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Mathematics Wolfram Language Optimization Numerical Computation

Can interpolating functions be used in NMinimize constraints?

Frank Kampas

Frank Kampas, Physicist at Large Consulting

Posted 6 years ago

In[1]:= int = Interpolation[Table[{i, (i - 5)^2}, {i, 10}]] Out[1]= InterpolatingFunction[{{1, 10}}, { 5, 3, 0, {10}, {4}, 0, 0, 0, 0, Automatic, {}, {}, False}, {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}}, {{16}, {9}, {4}, {1}, \ {0}, {1}, {4}, {9}, {16}, {25}}, {Automatic}] In[5]:= NMinimize[{o, o == int[x]}, {x, 1, 10}] During evaluation of In[5]:= InterpolatingFunction::dmval: Input value {0.652468} lies outside the range of data in the interpolating function. Extrapolation will be used. During evaluation of In[5]:= NMinimize::bcons: The following constraints are not valid: {o==InterpolatingFunction[{{1,10}},{5,3,0,{10},{4},0,0,0,0,Automatic,{},{},False},{{1,2,3,4,5,6,7,8,9,10}},{{16},{9},{4},{1},{0},{1},{4},{9},{16},{25}},{Automatic}][x]}. Constraints should be equalities, inequalities, or domain specifications involving the variables. Out[5]= NMinimize[{o, o == InterpolatingFunction[{{1, 10}}, { 5, 3, 0, {10}, {4}, 0, 0, 0, 0, Automatic, {}, {}, False}, {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}}, {{16}, {9}, {4}, {1}, {0}, {1}, { 4}, {9}, {16}, {25}}, {Automatic}][x]}, {x, 1, 10}]

In[1]:= int = Interpolation[Table[{i, (i - 5)^2}, {i, 10}]]

Out[1]= InterpolatingFunction[{{1, 10}}, {
 5, 3, 0, {10}, {4}, 0, 0, 0, 0, Automatic, {}, {}, 
  False}, {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}}, {{16}, {9}, {4}, {1}, \
{0}, {1}, {4}, {9}, {16}, {25}}, {Automatic}]

In[5]:= NMinimize[{o, o == int[x]}, {x, 1, 10}]

During evaluation of In[5]:= InterpolatingFunction::dmval: Input value {0.652468} lies outside the range of data in the interpolating function. Extrapolation will be used.

During evaluation of In[5]:= NMinimize::bcons: The following constraints are not valid: {o==InterpolatingFunction[{{1,10}},{5,3,0,{10},{4},0,0,0,0,Automatic,{},{},False},{{1,2,3,4,5,6,7,8,9,10}},{{16},{9},{4},{1},{0},{1},{4},{9},{16},{25}},{Automatic}][x]}. Constraints should be equalities, inequalities, or domain specifications involving the variables.

Out[5]= NMinimize[{o, 
  o == InterpolatingFunction[{{1, 10}}, {
     5, 3, 0, {10}, {4}, 0, 0, 0, 0, Automatic, {}, {}, False}, {{1, 
     2, 3, 4, 5, 6, 7, 8, 9, 10}}, {{16}, {9}, {4}, {1}, {0}, {1}, {
     4}, {9}, {16}, {25}}, {Automatic}][x]}, {x, 1, 10}]

POSTED BY: Frank Kampas

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Frank Kampas

Frank Kampas, Physicist at Large Consulting

Posted 6 years ago

Giving numbers after the variable in NMinimize gives a suggested search range, not bounds on the variables. I think I've found a work-around In[13]:= int = Interpolation[Table[{i, (i - 5)^2}, {i, 10}]]; In[14]:= NMinimize[{o, o == int[x]}, {o, x}] // Quiet Out[14]= {\[Infinity], {o -> Indeterminate, x -> Indeterminate}} In[15]:= NMinimize[{o, int[x] <= o <= int[x]}, {o, x}] // Quiet Out[15]= {-7.4505910^-9, {o -> -7.4505910^-9, x -> 5.}}

Giving numbers after the variable in NMinimize gives a suggested search range, not bounds on the variables.

