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1、.2016 年大连理工大学优化方法上机大作业学院：专业：班级：学号：姓名：上机大作业 1：1.最速下降法：.function f=fun(x)f=(1-x(1)2+100*(x(2)-x(1)2)2;endfunction g=grad(x)g=zeros(2,1);g(1)=2*(x(1)-1)+400*x(1)*(x(1)2-x(2);g(2)=200*(x(2)-x(1)2);.endfunction x_star=steepest(x0,eps)gk=grad(x0);res=norm(gk);k=0;while res eps&k f0+0.1*ak*slope ak=ak/4;xk
2、=x0+ak*dk;f1=fun(xk);end k=k+1;x0=xk;gk=grad(xk);res=norm(gk);fprintf(-The%d-th iter,the residual is%fn,k,res);end x_star=xk;end.clear x0=0,0;eps=1e-4;x=steepest(x0,eps).2.牛顿法：function f=fun(x)f=(1-x(1)2+100*(x(2)-x(1)2)2;endfunction g=grad2(x)g=zeros(2,2);.g(1,1)=2+400*(3*x(1)2-x(2);g(1,2)=-400*x(1
3、);g(2,1)=-400*x(1);g(2,2)=200;endfunction g=grad(x)g=zeros(2,1);g(1)=2*(x(1)-1)+400*x(1)*(x(1)2-x(2);g(2)=200*(x(2)-x(1)2);endfunction x_star=newton(x0,eps)gk=grad(x0);bk=grad2(x0)(-1);res=norm(gk);k=0;while res eps&k clear x0=0,0;eps=1e-4;x1=newton(x0,eps)-The 1-th iter,the residual is 447.213595-T
4、he 2-th iter,the residual is 0.000000 x1=1.0000 1.0000.3.BFGS 法：function f=fun(x)f=(1-x(1)2+100*(x(2)-x(1)2)2;end.function g=grad(x)g=zeros(2,1);g(1)=2*(x(1)-1)+400*x(1)*(x(1)2-x(2);g(2)=200*(x(2)-x(1)2);endfunction x_star=bfgs(x0,eps)g0=grad(x0);gk=g0;res=norm(gk);Hk=eye(2);k=0;while res eps&k f0+0
5、.1*ak*slope ak=ak/4;xk=x0+ak*dk;f1=fun(xk);end k=k+1;fa0=xk-x0;x0=xk;.go=gk;gk=grad(xk);y0=gk-g0;Hk=(eye(2)-fa0*(y0)/(fa0)*(y0)*(eye(2)-(y0)*(fa0)/(fa0)*(y0)+(fa0*(fa0)/(fa0)*(y0);res=norm(gk);fprintf(-The%d-th iter,the residual is%fn,k,res);end x_star=xk;End clear x0=0,0;eps=1e-4;x=bfgs(x0,eps).4.共
6、轭梯度法：.function f=fun(x)f=(1-x(1)2+100*(x(2)-x(1)2)2;endfunction g=grad(x)g=zeros(2,1);g(1)=2*(x(1)-1)+400*x(1)*(x(1)2-x(2);g(2)=200*(x(2)-x(1)2);endfunction x_star=CG(x0,eps)gk=grad(x0);res=norm(gk);k=0;dk=-gk;while res eps&k f0+0.1*ak*slope ak=ak/4;xk=x0+ak*dk;f1=fun(xk);end.k=k+1;x0=xk;g0=gk;gk=gr
7、ad(xk);res=norm(gk);p=(gk/g0)2;dk1=dk;dk=-gk+p*dk1;fprintf(-The%d-th iter,the residual is%fn,k,res);end x_star=xk;end clear x0=0,0;eps=1e-4;x=CG(x0,eps).上机大作业 2：.function f=obj(x)f=4*x(1)-x(2)2-12;endfunction h,g=constrains(x)h=x(1)2+x(2)2-25;g=zeros(3,1);.g(1)=-10*x(1)+x(1)2-10*x(2)+x(2)2+34;g(2)=-
