最优化报告

最优控制及最优化实验

姓名：李佳乐

学号：40950104

班级：电091 班

学院：自动化学院

同组：张蕾陈瑜

最优化报告

实验一：线性规划

实验目的：要求学生能用 Matlab Optimization Toolbox求解线性规划问题。实验内容：练习使用 linprog求解线性规划问题。

【练习1】将例1 中的最优化问题的前三个不等式约束变为等式约束（如下所示），重新求解，判断解的存在性。（注：需要调用

[x,fval,exitflag,output,lambda] =linprog(...)，通过exitflag的值判断解的存在性）

min?5x1?4x2?6x3

s..tx1?x2?x3?20

3x1?2x2?4x3?42

3x1?2x2?30

x1,x2,x3?0

程序：

f=[-5;-4;-6];

A=[];b=[];

Aeq=[1,-1,1;3,2,4;3,2,0];

beq=[20;42;30];

lb=zeros(3,1);

[x,fval,exitflag,output,lambda]=linprog(f,A,b,Aeq,beq,lb,[])

结果：

Exiting: One or more of the residuals, duality gap, or total relative error

has grown 100000 times greater than its minimum value so far:

the primal appears to be infeasible (and the dual unbounded).

(The dual residual < TolFun=1.00e-008.)

x =

16.3148

0.0000

6.9098

fval =

-123.0326

exitflag =

-2

output =

iterations: 4

algorithm: 'large-scale: interior point'

cgiterations: 0

message: [1x258 char]

lambda =

ineqlin: [0x1 double]

eqlin: [3x1 double]

upper: [3x1 double]

lower: [3x1 double]

因为exitflag = -2<0，所以没有解。

【练习2】建一个.m文件，求解下面的线性规划问题。

min?2x1?x2?4x3?3x4?x5

s..2tx2?x3?4x4?2x5?54

3x1?4x2?5x3?x4?x5?62

x1,x2?0,x3?3.32,x4?0.678,x5?2.57

程序：

f=[-2;-1;-4;-3;-1];

A=[0 2 1 4 2;3 4 5 -1 -1];

b=[54;62];

Aeq=[];

beq=[];

lb=[0;0;3.32;0.678;2.57];

[x,fval,exitflag,output,lambda]=linprog(f,A,b,Aeq,beq,lb,[])

结果：

Optimization terminated.

x =

19.7850

0.0000

3.3200

11.3850

2.5700

fval =

-89.5750

exitflag =

output =

iterations: 6

algorithm: 'large-scale: interior point' cgiterations: 0

message: 'Optimization terminated.'

lambda =

ineqlin: [2x1 double]

eqlin: [0x1 double]

upper: [5x1 double]

lower: [5x1 double]

exitflag =1>0，所以收敛为一个解。 4 因为

实验二：二次规划问题的求解

实验目的：要求学生能用 Optimization toolbox求解二次规划问题。实验内容：使用quadprog()函数求解二次规划问题

min1xTHx?fT

二次规划问题的数学表示为：

s..tAx?b

Aeqx?beq

lb?x?ub

调用格式：

x = quadprog(H,f,A,b)

x = quadprog(H,f,A,b,Aeq,beq)

x = quadprog(H,f,A,b,Aeq,beq,lb,ub)

[x,fval] = quadprog(...)

[x,fval,exitflag,output,lambda] = quadprog(...)

【练习3】试求解下面的二次规划问题

min(x22

1?1)?(x2?2)2?(x3?3)2?(x4?4)

s..tx1?x2?x3?x4?5

1?3x2?2x3?x4?10

x1,x2,x3,x4?0

程序：

H=diag([1 1 1 1]);

f=[-2;-4;-6;-8];

A=[1,1,1,1;3,3,2,1];

b=[5,10];

lb=zeros(4,1);

[x,fval,exitflag,output,lambda]=quadprog(H,f,A,b)

结果：

x =

-1.7500

0.2500

2.2500

4.2500

fval =

-31.8750

exitflag =

output =

iterations: 2

algorithm: 'medium-scale: active-set' firstorderopt: []

cgiterations: []

message: 'Optimization terminated.'

lambda =

lower: [4x1 double]

upper: [4x1 double]

eqlin: [0x1 double]

ineqlin: [2x1 double]

exitflag =1>0，所以收敛为一个解。 6 因为

实验三：用加权法求解多目标优化问题

实验目的：掌握用加权法求解多目标优化问题

实验内容：用加权法解多目标线性优化问题min Cx st. Ax>=b，即最小化两个目

标函数 1 1 2 f (x) = 3x + x 和2 1 2 f (x) = ?x ? 2x st. Ax >=b。

实验方法：使用加权法思想，将两个目标的优化问题转换成一个目标的优化问

题。分别设置两个目标的权重为w1 和w2, 将两个目标加权

w1f1(x)+w2f(x)，然后优化。即minw1f1(x)+w2f(x) st. Ax>=b. 通

过取不同的权重，可以得到多个解。令1>=w1,w2>=0, w2=1-w1, 分

别取w1=0,0.2，0.4，0.6，0.8，1求解w1f1(x)+w2f(x)，给出最优

解，并计算出两个目标函数的值。

程序：

for W1=0:0.2:1

W2=1-W1

f=[3*W1-W2;W1-2*W2];