I think I've found a work-around

In[13]:= int = Interpolation[Table[{i, (i - 5)^2}, {i, 10}]];

In[14]:= NMinimize[{o, o == int[x]}, {o, x}] // Quiet

Out[14]= {\[Infinity], {o -> Indeterminate, x -> Indeterminate}}

In[15]:= NMinimize[{o, int[x] <= o <= int[x]}, {o, x}] // Quiet

Out[15]= {-7.45059*10^-9, {o -> -7.45059*10^-9, x -> 5.}}

POSTED BY: Frank Kampas

Michael Rogers

Michael Rogers, Emory University

Posted 6 years ago

There are a couple of issues. The major one, I think, is the `o` and not the interpolating function: NMinimize[{o , o == (x - 5)^2}, {x, 1, 10}] (* NMinimize::bcons: The following constraints are not valid: {o==(-5+x)^2}. Constraints should be equalities, inequalities, or domain specifications involving the variables. ) ( Out[.]= NMinimize[{o, o == (-5 + x)^2}, {x, 1, 10}] ) A potential one is that a constraint of the form `<some function of x> == int[x]` results in a discrete set of points, which I don't think `NMinimize` can handle: NMinimize[{x , x == int[x]}, {x, 1, 10}] ( InterpolatingFunction::dmval: Input value {0.652468} lies outside the range of data in the interpolating function. Extrapolation will be used. ) ( NMinimize::nsol: There are no points that satisfy the constraints {}. ) ( Out[.]= {?, {x -> Indeterminate}} ) A trivial one: I don't know what the `1` and `10` do in the second argument `{x, 1, 10}`, but they do not constrain `x` to be in the domain of `int[x]`: NMinimize[{x , 1 < int[x]}, {x, 1, 10}] ( InterpolatingFunction::dmval: Input value {0.652468} lies outside the range of data in the interpolating function. Extrapolation will be used. ... <more messages> ... ) ( Out[.]= {-26.7218, {x -> -26.7218}} <-- solution not in the domain ) Mariusz has already given examples that work. Here is another: NMinimize[{(x - 4)^2 , x < int[x]}, {x} ? Cuboid @@ Transpose@ int["Domain"]] ( InterpolatingFunction::dmval: Input value {0.665389} lies outside the range of data in the interpolating function. Extrapolation will be used. ) ( Out[.]= {0.626136, {x -> 3.20871}} *) Constraints are enforced with a penalty function, I believe, which means that `NMinimize` is not guaranteed to sample strictly within the constraints; however, the solution should satisfy the constraints.

There are a couple of issues. The major one, I think, is the o and not the interpolating function:

NMinimize[{o , o == (x - 5)^2}, {x, 1, 10}]

(*  NMinimize::bcons: The following constraints are not valid: {o==(-5+x)^2}. 
    Constraints should be equalities, inequalities, or domain specifications involving 
    the variables.  *)

(*  Out[.]= NMinimize[{o, o == (-5 + x)^2}, {x, 1, 10}]  *)

A potential one is that a constraint of the form <some function of x> == int[x] results in a discrete set of points, which I don't think NMinimize can handle:

NMinimize[{x , x == int[x]}, {x, 1, 10}]

(*  InterpolatingFunction::dmval: Input value {0.652468} lies outside the range of data 
     in the interpolating function. Extrapolation will be used.  *)

(* NMinimize::nsol: There are no points that satisfy the constraints {}. *)

(*  Out[.]= {?, {x -> Indeterminate}}  *)

A trivial one: I don't know what the 1 and 10 do in the second argument {x, 1, 10}, but they do not constrain x to be in the domain of int[x]:

NMinimize[{x , 1 < int[x]}, {x, 1, 10}]

(*  InterpolatingFunction::dmval: Input value {0.652468} lies outside the range of data 
    in the interpolating function. Extrapolation will be used.
    ... <more messages> ... *)

(*  Out[.]= {-26.7218, {x -> -26.7218}}   <-- solution not in the domain  *)

Mariusz has already given examples that work. Here is another:

NMinimize[{(x - 4)^2 , x < int[x]}, {x} ? Cuboid @@ Transpose@ int["Domain"]]

(*  InterpolatingFunction::dmval: Input value {0.665389} lies outside the range of data 
    in the interpolating function. Extrapolation will be used.  *)

(*  Out[.]= {0.626136, {x -> 3.20871}}  *)

Constraints are enforced with a penalty function, I believe, which means that NMinimize is not guaranteed to sample strictly within the constraints; however, the solution should satisfy the constraints.

