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  • 网络存储系统NSS基准性能测试(7.17)

    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(balanced job bounds.m)

    clear;
     
    % input
    N = input('Simultaneous browser connections N = ');                           
     
    %initializtion
    %Interarrival Times in Seconds
    z1 = 30.0;
    z2 = 96.1;
    z3 = 246.7;
    z4 = 735.7;
     
    %Average * Time in Seconds
    t1 = 0.007229;
    t2 = 0.00036;
    t3 = 0.0000618;
    t4 = 0.2070;
    t5 = 0.6776;
    t6 = 0.506;
    t7 = 0.005294;
    t8 = 79.6;
    t9 = 118.8;
     
    %File Size in MB
    fs1 = 1.2;
    fs2 = 19.6;
    fs3 = 78.9;
    fs4 = 220.6;
     
    %Block Size in MB
    bs1 = 0.016;
    bs2 = 0.015;
     
    p1 = 0;
    p2 = 1;
    p3 = 0;
    p4 = 0;
    p5 = 0.8;
     
    STR = 15;
     
    %Average Seek in Seconds
    seek1 = 0.016;
    seek2 = 0.012;
    seek3 = 0.012;
    seek4 = 0.0115;
     
    %Average Latency in Seconds
    latency1 = 0.00833;
    latency2 = 0.00687;
    latency3 = 0.00687;
    latency4 = 0.00555;
     
    %Average Number of Seeks per Class
    n1 = 6;
    n2 = 11;
    n3 = 26;
    n4 = 60;
     
    %Transfer Rate in MB/s
    trate1 = 6.00;
    trate2 = 4.67;
    trate3 = 9.34;
    trate4 = 9.34;
     
    % response time of BJB
    D(1) = t1+t2+t3*(fs4/bs1)+t4+t5+(1-p1)*(t6+t7*(fs4/bs2));
    D(2) = p1*(1/STR)*((seek1+latency1)*n1+(bs1/trate1)*(fs4/bs1));
    D(3) = (5/2)*(seek1+latency1+bs1/trate1)*(1-p3)+(1/2)*(seek1+latency1+bs1/trate1)*(1-p1)*(1-p4)+(1/3)*(seek1+latency1+bs1/trate1)*(1-p1)*(1-p4);
    D(4) = (1-p1)*(1/4)*p5*t8+(1-p1)*(1-p5)*t9;
    Dmax = max(D(1:4));                                         
    Dsum = D(1)+D(2)+D(3)+D(4);                    
    Davg = Dsum/4;                                  
    for n = 1:N
        Rmin(n) = max(n*Dmax-z1, Dsum+((n-1)*Davg*Dsum/(Dsum+z1))); % lower bound of response time 
        Rmax(n) = Dsum+((n-1)*Dmax*(n-1)*Dsum/(((n-1)*Dsum)+z1));   % upper bound of response time    
    end
     
    % response time of MVA
    for m = 1:4
        L(m) = 0;                    
    end    
    for n = 1:N
        R(1) = D(1);
        for m = 1:4
            R(m) = D(m) * (1 + L(m));
        end
        Tau = n / sum(R(:));
        for m = 1:4
            L(m) = Tau * R(m);          
        end
        
        Rn(n,1) = D(1);
        for m = 1:4
            Ln(n,m) = L(m);
            Rn(n,m) = R(m);
        end
        Taun(n) = Tau;
    end
        
    for n = 1:N    
        RTn(n) = sum(Rn(n,1:4));      % average response time    
    end
     
    t = 1:N;
     
    % response time 
    figure(1), plot(t, Rmin, 'g', t, RTn, 'r', t, Rmax, 'b'),xlabel('Simultaneous browser connections'),ylabel('Response time(s)');

    三、NSS(网络存储系统)边界性能

     

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  • 原文地址:https://www.cnblogs.com/innosight/p/3271224.html
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