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i am trying to solve a general linear objective function with non-linear constraints. Can someone help me solve this. Here is an example of one problem i am trying to solve,

minimize $b$

subject to $b \geq 0$

$yi - xi*m -c \geq 0$ for all i

$yi - xi*m -c \leq b(\sqrt(1+m^2))$ for all i

(xi,yi) = (1,2), (3,5), (9,7), (6,8), (7,6.5), (3,5.8), (10,19), (4,6)

The above problem is an example of what I am trying to solve. The number of (xi,yi) values is in the range of thousands for the real problem.

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You can try the Lasserre moment relaxations of polynomial optimization problems, see the survey of Monique Laurent:

The code for your example in YALMIP (a MATLAB package) is:

clear b c m r
sdpvar b c m r

x = [1 2; 3 5; 9 7; 6 8; 7 6.5; 3 5.8; 10 19; 4 6]

constraints = set(b>=0) + set(r^2<=1+m^2) + set(m>=0)
for i = 1 : size(x,1)
constraints = constraints + set(0 <= x(i,2)-x(i,1)*m-c <= b*r)




This gives the following (at least numerically) optimal solution:

b = 4.7548
c = -9.9939
m = 1.8874
r = 2.1350

For this to work you need an SDP solver like SeDuMi. Implementations of the Lasserre moment approach other than YALMIP are SOSTools, GloptiPoly and SparsePOP. I have also some slides which might be helpful:

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I have edited the question, I hope the question is clearer now. I am reading through your slides. It is very informational. Thanks. – Sainath adapa Oct 23 '12 at 2:33
if i square the constraints, i can solve through sequential quadratic programming, isn't it?Is there efficient way to solve this problem other than SQP? – Sainath adapa Oct 23 '12 at 2:55
I just remarked that I accidentally added the constraint "set(m>=0)". However, without this constraint YALMIP does not find (at least at this relatively low relaxation degree of 4) an optimal solution. And moreover, if you add the constraint "set(m<=0)", then YALMIP gives a lower bound of 11.7390 (at the same relaxation degree) for the true optimal value of the problem with the additional constraint m<=0. Henceforth, the solution I gave above should nevertheless be correct (numerically). – Markus Schweighofer Oct 24 '12 at 7:37

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