Data 604 - Discussion #2

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 \(\rho\).

Known values:

• Mean interarrival time: 1.25 minutes
• Mean service time: 1 minute
• No. of servers: 1

# mean interarrival time (minutes)
mit <- 1.25

# mean service time (minute)
mst <- 1 

# number of servers 
c <- 1

# arrival rate 
lambda <- 1/mit

# service rate
mu <- 1/mst

# utilization of servers
rho <- lambda/(c * mu)

# avg no. of entities in queue
L_q <- (rho)**2/(1-rho)

# avg no. of entities in system
L <- (rho)/(1-rho)

# avg time in queue
W_q <- rho/(mu * (1-rho))

# avg time in system
W <- 1/(mu * (1-rho))

	M/M/1	Value
\(\lambda\)	\(\frac{1}{E(interarrival-time)}\)	0.8
\(\mu\)	\(\frac{1}{E(service-time)}\)	1
\(\rho\)	\(\frac{\lambda}{c\cdot\mu}\)	0.8
\(W_q\)	\(\frac{\rho}{\mu\cdot(1-\rho)}\)	4 minutes
\(W\)	\(\frac{1}{\mu\cdot(1-\rho)}\)	5 minutes
\(L_q\)	\(\frac{\rho^2}{1-\rho}\)	3.2 entities
\(L\)	\(\frac{\rho}{1-\rho}\)	4 entities

For each, interpret in words

This queue system has single server with exponential interarrival and service times.

• \(\rho\): With mean interarrival of 1.25 minutes and mean service time of 1 minute, the server has 80% chance of being busy.

• \(W_q\): The average time an entity spends waiting in queue is 4 minutes.

• \(W\) : The average time an entity spends both in queue and in service is 5 minutes

• \(L_q\): The average number of entities waiting in queue are 3.

• \(L\): The average number of entities waiting in queue and in system are 4.

Simulate in Simio

Simulation Setup

Results

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