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資源簡(jiǎn)介

乘子法核心代碼,fun、dfun為目標(biāo)函數(shù)和梯度;hf、dhf為等式約束函數(shù)及其Jacobi矩陣的轉(zhuǎn)置;gf、dgf為不等式約束函數(shù)及其Jacobi矩陣的轉(zhuǎn)置;x0為初始點(diǎn)

資源截圖

代碼片段和文件信息

function?[xvalk]=bfgs(fungfunx0varargin)
%功能:?用BFGS算法求解無約束問題:??min?f(x)
%輸入:?x0是初始點(diǎn)?fun?gfun分別是目標(biāo)函數(shù)及其梯度;
%?varargin是輸入的可變參數(shù)變量?簡(jiǎn)單調(diào)用bfgs時(shí)可以忽略它
%?但若其它程序循環(huán)調(diào)用該程序時(shí)將發(fā)揮重要的作用
%輸出:??x?val分別是近似最優(yōu)點(diǎn)和最優(yōu)值??k是迭代次數(shù).
maxk=500;???%給出最大迭代次數(shù)
rho=0.55;?sigma1=0.4;?epsilon1=1e-5;?
k=0;???n=length(x0);?
Bk=eye(n);???%Bk=feval(‘Hess‘x0);?
while(k????gk=feval(gfunx0varargin{:});?%計(jì)算梯度
????if(norm(gk)????dk=-Bk\gk;??%解方程組?計(jì)算搜索方向
????m=0;?mk=0;
????while(m<20)???%?用Armijo搜索求步長(zhǎng)?
????????newf=feval(funx0+rho^m*dkvarargin{:});
????????oldf=feval(funx0varargin{:});
????????if(newf????????????mk=m;?break;
????????end
????????m=m+1;
????end
????%BFGS校正
????x=x0+rho^mk*dk;??
????sk=x-x0;??yk=feval(gfunxvarargin{:}

?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----
?????文件????????1083??2009-10-15?20:56??乘子法程序\bfgs.m
?????文件??????????57??2009-10-15?20:26??乘子法程序\df1.m
?????文件??????????53??2009-10-15?21:49??乘子法程序\dg1.m
?????文件??????????42??2009-10-15?20:26??乘子法程序\dh1.m
?????文件?????????311??2009-10-15?17:55??乘子法程序\dmpsi.m
?????文件??????????46??2009-10-15?20:26??乘子法程序\f1.m
?????文件??????????44??2009-10-15?21:50??乘子法程序\g1.m
?????文件??????????40??2009-10-15?20:26??乘子法程序\h1.m
?????文件?????????365??2009-10-15?17:51??乘子法程序\mpsi.m
?????文件????????1669??2009-10-15?20:36??乘子法程序\multphr.m

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