Problem Description

Consider the electrostatics problem of placing 10 electrons in a conducting body. The electrons will arrange themselves so as to minimize their total potential energy, subject to the constraint of lying inside the body. It is well known that all the electrons will be on the boundary of the body at a minimum. The electrons are indistinguishable, so there is no unique minimum for this problem (permuting the electrons in one solution gives another valid solution). This example was inspired by Dolan, Moré, and Munson [58].

This example is a conducting body defined by the following inequalities:

$$x^2+y^2+(z+1{)}^2\le 1$$.

This body looks like a pyramid on a sphere.

[X,Y] = meshgrid(-1:.01:1);
Z1 = -abs(X) - abs(Y);
Z2 = -1 - sqrt(1 - X.^2 - Y.^2);
Z2 = real(Z2);
W1 = Z1; W2 = Z2;
W1(Z1 < Z2) = nan; % only plot points where Z1 > Z2
W2(Z1 < Z2) = nan; % only plot points where Z1 > Z2
hand = figure; % handle to the figure, since we'll plot more later
set(gcf,'Color','w') % white background
surf(X,Y,W1,'LineStyle','none');
hold on
surf(X,Y,W2,'LineStyle','none');
view(-44,18)

There is a slight gap between the upper and lower surfaces of the figure. This is an artifact of the general plotting routine used to create the figure. This routine erases any rectangular patch on one surface that touches the other surface.

Create Variables

Generate a symbolic vector x as a 30-by-1 vector composed of real symbolic variables xij, i between 1 and 10, and j between 1 and 3. These variables represent the three coordinates of electron i: xi1 corresponds to the $$x$$ coordinate, xi2 corresponds to the $$y$$ coordinate, and xi3 corresponds to the $$z$$ coordinate.

x = cell(3, 10);
for i = 1:10
    for j = 1:3
        x{j,i} = sprintf('x%d%d',i,j);
    end
end
x = x(:); % now x is a 30-by-1 vector
x = sym(x, 'real');

Examine the vector x.

x

x = 
$$\left(\begin{array}{c}{x}_{11}\\ x_{12}\\ x_{13}\\ x_{21}\\ x_{22}\\ x_{23}\\ x_{31}\\ x_{32}\\ x_{33}\\ x_{41}\\ x_{42}\\ x_{43}\\ x_{51}\\ x_{52}\\ x_{53}\\ x_{61}\\ x_{62}\\ x_{63}\\ x_{71}\\ x_{72}\\ x_{73}\\ x_{81}\\ x_{82}\\ x_{83}\\ x_{91}\\ x_{92}\\ x_{93}\\ x_{101}\\ x_{102}\\ x_{103}\end{array}\right)$$

Include Linear Constraints

Write the linear constraints

as a set of four linear inequalities for each electron:

xi3 - xi1 - xi2 ≤ 0
xi3 - xi1 + xi2 ≤ 0
xi3 + xi1 - xi2 ≤ 0
xi3 + xi1 + xi2 ≤ 0

Therefore there are a total of 40 linear inequalities for this problem.

Write the inequalities in a structured way:

B = [1,1,1;-1,1,1;1,-1,1;-1,-1,1];

A = zeros(40,30);
for i=1:10
    A(4*i-3:4*i,3*i-2:3*i) = B;
end

b = zeros(40,1);

You can see that A*x ≤ b represents the inequalities:

disp(A*x)

$$\left(\begin{array}{c}{x}_{11}+x_{12}+x_{13}\\ x_{12}-x_{11}+x_{13}\\ x_{11}-x_{12}+x_{13}\\ x_{13}-x_{12}-x_{11}\\ x_{21}+x_{22}+x_{23}\\ x_{22}-x_{21}+x_{23}\\ x_{21}-x_{22}+x_{23}\\ x_{23}-x_{22}-x_{21}\\ x_{31}+x_{32}+x_{33}\\ x_{32}-x_{31}+x_{33}\\ x_{31}-x_{32}+x_{33}\\ x_{33}-x_{32}-x_{31}\\ x_{41}+x_{42}+x_{43}\\ x_{42}-x_{41}+x_{43}\\ x_{41}-x_{42}+x_{43}\\ x_{43}-x_{42}-x_{41}\\ x_{51}+x_{52}+x_{53}\\ x_{52}-x_{51}+x_{53}\\ x_{51}-x_{52}+x_{53}\\ x_{53}-x_{52}-x_{51}\\ x_{61}+x_{62}+x_{63}\\ x_{62}-x_{61}+x_{63}\\ x_{61}-x_{62}+x_{63}\\ x_{63}-x_{62}-x_{61}\\ x_{71}+x_{72}+x_{73}\\ x_{72}-x_{71}+x_{73}\\ x_{71}-x_{72}+x_{73}\\ x_{73}-x_{72}-x_{71}\\ x_{81}+x_{82}+x_{83}\\ x_{82}-x_{81}+x_{83}\\ x_{81}-x_{82}+x_{83}\\ x_{83}-x_{82}-x_{81}\\ x_{91}+x_{92}+x_{93}\\ x_{92}-x_{91}+x_{93}\\ x_{91}-x_{92}+x_{93}\\ x_{93}-x_{92}-x_{91}\\ x_{101}+x_{102}+x_{103}\\ x_{102}-x_{101}+x_{103}\\ x_{101}-x_{102}+x_{103}\\ x_{103}-x_{102}-x_{101}\end{array}\right)$$

