优化方法作业第二版(共19页).doc

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1、精选优质文档-倾情为你奉上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：指导教师:肖现涛第一题：1. 最速下降法源程序如下：function x_star = ZSXJ(x0,eps) gk = grad(x0); res = norm(gk); k = 0; while res eps & k f0 + 0.0001*ak*slope ak = ak/2; xk = x0 + ak*dk; f1 = fun(xk); end k = k+1; x0 = xk; gk = grad(xk);res = norm(gk); fprintf(-The %d-th iter,

2、the residual is %fn,k,res); end x_star = xk; end 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 运行结果： x0=0,0; esp=1e-4; xk=ZSXJ(x0,eps)-The 1-th iter, the residual is 13.-The 2-th iter,

3、the residual is 12.-The 3-th iter, the residual is 11.-The 9144-th iter, the residual is 0.-The 9145-th iter, the residual is 0.-The 9146-th iter, the residual is 0.xk = 0.99990.9998MATLAB截屏：2. 牛顿法源程序如下：function x_star = NEWTON(x0,eps) gk = grad(x0); bk = grad2(x0)(-1); res = norm(gk); k = 0; while

4、res eps & k x0=0,0;eps=1e-4; xk=NEWTON(x0,eps)-The 1-th iter, the residual is 447.-The 2-th iter, the residual is 0.xk = 1.00001.0000MATALB截屏;3. BFGS方法源程序如下：function x_star = Bfgs(x0,eps) g0 = grad(x0); gk=g0; res = norm(gk); Hk=eye(2); k = 0; while res eps & k f0 + 0.1*ak*slope ak = ak/2; xk = x0 +

5、 ak*dk; f1 = fun(xk); end k = k+1; fa0=xk-x0; x0 = xk; g0=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; endfunction f=fun(x) f=(1-x(1)2 + 100*(x(2)-x(1

6、)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 运行结果： x0=0,0; esp=1e-4; xk=Bfgs(x0,eps)-The 1-th iter, the residual is 3.-The 2-th iter, the residual is 2.-The 3-th iter, the residual is 3.-The 1516-th iter, the residual is 0.-The 1517

7、-th iter, the residual is 0.xk = 1.0001 1.0002MATLAB截屏：4. 共轭梯度法源程序如下：function x_star =Conj (x0,eps) gk = grad(x0);res = norm(gk); k = 0; dk = -gk; while res eps & k f0 + 0.1*ak*slope ak = ak/2; xk = x0 + ak*dk; f1 = fun(xk); end d0=dk;g0=gk; k=k+1; x0=xk;gk=grad(xk);f=(norm(gk)/norm(g0)2; res=norm(g

8、k);dk=-gk+f*d0;fprintf(-The %d-th iter, the residual is %fn,k,res); end x_star = xk; end 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)=400*x(1)3-400*x(1)*x(2)+2*x(1)-2;g(2)=-200*x(1)2+200*x(2);end 运行结果： x0=0,0; eps=1e-4; xk=Conj(x0,eps)-The 1-th iter, the resid

9、ual is 3.-The 2-th iter, the residual is 1.-The 3-th iter, the residual is 4.-The 4-th iter, the residual is 0.-The 73-th iter, the residual is 0.-The 74-th iter, the residual is 0.-The 75-th iter, the residual is 0.-The 76-th iter, the residual is 0.xk = 0.9999 0.9999MATLAB截屏：第二题：解 ：目标函数文件 f1.mfunc

10、tion f=f1(x)f=4*x(1)-x(2)2-12;等式约束函数文件 h1.mfunction he=h1(x)he=25-x(1)2-x(2)2;不等式约束函数文件 g1.mfunction gi=g1(x)gi=10*x(1)-x(1)2+10*x(2)-x(2)2-34;目标函数的梯度文件 df1.mfunction g=df1(x)g = 4, 2.0*x(2);等式约束（向量）函数的Jacobi矩阵（转置）文件dh1.mfunction dhe=dh1(x)dhe = -2*x(1), -2*x(2);不等式约束（向量）函数的Jacobi矩阵（转置）文件dg1.mfuncti

11、on dgi=dg1(x)dgi = 10-2*x(1), 10-2*x(2);然后在 Matlab 命令窗口输入如下命令:x0=0,0;x,mu,lambda,output=multphr(f1,h1,g1,df1,dh1,dg1,x0);得到如下输出:x =4.8211.571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_beginvariable x(n)minimize( x*a*x-b*x)subject toc * x =0cvx_end运算结果:Calling SDPT

12、3 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: Infeasible path-following algorithms* version predcorr gam expon scale_data NT 1 0.000 1 0 it pste

13、p dstep pinfeas dinfeas gap prim-obj dual-obj cputime- 0|0.000|0.000|8.0e-001|6.5e+000|3.1e+002| 1.e+001 0.e+000| 0:0:00| chol 1 1 1|1.000|0.987|4.3e-007|1.5e-001|1.6e+001| 9.e+000 -2.e-001| 0:0:01| chol 1 1 2|1.000|1.000|2.6e-007|7.6e-003|1.4e+000| 1.e+000 -5.e-002| 0:0:01| chol 1 1 3|1.000|1.000|2

