intfmincon Solves a constrainted multi-variable mixed integer non linear programming problem xopt = intfmincon(f,x0,intcon,A,b) xopt = intfmincon(f,x0,intcon,A,b,Aeq,beq) xopt = intfmincon(f,x0,intcon,A,b,Aeq,beq,lb,ub) xopt = intfmincon(f,x0,intcon,A,b,Aeq,beq,lb,ub,nlc) xopt = intfmincon(f,x0,intcon,A,b,Aeq,beq,lb,ub,nlc,options) [xopt,fopt] = intfmincon(.....) [xopt,fopt,exitflag]= intfmincon(.....) [xopt,fopt,exitflag,gradient]=intfmincon(.....) [xopt,fopt,exitflag,gradient,hessian]=intfmincon(.....) f : A function, representing the objective function of the problem. x0 : A vector of doubles, containing the starting values of variables of size (1 X n) or (n X 1) where 'n' is the number of variables. intcon : A vector of integers, representing the variables that are constrained to be integers. A : A matrix of doubles, containing the coefficients of linear inequality constraints of size (m X n) where 'm' is the number of linear inequality constraints. b : A vector of doubles, related to 'A' and represents the linear coefficients in the linear inequality constraints of size (m X 1). Aeq : A matrix of doubles, containing the coefficients of linear equality constraints of size (m1 X n) where 'm1' is the number of linear equality constraints. beq : A vector of double, vector of doubles, related to 'Aeq' and represents the linear coefficients in the equality constraints of size (m1 X 1). lb : A vector of doubles, containing the lower bounds of the variables of size (1 X n) or (n X 1) where 'n' is the number of variables. ub : A vector of doubles, containing the upper bounds of the variables of size (1 X n) or (n X 1) where 'n' is the number of variables. nlc : A function, representing the Non-linear Constraints functions(both Equality and Inequality) of the problem. It is declared in such a way that non-linear inequality constraints (c), and the non-linear equality constraints (ceq) are defined as separate single row vectors. options : A list, containing the option for user to specify. See below for details. xopt : A vector of doubles, containing the the computed solution of the optimization problem. fopt : A double, containing the value of the function at xopt. exitflag : An integer, containing the flag which denotes the reason for termination of algorithm. See below for details. gradient : a vector of doubles, containing the Objective's gradient of the solution. hessian : a matrix of doubles, containing the Objective's hessian of the solution. Search the minimum of a mixed integer constrained optimization problem specified by : Find the minimum of f(x) such that \begin{eqnarray} &\mbox{min}_{x} & f(x) \\ & \text{Subjected to:} & A \boldsymbol{\cdot} x \leq b \\ & & Aeq \boldsymbol{\cdot} x \ = beq\\ & & c(x) \leq 0\\ & & ceq(x) \ = 0\\ & & lb \leq x \leq ub \\ & & x_{i} \in \!\, \mathbb{Z}, i \in \!\, I \end{eqnarray} intfmincon calls Bonmin, an optimization library written in C++, to solve the Constrained Optimization problem. The options allow the user to set various parameters of the Optimization problem. The syntax for the options is given by: options= list("IntegerTolerance", [---], "MaxNodes",[---], "MaxIter", [---], "AllowableGap",[---] "CpuTime", [---],"gradobj", "off", "hessian", "off" ); IntegerTolerance : A Scalar, a number with that value of an integer is considered integer. MaxNodes : A Scalar, containing the maximum number of nodes that the solver should search. CpuTime : A scalar, specifying the maximum amount of CPU Time in seconds that the solver should take. AllowableGap : A scalar, that specifies the gap between the computed solution and the the objective value of the best known solution stop, at which the tree search can be stopped. MaxIter : A scalar, specifying the maximum number of iterations that the solver should take. gradobj : A string, to turn on or off the user supplied objective gradient. hessian : A scalar, to turn on or off the user supplied objective hessian. The default values for the various items are given as: options = list('integertolerance',1d-06,'maxnodes',2147483647,'cputime',1d10,'allowablegap',0,'maxiter',2147483647,'gradobj',"off",'hessian',"off") The exitflag allows to know the status of the optimization which is given back by Ipopt. 