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Merge pull request '第一周' (#1) from 姚安欣 into main

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p3itgm2rp 4 months ago
parent 1408686570 d455ce1565
commit 371c0c70c3
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40
LPej1_3.m
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clc; clear; close all; format long g;
prob=optimproblem;
x=optimvar('x',9,1,'LowerBound',0,'Type','integer');
p=[0.25,0.35,0.50];
r=[1.25,2.00,2.80];
ot=[5,10,0;7,9,12;6,8,0;4,0,11;7,0,0];
total=1000:200:3000;
odp=[300/6000,321/10000,250/4000,783/7000,200/4000];
X=[];
Q=[];
for i=1:length(total)
% prob.Objective=(ot(1,1)*ones(1,3)*x(1:3)+ot(1,2)*x(7))*odp(1)+(ot(2,1)*ones(1,3)*x(4:6)+ot(2,2)*x(8)+ot(2,3)*x(9))*odp(2)+(ot(3,1)*(x(1)+x(4))+ot(3,2)*(x(7)+x(8)))*odp(3)+(ot(4,1)*(x(2)+x(5))+ot(4,3)*x(9))*odp(4)+(ot(5,1)*(x(3)+x(6)))*odp(5)-(r(1)-p(1))*ones(1,6)*x(1:6)-(r(2)-p(2))*ones(1,2)*x(7:8)-(r(3)-p(3))*x(9);
dp1=(ot(1,1)*ones(1,3)*x(1:3)+ot(1,2)*x(7))*odp(1);
dp2=(ot(2,1)*ones(1,3)*x(4:6)+ot(2,2)*x(8)+ot(2,3)*x(9))*odp(2);
dp3=(ot(3,1)*(x(1)+x(4))+ot(3,2)*(x(7)+x(8)))*odp(3);
dp4=(ot(4,1)*(x(2)+x(5))+ot(4,3)*x(9))*odp(4);
dp5=(ot(5,1)*(x(3)+x(6)))*odp(5);
dp=dp1+dp2+dp3+dp4+dp5;
prob.Objective=dp-(r(1)-p(1))*ones(1,6)*x(1:6)-(r(2)-p(2))*ones(1,2)*x(7:8)-(r(3)-p(3))*x(9);
prob.Constraints.con1=p(1)*ones(1,6)*x(1:6)+p(2)*ones(1,2)*x(7:8)+p(3)*x(9)+dp<=total(i);%总费用
prob.Constraints.con2=ot(1,1)*ones(1,3)*x(1:3)+ot(1,2)*x(7)<=6000;%A1
prob.Constraints.con3=ot(2,1)*ones(1,3)*x(4:6)+ot(2,2)*x(8)+ot(2,3)*x(9)<=10000;
prob.Constraints.con4=ot(3,1)*(x(1)+x(4))+ot(3,2)*(x(7)+x(8))<=4000;
prob.Constraints.con5=ot(4,1)*(x(2)+x(5))+ot(4,3)*x(9)<=7000;
prob.Constraints.con6=ot(5,1)*(x(3)+x(6))<=4000;
[sol,fval,flag,out]=solve(prob);
xx=sol.x;
X=[X,xx];
Q=[Q,-fval];
end
plot(total,Q,'*-');

61
Mengtkl.m
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%27页例题
clc; clear; close all;
n = 10000; % 使用较小的 n 值以便更容易可视化
x = unifrnd(0, 12, [1, n]);
y = unifrnd(0, 9, [1, n]);
ans=sum(y < x.^2 & x <= 3)+sum(y < 12 - x & x >= 3);
ans=ans/n;
% 找出满足条件的点
condition1 = y < x.^2 & x <= 3;
condition2 = y < 12 - x & x >= 3;
condition_met = condition1 | condition2; % 满足任一条件的点
condition_not_met = ~condition_met; % 不满足任何条件的点
% 创建图形窗口
figure;
hold on;%在同一张图上绘图
% 绘制不满足任何条件的点