POSTED BY: Michael Rogers

Frank Kampas

Frank Kampas, Physicist at Large Consulting

Posted 6 years ago

That works for this example problem, but the problem I'm actually working with needs the Interpolation function in the constraints.

POSTED BY: Frank Kampas

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 6 years ago

Ok maybe so: For 1D: int = Interpolation[Table[{i, (i - 5)^2}, {i, 10}]]; {NMinimize[{x^4 - 3 x^2 - x, (x - 5)^2 < 8}, x] // Quiet, NMinimize[{x^4 - 3 x^2 - x, int[x] < 8}, x] // Quiet, NMinimize[{x^4 - 3 x^2 - x, int[x] < 8 && 1 < x < 10}, x] // Quiet} (* {{5.91934, {x -> 2.17157}}, {5.91934, {x -> 2.17157}}, {5.91934, {x ->2.17157}}}) For 2D:* f = Interpolation[ Flatten[Table[{{x, y}, x^2 + y^2 + x}, {x, -20, 20}, {y, -20, 20}], 1]](Generating fake data) NMinimize[{x^2 - (y - 1)^2, f[x, y] <= 4}, {x, y}] (* {-9.22366, {x -> -0.299019, y -> -2.05173}} ) NMinimize[{x^2 - (y - 1)^2, x^2 + y^2 + x <= 4}, {x, y}]( Check ) ( {-9.22366, {x -> -0.299019, y -> -2.05173}} ) (OK*)

Ok maybe so:

For 1D:

     int = Interpolation[Table[{i, (i - 5)^2}, {i, 10}]];

    {NMinimize[{x^4 - 3 x^2 - x, (x - 5)^2 < 8}, x] // Quiet, 
     NMinimize[{x^4 - 3 x^2 - x, int[x] < 8}, x] // Quiet, 
     NMinimize[{x^4 - 3 x^2 - x, int[x] < 8 && 1 < x < 10}, x] // Quiet}

     (* {{5.91934, {x -> 2.17157}}, {5.91934, {x -> 2.17157}}, {5.91934, {x ->2.17157}}}*)

For 2D:

    f = Interpolation[
      Flatten[Table[{{x, y}, x^2 + y^2 + x}, {x, -20, 20}, {y, -20, 20}], 
       1]](*Generating fake data*)
    NMinimize[{x^2 - (y - 1)^2, f[x, y] <= 4}, {x, y}]

   (* {-9.22366, {x -> -0.299019, y -> -2.05173}} *)

  NMinimize[{x^2 - (y - 1)^2, x^2 + y^2 + x <= 4}, {x, y}](* Check *)

    (* {-9.22366, {x -> -0.299019, y -> -2.05173}} *) (*OK*)

POSTED BY: Mariusz Iwaniuk

Mariusz Iwaniuk

Mariusz Iwaniuk, engineer, math hobbyist.

Posted 6 years ago

Try: int = Interpolation[Table[{i, (i - 5)^2}, {i, 10}]]; NMinimize[{int[x], 1 < x < 10}, x] (* {6.552410^-20, {x -> 5.}} ) FindMinimum[{int[x], 1 < x < 10}, {x, 2}] (* {0., {x -> 5.}} ) Using FunctionInterpolation*: int2 = FunctionInterpolation[Cos[x], {x, 0, 6}]; {NMinimize[{int2[x], 0 < x < 6}, x], Plot[int2[x], {x, 0, 6}]}

Try:

int = Interpolation[Table[{i, (i - 5)^2}, {i, 10}]];
NMinimize[{int[x], 1 < x < 10}, x]

(* {6.5524*10^-20, {x -> 5.}} *)

FindMinimum[{int[x], 1 < x < 10}, {x, 2}]

(* {0., {x -> 5.}} *)

Using FunctionInterpolation:

int2 = FunctionInterpolation[Cos[x], {x, 0, 6}];
{NMinimize[{int2[x], 0 < x < 6}, x], Plot[int2[x], {x, 0, 6}]}

POSTED BY: Mariusz Iwaniuk

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