任务数与最大T延时的关系：仿真实验

f0 = 1000000000; f1 = 5000000000; C = 2000000; z = 0; Q = [5 10 15 20 25]; % 任务数作为自变量，确定随机数的生成个数 Tc = zeros(1, length(Q)); % 初始化每个Q对应的最大T延时 for s = 1:5 nums = randi([10000 50000], 1, Q(s)); % 随机数的生成，任务输入大小dn sorted_nums = sort(nums); max_T0 = 0; % 从小到大排序 for i = 1:Q(s) T0 = 0; T1 = 0; for m = 1:i c(m) = 1000 * sorted_nums(m); T0 = T0 + c(m) / f0; end for n = i + 1:Q(s) c(n) = 1000 * sorted_nums(n); d(n) = sorted_nums(n); T1 = T1 + d(n) / C + c(n) / f1; end if T0 >= T1 k = i; break; end end % 确定了断点，要进行补偿 for m = 1:k T0 = T0 - c(m) / f0; T1 = T1 + d(m) / C + c(m) / f1; if T0 <= T1 break; end max_T0 = T1; % 更新当前Q对应的最大T end Tc(s) = max_T0; % 记录当前Q对应的最大T end plot(Q, Tc); xlabel('任务数'); ylabel('最大T延时'); title('任务数与最大T延时的关系'); grid on;