Model formulation: a conclusion about the model before solving it
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I have found this simple model in a paper discussing robust optimization [1]
$$max vec{c}^{T}vec{x}$$
s.t.
$$sum_j a_{ij}x_j + sum_j tilde{a}_{ij}y_j leq b_i ;;;forall i$$
$$ -y_j leq x_jleq y_j ;;;forall j$$
$$vec{l} leq vec{x} leq vec{u} $$
$$vec{y} geq 0$$
$a_{ij}$ and $tilde{a}_{ij}$ are such that they form an interval $[a_{ij}-tilde{a}_{ij}, a_{ij}+tilde{a}_{ij}]$ (but I don't know whether it is relevant to what follows).
Let $vec{x}^{*}$ be the optimal solution. The authors state "At optimality, clearly, $y_j=|x_j^{*}|$". Unfortunately, I am not able to understand this conclusion. Would you please help me understand?
Thanks
[1] Bertsimas, Sim, 2004, The Price of Robustness. Operations Research
optimization linear-programming mathematical-modeling
asked Dec 4 at 15:58
Libra
361210
add a comment |
up vote
1
down vote
favorite
I have found this simple model in a paper discussing robust optimization [1]
$$max vec{c}^{T}vec{x}$$
s.t.
$$sum_j a_{ij}x_j + sum_j tilde{a}_{ij}y_j leq b_i ;;;forall i$$
$$ -y_j leq x_jleq y_j ;;;forall j$$
$$vec{l} leq vec{x} leq vec{u} $$
$$vec{y} geq 0$$
$a_{ij}$ and $tilde{a}_{ij}$ are such that they form an interval $[a_{ij}-tilde{a}_{ij}, a_{ij}+tilde{a}_{ij}]$ (but I don't know whether it is relevant to what follows).
Let $vec{x}^{*}$ be the optimal solution. The authors state "At optimality, clearly, $y_j=|x_j^{*}|$". Unfortunately, I am not able to understand this conclusion. Would you please help me understand?
Thanks
[1] Bertsimas, Sim, 2004, The Price of Robustness. Operations Research
optimization linear-programming mathematical-modeling
asked Dec 4 at 15:58
Libra
361210
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have found this simple model in a paper discussing robust optimization [1]
$$max vec{c}^{T}vec{x}$$
s.t.
$$sum_j a_{ij}x_j + sum_j tilde{a}_{ij}y_j leq b_i ;;;forall i$$
$$ -y_j leq x_jleq y_j ;;;forall j$$
$$vec{l} leq vec{x} leq vec{u} $$
$$vec{y} geq 0$$
$a_{ij}$ and $tilde{a}_{ij}$ are such that they form an interval $[a_{ij}-tilde{a}_{ij}, a_{ij}+tilde{a}_{ij}]$ (but I don't know whether it is relevant to what follows).
Let $vec{x}^{*}$ be the optimal solution. The authors state "At optimality, clearly, $y_j=|x_j^{*}|$". Unfortunately, I am not able to understand this conclusion. Would you please help me understand?
Thanks
[1] Bertsimas, Sim, 2004, The Price of Robustness. Operations Research
optimization linear-programming mathematical-modeling
asked Dec 4 at 15:58
Libra
361210
I have found this simple model in a paper discussing robust optimization [1]
$$max vec{c}^{T}vec{x}$$
s.t.
$$sum_j a_{ij}x_j + sum_j tilde{a}_{ij}y_j leq b_i ;;;forall i$$
$$ -y_j leq x_jleq y_j ;;;forall j$$
$$vec{l} leq vec{x} leq vec{u} $$
$$vec{y} geq 0$$
$a_{ij}$ and $tilde{a}_{ij}$ are such that they form an interval $[a_{ij}-tilde{a}_{ij}, a_{ij}+tilde{a}_{ij}]$ (but I don't know whether it is relevant to what follows).
Let $vec{x}^{*}$ be the optimal solution. The authors state "At optimality, clearly, $y_j=|x_j^{*}|$". Unfortunately, I am not able to understand this conclusion. Would you please help me understand?
Thanks
[1] Bertsimas, Sim, 2004, The Price of Robustness. Operations Research
optimization linear-programming mathematical-modeling
optimization linear-programming mathematical-modeling
asked Dec 4 at 15:58
Libra
361210
asked Dec 4 at 15:58
Libra
361210
asked Dec 4 at 15:58
Libra
361210
asked Dec 4 at 15:58
Libra
361210
asked Dec 4 at 15:58
Libra
361210
361210
add a comment |
add a comment |
1 Answer
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If the reformulated constraint $sum_j a_{ij}x_j + tilde{a}_{ij}|x_j| leq b_i$ is binding at optimality, a proof is easy (there is simply no other $y_j$ that is feasible). The conclusion is erroneous otherwise (although the solution remains optimal if you let $y_j = |x_j|$).
answered Dec 4 at 16:14
LinAlg
7,9241520
So, if I understand correctly, since it is a maximization problem under a budget constraint, the original constraint (involving $vec{y}$) should be binding at optimality. Further, since the objective involves only the $vec{x}$ I would set $vec{x}$ "as large as possible" and $vec{y}$ to $0$, but this is prevented by the second set of constraints where $vec{y}$ is greater than $vec{x}$. So, the best solution is to set $vec{x} = vec{y}$. Is this a correct reasoning? Still, I miss the absolute value.
