最优化——拟牛顿方法matlab程序

% BFGS
function [x, output] = bfgs(fun, dfun, x0, varargin)
% Step 1: initialization
epsi = 1.0e-6;
k = 0;
funcN = 0;
rho = 0.01; l = 0.15; u = 0.85;
x = x0;
f = feval(fun, x, varargin{:});
funcN = funcN + 1;
n = length(x0);
H = eye(n);
% Step 2: check termination condition
g = feval(dfun, x, varargin{:});
while norm(g) > epsi & k <= 100
    itercon = true;
    d = -H*g;
% Step 3: line search
    alpha_0 = 1.0;
    gd = g'*d;
    [alpha, funcNk, exitflag] = ...
        lines(fun, rho, l, u, alpha_0, f, gd, x, d, varargin{:});
    funcN = funcN + funcNk;
    if exitflag == -1
        itercon = false;
        restart = true;
        H = eye(n);
        gold = g;
    end
% Step 4: compute new point
    if itercon
        s = alpha * d;
        x = x + s;
        f = feval(fun, x, varargin{:});
        funcN = funcN + 1;
        gold = g;
        g = feval(dfun, x, varargin{:});
% Stept 5: update H
        y = g - gold;
        hy = H*y; sy = s'*y; yhy = y'*hy;
        if sy < 0.2*yhy
            theta = 0.8*yhy/(yhy-sy);
            s = theta*s + (1.0-theta)*hy;
            sy = 0.2*yhy;
        end
        v = sqrt(yhy)*(s/sy-hy/yhy);
        H = H + s*s'/sy-hy*hy'/yhy+v*v';
    end
    k = k + 1;
end
% Step 6: output
output.fval = f;
output.iteration = k;
output.funcount = funcN;
output.gnorm = norm(g);
end

测试

% f1.m
function f = f1(x)
% objective function
f = (x(2)-x(1)^2)^2+(1-x(1))^2;
end

% df1.m
function g = df1(x)
% grads function
g = [4.0*(x(1)^3-x(1)*x(2))+2*x(1)-2;2.0*(x(2)-x(1)^2)];
end

% Command
>> x0=[-1.9;2];
>> [x, output]=bfgs('f1', 'df1', x0)

结果

x =

    1.0000
    1.0000

output =

         fval: 6.3098e-017
    iteration: 12
     funcount: 28
        gnorm: 3.8353e-008