优化方法作业第二版..doc

1、优化方法上机大作业院系：化工与环境生命学部姓名：李翔宇学号： 31607007指导老师 :肖现涛第一题：1. 最速下降法源程序如下：function x_star =zsxj x0,eps gk = gradx0;res = normgk; k = 0;while res eps & k f0 + 0.0001*ak*slope ak = ak/2;xk = x0 + ak*dk; f1 = funxk; endk = k+1; x0 = xk;gk = gradxk;res = normgk;fprintf-the %d-th iter, the residual is %fn,k,res;

2、endx_star = xk; endfunction f = funxf = 1-x12 + 100*x2-x122;endfunction g = gradx g = zeros2,1;g1=2*x1-1+400*x1*x12-x2; g2 = 200*x2-x12;end运行结果： x0=0,0; esp=1e-4; xk= zsxj x0,eps-the 1-th iter, the residual is 13.372079-the 2-th iter, the residual is 12.079876-the 3-th iter, the residual is 11.05410

3、5-the 9144-th iter, the residual is 0.000105-the 9145-th iter, the residual is 0.000102-the 9146-th iter, the residual is 0.000100 xk =0.99990.9998matlab截屏：2. 牛顿法源程序如下：function x_star =newton x0,eps gk = gradx0;bk = grad2x0-1;res = normgk; k = 0;while res eps & k x0=0,0;eps=1e-4; xk= newton x0,eps-t

4、he 1-th iter, the residual is 447.213595-the 2-th iter, the residual is 0.000000 xk =1.00001.0000matalb截屏 ;3. bfgs 方法源程序如下：function x_star =bfgs x0,eps g0 = gradx0;gk=g0;res = normgk;hk=eye2;k = 0;while res eps & k f0 + 0.1*ak*slope ak = ak/2;xk = x0 + ak*dk; f1 = funxk; endk = k+1; fa0=xk-x0; x0 =

5、xk; g0=gk;gk = gradxk; y0=gk-g0;hk=eye2-fa0*y0/fa0*y0*eye2-y0*fa0/fa0*y0+fa0*fa0/fa0*y0; res = normgk;fprintf-the %d-th iter, the residual is %fn,k,res; endx_star = xk;endfunction f=funxf=1-x12 + 100*x2-x122;endfunction g = gradx g = zeros2,1;g1=2*x1-1+400*x1*x12-x2;g2 = 200*x2-x12;end运行结果： x0=0,0;

6、esp=1e-4; xk= bfgs x0,eps-the 1-th iter, the residual is 3.271712-the 2-th iter, the residual is 2.381565-the 3-th iter, the residual is 3.448742-the 1516-th iter, the residual is 0.000368-the 1517-th iter, the residual is 0.000099 xk =1.00011.0002matlab截屏：4. 共轭梯度法源程序如下：function x_star =conj x0,eps

7、gk = gradx0;res = normgk;k = 0;dk = -gk;while res eps & k f0 + 0.1*ak*slopeak = ak/2;xk = x0 + ak*dk; f1 = funxk; end d0=dk;g0=gk; k=k+1;x0=xk;gk=gradxk;f=normgk/normg02; res=normgk;dk=-gk+f*d0;fprintf-the %d-th iter, the residual is %fn,k,res;endx_star = xk; endfunction f=funxf=1-x12+100*x2-x122;en

8、dfunction g=gradx g=zeros2,1;g1=400*x13-400*x1*x2+2*x1-2; g2=-200*x12+200*x2;end运行结果： x0=0,0; eps=1e-4; xk=conjx0,eps-the 1-th iter, the residual is 3.271712-the 2-th iter, the residual is 1.380542-the 3-th iter, the residual is 4.527780-the 4-th iter, the residual is 0.850596-the 73-th iter, the re

9、sidual is 0.001532-the 74-th iter, the residual is 0.000402-the 75-th iter, the residual is 0.000134-the 76-th iter, the residual is 0.000057 xk =0.99990.9999matlab截屏：其次题：解 ：目标函数文件 f1.m function f=f1xf=4*x1-x22-12;等式约束函数文件h1.m function he=h1xhe=25-x12-x22;不等式约束函数文件g1.m function gi=g1xgi=10*x1-x12+10

10、*x2-x22-34;目标函数的梯度文件df1.m function g=df1xg = 4, 2.0*x2;等式约束（向量）函数的jacobi 矩阵（转置）文件dh1.m function dhe=dh1xdhe = -2*x1, -2*x2;不等式约束（向量）函数的jacobi 矩阵（转置）文件dg1.m function dgi=dg1xdgi = 10-2*x1, 10-2*x2;然后在 matlab命令窗口输入如下命令: x0=0,0;x,mu,lambda,output=multphrf1,h1,g1,df1,dh1,dg1,x0;得到如下输出 :x =4.898717426488

11、211.00128197198571算法编程利用程序调用格式第三题：1. 解：将目标函数改写为向量形式：x*a*x-b*x程序代码：n=2; a=0.5,0;0,1; b=2 4;c=1 1;cvx_begin variable xnminimize x*a*x-b*xsubject to c * x =0cvx_end运算结果 :calling sdpt3 4.0: 7 variables, 3 equality constraintsfor improved efficiency, sdpt3 is solving the dual problem.num. of constraints

