Formulating the dual of an exponential cone optimization problem
$begingroup$
I have an optimization problem that is written
begin{align}
min_{{s_kappa} }& frac{1}{2}sum_{kappa} eta_kappa left( exp(r_kappa s_kappa) + expleft( frac{s^2_kappa}{2} right) right) \
text{s.t.} & sum_{kappa} expleft( frac{s_kappa}{2} right) leq C
end{align}
where $r_kappa < 0$, $C> 0$ and ${ s_kappa}$ is a finite set of decision variables.
I believe then that the Lagrangian is given by
begin{align}
mathcal{L}(s_kappa,lambda) = frac{1}{2}sum_{kappa} eta_kappa left( exp(r_kappa s_kappa) + expleft( frac{s^2_kappa}{2} right) right) + lambdaleft( sum_{kappa} expleft( frac{s_kappa}{2} right) - Cright)
end{align}
and the KKT conditions then give $lambda geq 0$ (dual feasibility),
$0 = lambda left( sum_{kappa} expleft( frac{s_kappa}{2} right) - C right)$ (complimentary slackness), and
begin{align}
0 &= frac{1}{2} eta_kappa left( r_kappa exp(r_kappa s_kappa) + s_kappa expleft(frac{s^2_kappa}{2}right) right) + frac{lambda exp(frac{s_kappa}{2})}{2}, forall kappa
end{align}
I would like to write the dual problem for this, which should be $g(lambda) = inf_{{s_kappa} } mathcal{L}(s_kappa,lambda)$. I would like to this because I have from other sources some properties on $lambda$ which I would like to translate into properties on the optimal objective.
My problem is that though I can write $lambda$ in terms of $s_kappa$ for a given $s_kappa$
begin{align}
lambda = - eta_kappa left( r_kappa exp(r_kappa s_kappa) + s_kappa expleft( frac{s^2_kappa}{2} right) right)
end{align}
I am struggling to write $s_kappa$ in terms of $lambda$ in order to substitute back into the Lagrangian to get the dual objective function. Any suggestions?
optimization convex-optimization duality-theorems
asked Jan 15 at 0:10
NeedsToKnowMoreMathsNeedsToKnowMoreMaths
728
$endgroup$
add a comment |
$begingroup$
I have an optimization problem that is written
begin{align}
min_{{s_kappa} }& frac{1}{2}sum_{kappa} eta_kappa left( exp(r_kappa s_kappa) + expleft( frac{s^2_kappa}{2} right) right) \
text{s.t.} & sum_{kappa} expleft( frac{s_kappa}{2} right) leq C
end{align}
where $r_kappa < 0$, $C> 0$ and ${ s_kappa}$ is a finite set of decision variables.
I believe then that the Lagrangian is given by
begin{align}
mathcal{L}(s_kappa,lambda) = frac{1}{2}sum_{kappa} eta_kappa left( exp(r_kappa s_kappa) + expleft( frac{s^2_kappa}{2} right) right) + lambdaleft( sum_{kappa} expleft( frac{s_kappa}{2} right) - Cright)
end{align}
and the KKT conditions then give $lambda geq 0$ (dual feasibility),
$0 = lambda left( sum_{kappa} expleft( frac{s_kappa}{2} right) - C right)$ (complimentary slackness), and
begin{align}
0 &= frac{1}{2} eta_kappa left( r_kappa exp(r_kappa s_kappa) + s_kappa expleft(frac{s^2_kappa}{2}right) right) + frac{lambda exp(frac{s_kappa}{2})}{2}, forall kappa
end{align}
I would like to write the dual problem for this, which should be $g(lambda) = inf_{{s_kappa} } mathcal{L}(s_kappa,lambda)$. I would like to this because I have from other sources some properties on $lambda$ which I would like to translate into properties on the optimal objective.