Create the Nonlinear Constraints, Their Gradients and Hessians

The nonlinear constraints

are also structured. Generate the constraints, their gradients, and Hessians as follows.

c = sym(zeros(1,10));
i = 1:10;
c = (x(3*i-2).^2 + x(3*i-1).^2 + (x(3*i)+1).^2 - 1).';

gradc = jacobian(c,x).'; % .' performs transpose

hessc = cell(1, 10);
for i = 1:10
    hessc{i} = jacobian(gradc(:,i),x);
end

The constraint vector c is a row vector, and the gradient of c(i) is represented in the ith column of the matrix gradc. This is the correct form, as described in Nonlinear Constraints.

The Hessian matrices, hessc{1}, ..., hessc{10}, are square and symmetric. It is better to store them in a cell array, as is done here, than in separate variables such as hessc1, ..., hesssc10.

Use the .' syntax to transpose. The ' syntax means conjugate transpose, which has different symbolic derivatives.

Create the Objective Function, Its Gradient and Hessian

The objective function, potential energy, is the sum of the inverses of the distances between each electron pair:

$$energy=\sum _{i<j}\frac{1}{\left|x_i-x_j\right|}$$.

The distance is the square root of the sum of the squares of the differences in the components of the vectors.

Calculate the energy, its gradient, and its Hessian as follows.

energy = sym(0);
for i = 1:3:25
    for j = i+3:3:28
        dist = x(i:i+2) - x(j:j+2);
        energy = energy + 1/sqrt(dist.'*dist);
    end
end

gradenergy = jacobian(energy,x).';

hessenergy = jacobian(gradenergy,x);

Create the Objective Function File

The objective function should have two outputs, energy and gradenergy. Put both functions in one vector when calling matlabFunction to reduce the number of subexpressions that matlabFunction generates, and to return the gradient only when the calling function (fmincon in this case) requests both outputs. This example shows placing the resulting files in your current folder. Of course, you can place them anywhere you like, as long as the folder is on the MATLAB path.

currdir = [pwd filesep]; % You may need to use currdir = pwd 
filename = [currdir,'demoenergy.m'];
matlabFunction(energy,gradenergy,'file',filename,'vars',{x});

This syntax causes matlabFunction to return energy as the first output, and gradenergy as the second. It also takes a single input vector {x} instead of a list of inputs x11, ..., x103.

The resulting file demoenergy.m contains, in part, the following lines or similar ones:

function [energy,gradenergy] = demoenergy(in1)
%DEMOENERGY
%   [ENERGY,GRADENERGY] = DEMOENERGY(IN1)
...
x101 = in1(28,:);
...
energy = 1./t140.^(1./2) + ...;
if nargout > 1
    ...
    gradenergy = [(t174.*(t185 - 2.*x11))./2 - ...];
end

This function has the correct form for an objective function with a gradient; see Writing Scalar Objective Functions.

Create the Constraint Function File

Generate the nonlinear constraint function, and put it in the correct format.

filename = [currdir,'democonstr.m'];
matlabFunction(c,[],gradc,[],'file',filename,'vars',{x},...
    'outputs',{'c','ceq','gradc','gradceq'});

The resulting file democonstr.m contains, in part, the following lines or similar ones:

function [c,ceq,gradc,gradceq] = democonstr(in1)
%DEMOCONSTR
%    [C,CEQ,GRADC,GRADCEQ] = DEMOCONSTR(IN1)
...
x101 = in1(28,:);
...
c = [t417.^2 + ...];
if nargout > 1
    ceq = [];
end
if nargout > 2
    gradc = [2.*x11,...];
end
if nargout > 3
    gradceq = [];
end

This function has the correct form for a constraint function with a gradient; see Nonlinear Constraints.

Generate the Hessian Files

To generate the Hessian of the Lagrangian for the problem, first generate files for the energy Hessian and for the constraint Hessians.

