2016年大连理工大学优化方法上机大作业

1、精选优质文档-倾情为你奉上 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); end function x_star = steepest(x0,eps) gk = grad(x0); res = norm(gk); k = 0; wh

2、ile res > eps && k<=1000 dk = -gk; ak =1; f0 = fun(x0); f1 = fun(x0+ak*dk); slope = dot(gk,dk); while f1 > f0 + 0.1*ak*slope ak = ak/4; xk = 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); e

3、nd 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); g(2,1)=-400*x(1); g(2,2)=200; end function g = grad(x)

4、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); end function x_star = newton(x0,eps) gk = grad(x0); bk = grad2(x0)(-1); res = norm(gk); k = 0; while res > eps && k<=1000 dk=-bk*gk; xk=x0+dk; k = k+1; x0 = xk; gk = grad(xk); bk = grad2(xk)(-1); res = norm(

5、gk); fprintf('-The %d-th iter, the residual is %fn',k,res); end x_star = xk; end>> clear>> x0=0,0'>> eps=1e-4;>> x1=newton(x0,eps)-The 1-th iter, the residual is 447.-The 2-th iter, the residual is 0.x1 = 1.0000 1.00003.BFGS法：function f = fun(x) f = (1-x(1)2 + 100

6、*(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); end function x_star = bfgs(x0,eps) g0 = grad(x0); gk=g0; res = norm(gk); Hk=eye(2); k = 0; while res > eps && k<=1000 dk = -Hk*gk; ak =1; f0 = fun(x0); f1 = fun(x0+ak*

7、dk); slope = dot(gk,dk); while f1 > f0 + 0.1*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

8、('-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) = 2

9、00*(x(2)-x(1)2); end function x_star =CG(x0,eps) gk = grad(x0);res = norm(gk); k = 0; dk = -gk; while res > eps && k<=1000 ak =1; f0 = fun(x0); f1 = fun(x0+ak*dk); slope = dot(gk,dk); while f1 > f0 + 0.1*ak*slope ak = ak/4; xk = x0 + ak*dk; f1 = fun(xk); end k = k+1; x0 = xk; g0=gk;

10、 gk = grad(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(

11、2)2-25;g=zeros(3,1);g(1)=-10*x(1)+x(1)2-10*x(2)+x(2)2+34;g(2)=-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_

12、inequ h_inequ=h_inequ+(0.5/pena)*(max(0,(r_al(i)+pena*g(i).2-r_al(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)/pe

13、na);endh1 = max(h_inequ);f= max(abs(h_equ),h1); %sqrt(h_equ+h_inequ);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_

14、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_itera<max_itera 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'); break end %c判断函数下降度? if c

15、ompare(x_al)<cta*compareFlag pena=1*pena; %可以根据需要修改惩罚系数变量 else pena=min(1000,c_scale*pena); %乘法系数最大1000 disp('pena=2*pena'); end %?更新拉格朗日乘子 h,g=constrains(x_al); for i=1:1 %等式约束部分 r_al(i)= r_al0(i)+pena*h(i); end for i=1:3 %不等式约束部分 r_al(i+1)=max(0,(r_al0(i+1)+pena*g(i); end out_itera=out_

16、itera+1;end%+迭代结束+disp('the iteration number');k=out_itera;disp('the value of constrains'); compare(x_al)disp('the opt point'); x=x_al; fmin=obj(X);>> clear>> x_al=0,0;>> x,fmin,k=almain(x_al)上机大作业3： 1、>> clear all n=3; c=-3,-1,-3' A=2,1,1;1,2,3;2,

17、2,1;-1,0,0;0,-1,0;0,0,-1;b=2,5,6,0,0,0'cvx_begin variable x(n) minimize( c'*x) subject to A*x<=bcvx_end Calling SDPT3 4.0: 6 variables, 3 equality constraints- num. of constraints = 3 dim. of linear var = 6* SDPT3: Infeasible path-following algorithms* version predcorr gam expon scale_dat

18、a NT 1 0.000 1 0 it pstep dstep pinfeas dinfeas gap prim-obj dual-obj cputime- 0|0.000|0.000|1.1e+01|5.1e+00|6.0e+02|-7.e+01 0.e+00| 0:0:00| chol 1 1 1|0.912|1.000|9.4e-01|4.6e-02|6.5e+01|-5.e+00 -2.e+01| 0:0:01| chol 1 1 2|1.000|1.000|1.3e-07|4.6e-03|8.5e+00|-2.e+00 -1.e+01| 0:0:01| chol 1 1 3|1.00

