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INTRODUCTION TO NONELINEAR OPTIMIZATION Excise 5.2 Freudenstein and Roth Test Function

發布時間：2024/7/19 编程问答 27 豆豆

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Amir Beck’s INTRODUCTION TO NONELINEAR OPTIMIZATION Theory, Algorithms, and Applications with MATLAB Excise 5.2

INTRODUCTION TO NONELINEAR OPTIMIZATION Theory, Algorithms, and Applications with MATLAB. Amir Beck. 2014

本文主要涉及題目(ii)的MATLAB部分，
題目(i): https://blog.csdn.net/IYXUAN/article/details/121454946

實驗目的

掌握回溯法梯度下降、阻尼牛頓法和混合牛頓法；
學會使用MATLAB，通過上述三種方法求解優化問題；
了解這三種方法的優劣，并能夠分析結果產生的原因。

實驗環境

MATLAB R2021a

實驗內容

Consider the Freudenstein and Roth test function

$f(x)=f1(x)2+f2(x)2,x∈R2,f\left(x\right)=f_1\left(x\right)^2+f_2\left(x\right)^2,x\in \mathbb{R}^2,$

where

$f1(x)=?13+x1+((5?x2)x2?2)x2,f_1\left(x\right)=-13+x_1+\left(\left(5-x_2\right)x_2-2\right)x_2\ ,$

$f2(x)=?29+x1+((x2+1)x2?14)x2.f_2\left(x\right)=-29+x_1+\left(\left(x_2+1\right)x_2-14\right)x_2\ .$

Use MATLAB to employ the following three methods on the problem of minimizing $f$ :

the gradient method with backtracking and parameters

(s,α,β)=(1,0.5,0.5)\left(s,\alpha,\beta\right)=\left(1,0.5,0.5\right)

the hybrid Newton’s method with parameters

(s,α,β)=(1,0.5,0.5)\left(s,\alpha,\beta\right)=\left(1,0.5,0.5\right)

damped Gauss–Newton’s method with a backtracking line search strategy with parameters

(s,α,β)=(1,0.5,0.5)\left(s,\alpha,\beta\right)=\left(1,0.5,0.5\right)

All the algorithms should use the stopping criteria $∣∣?f(x)∣∣≤?,?=10?5\left|\left|\ \nabla f\left(x\right)\right|\right|\le\epsilon,\epsilon={10}^{-5}$ . Each algorithm should be employed four times on the following four starting points: $(?50,7)T,(20,7)T,(20,?18)T,(5,?10)T\left(-50,7\right)^T,\left(20,7\right)^T,\left(20,-18\right)^T,\left(5,-10\right)^T$ . For each of the four starting points, compare the number of iterations and the point to which each method converged. If a method did not converge, explain why.

算法描述

gradient_method_backtracking

回溯法梯度下降算法描述：

設定 $?\epsilon$ ；

初始化 $x0∈Rnx_0\in \mathbb{R}^n$ ；

FOR k = 0, 1, 2, …

選取下降方向

d_k=g(x_k)

；

選取步長

t_k

，使得

f(x_k+t_kd_k)<f(x_k)

；

設置

x_{k+1}=x_k+t_kd_k

；

∥g(xk+1)∥≤?\Vert g(x_{k+1}) \Vert \le \epsilon

THEN STOP，OUTPUT

x_{k+1}

；

在算法循環的 2. 步長選取中，對于定步長的梯度下降來說， $t_k$ 為一常量，即 $tk=tˉt_k=\bar{t}$ . 而對于采用回溯法的非精確線搜索來說，令步長的初值 $t_k=s, s>0$ ，更新步長 $tk←βtk,β∈(0,1)t_k \gets \beta t_k, \beta \in (0,1)$ ，直到滿足 $f(xk)?f(xk+tkdk)≥?αtk?f(xk)Tdk,α∈(0,1)f(x_k)-f(x_k+t_kd_k)\ge -\alpha t_k\nabla f(x_k)^Td_k, \alpha \in (0,1)$ ，可以證明，當 $t∈[0,?]t\in [0,\epsilon]$ ，上述不等式總成立。

newton_backtracking

回溯牛頓算法描述：

回溯牛頓法和上述梯度下降相似，只需將1. 下降方向選取改為牛頓方向即可。

$dk=?(?2f(xk))?1?f(xk)d_k=-\left( \nabla^2f(x_k) \right )^{-1}\nabla f(x_k)$ .

