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diff --git a/demos/intfmincon.dem.sce b/demos/intfmincon.dem.sce
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+mode(1)
+//
+// Demo of intfmincon.sci
+//
+
+//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
+halt()   // Press return 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
+halt()   // Press return 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
+halt()   // Press return 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
+//========= E N D === O F === D E M O =========//
diff --git a/demos/intfminimax.dem.sce b/demos/intfminimax.dem.sce
new file mode 100644
index 0000000..db74b92
--- /dev/null
+++ b/demos/intfminimax.dem.sce
@@ -0,0 +1,57 @@
+mode(1)
+//
+// Demo of intfminimax.sci
+//
+
+// A basic case :
+// we provide only the objective function and the nonlinear constraint
+// function
+function f = myfun(x)
+f(1)= 2*x(1)^2 + x(2)^2 - 48*x(1) - 40*x(2) + 304;     //Objectives
+f(2)= -x(1)^2 - 3*x(2)^2;
+f(3)= x(1) + 3*x(2) -18;
+f(4)= -x(1) - x(2);
+f(5)= x(1) + x(2) - 8;
+endfunction
+// The initial guess
+x0 = [0.1,0.1];
+// The expected solution : only 4 digits are guaranteed
+xopt = [4 4]
+fopt = [0 -64 -2 -8 0]
+intcon = [1]
+maxfopt = 0
+// Run fminimax
+[x,fval,maxfval,exitflag] = intfminimax(myfun, x0,intcon)
+// Press ENTER to continue
+halt()   // Press return to continue
+ 
+// A case where we provide the gradient of the objective
+// functions and the Jacobian matrix of the constraints.
+// The objective function and its gradient
+function [f,G] = myfun(x)
+f(1)= 2*x(1)^2 + x(2)^2 - 48*x(1) - 40*x(2) + 304;
+f(2)= -x(1)^2 - 3*x(2)^2;
+f(3)= x(1) + 3*x(2) -18;
+f(4)= -x(1) - x(2);
+f(5)= x(1) + x(2) - 8;
+G = [ 4*x(1) - 48, -2*x(1), 1, -1, 1;
+2*x(2) - 40, -6*x(2), 3, -1, 1; ]'
+endfunction
+// The nonlinear constraints
+function [c,ceq,DC,DCeq] = confun(x)
+// Inequality constraints
+c = [1.5 + x(1)*x(2) - x(1) - x(2), -x(1)*x(2) - 10]
+// No nonlinear equality constraints
+ceq=[]
+DC= [x(2)-1, -x(2);
+x(1)-1, -x(1)]'
+DCeq = []'
+endfunction
+// Test with both gradient of objective and gradient of constraints
+minimaxOptions = list("GradObj","on","GradCon","on");
+// The initial guess
+x0 = [0,10];
+intcon = [2]
+// Run intfminimax
+[x,fval,maxfval,exitflag] = intfminimax(myfun,x0,intcon,[],[],[],[],[],[], confun, minimaxOptions)
+//========= E N D === O F === D E M O =========//