8、x(1);g(3)=-x(2);endfunction f=alobj(x)%拉格朗日增广函数%N_equ 等式约束个数?%N_inequ 不等式约束个数N_equ=1;N_inequ=3;global r_al pena;%全局变量h_equ=0;h_inequ=0;h,g=constrains(x);%等式约束部分?for i=1:N_equ h_equ=h_equ+h(i)*r_al(i)+(pena/2)*h(i).2;end%不等式约束部分for i=1:N_inequh_inequ=h_inequ+(0.5/pena)*(max(0,(r_al(i)+pena*g(i).2-r_a
9、l(i).2);end.%拉格朗日增广函数值f=obj(x)+h_equ+h_inequ;function f=compare(x)global r_al pena N_equ N_inequ;N_equ=1;N_inequ=3;h_inequ=zeros(3,1);h,g=constrains(x);%等式部分for i=1:1 h_equ=abs(h(i);end%不等式部分for i=1:3h_inequ=abs(max(g(i),-r_al(i+1)/pena);endh1=max(h_inequ);f=max(abs(h_equ),h1);%sqrt(h_equ+h_inequ);.
10、function x,fmin,k=almain(x_al)%本程序为拉格朗日乘子算法示例算法%函数输入：%x_al:初始迭代点%r_al：初始拉格朗日乘子 N-equ:等式约束个数 N_inequ:不等式约束个数?%函数输出%X：最优函数点 FVAL:最优函数值%=程序开始=global r_al pena ;%参数（全局变量）pena=10;%惩罚系数r_al=1,1,1,1;c_scale=2;%乘法系数乘数cta=0.5;%下降标准系数e_al=1e-4;%误差控制范围max_itera=25;out_itera=1;%迭代次数%=算法迭代开始=while out_iteramax_i
11、tera x_al0=x_al;r_al0=r_al;%判断函数?compareFlag=compare(x_al0);%无约束的拟牛顿法 BFGS X,fmin=fminunc(alobj,x_al0);x_al=X;%得到新迭代点%判断停止条件?if compare(x_al)e_al.disp(we get the opt point);breakend%c 判断函数下降度?if compare(x_al)clear x_al=0,0;x,fmin,k=almain(x_al).上机大作业 3：.1、clear alln=3;c=-3,-1,-3;A=2,1,1;1,2,3;2,2,1;
12、-1,0,0;0,-1,0;0,0,-1;b=2,5,6,0,0,0;cvx_beginvariable x(n).minimize(c*x)subject toA*x=bcvx_endCalling SDPT3 4.0:6 variables,3 equality constraints-num.of constraints=3dim.of linear var=6*SDPT3:Infeasible path-following algorithms*versionpredcorrgamexponscale_dataNT10.00010it pstep dstep pinfeas dinfe
13、as gapprim-objdual-objcputime-0|0.000|0.000|1.1e+01|5.1e+00|6.0e+02|-7.000000e+010.000000e+00|0:0:00|chol111|0.912|1.000|9.4e-01|4.6e-02|6.5e+01|-5.606627e+00-2.967567e+01|0:0:01|chol 112|1.000|1.000|1.3e-07|4.6e-03|8.5e+00|-2.723981e+00-1.113509e+01|0:0:01|chol 11.3|1.000|0.961|2.3e-08|6.2e-04|1.8e
14、+00|-4.348354e+00-6.122853e+00|0:0:01|chol114|0.881|1.000|2.2e-08|4.6e-05|3.7e-01|-5.255152e+00-5.622375e+00|0:0:01|chol 115|0.995|0.962|1.6e-09|6.2e-06|1.5e-02|-5.394782e+00-5.409213e+00|0:0:01|chol 116|0.989|0.989|2.7e-10|5.2e-07|1.7e-04|-5.399940e+00-5.400100e+00|0:0:01|chol117|0.989|0.989|5.3e-1
