Solving equation problem with constrains using the Optimization Toolbox

I have a problem with handful of variables and constrains, it look a lot like this one:

Set Up a Linear Program, Problem-Based - MATLAB & Simulink (mathworks.com)

However, instead of finding the minimum cost like the problem above, I already have a target cost, but find the variables that fits this target cost. So I look at the equation problems:

Create equation problem - MATLAB eqnproblem (mathworks.com)

But it seems that the equation problem wouldn't allow me to add constrains, what should I do?

Answer 1

[Matt J]

Use your first approach except specify a least squares objective, like below.

x=optimvar('x','lower',0);
y=optimvar('y','lower',0);
target=10; %target cost

con(1)=x>=y;
con(2)=x+y>=1;

solve( optimproblem('Objective', (x+2*y-target).^2,'Constraints',con) )

Solving problem using lsqlin.

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.

ans = struct with fields:
    x: 3.8238
    y: 3.0881

Be mindful, however, that the solution will probably be non-unique. For example, another solution to the example problem above is x=10, y=0.

[MENGZE WU]

Thank you for your help! But how can I get all the possible solution, not just one, maybe in the form of plot?

[Matt J]

Is it a 2D problem? You can't plot a 4+ dimensional polygon.

[MENGZE WU]

It is acutally a 3D problem.

[Walter Roberson]

Are all of the variables integer constrained? If so often to get "all" the solutions, using exhaustive search can be the easiest way -- construct all of the potential combinations of the variables, filter out the combinations that do not meet constraints, evaluate the function at each remaining combination.

[Matt J]

All points meeting the target cost satisfy the equation

dot(f,x)=targetCost

which is a line in 2D and a plane in 3D. The set of solutions will be the intersection of this line/plane with the convex region specified by your other constraints. You can compute the vertices of this intersection using intersectionHull() in this FEX package:

https://www.mathworks.com/matlabcentral/fileexchange/30892-analyze-n-dimensional-convex-polyhedra

For example, applying this to the toy 2D problem above,

Aineq=[-1,1;

-1 -1;

-eye(2)]; bineq=[0;-1; 0 ;0]; %inequality constraints

f=[1 2];

targetCost=10;

I=intersectionHull('lcon',Aineq,bineq,[],[],'lcon',[],[],f,targetCost);

I.vert, %vertices

ans =

10.0000 0

3.3333 3.3333

In other words, the set of solutions are all points on the line segment from (x,y)=[3.33,3.33] to (x,y)=[10,0]. In 3D, if you had only inequality constraints, you would get vertices of a polygon floating in the plane dot(f,x)=targetCost. If you wanted to plot this region, you could feed the vertices to patch(), similar to the figure below.

[Walter Roberson]

If the set of solutions is all points on a particular line or particular plane, then you cannot return them all -- not unless you want to iterate over each distinguishable floating point number in the range.

That is why I was suggesting that the system needed to be integer constrained variables: for those it is possible to list all of the solutions (but it might need a lot of memory if the set of points is large enough.)

[Matt J]

If the set of solutions is all points on a particular line or particular plane, then you cannot return them all

No, you can't, but it will be a convex polyhedral region and, if bounded, you can use its vertices both to plot the region and to sample an arbitrary number of points from its interior.