0 : Optimal Solution Found 1 : InFeasible Solution. 2 : Objective Function is Continuous Unbounded. 3 : Limit Exceeded. 4 : User Interrupt. 5 : MINLP Error. For more details on exitflag, see the Bonmin documentation which can be found on http://www.coin-or.org/Bonmin A few examples displaying the various functionalities of intfmincon have been provided below. You will find a series of problems and the appropriate code snippets to solve them. Here we solve a simple objective function, subjected to three linear inequality constraints. Find x in R^2 such that it minimizes: \begin{eqnarray} \mbox{min}_{x}\ f(x) = x_{1}^{2} - x_{1} \boldsymbol{\cdot} x_{2}/3 + x_{2}^{2} \end{eqnarray} \\\text{Subjected to:}\\ \begin{eqnarray} \hspace{70pt} &x_{1} + x_{2}&\leq 2\\ \hspace{70pt} &x_{1} + \dfrac{x_{2}}{4}&\leq 1\\ \hspace{70pt} &-x_{1} + x_{2}&\geq -2\\ \end{eqnarray}\\ \text{With integer constraints as: } \\ \begin{eqnarray} \begin{array}{c} [1] \\ \end{array} \end{eqnarray} Here we build up on the previous example by adding linear equality constraints. We add the following constraints to the problem specified above: \begin{eqnarray} &x_{1} - x_{2}&= 1 \\&2x_{1} + x_{2}&= 2 \\\end{eqnarray} In this example, we proceed to add the upper and lower bounds to the objective function. Find x in R^2 such that it minimizes: \begin{eqnarray} -1 &\leq x_{1} &\leq \infty\\ -\infty &\leq x_{2} &\leq 1 \end{eqnarray} Finally, we add the non-linear constraints to the problem. Note that there is a notable difference in the way this is done as compared to defining the linear constraints. \begin{eqnarray} \mbox{min}_{x}\ f(x) = x_{1} \boldsymbol{\cdot} x_{2} + x_{2} \boldsymbol{\cdot} x_{3} \end{eqnarray} \\\text{Subjected to:}\\ \begin{eqnarray} \hspace{70pt} &x_{1}^{2} - x_{2}^{2} + x_{3}^{2}&\leq 2\\ \hspace{70pt} &x_{1}^{2} + x_{2}^{2} + x_{3}^{2}&\leq 10\\ \end{eqnarray}\\ \text{With integer constraints as: }\\ \begin{eqnarray} \begin{array}{c} [2] \\ \end{array} \end{eqnarray} We can further enhance the functionality of intfmincon by setting input options. We can pre-define the gradient of the objective function and/or the hessian of the lagrange function and thereby improve the speed of computation. This is elaborated on in example 5. We take the following problem and add simple non-linear constraints, specify the gradients and the hessian of the Lagrange Function. We also set solver parameters using the options. Infeasible Problems: Find x in R^3 such that it minimizes: \begin{eqnarray} f(x) = x_{1} \boldsymbol{\cdot} x_{2} + x_{2} \boldsymbol{\cdot} x_{3} \end{eqnarray} \\\text{Subjected to:}\\ \begin{eqnarray} \hspace{70pt} &x_{1}^{2} &\leq 1\\ \hspace{70pt} &x_{1}^{2} + x_{2}^{2}&\leq 1\\ \hspace{70pt} &x_{3}^{2}&\leq 1\\ \hspace{70pt} &x_{1}^{3}&\leq 0.5\\ \hspace{70pt} &x_{2}^{2} + x_{3}^{2}&\leq 0.75\\ \end{eqnarray}\\ \text{With variable bounds as: }\\ \begin{eqnarray} \hspace{70pt} 0 &\leq x_{1} &\leq 0.6\\ \hspace{70pt} 0.2 &\leq x_{2} &\leq \infty\\ \end{eqnarray}\\ \text{With integer constraints as: } \\ \begin{eqnarray} \begin{array}{c} [2] \\ \end{array} \end{eqnarray} Unbounded Problems: Find x in R^3 such that it minimizes: \begin{eqnarray} \mbox{min}_{x}\ f(x) = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}\\ \end{eqnarray}\\ \text{With variable bounds as: }\\ \begin{eqnarray} -\infty &\leq x_{1} &\leq 0\\ -\infty &\leq x_{2} &\leq 0\\ -\infty &\leq x_{3} &\leq 0\\ \end{eqnarray}\\ \text{With integer constraints as: } \\ \begin{eqnarray} \begin{array}{c} [3] \\ \end{array} \end{eqnarray} Harpreet Singh