How to define objective for this optimization problem?

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Barbab
Barbab 2021-12-9
评论： Alan Weiss 2021-12-9
Ran in:
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 个评论
Walter Roberson
Walter Roberson 2021-12-9
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 2021-12-9
Ran in:
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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回答（1 个）

Walter Roberson
Walter Roberson 2021-12-9
https://www.mathworks.com/help/optim/ug/supported-operations-on-optimization-variables-expressions.html
  • 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 个评论
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Alan Weiss
Alan Weiss 2021-12-9
Yes, indeed. See the Note on the page Walter linked.
Alan Weiss
MATLAB mathematical toolbox documentation

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