""" 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], )