Compute utilization, response time, average number of requests in service and throughput for a M/M/m queue, a queueing system with m identical servers connected to a single FCFS queue.
INPUTS
lambda
Arrival rate (lambda>0
).
mu
Service rate (mu>lambda
).
m
Number of servers (m ≥ 1
).
Default is m=1
.
k
Number of requests in the system (k ≥ 0
).
OUTPUTS
U
Service center utilization, U = \lambda / (m \mu).
R
Service center mean response time
Q
Average number of requests in the system
X
Service center throughput. If the system is ergodic,
we will always have X = lambda
p0
Steady-state probability that there are 0 requests in the system
pm
Steady-state probability that an arriving request has to wait in the queue
pk
Steady-state probability that there are k requests in the system (including the one being served).
If this function is called with less than four parameters, lambda, mu and m can be vectors of the same size. In this case, the results will be vectors as well.
REFERENCES
See also: erlangc,qsmm1,qsmminf,qsmmmk.
The following code
# This is figure 6.4 on p. 220 Bolch et al. rho = 0.9; ntics = 21; lambda = 0.9; m = linspace(1,ntics,ntics); mu = lambda./(rho .* m); [U R Q X] = qsmmm(lambda, mu, m); qlen = X.*(R-1./mu); plot(m,Q,"o",qlen,"*"); axis([0,ntics,0,25]); legend("Jobs in the system","Queue Length","location","northwest"); legend("boxoff"); xlabel("Number of servers (m)"); title("M/M/m system, \\lambda = 0.9, \\mu = 0.9");
Produces the following figure
Figure 1 |
---|
The following code
## Given a M/M/m queue, compute the steady-state probability pk of ## having k jobs in the systen. lambda = 0.5; mu = 0.15; m = 5; k = 0:10; pk = qsmmm(lambda, mu, m, k); plot(k, pk, "-o", "linewidth", 2); xlabel("N. of jobs (k)"); ylabel("P_k"); title(sprintf("M/M/%d system, \\lambda = %g, \\mu = %g", m, lambda, mu));
Produces the following figure
Figure 1 |
---|
Package: queueing