How to define objective for this optimization problem?

7 vues (au cours des 30 derniers jours)
Barbab
Barbab le 9 Déc 2021
Commenté : Alan Weiss le 9 Déc 2021
My objective is:
subject to:
where k and h are two given vectors of size N. For this objective I tried:
x = optimvar('x',1,2,3,'LowerBound',0)
x =
1×2×3 OptimizationVariable array with properties: Array-wide properties: Name: 'x' Type: 'continuous' IndexNames: {{} {} {}} Elementwise properties: LowerBound: [1×2×3 double] UpperBound: [1×2×3 double] See variables with show. See bounds with showbounds.
k = rand(100,1);
h = rand(100,1);
obj = sum((x(1) + x(2)/x(3) * (k.^(-x(3))-1) - h).^2,'all')
Error using optim.internal.problemdef.operator.PowerOperator
Exponent must be a finite real numeric scalar.

Error in optim.internal.problemdef.Power

Error in .^
What is the proper way to define the objective? What's the best solver?
  2 commentaires
Walter Roberson
Walter Roberson le 9 Déc 2021
You appear to be using Problem Based Optimization rather than Solver Based.
We do not know what your z(3) is, but the context of the error message hints it might be one of your optimization variables.
Barbab
Barbab le 9 Déc 2021
Sorry, I edited my question, the problem is not connected to the optimization variables:
x = optimvar('x',1,2,3,'LowerBound',0)
x =
1×2×3 OptimizationVariable array with properties: Array-wide properties: Name: 'x' Type: 'continuous' IndexNames: {{} {} {}} Elementwise properties: LowerBound: [1×2×3 double] UpperBound: [1×2×3 double] See variables with show. See bounds with showbounds.
k = rand(100,1);
h = rand(100,1);
obj = sum((x(1) + x(2)/x(3) * (k.^(-x(3))-1) - h).^2,'all')
Error using optim.internal.problemdef.operator.PowerOperator
Exponent must be a finite real numeric scalar.

Error in optim.internal.problemdef.Power

Error in .^

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Réponses (1)

Walter Roberson
Walter Roberson le 9 Déc 2021
  • x and y represent optimization arrays of arbitrary size (usually the same size).
  • a is a scalar numeric constant.
Supported operations
  • Pointwise power x.^a
Look through that list and notice that a.^x is not one of the supported operations. You cannot use problem-based optimization to take a constant to a power that is an optimization variable.
  3 commentaires
Walter Roberson
Walter Roberson le 9 Déc 2021
That just might work.
Alan Weiss
Alan Weiss le 9 Déc 2021
Yes, indeed. See the Note on the page Walter linked.
Alan Weiss
MATLAB mathematical toolbox documentation

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