12、= 3dim. of socpvar = 4,num. of socp blk = 1 dim. of linear var = 3*sdpt3: infeasible path-following algorithms* version predcorr gam expon scale_datant10.00010it pstep dstep pinfeas dinfeas gapprim-objdual-objcputime0|0.000|0.000|8.0e-001|6.5e+000|3.1e+002| 1.000000e+001 0.000000e+000| 0:0:00| chol

13、1 11|1.000|0.987|4.3e-007|1.5e-001|1.6e+001| 9.043148e+000 -2.714056e-001| 0:0:01| chol 1 12|1.000|1.000|2.6e-007|7.6e-003|1.4e+000| 1.234938e+000 -5.011630e-002| 0:0:01| chol 1 13|1.000|1.000|2.4e-007|7.6e-004|3.0e-001| 4.166959e-0011.181563e-001| 0:0:01| chol114|0.892|0.877|6.4e-008|1.6e-004|5.2e-

14、002| 2.773022e-0012.265122e-001| 0:0:01| chol115|1.000|1.000|1.0e-008|7.6e-006|1.5e-002| 2.579468e-0012.427203e-001| 0:0:01| chol116|0.905|0.904|3.1e-009|1.4e-006|2.3e-003| 2.511936e-0012.488619e-001| 0:0:01| chol117|1.000|1.000|6.1e-009|7.7e-008|6.6e-004| 2.503336e-0012.496718e-001| 0:0:01| chol118

15、|0.903|0.903|1.8e-009|1.5e-008|1.0e-004| 2.500507e-0012.499497e-001| 0:0:01| chol119|1.000|1.000|4.9e-010|3.5e-010|2.9e-005| 2.500143e-0012.499857e-001| 0:0:01| chol1110|0.904|0.904|5.7e-011|1.3e-010|4.4e-006| 2.500022e-0012.499978e-001| 0:0:01| chol2211|1.000|1.000|5.2e-013|1.1e-011|1.2e-006| 2.500

16、006e-0012.499994e-001| 0:0:01| chol2212|1.000|1.000|5.9e-013|1.0e-012|1.8e-007| 2.500001e-0012.499999e-001| 0:0:01| chol2213|1.000|1.000|1.7e-012|1.0e-012|4.2e-008| 2.500000e-0012.500000e-001| 0:0:01| chol2214|1.000|1.000|2.3e-012|1.0e-012|7.3e-009| 2.500000e-0012.500000e-001| 0:0:01|stop: maxrelati

17、ve gap, infeasibilities 1.49e-008number of iterations = 14primal objective value = 2.50000004e-001 dual objective value = 2.49999996e-001 gap := tracexz = 7.29e-009relative gap= 4.86e-009 actual relative gap= 4.86e-009rel. primal infeas scaled problem= 2.33e-012rel. dual= 1.00e-012rel. primal infeas

18、 unscaled problem = 0.00e+000 rel. dual = 0.00e+000normx, normy, normz = 3.2e+000, 1.5e+000, 1.9e+000 norma, normb, normc = 3.9e+000, 4.2e+000, 2.6e+000 total cpu time secs = 0.99cpu time per iteration = 0.07termination code= 0dimacs: 3.3e-012 0.0e+000 1.3e-012 0.0e+000 4.9e-009 4.9e-009status: solv

19、edoptimal value cvx_optval: -32. 程序代码：n=3;a=-3 -1 -3;b=2;5;6;c=2 1 1;1 2 3;2 2 1;cvx_begin variable xn minimize a*x subject toc * x =0cvx_end运行结果：calling sdpt3 4.0: 6 variables, 3 equality constraintsnum. of constraints = 3 dim. of linear var = 6*sdpt3: infeasible path-following algorithms* version

20、predcorr gam expon scale_datant10.00010it pstep dstep pinfeas dinfeas gapprim-objdual-objcputime0|0.000|0.000|1.1e+001|5.1e+000|6.0e+002|-7.000000e+001 0.000000e+000| 0:0:00| chol 1 11|0.912|1.000|9.4e-001|4.6e-002|6.5e+001|-5.606627e+000 -2.967567e+001| 0:0:00| chol112|1.000|1.000|1.3e-007|4.6e-003

21、|8.5e+000|-2.723981e+000 -1.113509e+001| 0:0:00| chol113|1.000|0.961|2.3e-008|6.2e-004|1.8e+000|-4.348354e+000 -6.122853e+000| 0:0:00| chol114|0.881|1.000|2.2e-008|4.6e-005|3.7e-001|-5.255152e+000 -5.622375e+000| 0:0:00| chol115|0.995|0.962|1.6e-009|6.2e-006|1.5e-002|-5.394782e+000 -5.409213e+000| 0:0:00| chol116|0.989|0.989|2.7e-010|5.2e-007|1.7e-004|-5.399940e+000 -5.400100e+000| 0:0:00| chol117|0.989|0.989|5.3e-011|5.8e-009|1.8e-006|-5.399999e+000 -5.400001e+000| 0:0:00| chol8|1.000|0.