优化设计方法

《优化设计方法》姓名：陈辰学号：10104013院系：工学院专业：机械设计制造及其自动化班级：10机制一班第四章例题functiony=OPT_fun0(x)y=3*x(1)*x(1)-2*x(1)*x(2)+x(2)*x(2);formatlong;x0=[1,1];[x,f_opt,c,d]=fminsearch(@OPT_fun0,x0);x,f_opt结果：x=1.0e-004* 2-0.10965867079967 0.19889287750638f_opt=1.192541061251568e-009第四章课后习题4-1functiony=OPT_fun0(x)y=1.5*x(1)*x(1)+0.5*x(2)*x(2)-x(1)*x(2)-2*x(1);formatlong;x0=[2,2];[x,f_opt,c,d]=fminsearch(@OPT_fun0,x0);x,f_opt,c,d结果：x=0.99999695404938 1.00001655313968f_opt=-0.99999999979866c=1d=iterations:393funcCount:77algorithm:'Nelder-Meadsimplexdirectsearch用Hession矩阵进行证明：已知f(X)=3/2x(i)*x(i)+1/2x(2)*x(2)-X(i)*x(2)-2x(i)对该函数进行求导：一阶导为：f(X1)=3X(1)-X(2)-2；二阶导为：卜(X1)=3; 4(x2)=1； f''(X1X2)=f''(x2X1)=-1;由Hession矩阵判定条件之，令一阶导数值为0，固有E3W=0; f(fx⑴=0；得:x(1)=1; x(2)=1该函数的Hession矩阵为：--H=由该Hession矩阵知，一阶主4子式为3>0,二阶主子式为:3*1-(-1)*(-1)=2>0.所以该Hession矩阵为正定。即可知X=[11]为该函数的极小值点。即可知Matlab软件所算的值是正确的。functiony=OPT_fun0(x)y=x(1)*x(1)+x(1)*x(2)+2*x(2)*x(2)+4*x(1)+6*x(2)+10;formatlong;x0=[0,0];[x,f_opt,c,d]=fminsearch(@OPT_fun0,x0);x,f_opt,c结果：x=-1.42857209403050-1.14288457948509f_opt=3.71428571580995c=1结论：由结果可知该函数的最优点为：X=[-1.43-1.14]故此函数的极值为：fopt=3.715functiony=a123(x)y=x(1)*x(1)*x(1)+x(1)*x(2)-3*x(2)*x(2)*x(2)+3*x(1)*x(1)+3*x(2)*x(2)-9*x(1);formatlong;x0=[0,0][x,f_opt,c,d]=fminsearch(@a123,x0);x,f_opt,c结果：x0=0 0x=1.01162643167122-0.13945566799890f_opt=-5.07378442202651c=16结论：由上述结果知该函数的最优解点为：X=[1.01-0.14]故此函数的极值为:f_opt=-5.07functiony=a123(x)y=x(1)*x(1)*x(1)*x(1)+2*x(1)*x(1)*x(2)+x(2)*x(2)+x(1)*x(1)-2*x(2)+5;formatlong;x0=[0,0][x,f_opt,c,d]=fminsearch(@a123,x0);x,f_opt,c结果：X0=0 0X=0.00001815690060 0.99998886868293f_opt=4.000000001112927C=1结论：由上述结果知该函数的最优解点为：X=[0 1]故此函数的极值为:f_opt=4.004-2functiony=OPT_fun0(x)y=4*x(1)*x(2)+4/x(1)+4/x(2);formatlong;x0=[0.5,0.5];[x,f_opt,c,d]=fminsearch(@OPT_fun0,x0);x,f_opt,c结果:结果:1.00001878336170 1.00000069475736f_opt=12.00000000146536结论：由上述结果知该函数的最优解点为：X=[1 1]故此函数的极值为:f_opt=12.004-3functiony=OPT_fun1(x)y=x(1)*x(1)-x(1)*x(2)+3*x(2)*x(2);formatlong;x0=[1,1][x,f_opt,c,d]=fminsearch(@OPT_fun1,x0);x,f_opt,c9结果：x0=11x=1.0e-004*0.02440977771285 0.30798296188731f_opt=2.776385560476180e-009c=1结论：有输出的结果可知，最终在x(0)=[0.020.31]处约束，且最优解值为：f(x)=2.78 104-4functiony=OPT_fun1(x)y=4+4.5*x(1)-4*x(2)+x(1)*x(1)+2*x(2)*x(2)-2*x(1)*x(2)+x(1)*x(1)*x(1)*x(1)-2*x(1)*x(1)*x(2);formatlong;x0=[-2,2][x,f_opt,c,d]=fminsearch(@OPT_fun1,x0);x,f_opt,c结果：X0=-2 