scatter(x(condition_not_met), y(condition_not_met), 'k.'); % k----黑色 .----绘制样式
%scatter绘制散点图
%x(condition_not_met) 会返回一个新的向量，其中只包含 x 中对应 condition_not_met 为 true 的元素。
% 绘制满足第一个条件的点
scatter(x(condition1), y(condition1), 'r.'); % 红色
% 绘制满足第二个条件的点
scatter(x(condition2), y(condition2), 'b.'); % 蓝色
% 添加图例和标签
legend('不满足任何条件的点', '满足 y < x^2 且 x <= 3 的点', '满足 y < 12 - x 且 x >= 3 的点');
xlabel('x');
ylabel('y');
title('随机生成的点和满足条件的点');
hold off;
%蒙特卡洛法求圆周率qw
clc;clear;close all;
n=10^5;
x=unifrnd(-1,1,[1,n]);
y=unifrnd(-1,1,[1,n]);
con1=x.^2+y.^2<=1;
con2=~con1;
ans=sum(x.^2+y.^2<=1);
ans=ans/n*4;
figure ;
hold on;
scatter(x(con1),y(con1),'r.');
scatter(x(con2),y(con2),'k.');
hold off;

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NLP.m
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%非线性规划
%P42例题
%求解
clc,clear,format long g,close all;
syms x1 x2;%符号变量(用于定义函数、求解方程、计算导数和积分)
f=(339-0.01*x1-0.003*x2)*x1+(399-0.004*x1-0.01*x2)*x2-(195*x1+225*x2+400000);
f=simplify(f);
f1=diff(f,x1),f2=diff(f,x2);
[x10,x20]=solve(f1,f2);%求驻点即使f1,f2为0的点
x10=round(double(x10));%四舍五入
x20=round(double(x20));
f0=subs(f,{x1,x2},{x10,x20});
f0=double(f0);
subplot(121);
fmesh(f,[0,10000,0,10000]);%3维
%xlabel('x1','Interpreter','latex');
%ylabel('y1','Interpreter','latex');
xlabel('x1');
ylabel('y1');
subplot(122);
L=fcontour(f,[0,10000,0,10000]);
contour(L.XData,L.YData,L.ZData,'ShowText','on');
xlabel('x1');
ylabel('y1');
p1=339-0.01*x10-0.003*x20;
p2=399-0.004*x10-0.01*x20;
c=195*x10+225*x20+400000;
rate=f0/c;
%灵敏性检验 分析 最优解 对 19村彩电价格下降幅度 的灵敏性
clc,clear,close all,format long g;
syms x1 x2 a;
X1=4735,X2=7043,AA=0.01,FF=553641;
f=(339-a*x1-0.003*x2)*x1+(399-0.004*x1-0.01*x2)*x2-(195*x1+225*x2+400000);
f1=diff(f,x1),f2=diff(f,x2);
[x10,x20]=solve(f1,f2);
subplot(121);
fplot(x10,[0.002,0.02]);
xlabel('a');
ylabel('x1');
subplot(122);
fplot(x20,[0.002,0.02]);
xlabel('a');
ylabel('x2');
dx1=diff(x10,a);
dx10=subs(dx1,a,0.01),dx10=double(dx10);
sx1a=dx10*AA/X1;
dx2=diff(x20,a);
dx20=subs(dx2,a,0.01),dx20=double(dx20);
sx2a=dx20*AA/X2;
F=subs(f,[x1,x2],[x10,x20]);
F=simplify(F);
figure,fplot(F,[0.002,0.02]);
xlabel('a');
ylabel('profit');
dpro=diff(F,a);
dpro0=subs(F,a,0.01);
sya=dpro0*AA/FF;
%当a提高10%
f3=subs(f,[x1,x2,a],[X1,X2,0.011]),f3=double(f3);
f4=subs(F,[a],0.011);
%相对误差
delta=(f4-f3)/f4;
delta=double(delta);
