%%  This script demonstrates the Erlang B and Erlang C functions.%%% First, a couple of easy ones.%disp('Consider a queueing system with lambda=10/minute, mu=4/minute,');disp('5 servers, and no waiting line.  Find the probability that an');disp('arriving customer will find all servers busy.');disp('pbusy=erlangb(5,10/4)');pbusy=erlangb(5,10/4)disp('Next, suppose that we have an infinite waiting line. What is the');disp('probability that an arriving customer will have to wait?');disp('pwait=erlangc(5,10/4)');pwait=erlangc(5,10/4)%% Now, a couple of hard ones.%disp('Consider a queueing system with lambda=900/minute, mu=1/minute,');disp('1000 servers, and no waiting line.  Find the probability that an');disp('arriving customer will find all servers busy.');disp('pbusy=erlangb(1000,900)');pbusy=erlangb(1000,900)disp('Next, suppose that we have an infinite waiting line. What is the');disp('probability that an arriving customer will have to wait?');disp('pwait=erlangc(1000,900)');pwait=erlangc(1000,900)%% Now, demonstrate the inverse functions.  %disp('Consider a queueing system with lambda=100/minute, mu=4/minute,');disp('n servers, and no waiting line.  Find the smallest n such that');disp('the probability of an arriving customer finding all servers busy');disp('is less than 0.001');disp('n=erlangbinv(0.001,100/4)');n=erlangbinv(0.001,100/4)disp('Next, suppose that we have an infinite waiting line. What is the');disp('number of servers required to reduce the probability of having to');disp('wait below 0.001 ');disp('n=erlangcinv(0.001,100/4)');n=erlangcinv(0.001,100/4)%% Demonstrate the findrho functions.%disp('Consider a queueing system with n=1000 servers and no waiting line.');  disp('How much traffic can be handled while keeping the probability that');disp('an arriving customer will be blocked below 1%?');disp('rhob=findrhob(1000,0.01)');rhob=findrhob(1000,0.01)disp('Consider a queueing system with n=1000 servers and an infinite waiting');disp('line.  How much traffic can be handled while keeping the probability');disp('that an arriving customer will wait below 1%?');disp('rhoc=findrhoc(1000,0.01)');rhoc=findrhoc(1000,0.01)%% Finally demonstrate the engset function.%disp('Consider a queueing system with a population of m=200');disp('customers, n=150 servers, and rho=0.5.  What is the');disp('blocking probability?');disp('E=engset(200,150,0.50)');E=engset(200,150,0.50)