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  • 网络存储系统NSS基准性能测试(7.24)

    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;
     
    N = input('Simultaneous browser connections N = ');
    %model M/M/1
    Z = 0.003;
    %host
    E1 = 1/(64*33/8)+0.0002;
    for n = 1 : N
        u(n) = (sqrt(n*n+4*n)-n)/2;    %utilization in percent
        rate(1,n) = u(n)/E1;   %I/O request rate in IOPS
     
    end
        arate1 = sum(rate(1,n))/n;
        D(1) = E1 + arate1*(E1^2)/(2*(1-arate1*E1));
     
    %fcf
    E2 = 1/1062.5;
    for n = 1 : N
        arate2 = rate(1,n);
        D(2) = arate2*(E2^2)/(2*(1-arate2*E2));
    end
     
    %dacc
    h = 2.283;
    v = 2.798;
    E3 = 1/(64*66/8)+0.00016;
    E4 = 1/(64*66/8)+0.00018;
    ts = E3+(E3^2)*sqrt(1.39794);
    for n = 1 : N
        u(n) = (sqrt(n*n+4*n)-n)/2;    %utilization in percent
        rate(3,n) = u(n)/E3;   %I/O request rate in IOPS 
        rate(4,n) = u(n)/E4;   %I/O request rate in IOPS 
        arate3 = sum(rate(3,n))/n;
        arate4 = sum(rate(4,n))/n;
        D(3) = ts+((0.5*(h+(4*v-3*h-1)*0.5)+2*(1+h-2*v)*0.25)*(1/(E3^2))*arate3*E3*(2*(E3^2)))/((1/E3)-arate3);
        D(4) = ts+((0.5*(h+(4*v-3*h-1)*0.5)+2*(1+h-2*v)*0.25)*(1/(E4^2))*arate4*E3*(2*(E4^2)))/((1/E3)-arate4);
    end
     
    Dmax = max(D(1:4));                             % maximum service demand per code              
    Dsum = D(1)+D(2)+D(3)+D(4);           % sum of total service demands
    Davg = Dsum/4;                                  % average service demand per queue
     
    for n = 1:N
        Rmin(n) = max(n * Dmax - Z, Dsum + ((n-1)*Davg*Dsum/(Dsum+Z))); % lower bound of response time 
        Rmax(n) = Dsum + ((n-1)*Dmax*(n-1)*Dsum/(((n-1)*Dsum)+Z));      % 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/3271223.html
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