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(.....)
a function, representing the objective function of the problem
a vector of doubles, containing the starting values of variables.
a vector of integers, represents which variables are constrained to be integers
a matrix of double, represents the linear coefficients in the inequality constraints A⋅x ≤ b.
a vector of double, represents the linear coefficients in the inequality constraints A⋅x ≤ b.
a matrix of double, represents the linear coefficients in the equality constraints Aeq⋅x = beq.
a vector of double, represents the linear coefficients in the equality constraints Aeq⋅x = beq.
Lower bounds, specified as a vector or array of double. lb represents the lower bounds elementwise in lb ≤ x ≤ ub.
Upper bounds, specified as a vector or array of double. ub represents the upper bounds elementwise in lb ≤ x ≤ ub.
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 are defined first as a single row vector (c), followed by non-linear equality constraints as another single row vector (ceq). Refer Example for definition of Constraint function.
a list, containing the option for user to specify. See below for details.
a vector of doubles, containing the the computed solution of the optimization problem.
a scalar of double, containing the the function value at x.
a scalar of integer, containing the flag which denotes the reason for termination of algorithm. See below for details.
a vector of doubles, containing the Objective's gradient of the solution.
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
The routine calls Bonmin for solving the Bounded Optimization problem, Bonmin is a library written in C++.
The options allows the user to set various parameters of the Optimization problem. It should be defined as type "list" and contains the following fields.
The exitflag allows to know the status of the optimization which is given back by Ipopt.
For more details on exitflag see the Bonmin documentation, go to http://www.coin-or.org/Bonmin
//Find x in R^2 such that it minimizes: //f(x)= -x1 -x2/3 //x0=[0,0] //constraint-1 (c1): x1 + x2 <= 2 //constraint-2 (c2): x1 + x2/4 <= 1 //constraint-3 (c3): x1 - x2 <= 2 //constraint-4 (c4): -x1/4 - x2 <= 1 //constraint-5 (c5): -x1 - x2 <= -1 //constraint-6 (c6): -x1 + x2 <= 2 //constraint-7 (c7): x1 + x2 = 2 //Objective function to be minimised function [y, dy]=f(x) y=-x(1)-x(2)/3; dy= [-1,-1/3]; endfunction //Starting point, linear constraints and variable bounds x0=[0 , 0]; intcon = [1] A=[1,1 ; 1,1/4 ; 1,-1 ; -1/4,-1 ; -1,-1 ; -1,1]; b=[2;1;2;1;-1;2]; Aeq=[1,1]; beq=[2]; lb=[]; ub=[]; nlc=[]; //Options options=list("GradObj", "on"); //Calling Ipopt [x,fval,exitflag,grad,hessian] =intfmincon(f, x0,intcon,A,b,Aeq,beq,lb,ub,nlc,options) // Press ENTER to continue |
//Find x in R^3 such that it minimizes: //f(x)= x1*x2 + x2*x3 //x0=[0.1 , 0.1 , 0.1] //constraint-1 (c1): x1^2 - x2^2 + x3^2 <= 2 //constraint-2 (c2): x1^2 + x2^2 + x3^2 <= 10 //Objective function to be minimised function [y, dy]=f(x) y=x(1)*x(2)+x(2)*x(3); dy= [x(2),x(1)+x(3),x(2)]; endfunction //Starting point, linear constraints and variable bounds x0=[0.1 , 0.1 , 0.1]; intcon = [2] A=[]; b=[]; Aeq=[]; beq=[]; lb=[]; ub=[]; //Nonlinear constraints function [c, ceq, cg, cgeq]=nlc(x) c = [x(1)^2 - x(2)^2 + x(3)^2 - 2 , x(1)^2 + x(2)^2 + x(3)^2 - 10]; ceq = []; cg=[2*x(1) , -2*x(2) , 2*x(3) ; 2*x(1) , 2*x(2) , 2*x(3)]; cgeq=[]; endfunction //Options options=list("MaxIter", [1500], "CpuTime", [500], "GradObj", "on","GradCon", "on"); //Calling Ipopt [x,fval,exitflag,output] =intfmincon(f, x0,intcon,A,b,Aeq,beq,lb,ub,nlc,options) // Press ENTER to continue |
//The below problem is an unbounded problem: //Find x in R^3 such that it minimizes: //f(x)= -(x1^2 + x2^2 + x3^2) //x0=[0.1 , 0.1 , 0.1] // x1 <= 0 // x2 <= 0 // x3 <= 0 //Objective function to be minimised function y=f(x) y=-(x(1)^2+x(2)^2+x(3)^2); endfunction //Starting point, linear constraints and variable bounds x0=[0.1 , 0.1 , 0.1]; intcon = [3] A=[]; b=[]; Aeq=[]; beq=[]; lb=[]; ub=[0,0,0]; //Options options=list("MaxIter", [1500], "CpuTime", [500]); //Calling Ipopt [x,fval,exitflag,grad,hessian] =intfmincon(f, x0,intcon,A,b,Aeq,beq,lb,ub,[],options) // Press ENTER to continue |
//The below problem is an infeasible problem: //Find x in R^3 such that in minimizes: //f(x)=x1*x2 + x2*x3 //x0=[1,1,1] //constraint-1 (c1): x1^2 <= 1 //constraint-2 (c2): x1^2 + x2^2 <= 1 //constraint-3 (c3): x3^2 <= 1 //constraint-4 (c4): x1^3 = 0.5 //constraint-5 (c5): x2^2 + x3^2 = 0.75 // 0 <= x1 <=0.6 // 0.2 <= x2 <= inf // -inf <= x3 <= 1 //Objective function to be minimised function [y, dy]=f(x) y=x(1)*x(2)+x(2)*x(3); dy= [x(2),x(1)+x(3),x(2)]; endfunction //Starting point, linear constraints and variable bounds x0=[1,1,1]; intcon = [2] A=[]; b=[]; Aeq=[]; beq=[]; lb=[0 0.2,-%inf]; ub=[0.6 %inf,1]; //Nonlinear constraints function [c, ceq, cg, cgeq]=nlc(x) c=[x(1)^2-1,x(1)^2+x(2)^2-1,x(3)^2-1]; ceq=[x(1)^3-0.5,x(2)^2+x(3)^2-0.75]; cg = [2*x(1),0,0;2*x(1),2*x(2),0;0,0,2*x(3)]; cgeq = [3*x(1)^2,0,0;0,2*x(2),2*x(3)]; endfunction //Options options=list("MaxIter", [1500], "CpuTime", [500], "GradObj", "on","GradCon", "on"); //Calling Ipopt [x,fval,exitflag,grad,hessian] =intfmincon(f, x0,intcon,A,b,Aeq,beq,lb,ub,nlc,options) // Press ENTER to continue |