My problem is that though I can write $lambda$ in terms of $s_kappa$ for a given $s_kappa$
begin{align}
lambda = - eta_kappa left( r_kappa exp(r_kappa s_kappa) + s_kappa expleft( frac{s^2_kappa}{2} right) right)
end{align}
I am struggling to write $s_kappa$ in terms of $lambda$ in order to substitute back into the Lagrangian to get the dual objective function. Any suggestions?
optimization convex-optimization duality-theorems
asked Jan 15 at 0:10
NeedsToKnowMoreMathsNeedsToKnowMoreMaths
728
$endgroup$
1
$begingroup$
Given the presence of the term $s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)$, you'd likely need something like the $W$-Lambert function, or, more likely, you won't have a closed form expression for $s_{kappa}$. As an alternative to obtaining an explicit dual objective, you could just form the Wolfe Dual.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 0:49
$begingroup$
An alternative to that term is to introduce a constraint of the form $$ s^2_kappa leq alpha, forall kappa$$ and then the $eta_kappa$ term becomes $eta_kappa r_kappa exp(r_kappa s_kappa)$. The problem here is then that there would be a new lagrange multiplier $mu_kappa$ associated with each of these constraints. Would something like this be more ameanable?
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 1:53
$begingroup$
I'm afraid I don't really follow you. If you introduce those constraints into the primal, you now need to solve $$frac{1}{2}eta_{kappa}left[r_{kappa}exp(r_{kappa}s_{kappa})+s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)right]+frac{1}{2}lambdaexpleft(frac{s_{kappa}}{2}right)+2s_{kappa}mu_{kappa}=0$$ for $s_{kappa}$ in terms of both $lambda$ and $mu_{kappa}$, which seems more difficult rather than easier. I don't see how the $eta_{kappa}$ term changes.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 2:11
$begingroup$
Sorry, I meant that I could change the objective to $$min sum_kappa eta_kappa exp(r_kappa s_kappa)$$ by adding in the additional constraints. The $s^2_kappa$ term came from trying to regularize to avoid the need for the constraints.
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 15:05
add a comment |
$begingroup$
I have an optimization problem that is written
begin{align}
min_{{s_kappa} }& frac{1}{2}sum_{kappa} eta_kappa left( exp(r_kappa s_kappa) + expleft( frac{s^2_kappa}{2} right) right) \
text{s.t.} & sum_{kappa} expleft( frac{s_kappa}{2} right) leq C
end{align}
where $r_kappa < 0$, $C> 0$ and ${ s_kappa}$ is a finite set of decision variables.
I believe then that the Lagrangian is given by
begin{align}
mathcal{L}(s_kappa,lambda) = frac{1}{2}sum_{kappa} eta_kappa left( exp(r_kappa s_kappa) + expleft( frac{s^2_kappa}{2} right) right) + lambdaleft( sum_{kappa} expleft( frac{s_kappa}{2} right) - Cright)
end{align}
and the KKT conditions then give $lambda geq 0$ (dual feasibility),
$0 = lambda left( sum_{kappa} expleft( frac{s_kappa}{2} right) - C right)$ (complimentary slackness), and
begin{align}
0 &= frac{1}{2} eta_kappa left( r_kappa exp(r_kappa s_kappa) + s_kappa expleft(frac{s^2_kappa}{2}right) right) + frac{lambda exp(frac{s_kappa}{2})}{2}, forall kappa
end{align}
I would like to write the dual problem for this, which should be $g(lambda) = inf_{{s_kappa} } mathcal{L}(s_kappa,lambda)$. I would like to this because I have from other sources some properties on $lambda$ which I would like to translate into properties on the optimal objective.
My problem is that though I can write $lambda$ in terms of $s_kappa$ for a given $s_kappa$
begin{align}
lambda = - eta_kappa left( r_kappa exp(r_kappa s_kappa) + s_kappa expleft( frac{s^2_kappa}{2} right) right)
end{align}
I am struggling to write $s_kappa$ in terms of $lambda$ in order to substitute back into the Lagrangian to get the dual objective function. Any suggestions?
optimization convex-optimization duality-theorems
asked Jan 15 at 0:10
NeedsToKnowMoreMathsNeedsToKnowMoreMaths
728
$endgroup$
I have an optimization problem that is written
begin{align}
min_{{s_kappa} }& frac{1}{2}sum_{kappa} eta_kappa left( exp(r_kappa s_kappa) + expleft( frac{s^2_kappa}{2} right) right) \
text{s.t.} & sum_{kappa} expleft( frac{s_kappa}{2} right) leq C
end{align}
where $r_kappa < 0$, $C> 0$ and ${ s_kappa}$ is a finite set of decision variables.
