a03共轭梯度法编程

1、MK7FCCK282Z3SVMK6CZBW26Q6LZ9 MKHMNXTJJQ4S83 MKMDE2RESG33HSP90 第二章 11( 2)用大M法求解min w=2x1+x2-x3-x4s.t x1-x2+2x3-x4=2 2x1+x2-3x3+x4=6 x1+x2+x3+x4=7xi 0, i = 1,2, 3, 4 用 matlab 求解如下： f=2,1,-1,-1,200,200,200;a=1 -1 2 -1 1 0 0;2 1 -3 1 0 1 0;1 1 1 1 0 0 1; b=2 6 7;lb=zeros(7,1); x,fval,exitflag,output,lam

2、bda=linprog(f,a,b,lb) Optimization terminated.运行结果如下：x =3.00000.00001.00003.00000.00000.00000.0000 fval =2.0000exitflag =1output =iterations: 7 algorithm: large-scale: interior point cgiterations: 0message: Optimization terminated. lambda =f*=2 。ineqlin: 0x1 double eqlin: 3x1 double upper: 7x1 doubl

3、e lower: 7x1 double 从上述运行结果可以得出：最优解为 x= ，最小值约为P151第三章 26用共轭梯度算法求f(x) = (x1-1)A2+5*(x2-x1A2)A2的极小点，取初始点x0=。用 matlab 求解如下：function mg=MG()%共轭梯度法求解习题三第26 题%clc;clear;n=2;x=2 0;max_k=100;count_k=1;trace(1,1)=x(1);trace(2,1)=x(2);trace(3,1)=f_fun(x);k=0;g1=f_dfun(x);s=-g1;while count_k=max_kif k=ng0=f_df

4、un(x);s=-g0;k=0;elser_min=fminbnd(t) f_fun(x+t*s),-100,100); x=x+r_min*s;g0=g1;g1=f_dfun(x);if norm(g1)0.001 & k300)if k=0p=-H0*g1;elsevk=sk/(sk*yk)-(H0*yk)/(yk*H0*yk); w1=(yk*H0*yk)*vk*vk; H1=H0-(H0*yk*yk*H0)/(yk*H0*yk)+(sk*sk)/(sk*yk)+w1;p=-H1*g1;H0=H1;endx00=x0;result=Usearch1(f,x1,x2,df,x0,p);ar

5、f=result(1);x0=x0+arf*p;g0=g1;g1=subs(df,x1,x2,x0(1,1),x0(2,1);p0=p;yk=g1-g0; sk=x0-x00;k=k+1;end;kx0f0=subs(f,x1,x2,x0(1,1),x0(2,1)function result=Usearch1(f,x1,x2,df,x0,p) mu=0.001;sgma=0.99;a=0;b=inf;arf=1; pk=p;x3=x0;x4=x3+arf*pk;f1=subs(f,x1,x2,x3(1,1),x3(2,1);f2=subs(f,x1,x2,x4(1,1),x4(2,1); g

6、k1=subs(df,x1,x2,x3(1,1),x3(2,1); gk2=subs(df,x1,x2,x4(1,1),x4(2,1);while (f1-f20)if(gk2*pksgma*gk1*pk)a=arf;a=min(2*arf,(a+b)/2);x4=x3+arf*pk;f2=subs(f,x1,x2,x4(1,1),x4(2,1);gk2 =subs(df,x1,x2,x4(1,1),x4(2,1);while (f1-f2 0g2(x) = x1 0用 matlab 求解如下： function sumt=SUMT(x0,e0,max_k0) %SUM内点法 global x

7、 s Mx=x0;e=e0;max_k=max_k0;trace(1,1)=x(1);trace(2,1)=x(2);M=100;c=3;e_FR=10A-10;max_FR=200;for k=0:max_kx=FR(x,e_FR,max_FR);trace(1,k+2)=x(1);trace(2,k+2)=x(2);if f_pfun(x)ebreak;endM=c*M;end f=f_fun(x)ktracefunction f=FR(x0,e,max_k)global x s;count_k=1;k=0;x=x0;g1=f_dfun(x);s=-g1;while count_k=max

8、_kif k=ng0=f_dfun(x);s=-g0;k=0;elser_min=fminbnd(t) f_fun(x+t*s),-100,100); x=x+r_min*s;%g0=g1;g1=f_dfun(x)if norm(g1)=0&b=0g(1,1)=1;g(2,1)=1;elseif a=0g(1,1)=1+2*M*(-x(1)A2+x(2)*(-2*x(1); g(2,1)=1+2*M*(-x(1)A2+x(2);elseif a=0&b=0&b=0p=0;elseif a=0&b0p=bA2;elseif a=0p=aA2;elsep=aA2+bA2;end懑 8(1)运行结果

9、如下：x =1.0e-004 *-0.2058-0.2058f =-2.0576e-005k =5trace =100.0000 -0.0050 -0.0017 -0.0006 -0.0002 -0.0001 -0.0000 3.0000 -0.0050 -0.0017 -0.0006 -0.0002 -0.0001 -0.0000 从上述运行结果可以得出：最优解为 x= ，最小值约为 f*=0 。(2) min f(x) = x1A2+x2A2S.t 2x1+x2-2 w 0-x2+1 w 0用 matlab 求解如下： 主程序与( 1 )相同； 注意：以下三个函数体在运行计算习题四第 8(

