Kelton, Chapter 2.6, Problem 9

Given:
System type: M/D/1 queue, deterministic distribution
Arrival rate: \(\lambda=1\) Service rate: \(\mu=1/0.9\)



Find:
\(W_q\)
\(W\)
\(L_q\)
\(L\)
\(\rho\)


Mean arrival rate:

\[\lambda=1\]



Mean service rate:

\[\mu=\frac{1}{meanservicetime}\]

\[\mu=1/0.9\]



Server utilization:

\[\rho=\frac{\lambda}{\mu}\]

\[\rho=\frac{1}{1/0.9}=0.9\]



Customers in queue:

\[L_q = (1/2)*\frac{\rho^2}{1 - \rho}\]

\[L_q = (1/2) * \frac{0.9^2}{1 - 0.9} = 4.05 \quad customers in queue\]



Customers in system:

\[L= \rho+(1/2)*\frac{\rho^2}{1 - \rho}\]

\[L = 0.9+(1/2) * \frac{0.9^2}{1 - 0.9} = 4.95 \quad customers in system\]



Total time in system:
\[W = \frac{1}{\mu} +\frac{p}{2*\mu*(1-\rho)}\]

\[W = \frac{1}{1/0.9} +\frac{0.9}{2*(1/0.9)*(1-0.9)} = 4.95 \quad mins\]

Time in queue:
\[W_q = W-E(s) = W - (1/\mu)\]

\[W_q = 4.95-(1/(1/0.9)) = 4.05 \quad mins\]

Kelton, Chapter 4, Problem 15

Build a Simio model for the previous problem and compute the relevant metrics.


Model setup

Model setup



Results, 24 hours

Results, 24 hours



Experiment, 48 hours, 10 replications

Experiment, 48 hours, 10 replications



Experiment, 48 hours, 10 replications

Experiment, 48 hours, 10 replications

We see that a 48-hour simulation with a 10-hour warm up period returns closer results to the calculated results. The simulation returns 4.5 for average number in system, compared to the calculated 4.95.