I believe then that the Lagrangian is given by
begin{align}
mathcal{L}(s_kappa,lambda) = frac{1}{2}sum_{kappa} eta_kappa left( exp(r_kappa s_kappa) + expleft( frac{s^2_kappa}{2} right) right) + lambdaleft( sum_{kappa} expleft( frac{s_kappa}{2} right) - Cright)
end{align}
and the KKT conditions then give $lambda geq 0$ (dual feasibility),
$0 = lambda left( sum_{kappa} expleft( frac{s_kappa}{2} right) - C right)$ (complimentary slackness), and
begin{align}
0 &= frac{1}{2} eta_kappa left( r_kappa exp(r_kappa s_kappa) + s_kappa expleft(frac{s^2_kappa}{2}right) right) + frac{lambda exp(frac{s_kappa}{2})}{2}, forall kappa
end{align}
I would like to write the dual problem for this, which should be $g(lambda) = inf_{{s_kappa} } mathcal{L}(s_kappa,lambda)$. I would like to this because I have from other sources some properties on $lambda$ which I would like to translate into properties on the optimal objective.
My problem is that though I can write $lambda$ in terms of $s_kappa$ for a given $s_kappa$
begin{align}
lambda = - eta_kappa left( r_kappa exp(r_kappa s_kappa) + s_kappa expleft( frac{s^2_kappa}{2} right) right)
end{align}
I am struggling to write $s_kappa$ in terms of $lambda$ in order to substitute back into the Lagrangian to get the dual objective function. Any suggestions?
optimization convex-optimization duality-theorems
optimization convex-optimization duality-theorems
asked Jan 15 at 0:10
NeedsToKnowMoreMathsNeedsToKnowMoreMaths
728
asked Jan 15 at 0:10
NeedsToKnowMoreMathsNeedsToKnowMoreMaths
728
asked Jan 15 at 0:10
NeedsToKnowMoreMathsNeedsToKnowMoreMaths
728
asked Jan 15 at 0:10
NeedsToKnowMoreMathsNeedsToKnowMoreMaths
728
asked Jan 15 at 0:10
NeedsToKnowMoreMathsNeedsToKnowMoreMaths
728
728
1
$begingroup$
Given the presence of the term $s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)$, you'd likely need something like the $W$-Lambert function, or, more likely, you won't have a closed form expression for $s_{kappa}$. As an alternative to obtaining an explicit dual objective, you could just form the Wolfe Dual.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 0:49
$begingroup$
An alternative to that term is to introduce a constraint of the form $$ s^2_kappa leq alpha, forall kappa$$ and then the $eta_kappa$ term becomes $eta_kappa r_kappa exp(r_kappa s_kappa)$. The problem here is then that there would be a new lagrange multiplier $mu_kappa$ associated with each of these constraints. Would something like this be more ameanable?
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 1:53
$begingroup$
I'm afraid I don't really follow you. If you introduce those constraints into the primal, you now need to solve $$frac{1}{2}eta_{kappa}left[r_{kappa}exp(r_{kappa}s_{kappa})+s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)right]+frac{1}{2}lambdaexpleft(frac{s_{kappa}}{2}right)+2s_{kappa}mu_{kappa}=0$$ for $s_{kappa}$ in terms of both $lambda$ and $mu_{kappa}$, which seems more difficult rather than easier. I don't see how the $eta_{kappa}$ term changes.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 2:11
$begingroup$
Sorry, I meant that I could change the objective to $$min sum_kappa eta_kappa exp(r_kappa s_kappa)$$ by adding in the additional constraints. The $s^2_kappa$ term came from trying to regularize to avoid the need for the constraints.
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 15:05
add a comment |
1
$begingroup$
Given the presence of the term $s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)$, you'd likely need something like the $W$-Lambert function, or, more likely, you won't have a closed form expression for $s_{kappa}$. As an alternative to obtaining an explicit dual objective, you could just form the Wolfe Dual.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 0:49
$begingroup$
An alternative to that term is to introduce a constraint of the form $$ s^2_kappa leq alpha, forall kappa$$ and then the $eta_kappa$ term becomes $eta_kappa r_kappa exp(r_kappa s_kappa)$. The problem here is then that there would be a new lagrange multiplier $mu_kappa$ associated with each of these constraints. Would something like this be more ameanable?
