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 = '); Z = input('Think time Z = '); Tmpro = input('Tmpro = '); Sm = input('Sm = '); STUm = input('STUm = '); Pm = input('Pm = '); Tmp = input('Tmp = '); IPnum = input('IPnum = '); STPm = input('STPm = '); Tsnpro = input('Tsnpro = '); Ssn = input('Ssn = '); Tsnp = input('Tsnp = '); IPsnnum = input('IPsnnum = '); Frames = input('Frames = '); TRate = input('TRate = '); Smr5 = input('Smr5 = '); Psn = input('Psn = '); seek = input('seek = '); latency = input('latency = '); % response time of BJB D(1) = Tmpro*Sm/STUm+(1-Pm)*Tmp*IPnum; D(2) = (1-Pm)/STPm*(Tsnpro*Ssn/STUm+Tsnp*IPsnnum); D(3) = (1-Pm)*IPnum*Frames/TRate; D(4) = (1-Pm)/STPm*IPsnnum*Frames/TRate; D(5) = (1-Pm)*Smr5/TRate; D(6) = (1-Pm)/STPm*(1-Psn)*(seek+latency+STUm/TRate)*Ssn/STUm; Dmax = max(D(1:6)); % maximum service demand per code Dsum = D(1)+D(2)+D(3)+D(4)+D(5)+D(6); % sum of total service demands Davg = Dsum/6; % 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 = 1:6 L(m) = 0; end for n = 1:N R(1) = D(1); for m = 1:6 R(m) = D(m) * (1 + L(m)); end Tau = n / sum(R(:)); for m = 1:6 L(m) = Tau * R(m); end Rn(n,1) = D(1); for m = 1:6 Ln(n,m) = L(m); Rn(n,m) = R(m); end Taun(n) = Tau; end for n = 1:N RTn(n) = sum(Rn(n,1:6)); % 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