15、1|5.8e-09|1.8e-06|-5.399999e+00-5.400001e+00|0:0:01|chol118|1.000|0.994|2.8e-13|4.3e-11|2.7e-08|-5.400000e+00-5.400000e+00|0:0:01|stop:max(relative gap,infeasibilities)clear alln=2;c=-2,-4;G=0.5,0;0,1;A=1,1;-1,0;0,-1;b=1,0,0;cvx_beginvariable x(n)minimize(x*G*x+c*x)subject toA*x=bcvx_endCalling SDPT
16、3 4.0:7 variables,3 equality constraintsFor improved efficiency,SDPT3 is solving the dual problem.-num.of constraints=3dim.of socpvar=4,num.of socp blk=1dim.of linear var=3*SDPT3:Infeasible path-following algorithms*versionpredcorrgamexponscale_data.NT10.00010it pstep dstep pinfeas dinfeas gapprim-o
17、bjdual-objcputime-0|0.000|0.000|8.0e-01|6.5e+00|3.1e+02|1.000000e+010.000000e+00|0:0:00|chol111|1.000|0.987|4.3e-07|1.5e-01|1.6e+01|9.043148e+00-2.714056e-01|0:0:00|chol112|1.000|1.000|2.6e-07|7.6e-03|1.4e+00|1.234938e+00-5.011630e-02|0:0:00|chol3|1.000|1.000|2.4e-07|7.6e-04|3.0e-01|4.166959e-011.18
18、1563e-01|0:0:00|chol4|0.892|0.877|6.4e-08|1.6e-04|5.2e-02|2.773022e-012.265122e-01|0:0:00|chol5|1.000|1.000|1.0e-08|7.6e-06|1.5e-02|2.579468e-012.427203e-01|0:0:00|chol6|0.905|0.904|3.1e-09|1.4e-06|2.3e-03|2.511936e-012.488619e-01|0:0:00|chol7|1.000|1.000|6.1e-09|7.7e-08|6.6e-04|2.503336e-012.496718
19、e-01|0:0:00|chol8|0.903|0.903|1.8e-09|1.5e-08|1.0e-04|2.500507e-012.499497e-01|0:0:00|chol9|1.000|1.000|4.9e-10|3.5e-10|2.9e-05|2.500143e-012.499857e-01|0:0:00|chol10|0.904|0.904|4.7e-11|1.3e-10|4.4e-06|2.500022e-012.499978e-01|0:0:00|chol11|1.000|1.000|2.3e-12|9.4e-12|1.2e-06|2.500006e-012.499994e-
20、01|0:0:00|chol12|1.000|1.000|4.7e-13|1.0e-12|1.8e-07|2.500001e-012.499999e-01|0:0:00|chol13|1.000|1.000|2.0e-12|1.0e-12|4.2e-08|2.500000e-012.500000e-01|0:0:00|chol14|1.000|1.000|2.6e-12|1.0e-12|7.3e-09|2.500000e-012.500000e-01|0:0:00|stop:max(relative gap,infeasibilities)1.49e-08-number of iteratio
21、ns=14primal objective value=2.50000004e-01dualobjective value=2.49999996e-01gap:=trace(XZ)=7.29e-09relative gap=4.86e-09.111111111111111122222222actual relative gap=4.86e-09rel.primal infeas(scaled problem)=2.63e-12rel.dual=1.00e-12rel.primal infeas(unscaled problem)=0.00e+00rel.dual=0.00e+00norm(X),norm(y),norm(Z)=3.2e+00,1.5e+00,1.9e+00norm(A),norm(b),norm(C)=3.9e+00,4.2e+00,2.6e+00Total CPU time(secs)=0.36CPU time per iteration=0.03termination code=0DIMACS:3.7e-120.0e+001.3e-120.0e+004.9e-09-Status:SolvedOptimal value(cvx_optval):-3.4.9e-09