A=[0 -1;-3 1;1 0;0 1];

b=[-3;-6;0;0];

[x,fval]=linprog(f,A,b,[],[],[],[])

[x,fval,exitflag,output,lambda] =linprog(f,A,b,[],[],[],[]);

f1=[3 1]*x

f2=[-1 -2]*x

end

运行结果：

W2 =

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

x =

3.7982

1.3663

fval =

-6.5308

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

f1 =

12.7609

f2 =

-6.5308

W2 =

0.8000

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

x =

1.7031

1.0469

fval =

-1.8063

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

f1 =

6.1563

f2 =

-3.7969

W2 =

0.6000

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

x =

2.1537

1.1969

fval =

0.3347

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

f1 =

7.6582

f2 =

-4.5476

W2 =

0.4000

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

x =

-0.1104

-6.3251

fval =

1.1105

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

f1 =

-6.6562

f2 =

12.7605

W2 =

0.2000

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

x =

0.4425

-4.2035

fval =

-0.7079

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

the primal appears to be infeasible and the dual unbounded since the dual objective > 1e+10

and the primal objective > -1e+6.

f1 =

-2.8760

f2 =

7.9646

W2 =

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

both the primal and the dual appear to be infeasible.

x =

0.4557

-12.2588

fval =

-10.8917

Exiting: One or more of the residuals, duality gap, or total relative error has stalled:

both the primal and the dual appear to be infeasible.

f1 =

-10.8917

f2 =

24.0619

实验四：用支持向量机方法实现数据分类

训练数据：下面的数据为DNA 分类数据（共10 个），其中前面5个数据为第一类，后面5为第二类。

0.29729731 0.13513514 ;

0.27027027 0.15315315 ;

0.27027027 0.06306306 ;

0.42342342 0.28828829 ;

0.23423424 0.10810811 ;

0.35454544 0.50000000 ;

0.32727272 0.50000000 ;

0.25454546 0.51818180 ;

0.30000000 0.50000000 ;

0.29090910 0.64545456 ;

未知分类信息的数据

0.35135136 0.12612613;

0.35135136 0.18918919;

0.27927927 0.18918919;

0.20909090 0.15454545;

0.18181818 0.13636364;

0.36363636 0.46363636;

0.35454544 0.26363636;

0.29090910 0.50000000;

0.21818182 0.56363636;

0.20000000 0.56363636;

要求：（1）画出训练集数据点，其中第一类用“o”表示，第二类用“*”来表示。（2）求解二次规划问题，给出分类超平面的参数w,b并在图中画出分类超平面（可画一段线段）。（3）对未知数据进行分类，分类结果需在图中展现。属于第一类的请用实心的“.”来表示，属于第二类的请用“+”来表示。（4）给出全部程序代码。

程序：

x1=[0.29729731 0.27027027 0.27027027 0.42342342 0.23423424];

y1=[0.13513514 0.15315315 0.06306306 0.28828829 0.10810811];

x2=[0.35454544 0.32727272 0.25454546 0.30000000 0.29090910];

y2=[0.50000000 0.50000000 0.51818180 0.50000000 0.64545456];

plot(x1,y1,'o')

hold on

plot(x2,y2,'*')

H=[1 0 0;0 1 0;0 0 0];

A=[-0.29729731 -0.13513514 -1;-0.27027027 -0.15315315 -1;-0.27027027 -0.06306306 -1;-0.42342342 -0.28828829 -1;-0.23423424 -0.10810811 -1;0.35454544 0.50000000 1;0.32727272 0.50000000 1;0.25454546 0.51818180 1;0.30000000 0.50000000

1;0.29090910 0.64545456 1];

b=[-1;-1;-1;-1;-1;-1;-1;-1;-1;-1];

[x,fval]=quadprog(H,[],A,b,[],[],[],[])

hold on

x=0:1;

plot(x,(2.7792*x+2.2859)/8.5429,'y' )

hold on

plot(x,(2.7792*x+3.2859)/8.5429 )

hold on

plot(x,(2.7792*x+1.2859)/8.5429 )

w=[2.7792;-8.5426];

x2=[0.35135136 0.35135136 0.27927927 0.20909090 0.18181818 0.36363636 0.35454544 0.29090910 0.21818182 0.20000000];

y2=[0.12612613 0.18918919 0.18918919 0.15454545 0.13636364 0.46363636 0.26363636 0.50000000 0.56363636 0.56363636];

for n=1:10;

if([x2(n) y2(n)]*w+2.2859) >= 0;

hold on

plot(x2(n),y2(n),'g.')

else

hold on

plot(x2(n),y2(n),'g+')

end

结果： x =

2.7792 -8.5426

2.2859

fval =

40.3502

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