10、2) 题时使用 %* function g=f_dfun(x) global M;a=-2*x(1)-x(2)+2; b=x(2)-1;if a=0&b=0 g(1,1)=2*x(1); g(2,1)=2*x(2);elseif a=0 g(1,1)=2*x(1)+2*M*(-2*x(1)-x(2)+2)*(-2); g(2,1)=2*x(2)+2*M*(-2*x(1)-x(2)+2)*(-1);elseif a=0&b=0&b=0p=0;elseif a=0&b0 p=bA2;elseif a=0p=aA2; elsep=aA2+bA2; end%题 8(2) 运行结果如下：x = -0.0

11、000 1.00001.00006 trace =100.0000-0.0000-0.0000-0.0000 -0.0000-0.0000-0.0000-0.00003.0000 0.99010.99670.9989 0.99960.9999 1.0000 1.0000从上述运行结果可以得出：最优解为 x= ，最小值约为 f*=1 。P232 第四章 25 用梯度投影法求解下列线性约束优化问题min f(x) = x2+x1*x2+2*x2A2+2*x3A2+2*x2*x3+4*x1+6*x2+12*x3 s.t x1 + x2 + x3 2xi 0 , i = 1,2, 3取 x1= ,用

12、matlab 求解如下：f= x1A2+x1*x2+2*x2A2+2*x3A2+2*x2*x3+4*x1+6*x2+12*x3a=1 1 1;1 1 -2;b=6;-2;l=zeros(3,1);x0=1 1 3;x,fval,exitflag,output,lambda,grad,hessian=fmincon(f,x0,a,b,l,)运行结果如下：x =0 0 1fval =14exitflag =1output =iterations: 2funcCount: 14stepsize: 1algorithm: medium-scale: SQP, Quasi-Newton, line-se

13、arch firstorderopt: 0 cgiterations: message: 1x144 char lambda =lower: 3x1 doubleupper: 3x1 double eqlin: 0x1 double eqnonlin: 0x1 doubleineqlin: 2x1 doubleineqnonlin: 0x1 doublegrad =4.00008.000016.0000hessian =1.11460.67710.60420.67713.36462.47920.60422.47923.4583从上述运行结果可以得出：最优解为 x= ，最小值约为 f*=14 。

14、P234 第四章 34 （ 1）用乘子法求解max f(x) = 10x1+4.4x22 +2x3s.t x1+4x2+5x3 3x1 2, x2 0, x3 0 用 matlab 求解如下：function x,minf = ymh434(l) format long;syms x1 x2 x3 f=10*(x1)+4.4*(x2)A2+2*(x3); h=-x1-4*x2-5*x3+32,-x1-3*x2-2*x3+29,(x3)A2)/2+(x2)A2-3,x1-2,x2,x3;x0=2 2 0; v=1 0 0 0 0 0;M=2; alpha=2; gama=0.25; var=x1

15、 x2 x3; eps=1.0e-4;if nargin = 8eps = 1.0e-4;endm1 = transpose(x0); m2 = inf;while lFE = 0;u=subs(h,x1,x2,x3,m1(1),m1(2),m1(3);for i=1:length(h)if (v(i)+M*u(i)=0FE = FE +(v(i)+M*u(i)A2-(v(i)A2);elseFE=FE+(v(i)A2;endendSumF = f + (1/(2*M)*FE;m2,minf = minNT(SumF,transpose(m1),var,eps);Hm2 =subs(h,x1,

16、x2,x3,m2(1),m2(2),m2(3);Hm1 =subs(h,x1,x2,x3,m1(1),m1(2),m1(3);Hx2 = Funval(h,var,x2);Hx1 = Funval(h,var,x1);if norm(Hx2) = gamaM = alpha*M; x1 = x2;elsev = v - M*transpose(Hx2);x1 = x2;endendendminf = Funval(f,var,x);format short;运行结果如下：x =19.75 0 2.45minf =-202.42maxf1 =202.42f*=-202.42 。从上述运行结果可以

17、得出： 最优解为 x= ，最大值约为 f*=202.42, 最小值约为P235 第四章 35( 2)用序列二次规划法求解 min f(x)=-5x1-5x2-4x3-x1x3-6x4-5x5/(1+x5)-8x6/(1+x6)-10(1-2eA-x7+eA-2x7)s.t g1(x)=2x4+x5+0.8x6+x7-5=0g2(x)=x2A2+x3A2+x5A2+x6A2-5=0g3(x)=x1+x2+x3+x4+x5+x6+x710g4(x)=x1+x2+x3+x4 5g5(x)=x1+x3+x5+x6A2-x7A2-5 0, i=1,2,7用 matlab 求解如下： function f

18、=objfun35(x)f=-5*x(1)-5*x(2)-4*x(3)-x(1)*x(3)-6*x(4)-5*x(5)/(1+x(5)-8*x(6)/(1+x(6)-10*(1-2 *exp(-x(7)+exp(-2*x(7);function c,ceq = confun35(x)ceq=x(2)A2+x(3)A2+x(5)A2+x(6)A2-5; c=x(1)+x(3)+x(5)+x(6)A2-x(7)A2-5;clcclearx0=1,1,1,1,1,1,1;aeq=0,0,0,2,1,0.8,1;beq=5;a=1,1,1,1,1,1,1;1,1,1,1,0,0,0;b=10,5;lb

19、=0,0,0,0,0,0,0;ub=;x,fval,exitflag,output,lambda,grad,hessian=fmincon(objfun35,x0,a,b,aeq,beq,l b,ub,confun35,options)c,ceq=confun35(x)运行结果如下：Iter F-count f(x) constraint Step-size derivative optimalityProcedure08-31.49581Infeasible start point117-41.71750.86421-6.31.62226-43.03090.53581-1.090.924335-44.51990.83481-1.40.438444-44.45430.0462110.1260.247553-44.46360.004