# M/M/1/4 vs M/M/2/4 queue

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## M/M/1/4 vs M/M/2/4 queue

 Here is the code I wrote which is explained in comments. % system M/M/1/4 % when there are 3 clients in the system, the capability of the server doubles. clc; clear all; close all; lambda = 4; mu = 5; states = [0,1,2,3,4]; % system with capacity 4 states % the initial state of the system. The system is initially empty. initial_state = [1,0,0,0,0]; % define the birth and death rates between the states of the system. births_B = [lambda,lambda,lambda,lambda]; deaths_D = [mu,mu,2*mu,2*mu]; % get the transition matrix of the birth-death process transition_matrix = ctmcbd(births_B,deaths_D); % get the ergodic probabilities of the system P = ctmc(transition_matrix); % plot the ergodic probabilities (bar for bar chart) figure(1); bar(states,P,"r",0.5); % transient probability of state 0 until convergence to ergodic probability. Convergence takes place P0 and P differ by 0.01 index = 0; for T=0:0.01:50   index = index + 1;   P0 = ctmc(transition_matrix,T,initial_state);   Prob0(index) = P0(1);   if P0-P < 0.01     break;   endif endfor T = 0:0.01:T; figure(2); plot(T,Prob0,"r","linewidth",1.3); If I had 2 servers thus a M/M/2/4 queue, what would have changed? How 2 servers would have affected the code ?  In my opinion, it would make any difference, but is that true? -- Sent from: http://octave.1599824.n4.nabble.com/Octave-General-f1599825.html