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 1:53
$begingroup$
I'm afraid I don't really follow you. If you introduce those constraints into the primal, you now need to solve $$frac{1}{2}eta_{kappa}left[r_{kappa}exp(r_{kappa}s_{kappa})+s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)right]+frac{1}{2}lambdaexpleft(frac{s_{kappa}}{2}right)+2s_{kappa}mu_{kappa}=0$$ for $s_{kappa}$ in terms of both $lambda$ and $mu_{kappa}$, which seems more difficult rather than easier. I don't see how the $eta_{kappa}$ term changes.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 2:11
$begingroup$
Sorry, I meant that I could change the objective to $$min sum_kappa eta_kappa exp(r_kappa s_kappa)$$ by adding in the additional constraints. The $s^2_kappa$ term came from trying to regularize to avoid the need for the constraints.
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 15:05
1
1
$begingroup$
Given the presence of the term $s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)$, you'd likely need something like the $W$-Lambert function, or, more likely, you won't have a closed form expression for $s_{kappa}$. As an alternative to obtaining an explicit dual objective, you could just form the Wolfe Dual.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 0:49
$begingroup$
Given the presence of the term $s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)$, you'd likely need something like the $W$-Lambert function, or, more likely, you won't have a closed form expression for $s_{kappa}$. As an alternative to obtaining an explicit dual objective, you could just form the Wolfe Dual.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 0:49
$begingroup$
An alternative to that term is to introduce a constraint of the form $$ s^2_kappa leq alpha, forall kappa$$ and then the $eta_kappa$ term becomes $eta_kappa r_kappa exp(r_kappa s_kappa)$. The problem here is then that there would be a new lagrange multiplier $mu_kappa$ associated with each of these constraints. Would something like this be more ameanable?
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 1:53
$begingroup$
An alternative to that term is to introduce a constraint of the form $$ s^2_kappa leq alpha, forall kappa$$ and then the $eta_kappa$ term becomes $eta_kappa r_kappa exp(r_kappa s_kappa)$. The problem here is then that there would be a new lagrange multiplier $mu_kappa$ associated with each of these constraints. Would something like this be more ameanable?
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 1:53
$begingroup$
I'm afraid I don't really follow you. If you introduce those constraints into the primal, you now need to solve $$frac{1}{2}eta_{kappa}left[r_{kappa}exp(r_{kappa}s_{kappa})+s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)right]+frac{1}{2}lambdaexpleft(frac{s_{kappa}}{2}right)+2s_{kappa}mu_{kappa}=0$$ for $s_{kappa}$ in terms of both $lambda$ and $mu_{kappa}$, which seems more difficult rather than easier. I don't see how the $eta_{kappa}$ term changes.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 2:11
$begingroup$
I'm afraid I don't really follow you. If you introduce those constraints into the primal, you now need to solve $$frac{1}{2}eta_{kappa}left[r_{kappa}exp(r_{kappa}s_{kappa})+s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)right]+frac{1}{2}lambdaexpleft(frac{s_{kappa}}{2}right)+2s_{kappa}mu_{kappa}=0$$ for $s_{kappa}$ in terms of both $lambda$ and $mu_{kappa}$, which seems more difficult rather than easier. I don't see how the $eta_{kappa}$ term changes.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 2:11
$begingroup$
Sorry, I meant that I could change the objective to $$min sum_kappa eta_kappa exp(r_kappa s_kappa)$$ by adding in the additional constraints. The $s^2_kappa$ term came from trying to regularize to avoid the need for the constraints.
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 15:05
$begingroup$
Sorry, I meant that I could change the objective to $$min sum_kappa eta_kappa exp(r_kappa s_kappa)$$ by adding in the additional constraints. The $s^2_kappa$ term came from trying to regularize to avoid the need for the constraints.
