![]() If you write objective or constraint functions that assume feasibility, these functions can error or give unexpected results.įor example, if you take a square root or logarithm of x, and x 2 Intermediate iterations can lead to points that are infeasible (do not satisfy constraints). Iterations Can Violate Constraintsīe careful when writing your objective and constraint functions. For best results, use the lowest numbered constraints possible:įor example, with a constraint 5 x ≤ 20, use a bound x ≤ 4 instead of a linear inequality or nonlinear inequality.įor information on how to pass extra parameters to constraint functions, see Passing Extra Parameters. You can sometimes write constraints in several ways. For more information, see Linear Inequality Constraints and Nonlinear Constraints. x ≥ b is equivalent to the constraint –A.For example, a constraint of the form c i( x) ≥ 0 is equivalent to the constraint –c i( x) ≤ 0. Express greater-than constraints as less-than constraints by multiplying them by –1. Optimization Toolbox functions assume that inequality constraints are of the form c i( x) ≤ 0 or A x ≤ b. Both c and ceq are scalars or vectors representing several constraints. Nonlinear Constraints - c( x) ≤ 0 and ceq( x) = 0. Equality constraints have the same form as inequality constraints. A is an m-by- n matrix, which represents m constraints for an n-dimensional vector x. ![]() Optimization Toolbox solvers have special forms for constraints:īound Constraints - Lower and upper bounds on individual components: x ≥ l and x ≤ u. X = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon,options) X = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon) ![]() f( x), c( x), and ceq( x) can be nonlinear functions. X, b, beq, lb, and ub are vectors, A and Aeq are matrices, c( x) and ceq( x) are functions that return vectors, and f( x) is a function that returns a scalar. X =, constraint is g(x)-f(x)-1 (x-c)^4 -0.1(x-5)^2 -0.1(x-5)^2 (x-c)^4 -0.1(x-5)^2 -0.Find minimum of constrained nonlinear multivariable functionįinds the minimum of a problem specified by I updated your optimization function and the solution c is as follow Another issue A = int(f,)-int(g,) = 55/6 - c^5/5 instead ofĪ = 2*(int(f,)-int(g,)) = 55/3 - (2*c^5)/5įactor 2 is used whether for even whether for odd function (like cosine or since).Įven for those kind of function the integration interval is reduced by half Technically you need to know the optimum c before solving Your problem is more related to calculus than matlab toolĬ= are just ignored by fmincon, because they are not logical Does anyone know why this happens and why MATLAB does not respect the constraints when I put them up for the whole range ? But when I delete the unneccessary constraints in the range ]2 5] function =cons_Q6(x) Now, when I solve it like this I get a solution that doesn't fit all the constraints. = constraints file looks like this (I inserted a lot of values for x with an increment of 0.1): function =cons_Q6(x)Īs you can see, I've set the bounds for the unknown variable so that x(1)= and I have set the constraints in the range although I would only need them in the range of because of the bounds for x(1). Minimize A=2*(int(f,)-int(g,))=55/3 - (2*c^5)/5Īlso I have the constraint that f(x) must always be 1 unit above the functions g(x) and h.įrom the graph I know that the variable c must be between 0 and 2 (just a range for the fmincon function). I want to minimize the Area between the function f(x) and the two connected functions g(x) and h, in the interval I want to use MATLABs fmincon function to solve a non-linear problem of which I know that it can be solved in a different way very easily but I want to use fmincon (you might not need the following detailed information about the problem but I provided them in case you need):įunction f(x) is a quadratic function with its vertex at point (5|1).įunction g(x) is a polynom of order 4 with its vertex at Point (c|0).
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