matlab实现mm1排队系统的方法（代码）.rar

%% %基于排队论mm1m模型的仿真，五种业务，五种不同的到达率和服务率，运行结果：五种业务被服务的客户总数与member对应统计的不相符， %% clear all;close all;clc; %% % 初始参数设置： sim_time = 50; %总仿真时间 m = 5; %缓存队列最大长度 %可能到达的顾客数 arr_num = 200; sim_num = arr_num*5; %五种业务 %五种业务五种不同的到达率 lambda1 = 9/1;lambda2 = 5/1; lambda3 = 7/1; lambda4 = 3/1;lambda5 = 6/1; %lambda = lambda1 + lambda2 + lambda3 + lambda4 + lambda5;%相对于系统而言，将所有种类的业务视为一类，所以其到达速率为所有业务到达速率之和。 %五种业务五种不同的服务率 mu1 = 4;mu2 = 3;mu3 = 6;mu4 = 2;mu5 = 5; %平均到达时间 %arr_mean = 1/lambda; arr_mean1 = 1/lambda1;arr_mean2 = 1/lambda2;arr_mean3 = 1/lambda3;arr_mean4 = 1/lambda4;arr_mean5 = 1/lambda5;%不同业务的不同到达率 %平均服务时间 %ser_mean = 1/mu; %平均服务时间 ser_mean1 = 1/mu1;ser_mean2 = 1/mu2;ser_mean3 = 1/mu3;ser_mean4 = 1/mu4;ser_mean5 = 1/mu5;%不同业务有不同的服务率 %按负指数分布产生各顾客达到时间间隔 arr_int(1,:) = exprnd(arr_mean1,1,arr_num); arr_int(2,:) = exprnd(arr_mean2,1,arr_num); arr_int(3,:) = exprnd(arr_mean3,1,arr_num); arr_int(4,:) = exprnd(arr_mean4,1,arr_num); arr_int(5,:) = exprnd(arr_mean5,1,arr_num); %各顾客的到达时刻等于时间间隔的累积和 info_arr(1,:) = cumsum(arr_int(1,:)); info_arr(2,:) = cumsum(arr_int(2,:)); info_arr(3,:) = cumsum(arr_int(3,:)); info_arr(4,:) = cumsum(arr_int(4,:)); info_arr(5,:) = cumsum(arr_int(5,:)); %按负指数分布产生各顾客服务时间 ser_int(1,:) = exprnd(ser_mean1,1,arr_num); ser_int(2,:) = exprnd(ser_mean2,1,arr_num); ser_int(3,:) = exprnd(ser_mean3,1,arr_num); ser_int(4,:) = exprnd(ser_mean4,1,arr_num); ser_int(5,:) = exprnd(ser_mean5,1,arr_num); %% %声明一些需要用到的数组 info_rej = zeros(5,arr_num); info_left = zeros(5,arr_num); info_accept = zeros(5,arr_num); info_wait = zeros(5,arr_num); %% %信息初始化 total = zeros(5,sim_num); x = [info_arr(1,1:arr_num),info_arr(2,1:arr_num),info_arr(3,1:arr_num),info_arr(4,1:arr_num),info_arr(5,1:arr_num)]; total(1,:) = sort(x,'ascend'); %统计物种业务各顾客到达时刻，将所有业务客户的到达时间按升序排列 len_sim = sum(total(1,:)<= sim_time) %计算仿真顾客个数，即到达时刻在仿真时间内的顾客数 %% %第一个顾客的信息 for i = 1:5 if info_arr(i,1)==total(1,1) info_left(i,1) = info_arr(i,1) + ser_int(i,1); %该顾客离开时刻 info_accept(i,1) = 1; %该业务第一个顾客被接纳 info_wait(i,1) = 0; %该顾客无需等待 end end total(5,1) = 1; %该顾客肯定被接纳，此时系统内共有1 个顾客，故标志位置1 member = [1]; %其进入系统后，系统内已有成员序号为1 %% %第i个顾客的信息 for i = 2:len_sim if total(1,i)>sim_time %第i个顾客的到达时间超过了仿真时间，跳出循环 for j = 1:5 for k = 1:arr_num if info_arr(j,k) == total(1,i) info_wait(j,k) = 0; info_accept(j,k) = 0; end end end break; else %第i个顾客的到达时间未超过仿真时间，则计算在其到达时刻系统中已有的顾客个数 number = sum(total(4,member) > total(1,i)); end if number >= m+1 %系统已满，则系统拒绝第i个顾客， total(5,i) = 0; elseif number == 0 %系统有空闲服务窗，则第i个顾客直接接受服务 for j = 1:5 for k = 1:arr_num if info_arr(j,k) == total(1,i) info_wait(j,k) = 0; info_left(j,k) = info_arr(j,k) + ser_int(j,k); info_accept(j,k) = 1; total(4,i) = total(1,i) + ser_int(j,k); end end end total(5,i) = 1; %进入系统后，有1个顾客 member = [member,i]; else %系统有顾客正在接受服务，且等待队列未满，则第i个顾客进入系统 for j = 1:5 for k = 1:arr_num if info_arr(j,k) == total(1,i) %嵌套一个循环，与member(end)对应的数进行比较 for p = 1:5 for q = 1:arr_num if info_arr(p,q) == total(member(end)) info_wait(j,k) = info_left(p,q)-total(1,i); %info_arr(j,k);%等待时间=前1个顾客的离开时刻-该顾客到达时刻 info_left(j,k) = info_left(p,q)+ser_int(j,k); info_accept(j,k) = 1; total(4,i)=total(4,member(end))+ser_int(j,k);%离开时刻=前1个顾客的离开时刻+该顾客服务时间 end end end % end end end total(5,i) = number+1; member = [member,i]; end end %%分业务统计 num=zeros(7,1); for i=1:5 for j=1:arr_num if info_accept(i,j)==0 num(i,1) = num(i,1)+1; end end end num(6,1) = sum(num(1:5,1)); for i=1:len_sim if total(5,i) > 0 num(7,1)= num(7,1)+1; end end %% %输出结果 len_men = length(member) ls = mean(total(5,member)) num