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 15:05
add a comment |
1 Answer
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$begingroup$
After making the modifications you've allowed in the comments, the optimization problem becomes
$$begin{align*}
min_{{s_{kappa}}}quad&sum_{kappa}eta_{kappa}exp(r_{kappa}s_{kappa})\
text{s.t.}quad&sum_{kappa}expleft(frac{1}{2}s_{kappa}right)leq C\
&s_{kappa}^2leq alpha.
end{align*}$$
We can make the transformation $x_{kappa}=exp(frac{1}{2}s_{kappa})$ to render this more linear. Especially, we can change the $s_{kappa}^2leq alpha$ constriants into pairs of linear constraints $x_{kappa}leq exp(sqrt{alpha})$, and $x_{kappa}geq exp(-sqrt{alpha})$. This gives a new problem
$$begin{align*}
min_{{x_{kappa}}}quad&sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}\
text{s.t.}quad&sum_{kappa}x_{kappa}leq C\
&x_{kappa}leq exp(sqrt{alpha})\
&x_{kappa}geq exp(-sqrt{alpha})\
end{align*}$$
The KKT stationary-point constraint for the transformed problem is given by
$$2eta_{kappa}r_{kappa}(x_{kappa})^{2r_{kappa}-1}+lambda+mu_{kappa}-nu_{kappa}=0.$$
Then, it is a simple rearrangement for $x_{kappa}$ to give
$$x_{kappa}^*=left[frac{lambda+mu_{kappa}-nu_{kappa}}{2eta_{kappa}(-r_{kappa})}right]^{frac{1}{2r_{kappa}-1}}.$$
The Lagrangian for the transformed problem is then
$$mathcal{L}(x_{kappa},lambda,mu,nu)=sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}+lambdaleft[sum_{kappa}x_{kappa}-Cright]+sum_{kappa}mu_{k}[x_{kappa}-exp(sqrt{alpha})]-sum_{kappa}nu_{k}[x_{kappa}-exp(-sqrt{alpha})].$$
I will leave the cumbersome exercise of substituing $x_{kappa}^*$ into the Lagrangian, but hopefully this approach will help you progress.
answered Jan 15 at 20:35
nathan.j.mcdougallnathan.j.mcdougall
1,519818
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add a comment |
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1 Answer
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$begingroup$
After making the modifications you've allowed in the comments, the optimization problem becomes
$$begin{align*}
min_{{s_{kappa}}}quad&sum_{kappa}eta_{kappa}exp(r_{kappa}s_{kappa})\
text{s.t.}quad&sum_{kappa}expleft(frac{1}{2}s_{kappa}right)leq C\
&s_{kappa}^2leq alpha.
end{align*}$$
We can make the transformation $x_{kappa}=exp(frac{1}{2}s_{kappa})$ to render this more linear. Especially, we can change the $s_{kappa}^2leq alpha$ constriants into pairs of linear constraints $x_{kappa}leq exp(sqrt{alpha})$, and $x_{kappa}geq exp(-sqrt{alpha})$. This gives a new problem
$$begin{align*}
min_{{x_{kappa}}}quad&sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}\
text{s.t.}quad&sum_{kappa}x_{kappa}leq C\
&x_{kappa}leq exp(sqrt{alpha})\
&x_{kappa}geq exp(-sqrt{alpha})\
end{align*}$$
The KKT stationary-point constraint for the transformed problem is given by
$$2eta_{kappa}r_{kappa}(x_{kappa})^{2r_{kappa}-1}+lambda+mu_{kappa}-nu_{kappa}=0.$$
Then, it is a simple rearrangement for $x_{kappa}$ to give
$$x_{kappa}^*=left[frac{lambda+mu_{kappa}-nu_{kappa}}{2eta_{kappa}(-r_{kappa})}right]^{frac{1}{2r_{kappa}-1}}.$$
The Lagrangian for the transformed problem is then
$$mathcal{L}(x_{kappa},lambda,mu,nu)=sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}+lambdaleft[sum_{kappa}x_{kappa}-Cright]+sum_{kappa}mu_{k}[x_{kappa}-exp(sqrt{alpha})]-sum_{kappa}nu_{k}[x_{kappa}-exp(-sqrt{alpha})].$$
I will leave the cumbersome exercise of substituing $x_{kappa}^*$ into the Lagrangian, but hopefully this approach will help you progress.
answered Jan 15 at 20:35
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
add a comment |
$begingroup$
After making the modifications you've allowed in the comments, the optimization problem becomes
$$begin{align*}
min_{{s_{kappa}}}quad&sum_{kappa}eta_{kappa}exp(r_{kappa}s_{kappa})\
text{s.t.}quad&sum_{kappa}expleft(frac{1}{2}s_{kappa}right)leq C\
&s_{kappa}^2leq alpha.