The Hessian of the objective function, hessenergy, is a very large symbolic expression, containing over 150,000 symbols, as evaluating size(char(hessenergy)) shows. So it takes a substantial amount of time to run matlabFunction(hessenergy).

To generate a file hessenergy.m, run the following two lines:

filename = [currdir,'hessenergy.m'];
matlabFunction(hessenergy,'file',filename,'vars',{x});

In contrast, the Hessians of the constraint functions are small, and fast to compute:

for i = 1:10
    ii = num2str(i);
    thename = ['hessc',ii];
    filename = [currdir,thename,'.m'];
    matlabFunction(hessc{i},'file',filename,'vars',{x});
end

After generating all the files for the objective and constraints, put them together with the appropriate Lagrange multipliers in a file hessfinal.m, whose code appears at the end of this example..

Run the Optimization

Start the optimization with the electrons distributed randomly on a sphere of radius 1/2 centered at [0,0,–1].

rng default % for reproducibility
Xinitial = randn(3,10); % columns are normal 3-D vectors
for j=1:10
    Xinitial(:,j) = Xinitial(:,j)/norm(Xinitial(:,j))/2;
    % this normalizes to a 1/2-sphere
end
Xinitial(3,:) = Xinitial(3,:) - 1; % center at [0,0,-1]
Xinitial = Xinitial(:); % Convert to a column vector

Set the options to use the interior-point algorithm, and to use gradients and the Hessian.

options = optimoptions(@fmincon,'Algorithm','interior-point','SpecifyObjectiveGradient',true,...
    'SpecifyConstraintGradient',true,'HessianFcn',@hessfinal,'Display','final');

Call fmincon.

[xfinal fval exitflag output] = fmincon(@demoenergy,Xinitial,...
    A,b,[],[],[],[],@democonstr,options);

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

<stopping criteria details>

View the objective function value, exit flag, number of iterations, and number of function evaluations.

disp(fval)

   34.1365

disp(exitflag)

     1

disp(output.iterations)

    19

disp(output.funcCount)

    28

Even though the initial positions of the electrons were random, the final positions are nearly symmetric.

for i = 1:10
    plot3(xfinal(3*i-2),xfinal(3*i-1),xfinal(3*i),'r.','MarkerSize',25);
end

rotate3d
figure(hand)

Compare to Optimization Without Gradients and Hessians

The use of gradients and Hessians makes the optimization run faster and more accurately. To compare with the same optimization using no gradient or Hessian information, set the options not to use gradients and Hessians:

options = optimoptions(@fmincon,'Algorithm','interior-point',...
    'Display','final');
[xfinal2 fval2 exitflag2 output2] = fmincon(@demoenergy,Xinitial,...
    A,b,[],[],[],[],@democonstr,options);

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.

<stopping criteria details>

Examine the same data as before.

disp(fval2)

   34.1365

disp(exitflag2)

     1

disp(output2.iterations)

    78

disp(output2.funcCount)

        2463

Compare the number of iterations and number of function evaluations with and without derivative information.

tbl = table([output.iterations;output2.iterations],[output.funcCount;output2.funcCount],...
    'RowNames',{'With Derivatives','Without Derivatives'},'VariableNames',{'Iterations','Fevals'})

tbl=2×2 table
                           Iterations    Fevals
                           __________    ______

    With Derivatives           19           28 
    Without Derivatives        78         2463 

Clear the Symbolic Variable Assumptions

The symbolic variables in this example have the assumption, in the symbolic engine workspace, that they are real. To clear this assumption from the symbolic engine workspace, it is not sufficient to delete the variables. Clear the variable assumptions by using syms:

syms x

Verify that the assumptions are empty.

assumptions(x)

 
ans =
 
Empty sym: 1-by-0
 

Helper Function

The following code creates the hessfinal function.

function H = hessfinal(X,lambda)
%
% Call the function hessenergy to start
H = hessenergy(X);

% Add the Lagrange multipliers * the constraint Hessians
H = H + hessc1(X) * lambda.ineqnonlin(1);
H = H + hessc2(X) * lambda.ineqnonlin(2);
H = H + hessc3(X) * lambda.ineqnonlin(3);
H = H + hessc4(X) * lambda.ineqnonlin(4);
H = H + hessc5(X) * lambda.ineqnonlin(5);
H = H + hessc6(X) * lambda.ineqnonlin(6);
H = H + hessc7(X) * lambda.ineqnonlin(7);
H = H + hessc8(X) * lambda.ineqnonlin(8);
H = H + hessc9(X) * lambda.ineqnonlin(9);
H = H + hessc10(X) * lambda.ineqnonlin(10);
end