# Solve The Optimization Problem Compare the number of steps to solve an integer programming problem both with and without an initial feasible point. Intlinprog stopped because the objective value is within a gap tolerance of the optimal value, options. The intcon variables are integer within tolerance, options. Therefore, the problem variables have an implied matrix form.

The problem has eight integer variables and four linear equality constraints, and all variables are restricted to be positive. The uses to solve linear least-squares problems, see Least-Squares (Model Fitting) Algorithms.

For example, companies often want to minimize production costs or maximize revenue.

In manufacturing, it is often desirable to minimize the amount of material used to package a product with a certain volume.

Given \(100\) ft of wire fencing, determine the dimensions that would create a garden of maximum area. Solution Let \(x\) denote the length of the side of the garden perpendicular to the rock wall and \(y\) denote the length of the side parallel to the rock wall.

Then the area of the garden is \(A=x⋅y.\) We want to find the maximum possible area subject to the constraint that the total fencing is \(100\, ft.\) From Figure \(\Page Index\), the total amount of fencing used will be \(2x y.\) Therefore, the constraint equation is \(2x y=100.\) Solving this equation for \(y\), we have \(y=100−2x.\) Thus, we can write the area as \(A(x)=x⋅(100−2x)=100x−2x^2.\) Before trying to maximize the area function \(A(x)=100x−2x^2,\) we need to determine the domain under consideration.

Certainly, if we keep making the side lengths of the garden larger, the area will continue to become larger.

Local minimum found that satisfies the constraints. An object has an internal list of the variables used in its expressions.

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. Branch and Bound: nodes total num int integer relative explored time (s) solution fval gap (%) 10000 1.13 0 - - 18188 1.85 1 2.906000e 03 4.509804e 01 22039 2.30 2 2.073000e 03 2.270974e 01 24105 2.51 3 1.854000e 03 9.973046e 00 24531 2.56 3 1.854000e 03 1.347709e 00 24701 2.58 3 1.854000e 03 0.000000e 00 Optimal solution found. Each variable has a linear index in the expression, and a size.

In this section, we show how to set up these types of minimization and maximization problems and solve them by using the tools developed in this chapter.

The basic idea of the optimization problems that follow is the same.

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