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