14、.4e-007|7.6e-004|3.0e-001| 4.e-001 1.e-001| 0:0:01| chol 1 1 4|0.892|0.877|6.4e-008|1.6e-004|5.2e-002| 2.e-001 2.e-001| 0:0:01| chol 1 1 5|1.000|1.000|1.0e-008|7.6e-006|1.5e-002| 2.e-001 2.e-001| 0:0:01| chol 1 1 6|0.905|0.904|3.1e-009|1.4e-006|2.3e-003| 2.e-001 2.e-001| 0:0:01| chol 1 1 7|1.000|1.0

15、00|6.1e-009|7.7e-008|6.6e-004| 2.e-001 2.e-001| 0:0:01| chol 1 1 8|0.903|0.903|1.8e-009|1.5e-008|1.0e-004| 2.e-001 2.e-001| 0:0:01| chol 1 1 9|1.000|1.000|4.9e-010|3.5e-010|2.9e-005| 2.e-001 2.e-001| 0:0:01| chol 1 1 10|0.904|0.904|5.7e-011|1.3e-010|4.4e-006| 2.e-001 2.e-001| 0:0:01| chol 2 2 11|1.0

16、00|1.000|5.2e-013|1.1e-011|1.2e-006| 2.e-001 2.e-001| 0:0:01| chol 2 2 12|1.000|1.000|5.9e-013|1.0e-012|1.8e-007| 2.e-001 2.e-001| 0:0:01| chol 2 2 13|1.000|1.000|1.7e-012|1.0e-012|4.2e-008| 2.e-001 2.e-001| 0:0:01| chol 2 2 14|1.000|1.000|2.3e-012|1.0e-012|7.3e-009| 2.e-001 2.e-001| 0:0:01| stop: m

17、ax(relative gap, infeasibilities) 1.49e-008- number of iterations = 14 primal objective value = 2.e-001 dual objective value = 2.e-001 gap := trace(XZ) = 7.29e-009 relative gap = 4.86e-009 actual relative gap = 4.86e-009 rel. primal infeas (scaled problem) = 2.33e-012 rel. dual = 1.00e-012 rel. prim

18、al infeas (unscaled problem) = 0.00e+000 rel. dual = 0.00e+000 norm(X), norm(y), norm(Z) = 3.2e+000, 1.5e+000, 1.9e+000 norm(A), norm(b), norm(C) = 3.9e+000, 4.2e+000, 2.6e+000 Total CPU time (secs) = 0.99 CPU time per iteration = 0.07 termination code = 0 DIMACS: 3.3e-012 0.0e+000 1.3e-012 0.0e+000

19、 4.9e-009 4.9e-009- -Status: SolvedOptimal 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 x(n) minimize( a*x) subject to C * x =0cvx_end运行结果：Calling SDPT3 4.0: 6 variables, 3 equality constraints- num. of constraints = 3 dim. of linear var = 6* SDPT3: Inf

20、easible 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|1.1e+001|5.1e+000|6.0e+002|-7.e+001 0.e+000| 0:0:00| chol 1 1 1|0.912|1.000|9.4e-001|4.6e-002|6.5e+001|-5.e+000 -2.e+001| 0:0:00| chol 1

21、 1 2|1.000|1.000|1.3e-007|4.6e-003|8.5e+000|-2.e+000 -1.e+001| 0:0:00| chol 1 1 3|1.000|0.961|2.3e-008|6.2e-004|1.8e+000|-4.e+000 -6.e+000| 0:0:00| chol 1 1 4|0.881|1.000|2.2e-008|4.6e-005|3.7e-001|-5.e+000 -5.e+000| 0:0:00| chol 1 1 5|0.995|0.962|1.6e-009|6.2e-006|1.5e-002|-5.e+000 -5.e+000| 0:0:00

22、| chol 1 1 6|0.989|0.989|2.7e-010|5.2e-007|1.7e-004|-5.e+000 -5.e+000| 0:0:00| chol 1 1 7|0.989|0.989|5.3e-011|5.8e-009|1.8e-006|-5.e+000 -5.e+000| 0:0:00| chol 1 1 8|1.000|0.994|2.8e-013|4.3e-011|2.7e-008|-5.e+000 -5.e+000| 0:0:00| stop: max(relative gap, infeasibilities) 1.49e-008- number of itera

23、tions = 8 primal objective value = -5.e+000 dual objective value = -5.e+000 gap := trace(XZ) = 2.66e-008 relative gap = 2.26e-009 actual relative gap = 2.21e-009 rel. primal infeas (scaled problem) = 2.77e-013 rel. dual = 4.31e-011 rel. primal infeas (unscaled problem) = 0.00e+000 rel. dual = 0.00e+

24、000 norm(X), norm(y), norm(Z) = 4.3e+000, 1.3e+000, 1.9e+000 norm(A), norm(b), norm(C) = 6.7e+000, 9.1e+000, 5.4e+000 Total CPU time (secs) = 0.11 CPU time per iteration = 0.01 termination code = 0 DIMACS: 3.6e-013 0.0e+000 5.8e-011 0.0e+000 2.2e-009 2.3e-009- -Status: SolvedOptimal value (cvx_optval): -5.4专心-专注-专业