2x=-1.05275631816673 1.0277208586455611f_opt=-0.51340925137577c=1结论：由上述结果知该函数的最优解点为：X=[1.05 1.03]故此函数的极值为：f_opt=-0.514-5functiony=OPT_fun1(x)y=x(1)*x(1)+2*x(2)*x(2);formatlong;x0=[1,1][x,f_opt,c,d]=fminsearch(@OPT_fun1,x0);x,f_opt12结果：x0=11x=1.0e-004*0.03454644575964 0.17587500087544f_opt=6.305748878049377e-010c=1结论：由上述结果知该函数的最13优解点为：X=[0 0.18]故此函数的极值为：f_opt=6.314-6functiony=OPT_fun1(x)y=x(1)*x(1)-x(1)*x(2)+x(2)*x(2)+2*x(1)-4*x(2);formatlong;x0=[2,2][x,f_opt,c,d]=fminsearch(@OPT_fun1,x0);x,f_opt,c结果：X0=2 2x=0.00003131889914 2.0000196982037114f_opt=-3.99999999924803c=结论：由上述结果知该函数的最优解为：X=[02]故此函数的极值为：f_opt=-4.00第五章例题Aeq=[];Beq=[];f=[-60,-120];A=[9,4;3,10;4,5];B=[360;300;200];LB=zeros(2,1);UB=[];15[X,fopt,key]=linprog(f,A,B,Aeq,Beq,LB,UB);X,fopt,key>>A11Optimizationterminatedsuccessfully.X=20.000024.0000fopt=-4.0800e+003key=1第五章课后习题5-1 16①Aeq=[1,1,1,0;1,2,2.5,3];Beq=[4;5];f=[-1.1,-2.2,3.3,-4.4];A=[];B=[];LB=zeros(4,1);UB=[];[X,fopt,key]=linprog(f,A,B,Aeq,Beq,LB,UB);X,fopt,key>>cc2Optimizationterminatedsuccessfully.X=3.999999999914530.000000000085390.000000000000140.33333333330478fopt=-5.86666666663444key=1结论：由上述结果知该函数的最优点为：X=4.00.00.00.3故此函数的极小值为：fopt=-5.87②Aeq=[];Beq=[];f=[-7,-12];A=[9,4;4,5;3,10];B=[360;200;300];LB=zeros(2,1);UB=[];[X,fopt,key]=linprog(f,A,B,Aeq,Beq,LB,UB);X,fopt,key>>cc3Optimizationterminatedsuccessfully.X=1819.9999999999951323.99999999999991fopt=-4.279999999999648e+002key=1结论：由上述结果知该函数的最优解点为:X=20.024.0故此函数的最小值为：fopt=-4.28e+0025-2Aeq=[];Beq=[];f=[-7000,-12000]; 19A=[9,4;4,5;3,10];B=[360;200;300];LB=zeros(2,1);UB=[];[X,fopt,key]=linprog(f,A,B,Aeq,Beq,LB,UB);X,fopt,keyOptimizationterminatedsuccessfully.X=19.9999999999967323.99999999999963fopt=-4.279999999999726e+005key=1结论：由上述结果知该函数的最优解点为:X=20.024.0故此函数的最小值为：fopt=-4.28e+005…20f=[-0.30,-0.15];A=[-1,-1;1,1];B=[-600;1000];LB=zeros(2,1);UB=[800;1200];[X,fopt,key]=linprog(f,A,B,Aeq,Beq,LB,UB);X,fopt,keyX=7.999999999999761.99999999999523fopt=-2.699999999999213e+002key=1结论：由上述结果知该函数的最优解点为:X=8.02.0故此函数的最小值为：fopt=-2.70e+0025-6 21Aeq=[];Beq=[];f=[1,-2]；A=[1,1;-2,-1];B=[5;-3];LB=zeros(2,1);UB=[];[X,fopt,key]=linprog(f,A,B,Aeq,Beq,LB,UB);X,fopt,key>>cc4Optimizationterminatedsuccessfully.X=0.000000000167344.99999999277444fopt=-9.9999999853815322key=1结论：由上述结果知该函数的最优解点为:X=0.05.0故此函数的最小值为：fopt=-10.005-7Aeq=[1,2,1];Beq=[6];f=[-5,-4,-8];A=[5,3,0;2,-1,0];B=[15;4];LB=zeros(3,1);UB=[];[X,fopt,key]=linprog(f,A,B,Aeq,Beq,LB,UB);X,fopt,keyOptimizationterminatedsuccessfully.X=0.00000000000002 230.000000000000005.99999999999997fopt=-47.99999999999991key=