newton_hybrid

混合牛頓法描述：

在選取下降方向 $d_k$ 時，當 Hessian 矩陣 $?2f(xk)\nabla^2f(x_k)$ 非正定時，使用回溯法梯度下降處理，即 $dk=??f(xk)d_k=-\nabla f(x_k)$ 。當 $?2f(xk)\nabla^2f(x_k)$ 正定時， $d_k$ 可選用牛頓方向。在判定 Hessian 矩陣是否正定時，使用 Cholesky 分解作為判定條件。

實驗步驟

計算 $f (x), g (x), h (x)$

$f(x)=2x_1^2 + 12x_1x_2^2 - 32x_1x_2 - 84x_1 + 2x_2^6 - 8x_2^5 + 2x_2^4 - 80x_2^3 + 12x_2^2 + 864x_2 + 1010$

$g(x)=(12x22?32x2+4x1?84,24x2?32x1+24x1x2?240x22+8x23?40x24+12x25+864)Tg(x)=\left ( 12x_2^2 - 32x_2 + 4x_1 - 84,24x_2 - 32x_1 + 24x_1x_2 - 240x_2^2 + 8x_2^3 - 40x_2^4 + 12x_2^5 + 864\right )^T$

$h(x)=(424x2?3224x2?3260x24?160x23+24x22?480x2+24x1+24)h(x)=\begin{pmatrix} 4 &24x_2 - 32 \\ 24x_2 - 32 &60x_2^4 - 160x_2^3 + 24x_2^2 - 480x_2 + 24x_1 + 24 \end{pmatrix}$

clc;clear syms f1 f2 x1 x2 f g h f1=-13+x1+((5-x2)*x2-2)*x2; f2=-29+x1+((x2+1)*x2-14)*x2; f=expand(f1^2+f2^2) % f = 2*x1^2 + 12*x1*x2^2 - 32*x1*x2 - 84*x1 + 2*x2^6 - 8*x2^5 + 2*x2^4 - 80*x2^3 + 12*x2^2 + 864*x2 + 1010 g=gradient(f) % g = [12*x2^2 - 32*x2 + 4*x1 - 84; 24*x2 - 32*x1 + 24*x1*x2 - 240*x2^2 + 8*x2^3 - 40*x2^4 + 12*x2^5 + 864] h=hessian(f) % h = [4, 24*x2 - 32; 24*x2 - 32, 60*x2^4 - 160*x2^3 + 24*x2^2 - 480*x2 + 24*x1 + 24]

繪制函數的等高線和曲面圖

Figure 1. contour and surface plots of $f (x)$ around the global optimal solution $x^*=(5,4)$ .

Figure 2. contour and surface plots of $f (x)$ around the local optimal solution $x^*=(11.4128, -0.8968)$ .

clc;clear clc;clear clc;clear % local optimality x1 = linspace(11.40,11.42,50); x2 = linspace(-0.897,-0.8965,50); % global optimality % x1 = linspace(4.9,5.1,50); % x2 = linspace(3.9,4.1,50); [X1,X2] = meshgrid(x1,x2); f = 2*X1.^2 + 12*X1.*X2.^2 - 32*X1.*X2 - 84*X1 + 2*X2.^6 - 8*X2.^5 ...+ 2*X2.^4 - 80*X2.^3 + 12*X2.^2 + 864*X2 + 1010; createfigure(X1,X2,f) %% 由 surfc 函數生成： function createfigure(xdata1, ydata1, zdata1) %CREATEFIGURE(xdata1, ydata1, zdata1) % XDATA1: surface xdata % YDATA1: surface ydata % ZDATA1: surface zdata% Auto-generated by MATLAB on 08-Dec-2021 15:44:51% Create figure figure1 = figure;% Create axes axes1 = axes('Parent',figure1); hold(axes1,'on');% Create surf surf(xdata1,ydata1,zdata1,'Parent',axes1);% Create contour contour(xdata1,ydata1,zdata1,'ZLocation','zmin');view(axes1,[-113 13]); grid(axes1,'on'); axis(axes1,'tight'); hold(axes1,'off'); % Create colorbar colorbar(axes1);

