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The `solve`

function returns a solution as a structure, with each variable in the problem having
a field in the structure. To obtain numerical values of expressions in the problem
from this structure easily, use the `evaluate`

function.

For example, solve a linear programming problem in two variables.

x = optimvar('x'); y = optimvar('y'); prob = optimproblem; prob.Objective = -x -y/3; prob.Constraints.cons1 = x + y <= 2; prob.Constraints.cons2 = x + y/4 <= 1; prob.Constraints.cons3 = x - y <= 2; prob.Constraints.cons4 = x/4 + y >= -1; prob.Constraints.cons5 = x + y >= 1; prob.Constraints.cons6 = -x + y <= 2; sol = solve(prob)

sol = struct with fields: x: 0.6667 y: 1.3333

Suppose that you want the objective function value at the solution. You can rerun the problem, this time asking for the objective function value and the solution.

[sol,fval] = solve(prob)

sol = struct with fields: x: 0.6667 y: 1.3333 fval = -1.1111

Alternatively, for a time-consuming problem, save time by evaluating the objective
function at the solution using `evaluate`

.

fval = evaluate(prob.Objective,sol)

fval = -1.1111

To check whether the reported solution is accurate, you can review outputs from
`solve`

. Return all `solve`

outputs

[sol,fval,exitflag,output,lambda] = solve(prob);

Check the exit flag.

`exitflag = OptimalSolution`

generally means that`solve`

converged to the solution. For an explanation of the other`exitflag`

values, see`exitflag`

.Check the exit message at the command line or in the output structure. When the exit message states that the solver converged to a solution, then generally the solution is reliable. This message corresponds to

`exitflag = OptimalSolution`

.When you have integer constraints, check the absolute gap and the relative gap in the exit message or in the output structure. When these gaps are zero or nearly zero, the solution is reliable.

If `solve`

reports that your problem is infeasible (the exit flag
is `NoFeasiblePointFound`

), examine the problem infeasibility at a
variety of points to see which constraints might be overly restrictive. Suppose that
you have a single continuous optimization variable named `x`

that
has finite bounds on all components, and you have constraints
`constr1`

through `constr20`

.

N = 100; % check 100 points infeas = zeros(N,20); % allocate L = x.LowerBound; U = x.UpperBound; S = numel(L); pthist = cell(N); for k = 1:N pt = L + rand(size(L)).*(U-L); pthist{k} = pt; for j = 1:20 infeas(k,j) = infeasibility(['constr',num2str(j)],pt); end end

The result `infeas(a,b)`

has nonzero values wherever the
associated point `pt{a}`

is infeasible for constraint
`b`

.

If `solve`

takes a long time, there are a few possible causes and
remedies.

*Problem formulation is slow*. If you have defined objective or constraint expressions in nested loops, then`solve`

can take a long time to convert the problem internally to a matrix form. To speed the solution, try to formulate your expressions in a vectorized fashion. See Create Efficient Optimization Problems.*Mixed-integer linear programming solution is slow*. Sometimes you can speed up an integer problem by setting options. You can also reformulate the problem to make it faster to solve. See Tuning Integer Linear Programming.*Nonlinear programming solution is slow*. For suggestions, see Solver Takes Too Long. For further suggestions, see When the Solver Fails.*Solver Limit Exceeded*. To solve some problems,`solve`

can take more than the default number of solution steps. For problems with integer constraints, increase the number of allowed steps by increasing the`LPMaxIterations`

,`MaxNodes`

,`MaxTime`

, or`RootLPMaxIterations`

options to higher-than-default values. To set these options, use`optimoptions('intlinprog',...)`

. For non-integer problems, increase the`MaxIterations`

option using`optimoptions('linprog','MaxIterations',...)`

. See`options`

.

`evaluate`

| `infeasibility`

| `solve`