end{align*}$$
We can make the transformation $x_{kappa}=exp(frac{1}{2}s_{kappa})$ to render this more linear. Especially, we can change the $s_{kappa}^2leq alpha$ constriants into pairs of linear constraints $x_{kappa}leq exp(sqrt{alpha})$, and $x_{kappa}geq exp(-sqrt{alpha})$. This gives a new problem
$$begin{align*}
min_{{x_{kappa}}}quad&sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}\
text{s.t.}quad&sum_{kappa}x_{kappa}leq C\
&x_{kappa}leq exp(sqrt{alpha})\
&x_{kappa}geq exp(-sqrt{alpha})\
end{align*}$$
The KKT stationary-point constraint for the transformed problem is given by
$$2eta_{kappa}r_{kappa}(x_{kappa})^{2r_{kappa}-1}+lambda+mu_{kappa}-nu_{kappa}=0.$$
Then, it is a simple rearrangement for $x_{kappa}$ to give
$$x_{kappa}^*=left[frac{lambda+mu_{kappa}-nu_{kappa}}{2eta_{kappa}(-r_{kappa})}right]^{frac{1}{2r_{kappa}-1}}.$$
The Lagrangian for the transformed problem is then
$$mathcal{L}(x_{kappa},lambda,mu,nu)=sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}+lambdaleft[sum_{kappa}x_{kappa}-Cright]+sum_{kappa}mu_{k}[x_{kappa}-exp(sqrt{alpha})]-sum_{kappa}nu_{k}[x_{kappa}-exp(-sqrt{alpha})].$$
I will leave the cumbersome exercise of substituing $x_{kappa}^*$ into the Lagrangian, but hopefully this approach will help you progress.
answered Jan 15 at 20:35
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
add a comment |
$begingroup$
After making the modifications you've allowed in the comments, the optimization problem becomes
$$begin{align*}
min_{{s_{kappa}}}quad&sum_{kappa}eta_{kappa}exp(r_{kappa}s_{kappa})\
text{s.t.}quad&sum_{kappa}expleft(frac{1}{2}s_{kappa}right)leq C\
&s_{kappa}^2leq alpha.
end{align*}$$
We can make the transformation $x_{kappa}=exp(frac{1}{2}s_{kappa})$ to render this more linear. Especially, we can change the $s_{kappa}^2leq alpha$ constriants into pairs of linear constraints $x_{kappa}leq exp(sqrt{alpha})$, and $x_{kappa}geq exp(-sqrt{alpha})$. This gives a new problem
$$begin{align*}
min_{{x_{kappa}}}quad&sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}\
text{s.t.}quad&sum_{kappa}x_{kappa}leq C\
&x_{kappa}leq exp(sqrt{alpha})\
&x_{kappa}geq exp(-sqrt{alpha})\
end{align*}$$
The KKT stationary-point constraint for the transformed problem is given by
$$2eta_{kappa}r_{kappa}(x_{kappa})^{2r_{kappa}-1}+lambda+mu_{kappa}-nu_{kappa}=0.$$
Then, it is a simple rearrangement for $x_{kappa}$ to give
$$x_{kappa}^*=left[frac{lambda+mu_{kappa}-nu_{kappa}}{2eta_{kappa}(-r_{kappa})}right]^{frac{1}{2r_{kappa}-1}}.$$
The Lagrangian for the transformed problem is then
$$mathcal{L}(x_{kappa},lambda,mu,nu)=sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}+lambdaleft[sum_{kappa}x_{kappa}-Cright]+sum_{kappa}mu_{k}[x_{kappa}-exp(sqrt{alpha})]-sum_{kappa}nu_{k}[x_{kappa}-exp(-sqrt{alpha})].$$
I will leave the cumbersome exercise of substituing $x_{kappa}^*$ into the Lagrangian, but hopefully this approach will help you progress.
