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Re: FindMinimum numerical constraint functions

On 7/8/10 at 7:41 AM, Stuart.Nettleton at (Stuart Nettleton)

>Hi, I am experimenting with external numerical constraint functions
>in FindMinimum. Would anyone understand how the second FindMinimum
>might be made to work in the way of the first?

>FindMinimum[{Sin[x ] Sin[2 y], {x^2 + y^2 < 3, -x < 0, -y < 0}}, {x,
>Clear[obj, con]; vars = {x, y}; lenvars = Length[vars];

>obj[vars1_]:= Module[{subst = Thread[Array[m, lenvars] -> vars1]},
>ReplaceAll[Sin[m[1]] Sin[m[2]], subst]]/; VectorQ[vars1, NumberQ];

>con[vars2_] := Module[{subst = Thread[Array[n, lenvars] -> vars2]},
>ReplaceAll[{n[1]^2 + n[2]^2, -n[1], -n[2]}, subst]
>]/; VectorQ[vars2, NumberQ];

>FindMinimum[{obj, Thread[con[vars] < {3, 0, 0}]}, vars]

You have made both functions con and obj needlessly complex.
Additionally, you are requiring con to work for an array of
numbers. So, you will not get a list of constraints when this
function gets evaluated.

First, obj can be simplified to:


and con can be simplified to:

con[vars_] := Flatten@{Plus@@vars^2, -vars}

Since both of these are defined using SetDelayed a terminating
semicolon is not needed. And since they both localize the
supplied argument, it is not necessary to define them with
uniquely named arguments.

With these definitions, the FindMinimum problem can be done as follows:

In[95]:= FindMinimum[{obj[vars], Thread[con[vars] < {3, 0, 0}]}, vars]

Out[95]= {3.41113*10^-12,{x->0.00256066,y->1.33213*10^-9}}

and to verify this is the same answer as you would have gotten
by explicitly typing everything:

In[96]:= FindMinimum[{Sin[x] Sin[ y], {x^2 + y^2 < 3, -x < 0, -y <
     0}}, {x, y}]

Out[96]= {3.41113*10^-12,{x->0.00256066,y->1.33213*10^-9}}

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