執行函數

clc;clear%% init s = 1; alpha = .5; beta = .5; epsilon = 1e-5; p1 = [-50;7]; p2 = [20;7]; p3 = [20;-18]; p4 = [5;-10]; f = @(x) 2*x(1)^2 + 12*x(1)*x(2)^2 - 32*x(1)*x(2) - 84*x(1) ...+ 2*x(2)^6 - 8*x(2)^5 + 2*x(2)^4 - 80*x(2)^3 + 12*x(2)^2 ...+ 864*x(2) + 1010; g = @(x) [12*x(2)^2 - 32*x(2) + 4*x(1) - 84; 24*x(2) - 32*x(1)...+ 24*x(1)*x(2) - 240*x(2)^2 + 8*x(2)^3 - 40*x(2)^4 + 12*x(2)^5 + 864]; h = @(x) [4, 24*x(2) - 32; 24*x(2) - 32, 60*x(2)^4 - 160*x(2)^3 ...+ 24*x(2)^2 - 480*x(2) + 24*x(1) + 24];%% call func % gradient_method_backtracking [x_gb1,fun_val_gb1] = gradient_method_backtracking(f,g,p1,s,alpha,... beta,epsilon); [x_gb2,fun_val_gb2] = gradient_method_backtracking(f,g,p2,s,alpha,... beta,epsilon); [x_gb3,fun_val_gb3] = gradient_method_backtracking(f,g,p3,s,alpha,... beta,epsilon); [x_gb4,fun_val_gb4] = gradient_method_backtracking(f,g,p4,s,alpha,... beta,epsilon);% newton_backtracking x_nb1 = newton_backtracking(f,g,h,p1,alpha,beta,epsilon); x_nb2 = newton_backtracking(f,g,h,p2,alpha,beta,epsilon); x_nb3 = newton_backtracking(f,g,h,p3,alpha,beta,epsilon); x_nb4 = newton_backtracking(f,g,h,p4,alpha,beta,epsilon);% newton_hybrid x_nh1 = newton_hybrid(f,g,h,p1,alpha,beta,epsilon); x_nh2 = newton_hybrid(f,g,h,p2,alpha,beta,epsilon); x_nh3 = newton_hybrid(f,g,h,p3,alpha,beta,epsilon); x_nh4 = newton_hybrid(f,g,h,p4,alpha,beta,epsilon); function [x,fun_val]=gradient_method_backtracking(f,g,x0,s,alpha,... beta,epsilon)% Gradient method with backtracking stepsize rule % % INPUT %======================================= % f ......... objective function % g ......... gradient of the objective function % x0......... initial point % s ......... initial choice of stepsize % alpha ..... tolerance parameter for the stepsize selection % beta ...... the constant in which the stepsize is multiplied % at each backtracking step (0<beta<1) % epsilon ... tolerance parameter for stopping rule % OUTPUT %======================================= % x ......... optimal solution (up to a tolerance) % of min f(x) % fun_val ... optimal function valuex=x0; grad=g(x); fun_val=f(x); iter=0; while (norm(grad)>epsilon)iter=iter+1;t=s;while (fun_val-f(x-t*grad)<alpha*t*norm(grad)^2)t=beta*t;endx=x-t*grad;fun_val=f(x);grad=g(x);fprintf('iter_number = %3d norm_grad = %2.6f fun_val = %2.6f \n',...iter,norm(grad),fun_val); end function x=newton_backtracking(f,g,h,x0,alpha,beta,epsilon)% Newton’s method with backtracking % % INPUT %======================================= % f ......... objective function % g ......... gradient of the objective function % h ......... hessian of the objective function % x0......... initial point % alpha ..... tolerance parameter for the stepsize selection strategy % beta ...... the proportion in which the stepsize is multiplied % at each backtracking step (0<beta<1) % epsilon ... tolerance parameter for stopping rule % OUTPUT %======================================= % x ......... optimal solution (up to a tolerance) % of min f(x) % fun_val ... optimal function valuex=x0; gval=g(x); hval=h(x); d=hval\gval; iter=0; while ((norm(gval)>epsilon)&&(iter<10000))iter=iter+1;t=1;while(f(x-t*d)>f(x)-alpha*t*gval'*d)t=beta*t;endx=x-t*d;fprintf('iter= %2d f(x)=%10.10f\n',iter,f(x))gval=g(x);hval=h(x);d=hval\gval; end if (iter==10000)fprintf('did not converge\n') end function x=newton_hybrid(f,g,h,x0,alpha,beta,epsilon)% Hybrid Newton’s method % % INPUT %======================================= % f ......... objective function % g ......... gradient of the objective function % h ......... hessian of the objective function % x0......... initial point % alpha ..... tolerance parameter for the stepsize selection strategy % beta ...... the proportion in which the stepsize is multiplied % at each backtracking step (0<beta<1) % epsilon ... tolerance parameter for stopping rule % OUTPUT %======================================= % x ......... optimal solution (up to a tolerance) % of min f(x) % fun_val ... optimal function valuex=x0; gval=g(x); hval=h(x); [L,p]=chol(hval,'lower'); if (p==0)d=L'\(L\gval); elsed=gval; end iter=0; while ((norm(gval)>epsilon)&&(iter<10000))iter=iter+1;t=1;while(f(x-t*d)>f(x)-alpha*t*gval'*d)t=beta*t;endx=x-t*d;fprintf('iter= %2d f(x)=%10.10f\n',iter,f(x))gval=g(x);hval=h(x);[L,p]=chol(hval,'lower');if (p==0)d=L'\(L\gval);elsed=gval;end end if (iter==10000)fprintf('did not converge\n') end

