mm1排队模型matlab,基于排队论mm1m模型的仿真，代码和出现问题如正文

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%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，

%还请熟悉matlab编程的高手帮忙找找问题所在，小妹不胜感激

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clear all;close all;clc;

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% 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 %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