19、0|0.961|2.3e-08|6.2e-04|1.8e+00|-4.e+00 -6.e+00| 0:0:01| chol 1 1 4|0.881|1.000|2.2e-08|4.6e-05|3.7e-01|-5.e+00 -5.e+00| 0:0:01| chol 1 1 5|0.995|0.962|1.6e-09|6.2e-06|1.5e-02|-5.e+00 -5.e+00| 0:0:01| chol 1 1 6|0.989|0.989|2.7e-10|5.2e-07|1.7e-04|-5.e+00 -5.e+00| 0:0:01| chol 1 1 7|0.989|0.989|5.3e

20、-11|5.8e-09|1.8e-06|-5.e+00 -5.e+00| 0:0:01| chol 1 1 8|1.000|0.994|2.8e-13|4.3e-11|2.7e-08|-5.e+00 -5.e+00| 0:0:01| stop: max(relative gap, infeasibilities) < 1.49e-08- number of iterations = 8 primal objective value = -5.e+00 dual objective value = -5.e+00 gap := trace(XZ) = 2.66e-08 relative g

21、ap = 2.26e-09 actual relative gap = 2.21e-09 rel. primal infeas (scaled problem) = 2.77e-13 rel. dual " " " = 4.31e-11 rel. primal infeas (unscaled problem) = 0.00e+00 rel. dual " " " = 0.00e+00 norm(X), norm(y), norm(Z) = 4.3e+00, 1.3e+00, 1.9e+00 norm(A), norm(b), nor

22、m(C) = 6.7e+00, 9.1e+00, 5.4e+00 Total CPU time (secs) = 0.71 CPU time per iteration = 0.09 termination code = 0 DIMACS: 3.6e-13 0.0e+00 5.8e-11 0.0e+00 2.2e-09 2.3e-09- -Status: SolvedOptimal value (cvx_optval): -5.42、>> clear all n=2; c=-2,-4' G=0.5,0;0,1; A=1,1;-1,0;0,-1; b=1,0,0'cv

23、x_begin variable x(n) minimize( x'*G*x+c'*x) subject to A*x<=bcvx_end Calling SDPT3 4.0: 7 variables, 3 equality constraints For improved efficiency, SDPT3 is solving the dual problem.- num. of constraints = 3 dim. of socp var = 4, num. of socp blk = 1 dim. of linear var = 3* SDPT3: Infea

24、sible path-following algorithms* version predcorr gam expon scale_data NT 1 0.000 1 0 it pstep dstep pinfeas dinfeas gap prim-obj dual-obj cputime- 0|0.000|0.000|8.0e-01|6.5e+00|3.1e+02| 1.e+01 0.e+00| 0:0:00| chol 1 1 1|1.000|0.987|4.3e-07|1.5e-01|1.6e+01| 9.e+00 -2.e-01| 0:0:00| chol 1 1 2|1.000|1

25、.000|2.6e-07|7.6e-03|1.4e+00| 1.e+00 -5.e-02| 0:0:00| chol 1 1 3|1.000|1.000|2.4e-07|7.6e-04|3.0e-01| 4.e-01 1.e-01| 0:0:00| chol 1 1 4|0.892|0.877|6.4e-08|1.6e-04|5.2e-02| 2.e-01 2.e-01| 0:0:00| chol 1 1 5|1.000|1.000|1.0e-08|7.6e-06|1.5e-02| 2.e-01 2.e-01| 0:0:00| chol 1 1 6|0.905|0.904|3.1e-09|1.

26、4e-06|2.3e-03| 2.e-01 2.e-01| 0:0:00| chol 1 1 7|1.000|1.000|6.1e-09|7.7e-08|6.6e-04| 2.e-01 2.e-01| 0:0:00| chol 1 1 8|0.903|0.903|1.8e-09|1.5e-08|1.0e-04| 2.e-01 2.e-01| 0:0:00| chol 1 1 9|1.000|1.000|4.9e-10|3.5e-10|2.9e-05| 2.e-01 2.e-01| 0:0:00| chol 1 1 10|0.904|0.904|4.7e-11|1.3e-10|4.4e-06| 2.e-01 2.e-01| 0:0:00| chol 2 2 11|1.000|1.000|2.3e-12|9.4e-12|1.2e-06| 2.e-01 2.e-01| 0:0:00| chol 2 2 12|1.000|1.000|4.7e-13|1.0e-12|1