結果與分析

$p1=(?50,7)T,p2=(20,7)T,p3=(20,?18)T,p4=(5,?10)Tp_1=\left(-50,7\right)^T,p_2=\left(20,7\right)^T,p_3=\left(20,-18\right)^T,p_4=\left(5,-10\right)^T$

gradient_method_backtracking

x*fun_valiter

p1	5.0, 4.0	0.000000	-
p2	5.0, 4.0	0.000000	-
p3	11.4128, -0.8968	48.984254	-
p4	5.0, 4.0	0.000107	-

# p1 iter_number = 1 norm_grad = 16886.999242 fun_val = 11336.705024 iter_number = 2 norm_grad = 5600.459983 fun_val = 5803.984203 iter_number = 3 norm_grad = 1056.831987 fun_val = 4723.717473 iter_number = 4 norm_grad = 434.668354 fun_val = 4676.341028 iter_number = 5 norm_grad = 181.792854 fun_val = 4664.456052 iter_number = 6 norm_grad = 481.845230 fun_val = 4645.697772 iter_number = 7 norm_grad = 1450.431648 fun_val = 2784.316562 iter_number = 8 norm_grad = 626.401853 fun_val = 2597.516663 iter_number = 9 norm_grad = 220.402369 fun_val = 2565.305713 iter_number = 10 norm_grad = 116.512243 fun_val = 2561.054079 ...... iter_number = 20849 norm_grad = 0.000092 fun_val = 0.000000 iter_number = 20850 norm_grad = 0.000092 fun_val = 0.000000 iter_number = 20851 norm_grad = 0.000092 fun_val = 0.000000 iter_number = 20852 norm_grad = 0.000092 fun_val = 0.000000 ......# p2 iter_number = 1 norm_grad = 20020.154130 fun_val = 11955.275024 iter_number = 2 norm_grad = 8107.335439 fun_val = 3744.085975 iter_number = 3 norm_grad = 2993.085982 fun_val = 1129.842099 iter_number = 4 norm_grad = 949.370487 fun_val = 445.778197 iter_number = 5 norm_grad = 192.946640 fun_val = 320.464548 iter_number = 6 norm_grad = 86.882053 fun_val = 313.996463 iter_number = 7 norm_grad = 41.236184 fun_val = 311.880377 iter_number = 8 norm_grad = 223.087207 fun_val = 295.899225 iter_number = 9 norm_grad = 91.972368 fun_val = 287.479235 iter_number = 10 norm_grad = 40.293830 fun_val = 285.264643 ...... iter_number = 13961 norm_grad = 0.000151 fun_val = 0.000000 iter_number = 13962 norm_grad = 0.000151 fun_val = 0.000000 iter_number = 13963 norm_grad = 0.000151 fun_val = 0.000000 iter_number = 13964 norm_grad = 0.000151 fun_val = 0.000000 iter_number = 13965 norm_grad = 0.000151 fun_val = 0.000000 ......# p3 iter_number = 1 norm_grad = 10459534.883496 fun_val = 26901796.557585 iter_number = 2 norm_grad = 4328861.498929 fun_val = 9350549.111728 iter_number = 3 norm_grad = 1815010.486957 fun_val = 3306985.630845 iter_number = 4 norm_grad = 764085.159510 fun_val = 1178877.486706 iter_number = 5 norm_grad = 322588.854922 fun_val = 423840.683140 iter_number = 6 norm_grad = 136745.858345 fun_val = 154327.967236 iter_number = 7 norm_grad = 58355.714510 fun_val = 57260.204151 iter_number = 8 norm_grad = 25173.561268 fun_val = 21780.708212 iter_number = 9 norm_grad = 11034.676282 fun_val = 8504.776429 iter_number = 10 norm_grad = 4932.916085 fun_val = 3367.298679 ...... iter_number = 11759 norm_grad = 0.000034 fun_val = 48.984254 iter_number = 11760 norm_grad = 0.000034 fun_val = 48.984254 iter_number = 11761 norm_grad = 0.000034 fun_val = 48.984254 iter_number = 11762 norm_grad = 0.000034 fun_val = 48.984254 iter_number = 11763 norm_grad = 0.000034 fun_val = 48.984254 iter_number = 11764 norm_grad = 0.000034 fun_val = 48.984254 ......