$r = \lambda/\mu$$ $$L_q = \dfrac{r^2}{2(1 - r)}$$

### C.4.2 M/M/c Queue

\begin{aligned} \lambda_n & =\lambda \\ \mu_n & = \begin{cases} n \mu & 0 \leq n < c \\ c \mu & n \geq c \end{cases} \\ \rho & = \lambda/c\mu \quad r = \lambda/\mu \end{aligned}

$P_0 = \biggl[\sum\limits_{n=0}^{c-1} \dfrac{r^n}{n!} + \dfrac{r^c}{c!(1 - \rho)}\biggr]^{-1}$ $L_q = \biggl(\dfrac{r^c \rho}{c!(1 - \rho)^2}\biggl)P_0$

$P_n = \begin{cases} \dfrac{(r^n)^2}{n!} P_0 & 1 \leq n < c \\[1.5ex] \dfrac{r^n}{c!c^{n-c}} P_0 & n \geq c \end{cases}$

Results M/M/c $$\rho = \lambda/c \mu$$
$$c$$ $$P_0$$ $$L_q$$
1 $$1 - \rho$$ $$\dfrac{\rho^2}{1 - \rho}$$
2 $$\dfrac{1 - \rho}{1 + \rho}$$ $$\dfrac{2 \rho^3}{1 - \rho^2}$$
3 $$\dfrac{2(1 - \rho)}{2 + 4 \rho + 3 \rho^2}$$ $$\dfrac{9 \rho^4}{2 + 2 \rho - \rho^2 - 3 \rho^3}$$

### C.4.3 M/M/c/k Queue

\begin{aligned} \lambda_n & = \begin{cases} \lambda & n < k \\ 0 & n \geq k \end{cases} \\ \mu_n & = \begin{cases} n \mu & 0 \leq n < c \\ c \mu & c \leq n \leq k \end{cases} \\ \rho & = \lambda/c\mu \quad r = \lambda/\mu \\ \lambda_e & = \lambda (1 - P_k) \end{aligned} $P_0 = \begin{cases} \biggl[\sum\limits_{n=0}^{c-1} \dfrac{r^n}{n!} + \dfrac{r^c}{c!} \dfrac{1-\rho^{k-c+1}}{1 - \rho}\biggr]^{-1} & \rho \neq 1\\[1.5ex] \biggl[\sum\limits_{n=0}^{c-1} \dfrac{r^n}{n!} + \dfrac{r^c}{c!} (k-c+1)\biggl]^{-1} & \rho = 1 \end{cases}$ $P_n = \begin{cases} \dfrac{r^n}{n!} P_0 & 1 \leq n < c \\[1.5ex] \dfrac{r^n}{c!c^{n-c}} P_0 & c \leq n \leq k \end{cases}$

### C.4.4 M/G/c/c Queue

\begin{aligned} \lambda_n & = \begin{cases} \lambda & n < c \\ 0 & n \geq c \end{cases} \\ \mu_n & = \begin{cases} n \mu & 0 \leq n \leq c \\ 0 & n > c \end{cases} \\ \rho & = \lambda/c\mu \quad r = \lambda/\mu \\ \lambda_e & = \lambda (1 - P_k) \end{aligned}

$P_0 = \biggr[\sum\limits_{n=0}^c \dfrac{r^n}{n!}\biggl]^{-1}$ $\begin{array}{c} P_n = \dfrac{r^n}{n!} P_0 \\ 0 \leq n \leq c \end{array}$ $L_q = 0$

### C.4.5 M/M/1/k Queue

\begin{aligned} \lambda_n & = \begin{cases} (k - n)\lambda & 0 \leq n < k \\ 0 & n \geq k \end{cases} \\ \mu_n & = \begin{cases} (k - n)\lambda & 0 \leq n \leq k \\ 0 & n > k \end{cases} \\ r & = \lambda/\mu \quad \lambda_e = \lambda(k - L) \end{aligned} $P_0 = \biggl[\sum\limits_{n=0}^k \prod\limits_{j=0}^{n-1} \biggl(\dfrac{\lambda_j}{\mu_{j+1}}\biggr)\biggr]^{-1}$ $P_n = \binom{k}{n} n! r^n P_0$

$L_q = \begin{cases} \dfrac{\rho}{1 - \rho} - \dfrac{\rho (k \rho^k + 1)}{1 - \rho^{k+1}} & \rho \neq 1 \\ \dfrac{k(k - 1)}{2(k + 1)} & \rho = 1 \end{cases}$

### C.4.6 M/M/c/k Queue

\begin{aligned} \lambda_n & = \begin{cases} (k - n)\lambda & 0 \leq n < k \\ 0 & n \geq k \\ \end{cases} \\ \mu_n & = \begin{cases} n \mu & 0 \leq n < c \\ c \mu & n \geq c \\ \end{cases} \\ r & = \lambda/\mu \quad \lambda_e = \lambda(k - L) \end{aligned}

$P_0 = \biggl[\sum\limits_{n=0}^k \prod\limits_{j=0}^{n-1} (\dfrac{\lambda_j}{\mu_{j+1}})\biggr]^{-1}$ $P_n = \begin{cases} \binom{k}{n} r^n P_0 & 1 \leq n < c \\[2ex] \binom{k}{n} \dfrac{n!}{c^{n-c} c!} r^n P_0 & c \leq n \leq k \end{cases}$

$L_q = \begin{cases} \dfrac{P_0 r^c \rho}{c!(1 - \rho)^2}[1 - \rho^{k-c} - (k-c) \rho^{k-c} (1 - \rho)] & \rho < 1 \\[2ex] \dfrac{r^c (k - c)(k - c + 1)}{2c!} P_0 & \rho = 1 \end{cases}$

### C.4.7 M/M/1/k/k Queue

\begin{aligned} \lambda_n & = \begin{cases} (k - n)\lambda & 0 \leq n < k \\ 0 & n \geq k \\ \end{cases} \\ \mu_n & = \begin{cases} (k - n)\lambda & 0 \leq n \leq k \\ 0 & n > k \\ \end{cases} \\ r & = \lambda/\mu \quad \lambda_e = \lambda(k - L) \end{aligned} $P_0 = \biggl[\sum\limits_{n=0}^k \prod\limits_{j=0}^{n-1} \biggl(\dfrac{\lambda_j}{\mu_{j+1}}\biggr)\biggr]^{-1}$ $\begin{array}{c} P_n = \binom{k}{n} n! r^n P_0 \\[1.5ex] 0 \leq n \leq k \end{array}$

$L_q = k - \biggl(\dfrac{\lambda + \mu}{\lambda}\biggr)(1 - P_0)$

### C.4.8 M/M/c/k/k Queue

\begin{aligned} \lambda_n & = \begin{cases} (k - n)\lambda & 0 \leq n < k \\ 0 & n \geq k \\ \end{cases} \\ \mu_n & = \begin{cases} n \mu & 0 \leq n < c \\ c \mu & n \geq c \\ \end{cases} \\ r & = \lambda/\mu \quad \lambda_e = \lambda(k - L) \end{aligned} $P_0 = \biggl[\sum\limits_{n=0}^k \prod\limits_{j=0}^{n-1} (\dfrac{\lambda_j}{\mu_{j+1}})\biggr]^{-1}$

$P_n = \begin{cases} \binom{k}{n} r^n P_0 & 1 \leq n < c \\[2ex] \binom{k}{n} \dfrac{n!}{c^{n-c} c!} r^n P_0 & c \leq n \leq k \end{cases}$

$L_q = \sum\limits_{n=c}^k (n - c) P_n$