%相对误差很小，说明在19寸的价格下降幅度在一定范围内发生变化时，仍然按照不变时的最优解来确定生产方案，仍然可以获得几乎最优的利润。
%使用fmincon的简单例子
%蒙特卡罗确定初始值
clc,clear,close all;
n=10^7;
x1=unifrnd(0,12,[1,n]);
x2=unifrnd(0,12,[1,n]);
fmin=+inf;
for i=1:n
x=[x1(i),x2(i)];
if ((x1(i)-1)^2-x2(i)-6<=0) & (-2*x1(i)+3*x2(i)-6<=0)
temp=x(1)^2+x(2)^2-x(1)*x(2)-2*x(1)-5*x(2);
if temp<fmin
fmin=temp;
x0=x;
end
end
end
%[x,fval]= fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options)
A=[-2,3];
b=6;
[x,fval]=fmincon(@fun1,x0,A,b,[],[],[],[],@nonlfun1);
%option=optimoptions('fmincon','Algorithm','sqp');
%[x,fval]=fmincon(@fun1,x0,A,b,[],[],[],[],@nonlfun1,option);
disp('Thw optimal solution is:');
disp(x);
disp('The minimum value is:');
disp(fval);
figure;
fmesh(@(x1, x2) arrayfun(@(x1, x2) fun1([x1, x2]), x1, x2), [-4, 4, -4, 4]);%用法
xlabel('x1');
ylabel('x2');
zlabel('f(x1,x2)');
%非线性规划---料场选址
clc,clear,close all;
a=[1.25,8.75,0.5,5.75,3,7.25];
b=[1.25,0.75,4.75,5,6.5,7.25];
A=[1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0;0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0];
bb=[20;20];
Aeq=[eye(6),eye(6),zeros(6,4)];
beq=[3;5;4;7;6;11;];
lb=[zeros(12,1);-inf;-inf;-inf;-inf];
x0=[3,5,0,7,0,1,0,0,4,0,6,10,5,1,2,7];
[xans,fvans]=fmincon(@fun2,x0,A,bb,Aeq,beq,lb,[]);

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fun1.m
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function f=fun1(x)
f=x(1)*x(1)+x(2)*x(2)-x(1)*x(2)-2*x(1)-5*x(2);
end

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fun2.m
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function f=fun2(x)
a=[1.25,8.75,0.5,5.75,3,7.25];
b=[1.25,0.75,4.75,5,6.5,7.25];
f1 = 0; % 初始化f1
f2 = 0; % 初始化f2
for i=1:6
s=sqrt((a(i)-x(13))^2+(b(i)-x(14))^2);
f1=x(i)*s+f1;
end
for i=7:12
s=sqrt((a(i-6)-x(15))^2+(b(i-6)-x(16))^2);
f2=s*x(i)+f2;
end
f=f1+f2;
end

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intLP.m
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%整数规划
%P23 ej2.5
%optimproblem解法
clc; clear; close all; format long g;
prob=optimproblem;
x=optimvar('x',6,'LowerBound',0,'Type','integer');
prob.Objective=sum(x);
cnt=[35,40,50,45,55,30];
con=optimconstr(6);
con(1)=x(1)+x(6)>=35;
for i=1:5
con(i+1)=x(i)+x(i+1)>=cnt(i+1);
end
prob.Constraints.con=con;
[sol,fval,flag,out]=solve(prob);
X=sol.x;
%linprog解法
clc; clear; close all; format long g;
f=[1,1,1,1,1,1];
intcon=[1:6];
A=zeros(6,6);
A(1,1)=-1;
