平衡边界作业算法并发仿真测试基于三层架构的Web系统的基准性能

传送门 ☞ Android兵器谱 ☞ 转载请注明 ☞ http://blog.csdn.net/leverage_1229

一、实验参数列表


二、MATLAB脚本(appr_mva_bounds.m)

clear;

% input
N1 = input('Number of customers of class 1 N1 = ');   % number of customers of class 1
N2 = input('Number of customers of class 2 N2 = ');   %  number of customers of class 2
M = input('number of applicaion tiers M = ');         % number of applicaion tiers
 
for k = 2:(M+1)
    S1(k) = input('Average service time S1 = ');  % average service time per tier of class 1
    V1(k) = input('Visit number V1 = ');          % visit number per tier of class 1
end
Z1 = input('Think time Z1 = ');         % average think time of class 1
 
for k = 2:(M+1)
    S2(k) = input('Average service time S2 = ');  % average service time per tier of class 2
    V2(k) = input('Visit number V2 = ');          % visit number per tier of class 2
end
Z2 = input('Think time Z2 = ');         % average think time of class 2

% initialization
S1(1) = Z1;
V1(1) = 1;
for m = 1:(M+1)
    D1(m) = V1(m) * S1(m);              % service demand per tier of class 1
end

S2(1) = Z2;
V2(1) = 1;
for m = 1:(M+1)
    D2(m) = V2(m) * S2(m);              % service demand per tier of class 2
end

vN2 = [1 5:5:N2];

for n = 1:length(vN2)                   % up to a maximum number of customers
    N2 = vN2(n);

% APPROXIMATE MVA ALGORITHM
% initialization
    for m = 2:(M+1)
        Q1(m) = N1/M;
        Q2(m) = N2/M;
    end
    error = [1 1 1 1 1 1];

    R1(1) = D1(1);
    R2(1) = D2(1);

    while max(error) > 0.01
        for m = 2:(M+1)
            A1(m) = (N1-1)/N1*Q1(m) + Q2(m);        % average number of customers per tier of class 1
            A2(m) = (N2-1)/N2*Q2(m) + Q1(m);        % average number of customers per tier of class 2
        end
   
        for m = 2:(M+1)
            R1(m) = D1(m) * (1 + A1(m));            % average delay per tier of class 1
            R2(m) = D2(m) * (1 + A2(m));            % average delay per tier of class 2 
        end
    
        X1 = N1 / (sum(R1(:)));         % throughput of class 1
        X2 = N2 / (sum(R2(:)));         % throughput of class 2
    
        for m = 2:(M+1)
            Q1old(m) = Q1(m);           % old length of queue 1
            Q2old(m) = Q2(m);           % old length of queue 2
            Q1(m) = X1 * R1(m);         % new length of queue 1
            Q2(m) = X2 * R2(m);         % new length of queue 2
        end
    
        for m = 2:(M+1)
            error(m-1) = abs((Q1(m) - Q1old(m)) / Q1old(m));
            error(m-1+M) = abs((Q2(m) - Q2old(m)) / Q2old(m));
        end
    end

    
    for m = 2:(M+1)
        U1n(n,m) = X1 * S1(m) * V1(m);  % tier utilization of class 1
        U2n(n,m) = X2 * S2(m) * V2(m);  % tier utilization of class 2
    end
 
    X1n(n) = X1;                        % throughput of class 1
    X2n(n) = X2;                        % throughput of class 2
    
    RT1n(n) = sum(R1(2:(M+1)));         % response time of class 1
    RT2n(n) = sum(R2(2:(M+1)));         % response time of class 2
end


% BALANCED JOB BOUNDS
D1max = max(D1(2:(M+1)));               % maximum service demand per queue of class 1             
D2max = max(D2(2:(M+1)));               % maximum service demand per queue of class 2              
D1sum = sum(D1(2:(M+1)));               % sum of total service demands of class 1
D2sum = sum(D2(2:(M+1)));               % sum of total service demands of class 2
D1avg = D1sum/M;                        % average service demand per queue of class 1
D2avg = D2sum/M;                        % average service demand per queue of class 2
for n = 1:length(vN2)
    N = vN2(n) + N1;
    R2min(n) = max(N * D2max - Z2, D2sum + ((N-1)*D2avg*D2sum/(D2sum+Z2))); % lower bound of response time
    R2max(n) = D2sum + ((N-1)*D2max*(N-1)*D2sum/(((N-1)*D2sum)+Z2));        % upper bound of response time

    X2min(n) = vN2(n) / (Z2 + R2max(n));            % lower bound of throughput
    X2max(n) = vN2(n) / (Z2 + R2min(n));            % upper bound of throughput

     for m = 2:(M+1)
         U2minm(n,m) = X2min(n) * D2(m);            % lower bound of utilization per tier
         U2maxm(n,m) = X2max(n) * D2(m);            % upper bound of utilization per tier
     end
end

t = vN2;


% response time 
figure(1), plot(t, RT2n,'r', t, R2min, 'g', t, R2max, 'b'), xlabel('Simultaneous browser connections'), ylabel('Response time(s)'); 

% throughput
figure(2), plot(t, X2n,'r', t,X2min,'g', t, X2max, 'b'), xlabel('Simultaneous browser connections'), ylabel('Throughput(1/s)');
三、平衡作业边界预测响应时间和吞吐率

 




posted @ 2013-05-29 06:08  Innosight  阅读(171)  评论(0编辑  收藏  举报