answered Jan 15 at 20:35
nathan.j.mcdougallnathan.j.mcdougall
1,519818
$endgroup$
After making the modifications you've allowed in the comments, the optimization problem becomes
$$begin{align*}
min_{{s_{kappa}}}quad&sum_{kappa}eta_{kappa}exp(r_{kappa}s_{kappa})\
text{s.t.}quad&sum_{kappa}expleft(frac{1}{2}s_{kappa}right)leq C\
&s_{kappa}^2leq alpha.
end{align*}$$
We can make the transformation $x_{kappa}=exp(frac{1}{2}s_{kappa})$ to render this more linear. Especially, we can change the $s_{kappa}^2leq alpha$ constriants into pairs of linear constraints $x_{kappa}leq exp(sqrt{alpha})$, and $x_{kappa}geq exp(-sqrt{alpha})$. This gives a new problem
$$begin{align*}
min_{{x_{kappa}}}quad&sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}\
text{s.t.}quad&sum_{kappa}x_{kappa}leq C\
&x_{kappa}leq exp(sqrt{alpha})\
&x_{kappa}geq exp(-sqrt{alpha})\
end{align*}$$
The KKT stationary-point constraint for the transformed problem is given by
$$2eta_{kappa}r_{kappa}(x_{kappa})^{2r_{kappa}-1}+lambda+mu_{kappa}-nu_{kappa}=0.$$
Then, it is a simple rearrangement for $x_{kappa}$ to give
$$x_{kappa}^*=left[frac{lambda+mu_{kappa}-nu_{kappa}}{2eta_{kappa}(-r_{kappa})}right]^{frac{1}{2r_{kappa}-1}}.$$
The Lagrangian for the transformed problem is then
$$mathcal{L}(x_{kappa},lambda,mu,nu)=sum_{kappa}eta_{kappa}(x_{kappa})^{2r_{kappa}}+lambdaleft[sum_{kappa}x_{kappa}-Cright]+sum_{kappa}mu_{k}[x_{kappa}-exp(sqrt{alpha})]-sum_{kappa}nu_{k}[x_{kappa}-exp(-sqrt{alpha})].$$
I will leave the cumbersome exercise of substituing $x_{kappa}^*$ into the Lagrangian, but hopefully this approach will help you progress.
answered Jan 15 at 20:35
nathan.j.mcdougallnathan.j.mcdougall
1,519818
edited Jan 15 at 22:48
answered Jan 15 at 20:35
nathan.j.mcdougallnathan.j.mcdougall
1,519818
answered Jan 15 at 20:35
nathan.j.mcdougallnathan.j.mcdougall
1,519818
answered Jan 15 at 20:35
nathan.j.mcdougallnathan.j.mcdougall
1,519818
1,519818
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$begingroup$
Given the presence of the term $s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)$, you'd likely need something like the $W$-Lambert function, or, more likely, you won't have a closed form expression for $s_{kappa}$. As an alternative to obtaining an explicit dual objective, you could just form the Wolfe Dual.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 0:49
$begingroup$
An alternative to that term is to introduce a constraint of the form $$ s^2_kappa leq alpha, forall kappa$$ and then the $eta_kappa$ term becomes $eta_kappa r_kappa exp(r_kappa s_kappa)$. The problem here is then that there would be a new lagrange multiplier $mu_kappa$ associated with each of these constraints. Would something like this be more ameanable?
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 1:53
$begingroup$
I'm afraid I don't really follow you. If you introduce those constraints into the primal, you now need to solve $$frac{1}{2}eta_{kappa}left[r_{kappa}exp(r_{kappa}s_{kappa})+s_{kappa}expleft(frac{{s_{kappa}}^2}{2}right)right]+frac{1}{2}lambdaexpleft(frac{s_{kappa}}{2}right)+2s_{kappa}mu_{kappa}=0$$ for $s_{kappa}$ in terms of both $lambda$ and $mu_{kappa}$, which seems more difficult rather than easier. I don't see how the $eta_{kappa}$ term changes.
$endgroup$
– nathan.j.mcdougall
Jan 15 at 2:11
$begingroup$
Sorry, I meant that I could change the objective to $$min sum_kappa eta_kappa exp(r_kappa s_kappa)$$ by adding in the additional constraints. The $s^2_kappa$ term came from trying to regularize to avoid the need for the constraints.
$endgroup$
– NeedsToKnowMoreMaths
Jan 15 at 15:05