Problem 4.14.D

\(G/M/1\) queue system, continuous uniformly distributed interarrival time with minimum 1 and maximum 5, and service rate \(\mu\) = 0.4 per minute.

\(z = \int_{0}^{\infty}e^{-\mu t (1-z)}g(t)dt\)

\(g(t) = 1/(b-a) \textrm{, if } a \leq t \leq b\\\)

\(g(t) = \begin{cases} 1/(b-a) & \textrm{if } a \leq t \leq b\\ 0 & \textrm{otherwise} \end{cases}\)

\(z = \int_{0}^{\infty}e^{-\mu t (1-z)} \frac{1}{b-a}dt\)

\(z = -\frac{1}{\mu(b-a)(1-z)}(e^{-\mu(1-z)b} - e^{-\mu(1-z)a})\)

\(z = -\frac{1}{0.4(5-1)(1-z)}(e^{-0.4(1-z)5} - e^{-0.4(1-z)})\)

\(z = -\frac{1}{1.6(1-z)}(e^{-2(1-z)} - e^{-0.4(1-z)})\)

\(W = \frac{1}{\mu(1-z)}\)

\(1/\lambda = (a+b)/2 = (1 + 5)/2 = 3\)

\(\lambda = 1/3\)

eqn <- function (z) {
  return(z + ((1 / (1.6 * (1 - z))) * (exp(-2 * (1 - z)) - exp(-0.4 * (1-z)))))
}

uniroot(eqn, c(0, 1), f.upper = 1, tol = 1e-16)

## $root
## [1] 0.723485
## 
## $f.root
## [1] 0
## 
## $iter
## [1] 11
## 
## $init.it
## [1] NA
## 
## $estim.prec
## [1] 8.615043e-05

z <- uniroot(eqn, c(0, 1), f.upper = 1, tol = 1e-16)$root

mu <- 0.4
lambda <- 1/3

W = 1 / (mu * (1 - z))
rho <- lambda / mu
W_q <- (lambda / mu^2) / (2 * (1 - (lambda / mu)))
L_q <- lambda * W_q
W = W_q + (1 / mu)
L = lambda * W

knitr::kable(data.frame (z = z, W_q = W_q, L_q = L_q, W = W, L = L, rho = rho))

z	W_q	L_q	W	L	rho
0.723485	6.25	2.083333	8.75	2.916667	0.8333333

Results of single run of 52-week long simulation in Simio