# p4 iter_number = 1 norm_grad = 740484.580250 fun_val = 1125740.309089 iter_number = 2 norm_grad = 318800.867908 fun_val = 410873.876977 iter_number = 3 norm_grad = 135283.737154 fun_val = 147433.167712 iter_number = 4 norm_grad = 57286.724183 fun_val = 52647.763355 iter_number = 5 norm_grad = 24273.902315 fun_val = 18647.021114 iter_number = 6 norm_grad = 10264.843110 fun_val = 6442.533586 iter_number = 7 norm_grad = 4279.439838 fun_val = 2094.562330 iter_number = 8 norm_grad = 1694.354839 fun_val = 604.979471 iter_number = 9 norm_grad = 568.486420 fun_val = 158.093768 iter_number = 10 norm_grad = 98.759692 fun_val = 69.674565 ...... iter_number = 6072 norm_grad = 0.000107 fun_val = 0.000000 iter_number = 6073 norm_grad = 0.000107 fun_val = 0.000000 iter_number = 6074 norm_grad = 0.000107 fun_val = 0.000000 iter_number = 6075 norm_grad = 0.000107 fun_val = 0.000000 iter_number = 6076 norm_grad = 0.000107 fun_val = 0.000000 ......

newton_backtracking

x*fun_valiter

p1	5.0, 4.0	-0.000000	8
p2	5.0, 4.0	0.000000	11
p3	11.4128, -0.8968	48.984254	16
p4	11.4128, -0.8968	48.984254	13

# p1 iter= 1 f(x)=17489.2000310639 iter= 2 f(x)=3536.1276294897 iter= 3 f(x)=556.9651095482 iter= 4 f(x)=50.4029326519 iter= 5 f(x)=1.2248344195 iter= 6 f(x)=0.0013284549 iter= 7 f(x)=0.0000000018 iter= 8 f(x)=-0.0000000000# p2 iter= 1 f(x)=18114.8519548468 iter= 2 f(x)=3676.4191962662 iter= 3 f(x)=583.8090713510 iter= 4 f(x)=53.8298627069 iter= 5 f(x)=1.3684607619 iter= 6 f(x)=0.0016459572 iter= 7 f(x)=0.0000000027 iter= 8 f(x)=0.0000000007 iter= 9 f(x)=0.0000000002 iter= 10 f(x)=0.0000000000 iter= 11 f(x)=0.0000000000# p3 iter= 1 f(x)=21356364.5665758252 iter= 2 f(x)=5561714.7070109081 iter= 3 f(x)=1444932.1368662491 iter= 4 f(x)=374422.3650773597 iter= 5 f(x)=96843.5106842914 iter= 6 f(x)=25078.5606496656 iter= 7 f(x)=6556.5446423615 iter= 8 f(x)=1764.6276022580 iter= 9 f(x)=509.7515417783 iter= 10 f(x)=171.5021447876 iter= 11 f(x)=76.9343325176 iter= 12 f(x)=52.3964690617 iter= 13 f(x)=49.0499328705 iter= 14 f(x)=48.9842770185 iter= 15 f(x)=48.9842536792 iter= 16 f(x)=48.9842536792# p4 iter= 1 f(x)=695503.8833055196 iter= 2 f(x)=180005.8157903731 iter= 3 f(x)=46559.5630006172 iter= 4 f(x)=12100.5787363822 iter= 5 f(x)=3202.4785180644 iter= 6 f(x)=889.1211707798 iter= 7 f(x)=275.2063763915 iter= 8 f(x)=106.0983438392 iter= 9 f(x)=59.2021422471 iter= 10 f(x)=49.5512377643 iter= 11 f(x)=48.9860167506 iter= 12 f(x)=48.9842536959 iter= 13 f(x)=48.9842536792