A(1,6)=-1;
for i=1:5
A(i+1,i)=-1;
A(i+1,i+1)=-1;
end
lb=zeros(6,1);%注意不是lb=0
b=[-35;-40;-50;-45;-55;-30];
[x,fval]=intlinprog(f,intcon,A,b,[],[],lb);
%01规划
%背包问题
f=-[540,200,180,350,60,150,280,450,320,120];
intcon=1:10;
lb=zeros(10);
ub=ones(10);
A=[6,3,4,5,1,2,3,5,4,2];
b=30;
[x,fval]=intlinprog(f,intcon,A,b,[],[],lb,ub);

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mm1.m
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%投资利益与风险1998A题
%模型一 给定风险承受程度，求最大利益
f=[-0.05,-0.27,-0.19,-0.185,-0.185];
%A矩阵
%A=[0,0.025,0.015,0.055,0.026]; %错误
%b=[1,1,1,1,1];
A=[zeros(4,1),diag([0.025,0.015,0.055,0.026])];%不等式约束条件矩阵
%Aeq、beq
Aeq=[1,1.01,1.02,1.045,1.065];
beq=1;
%lb
%lb=0; %错误
lb=zeros(5,1);
%可承担风险率a
a=(0:0.001:0.05);
%保存最优解
Q=zeros(1,length(a));
xx=[];%空矩阵存放最优解对应x的值
for i=1:length(a)
b=a(i)*ones(4,1);
[x,y]=linprog(f,A,b,Aeq,beq,lb);
Q(i)=-y;%注意取负！！！
xx=[xx;x'];
end
plot(a,Q,'*r');
xlabel("风险率");
ylabel("最大收益");
%模型二 收益、风险按权重组合
% f0=[-0.05,-0.27,-0.19,-0.185,-0.185];
% w=(0:0.1:1);
% Aeq=[1,1.01,1.02,1.045,1.065,0];
% beq=1;
% lb=0;
% xx=[];
% Q=zeros(1,length(w));
% A=[zeros(5,1),diag([0.025,0.025,0.055,0.065,0])];
% b=ones(5,1);
% for i=1:length(w)
% f=[-w(i)*f0,1-w(i)];
% b=x(end)*b;
% [x,y]=linprog(f,A,b,Aeq,beq,lb);
% Q(i)=-y;
% xx=[xx,x'];
% end
% plot(w,Q,'*r');
%模型二 收益、风险按权重组合
clc; clear; close all; format long g;
M = 10000;
prob = optimproblem;
x = optimvar('x', 6, 1, 'LowerBound', 0);
r = [0.05, 0.28, 0.21, 0.23, 0.25];
p = [0, 0.01, 0.02, 0.045, 0.065];
q = [0, 0.025, 0.015, 0.055, 0.026]';
%w = 0:0.1:1;
w = 0.7:0.03:1;
V = [];
Q = [];
X = [];
prob.Constraints.con1 = (1 + p) * x(1:end-1) == M;
prob.Constraints.con2 = (q(2:end).* x(2:end-1))<= x(end); %下标从1开始
for i = 1:length(w)
prob.Objective = w(i) * x(end) - (1 - w(i)) * (r - p) * x(1:end-1); %注意大小写
[sol, fval, flag, out] = solve(prob);
xx = sol.x;
V = [V, max(q.* xx(1:end-1))];
Q = [Q, (r - p) * xx(1:end-1)];
X = [X, xx];
plot(V, Q, '*-');
grid on;
xlabel('风险');
ylabel('收益');
end
%模型三：达到一定盈利水平，极小化风险
clc; clear; close all; format long g;
M=10000;
k=1500:100:3000;
prob = optimproblem;
x = optimvar('x', 5, 1, 'LowerBound', 0);%下界为0
r = [0.05, 0.28, 0.21, 0.23, 0.25];
p = [0, 0.01, 0.02, 0.045, 0.065];
q = [0, 0.025, 0.015, 0.055, 0.026]';
V = [];
Q = [];
X = [];
t = optimvar('t', 'LowerBound', 0);
%prob.Objective=max(q.*x);%极小化风险
