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""" bcc.py
Queue with blocked customers cleared
Jobs (e.g messages) arrive randomly at rate 1.0 per minute at a
2-server system. Mean service time is 0.75 minutes and the
service-time distribution is (1) exponential, (2) Erlang-5, or (3)
hyperexponential with p=1/8,m1=2.0, and m2=4/7. However no
queue is allowed; a job arriving when all the servers are busy is
rejected.
Develop and run a simulation program to estimate the probability of
rejection (which, in steady-state, is the same as p(c)) Measure
and compare the probability for each service time distribution.
Though you should test the program with a trace, running just a few
jobs, the final runs should be of 10000 jobs without a trace. Stop
the simulation when 10000 jobs have been generated.
"""
from SimPy.Simulation import *
from random import seed, Random, expovariate, uniform
# Model components ------------------------
dist = ""
def bcc(lam, mu, s):
""" bcc - blocked customers cleared model
- returns p[i], i = 0,1,..s.
- ps = p[s] = prob of blocking
- lameff = effective arrival rate = lam*(1-ps)
See Winston 22.11 for Blocked Customers Cleared Model (Erlang B formula)
"""
rho = lam/mu
n = range(s+1)
p = [0]*(s+1)
p[0] = 1
sump = 1.0
for i in n[1:]:
p[i] = (rho/i)*p[i-1]
sump = sump + p[i]
p0 = 1.0/sump
for i in n:
p[i] = p[i]*p0
p0 = p[0]
ps = p[s]
lameff = lam*(1-ps)
L = rho*(1-ps)
return {'lambda': lam, 'mu': mu, 's': s,
'p0': p0, 'p[i]': p, 'ps': ps, 'L': L}
def ErlangVariate(mean, K):
""" Erlang random variate
mean = mean
K = shape parameter
g = rv to be used
"""
sum = 0.0
mu = K/mean
for i in range(K):
sum += expovariate(mu)
return (sum)
def HyperVariate(p, m1, m2):
""" Hyperexponential random variate
p = prob of branch 1
m1 = mean of exponential, branch 1
m2 = mean of exponential, branch 2
g = rv to be used
"""
if random() < p:
return expovariate(1.0/m1)
else:
return expovariate(1.0/m2)
def testHyperVariate():
""" tests the HyerVariate rv generator"""
ERR = 0
x = (1.0981, 1.45546, 5.7470156)
p = 0.0, 1.0, 0.5
g = Random(1113355)
for i in range(3):
x1 = HyperVariate(p[i], 1.0, 10.0, g)
# print p[i], x1
assert abs(x1 - x[i]) < 0.001, 'HyperVariate error'
def erlangB(rho, c):
""" Erlang's B formula for probabilities in no-queue
Returns p[n] list
see also SPlus and R version in que.q mmcK
que.py has bcc.
"""
n = range(c+1)
pn = range(c+1)
term = 1
pn[0] = 1
sum = 1
term = 1.0
i = 1
while i < (c+1):
term *= rho/i
pn[i] = term
sum += pn[i]
i += 1
for i in n:
pn[i] = pn[i]/sum
return(pn)
class JobGen(Process):
""" generates a sequence of Jobs
"""
def execute(self, JobRate, MaxJob, mu):
global NoInService, Busy
for i in range(MaxJob):
j = Job()
activate(j, j.execute(i, mu), delay=0.0)
t = expovariate(JobRate)
MT.tally(t)
yield hold, self, t
self.trace("Job generator finished")
def trace(self, message):
if JobGenTRACING:
print "%8.4f \t%s" % (now(), message)
class Job(Process):
""" Jobs that are either accepted or rejected
"""
def execute(self, i, mu):
""" Job execution, only if accepted"""
global NoInService, Busy, dist, NoRejected
if NoInService < c:
self.trace("Job %2d accepted b=%1d" % (i, Busy))
NoInService += 1
if NoInService == c:
Busy = 1
try:
BM.accum(Busy, now())
except:
"accum error BM=", BM
# yield hold,self,Job.g.expovariate(self.mu); dist= "Exponential"
yield hold, self, ErlangVariate(1.0/mu, 5)
dist = "Erlang "
# yield hold,self,HyperVariate(1.0/8,m1=2.0,m2=4.0/7,g=Job.g);
# dist= "HyperExpon "
NoInService -= 1
Busy = 0
BM.accum(Busy, now())
self.trace("Job %2d leaving b=%1d" % (i, Busy))
else:
self.trace("Job %2d REJECT b=%1d" % (i, Busy))
NoRejected += 1
def trace(self, message):
if JobTRACING:
print "%8.4f \t%s" % (now(), message)
# Experiment data -------------------------
c = 2
lam = 1.0 # per minute
mu = 1.0/0.75 # per minute
p = 1.0/8
m1 = 2.0
m2 = 4.0/7.0
K = 5
rho = lam/mu
NoRejected = 0
NoInService = 0
Busy = 0
JobRate = lam
JobMax = 10000
JobTRACING = 0
JobGenTRACING = 0
# Model/Experiment ------------------------------
seed(111333)
BM = Monitor()
MT = Monitor()
initialize()
jbg = JobGen()
activate(jbg, jbg.execute(1.0, JobMax, mu), 0.0)
simulate(until=20000.0)
# Analysis/output -------------------------
print 'bcc'
print "time at the end =", now()
print "now=", now(), " startTime ", BM.startTime
print "No Rejected = %d, ratio= %s" % (NoRejected, (1.0*NoRejected)/JobMax)
print "Busy proportion (%6s) = %8.6f" % (dist, BM.timeAverage(now()), )
print "Erlang pc (th) = %8.6f" % (erlangB(rho, c)[c], )