newton_hybrid

x*fun_valiter

p1	5.0, 4.0	-0.000000	8
p2	5.0, 4.0	0.000000	11
p3	11.4128, -0.8968	48.984254	16
p4	11.4128, -0.8968	48.984254	13

結果分析

回溯法梯度下降（gradient_method_backtracking）在 $f (x)$ 函數上的表現情況比較糟糕，在所有初始點（p1-p4）上均不能滿足結束條件 $∣∣?f(x)∣∣≤?,?=10?5\left|\left|\ \nabla f\left(x\right)\right|\right|\le\epsilon,\epsilon={10}^{-5}$ ，所以其迭代步數 $\to \infty$ . 但是可以發現，隨著迭代次數的不斷增加，最優解 $x^*$ 和函數值 $fun\_val$ 均收斂。在起始點為p1,p2,p4時，回溯梯度下降求得的最優解 $xˉ→(5.0,4.0),fun_val→0.0\bar{x} \to (5.0,4.0), fun\_val \to 0.0$ ，顯然是 $f (x)$ 的全局最優，但當以p3為起始點時，所得解為局部最優。此外，當以p4為起始點時， $fun_val→0.000107fun\_val \to 0.000107$ ，收斂效果不如p1,p2起始點。總之，回溯法梯度下降的結果說明，由梯度下降得到的 $f (x)$ 的最優解，無論是在全局最優還是局部最優，其梯度 $?f(x?)\nabla f\left(x^*\right)$ 難以收斂到 $0\mathscr{0}$ ，這也導致了梯度下降的發散。

回溯牛頓法（newton_backtracking）和混合牛頓法（newton_hybrid）得到的結果極為相似，二者在所有四個起始點上的收斂結果表現良好。當以p1,p2為起始點時，兩種方法均在10步左右收斂到全局最優 $(5.0, 4.0)$ 。當以p3,p4為起始點時，兩方法也能在16步以內收斂到局部最優 $(11.4128, ? 0.8968)$ . 觀察兩種方法在四個起始點上的輸出結果，不難發現，輸出結果完全相同，說明 $?2f(xk),k=1,2,…\nabla ^2f(x_k), k=1,2,\dots$ 是非正定的，所以混合牛頓法中的(a)步（純牛頓）永遠不執行，混合牛頓此時等價于回溯牛頓。

總之，在 $f (x)$ 上，當以 $∣∣?f(x)∣∣≤?,?=10?5\left|\left|\ \nabla f\left(x\right)\right|\right|\le\epsilon,\epsilon={10}^{-5}$ 為終止條件時，牛頓法的迭代次數明顯少于梯度下降法。兩種方法，當選取的起始點不同，可能會導致收斂不到全局最優解。

?? Sylvan Ding ??

總結

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编程问答

INTRODUCTION TO NONELINEAR OPTIMIZATION Excise 5.2 Freudenstein and Roth Test Function

Amir Beck’s INTRODUCTION TO NONELINEAR OPTIMIZATION Theory, Algorithms, and Applications with MATLAB Excise 5.2

實驗目的

實驗環境

實驗內容

算法描述

gradient_method_backtracking

newton_backtracking

newton_hybrid

實驗步驟

計算 f(x),g(x),h(x)f(x),g(x),h(x)f(x),g(x),h(x)

繪制函數的等高線和曲面圖

執行函數

結果與分析

gradient_method_backtracking

newton_backtracking

newton_hybrid

結果分析

總結

計算 $f (x), g (x), h (x)$