for i=1:length(k)
prob.Objective = t; % 极小化风险
prob.Constraints.con1 = (1 + p) * x == M;
prob.Constraints.con2=((r-p)*x>=k(i));%达到一定盈利水平
prob.Constraints.con3=(q.*x<=t);
[sol,fval,flag,out]=solve(prob);
if flag==1
xx=sol.x;
X=[X,xx];
Q=[Q,(r-p)*xx];
V=[V,fval];
else
xx=-1*ones(5,1);
X=[X,xx];
Q=[Q,-1];
V=[V,-1];
end
end
plot(k,V,'*-');

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mm2_graph.m
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%ej.4.10仓库选址
clc; clear; close all; format long g;
a=zeros(6);
a(1,[2,5])=[20,15];
a(2,[1,3,4,5])=[20,20,60,25];
a(3,[2,4,5])=[20,30,18];
a(4,[2,3])=[60,30];
a(5,[1,2,3,6])=[15,25,18,15];
a(6,5)=15;
s = cellstr(strcat('销售点',int2str([1:6]')));
G=graph(a,s);
P = plot(G,'layout','force','EdgeColor','k','NodeFontSize',12);
D=[];
for i=1:6
temp=0;
for j=1:6
[path,d]=shortestpath(G,i,j);
if d>temp;
temp=d;
end
end
D=[D,temp];
end
[minVal,minInd]=min(D);
disp('仓库应建在销售点');
disp(minInd);
%答案版
clc; clear; close all;
% 初始化邻接矩阵
a = zeros(6);
a(1, [2, 5]) = [20, 15];
a(2, 3:5) = [20, 60, 25];
a(3, [4, 5]) = [30, 18];
a(5, 6) = 15;
% 创建节点标签
s = cellstr(strcat('V', int2str([1:6]')));
% 创建图
G = graph(a, s, 'upper');
% 计算所有节点之间的最短路径距离矩阵
d = distances(G);
% 计算从每个节点到其他节点的最大最短路径距离
D = max(d, [], 2)'; %第三个参数 2 指定沿着矩阵的第二个维度（即列）进行操作。
% 找到最大最短路径距离中的最小值及其下标
[minD, minIndex] = min(D);
% 绘制图
plot(G, 'EdgeLabel', G.Edges.Weight, 'Layout', 'force');%设置边的标签
% 显示结果
fprintf('从每个节点到任意其他节点的最大最短路径距离: \n');
disp(D);
fprintf('最大最短路径距离中的最小值为: %f\n', minD);
fprintf('对应的节点为: V%d\n', minIndex);
%设备更新问题；
clc;clear;close all;
a=zeros(6);
a(1,[2:6])=[15,20,27,37,54];
a(2,[3:6])=[15,20,27,37];
a(3,[4:6])=[16,21,28];
a(4,[5:6])=[16,21];
a(5,6)=17;
s=cellstr(strcat(int2str((1:6)'),'年初'));
G=digraph(a,s);
P = plot(G,'layout','force','EdgeColor','k','NodeFontSize',12);
[path,d]=shortestpath(G,1,6);%注意参数个数
highlight(P,path,"EdgeColor","red",'LineWidth',3.5);
%最小生成树
clc;clear;
a=zeros(9);
a(1,[2:9])=[2 1 3 4 4 2 5 4]; % 顶点1到其他顶点的边的权重
a(2,[3 9])=[4 1]; % 顶点2到顶点3、顶点9的边的权重
a(3,4)=1; % 同上。因为写过1到3，和2到3的边的权重，无需重复设
a(4,5)=1;
a(5,6)=5;
a(6,7)=2;
a(7,8)=3;
a(8,9)=5;
s=cellstr(strcat('节点',int2str([1:9]')));
G=graph(a,s,'upper');
P=plot(G,'layout','force');
T=minspantree(G,'Method','sparse');
L=sum(T.Edges.Weight);
highlight(P,T,"EdgeColor","red",'LineWidth',2.5);
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