Hi, I'm trying to optimize a function which I know the results but matlab is giving me weird results. Here's what I'm trying to do:
max: f(x)= 1815×1+379×2
subject to: -1475×1-112013×2 >= -700000
x1,,x2<=80
x1,x2>=0
How would you do it?
MATLAB: Optimization giving unexpected results
optimization
Related Solutions
As Matt points out, the solution you found may have issues.
You can see the source of the problem by looking at your nonlinear constraint. There we see you have the difference (x1^2 - x2^2) in the denominator. So when x1 is approximately equal to x2, things go to hell.
You can fix part of that, by factoring the terms in the first radical.
There we see a fraction with
(x1^4 - x2^4)/(x1^2 - x2^2)this reduces to (x1^2 + x2^2), so no 0/0 there.
The second radical does have a divide by zero, when x1==x2. So again, this will be a problem. You will be able to get almost any value you want for that term, by TINY changes to x1 and x2.
Did the optimizer converge? Hard to tell. I'start the optimization at a different location though, since you still have a singular constraint.
If your goal is to minimize (x1^2-x2^2) you can do so by maximizing x2, as well as minimize x1. (Drop the constants out front! A minimization does not care about them, and at best, they may just introduce numerical problems.)
So your start point should be roughly:
[9.2 9.8 , ?]I put a question mark in there for x3, because x3 does not appear in the objective. So we can choose a start point for x3 by trying to satisfy the nonlinear constraint.
So now lets look more closely at the constraint. We know that x3 lies in [0.1, 18].
First, lets simplify the constraint.
K1 = 1.875^2/(300^2*2*pi)*sqrt(1.84e5/4/7.9e-9)K1 = 15.0018747945489K2 = 300^3/(3*1.84e5*pi/4)K2 = 62.2780212098721So we can write the entire nonlinear constraint in the form of a function handle:
CON = @(x123) K1*sqrt(x123(1)^2 + x123(2)^2)*sqrt(1 + K2*x123(3)/(x123(1)^4 - x123(2)^4)) - 230;Will this function ever be greater than zero, when x1 and x2 are at their best locations?
CON([9.2 9.8 .1])ans = -28.653992101973CON([9.2 9.8 18])ans = -93.8657086630407ezplot(@(x3) CON([9.2 9.8 x3]),[.1 18])
So at what I chose as the best point for x1 and x2, no value of x3 will allow this constraint to be positive. But if I choose other values for x1 and x2, as long as x1>x2, there are solutions that allow con to be positive.
ezplot(@(x3) CON([9.8 9.7 x3]),[.1 18])grid on
Things look very different when x1 is slightly less than x2 however.
ezplot(@(x3) CON([9.6 9.7 x3]),[.1 18])grid on
In fact, it appears there may be no solution at all when x1 is less than x2. I'd need to look more assiduously at the problem to know if that is true.
But it is clearly true that [9.8 9.7 2] is a feasible point. So what does fmincon find?
Define CONFUN as:
function [c, ceq] = confun(x123)K1 = 1.875^2/(300^2*2*pi)*sqrt(1.84e5/4/7.9e-9);K2 = 300^3/(3*1.84e5*pi/4);c = K1*sqrt(x123(1)^2+x123(2)^2)*sqrt(1+K2*x123(3)/(x123(1)^4-x123(2)^4))-230;c = -c;ceq = [];I negated CON in there.
LB=[9.2,6,0.1];UB=[15,9.8,18];A=[];b=[];Aeq=[];beq=[];x0=[9.8,9.7,2];[x,fval,exitflag] = fmincon(@(x123) x123(1)^2 - x123(2)^2,x0,A,b,Aeq,beq,LB,UB,@CONFUN)Local minimum possible. Constraints satisfied.fmincon stopped because the size of the current step is less thanthe default value of the step size tolerance and constraints are satisfied to within the default value of the constraint tolerance.<stopping criteria details>x = 9.53501396574518 9.53501396574518 2.80228092041589fval = 2.8421709430404e-14exitflag = 2Fmincon wants to push to the boundary where x1 == x2. The constraint is satisfied at x.
CON(x)ans = 1144008644.36073But, clearly, the solution is a bit of a singular one, since it lies right along the edge where x1==x2, so we have an effective divide by zero.
The construction of the A and b matrices should be,
A=-diff(speye(2000));b=-0.01*ones(2000,1);If you have done this, then there is nothing incorrect about the way you have built the constraints, assuming X(i+1)-X(i)>=0.01 are indeed the conditions you want. You should investigate whether the problem truly does have a feasible solution under these constraints and whatever other constraints you have imposed.
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