Data604 - Valerie Briot - Discussion # 2

Calculations

For an M/M/1 queue with mean interarrival time 1.25 minutes and mean service time 1 minute. Find all five of \({ W }_{ q }\), W, \({ L }_{ q }\), L, and \(\rho\). For each, interpret in words. Be sure to state all your units and the relevant time frame of operation. a

A M/M/1 queue signified the following: M, the arrival process is given by a Poisson distribution, in this case, since the mean interarrival time is 1.25, the average arrival rate \(\lambda\) = 1/1.25 arrivals/minute M, the service time is given by a exponential distribution, in this case, since the mean service time 1 minute, the average service rate \(\mu\) = 1/1.0 serviced/minute

There is only 1 server,

The queue is unlimited, the capacity of the system is unlimited as well, system is open, and the queueing rule = FIFO

\({ W }_{ q }\) denotes the steady-state average time in the queue,
W denotes the steady-state average time in the system,
\({ L }_{ q }\) denotes the steady-state average number of entities in the queue,
L denotes the steady-state average number of entities in the system, and \(\rho\) denotes the steady-state utilization of the server

These values are given by following equations (1): \({ W }_{ q }=\frac { \lambda }{ \mu (\mu -\lambda ) }\)
\(W=\frac { 1 }{ \mu -\lambda }\)
\({ L }_{ q }=\frac { { \lambda }^{ 2 } }{ \mu (\mu -\lambda }\)
\(L=\frac { \lambda }{ \mu -\lambda }\)
\(\rho =\frac { \lambda }{ \mu }\)

l_lambda <- 1/1.25
m_mu <- 1/1.0 

r_rho <- l_lambda/m_mu

W_q <- l_lambda/(m_mu*(m_mu - l_lambda))
W <- 1/(m_mu - l_lambda)
L <- l_lambda/(m_mu - l_lambda)
L_q <- (l_lambda^2)/(m_mu*(m_mu - l_lambda))

We will now display the results of the calculations in a tabular format:

results <- data.frame(t(c(W_q, W, L_q, L, r_rho)))
colnames(results) <- c("Wq", "W", "Lq", "L", "Rho")

results

##   Wq W  Lq L Rho
## 1  4 5 3.2 4 0.8

Data604 - Valerie Briot - Discussion # 2

Valerie Briot

September 6, 2017

Calculations

Simulation