1结论：由上述结果知该函数的最优解点为:X=0.00.06.0故此函数的最小值为：fopt=-48.00第六章已知约束优化问题：24minf(x)=4气-x;-12s.t.g(X)=x；+x；-25<0,g2(X)=-气<0,g3(X)=-%<0,试以Xo=E,1L,Xo=",1｝，Xo二晶]为复合型的初始点，用复合型法进行二次迭代计算。主程序：functionf=exefun(x)f=4*x(1)-x(2)*x(2)-12;function[c,ceq]=execonfun(x)c=[x(1)*x(1)+x(2)*x(2)-25;-x(1)；-x(2)];ceq=[];x0=[2,1];lb=[0,0]；ub=[];options=optimset('LargeScale','off,'display','iter','tolx',1e-6);[x,fval,exitflag,output]=fmincon('exefun',x0,[],[],[],[],lb,ub,'execonfun',options)25输出结果:maxDirectionalFirst-orderIterF-count f(x)constraintStep-sizederivativeoptimalityProcedureTOC\o"1-5"\h\z7 -21 0 1-20 611 -44.1111 7.111 1 -30.25.53Hessianmodifiedtwice

TOC\o"1-5"\h\z15 -37.3937 0.3937 1 6.320.409Hessianmodifiedtwice19 -37.0015 0.001526 10.391 0.123 -37 2.327e-008 10.00153 0.1Hessianmodified27 -37 4.416e-022 12.3e-008 4HessianmodifiedtwiceOptimizationterminatedsuccessfully:Searchdirectionlessthan2*options.TolXandmaximumconstraintviolationislessthanoptions.TolConActiveConstraints:326ActiveConstraints:326x=-0 5fval=-37exitflag=1output=iterations:6funcCount:27stepsize:1algorithm:'medium-scale:SQP,Quasi-Newton,line-searchfirstorderopt:4.0000cgiterations:[]

用外点法求解下列问题的最优解，并用MATLAB计算下列约束优化问题。minf(x)=气+七s.t.g1(X)=3-七<0,g2(X)=气20，主程序为：functionf=exefun(x)f=x(1)+x(2);function[c,ceq]=execonfun(x)27c=[3-x(2);27-x(1);x(2)];ceq=[];x0=[2,1];lb=[0,0];ub=[];options=optimset('LargeScale','off,'display','iter','tolx',1e-6);[x,fval,exitflag,output]=fmincon('exefun',x0,[],[],[],[],lb,ub,'execonfun',options)输出结果：maxDirectionalFirst-orderIterF-count f(x)constraintStep-sizederivativeoptimalityProcedureTOC\o"1-5"\h\z7 3.5 1.5 10.5 1infeasible11 3.5 1.5 1-8.59e-009 1Hessianmodifiedtwice;infeasibleOptimizationterminated:Nofeasiblesolutionfound.Searchdirectionlessthan2*options.TolXbutconstraintsarenotsatisfied. 28x=2 1.5fval=3.5000exitflag=-1output=iterations:2funcCount:11stepsize:1algorithm:'medium-scale:SQP,Quasi-Newton,line-searchfirstorderopt:1.00006-46-4用内点法求解下列问题的最优解，并用Matlab计算下列约束优化问题。minf(x)=气+气stg1(X)二一X2+X>0,,g2(X)=x1>0,主程序为：functionf=exefun(x)f=x(1)+