– Libra
Dec 4 at 17:44
@Libra maximization vs minimization is irrelevant (as you can always multiply $c$ with $-1$). If the objective causes the constraints to be binding (which can be with an "as small as possible" (negative) $x$ as well (depending on the sign of $a_{ij}$)), your reasoning is correct.
– LinAlg
Dec 4 at 17:53
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
If the reformulated constraint $sum_j a_{ij}x_j + tilde{a}_{ij}|x_j| leq b_i$ is binding at optimality, a proof is easy (there is simply no other $y_j$ that is feasible). The conclusion is erroneous otherwise (although the solution remains optimal if you let $y_j = |x_j|$).
answered Dec 4 at 16:14
LinAlg
7,9241520
So, if I understand correctly, since it is a maximization problem under a budget constraint, the original constraint (involving $vec{y}$) should be binding at optimality. Further, since the objective involves only the $vec{x}$ I would set $vec{x}$ "as large as possible" and $vec{y}$ to $0$, but this is prevented by the second set of constraints where $vec{y}$ is greater than $vec{x}$. So, the best solution is to set $vec{x} = vec{y}$. Is this a correct reasoning? Still, I miss the absolute value.
– Libra
Dec 4 at 17:44
@Libra maximization vs minimization is irrelevant (as you can always multiply $c$ with $-1$). If the objective causes the constraints to be binding (which can be with an "as small as possible" (negative) $x$ as well (depending on the sign of $a_{ij}$)), your reasoning is correct.
– LinAlg
Dec 4 at 17:53
add a comment |
up vote
0
down vote
accepted
If the reformulated constraint $sum_j a_{ij}x_j + tilde{a}_{ij}|x_j| leq b_i$ is binding at optimality, a proof is easy (there is simply no other $y_j$ that is feasible). The conclusion is erroneous otherwise (although the solution remains optimal if you let $y_j = |x_j|$).
answered Dec 4 at 16:14
LinAlg
7,9241520
So, if I understand correctly, since it is a maximization problem under a budget constraint, the original constraint (involving $vec{y}$) should be binding at optimality. Further, since the objective involves only the $vec{x}$ I would set $vec{x}$ "as large as possible" and $vec{y}$ to $0$, but this is prevented by the second set of constraints where $vec{y}$ is greater than $vec{x}$. So, the best solution is to set $vec{x} = vec{y}$. Is this a correct reasoning? Still, I miss the absolute value.
– Libra
Dec 4 at 17:44
@Libra maximization vs minimization is irrelevant (as you can always multiply $c$ with $-1$). If the objective causes the constraints to be binding (which can be with an "as small as possible" (negative) $x$ as well (depending on the sign of $a_{ij}$)), your reasoning is correct.
– LinAlg
Dec 4 at 17:53
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
If the reformulated constraint $sum_j a_{ij}x_j + tilde{a}_{ij}|x_j| leq b_i$ is binding at optimality, a proof is easy (there is simply no other $y_j$ that is feasible). The conclusion is erroneous otherwise (although the solution remains optimal if you let $y_j = |x_j|$).
answered Dec 4 at 16:14
LinAlg
7,9241520
If the reformulated constraint $sum_j a_{ij}x_j + tilde{a}_{ij}|x_j| leq b_i$ is binding at optimality, a proof is easy (there is simply no other $y_j$ that is feasible). The conclusion is erroneous otherwise (although the solution remains optimal if you let $y_j = |x_j|$).
answered Dec 4 at 16:14
LinAlg
7,9241520
answered Dec 4 at 16:14
LinAlg
7,9241520
answered Dec 4 at 16:14
LinAlg
7,9241520
answered Dec 4 at 16:14
LinAlg
7,9241520
7,9241520
So, if I understand correctly, since it is a maximization problem under a budget constraint, the original constraint (involving $vec{y}$) should be binding at optimality. Further, since the objective involves only the $vec{x}$ I would set $vec{x}$ "as large as possible" and $vec{y}$ to $0$, but this is prevented by the second set of constraints where $vec{y}$ is greater than $vec{x}$. So, the best solution is to set $vec{x} = vec{y}$. Is this a correct reasoning? Still, I miss the absolute value.
– Libra
Dec 4 at 17:44
@Libra maximization vs minimization is irrelevant (as you can always multiply $c$ with $-1$). If the objective causes the constraints to be binding (which can be with an "as small as possible" (negative) $x$ as well (depending on the sign of $a_{ij}$)), your reasoning is correct.
– LinAlg
Dec 4 at 17:53
add a comment |
So, if I understand correctly, since it is a maximization problem under a budget constraint, the original constraint (involving $vec{y}$) should be binding at optimality. Further, since the objective involves only the $vec{x}$ I would set $vec{x}$ "as large as possible" and $vec{y}$ to $0$, but this is prevented by the second set of constraints where $vec{y}$ is greater than $vec{x}$. So, the best solution is to set $vec{x} = vec{y}$. Is this a correct reasoning? Still, I miss the absolute value.
– Libra
Dec 4 at 17:44
@Libra maximization vs minimization is irrelevant (as you can always multiply $c$ with $-1$). If the objective causes the constraints to be binding (which can be with an "as small as possible" (negative) $x$ as well (depending on the sign of $a_{ij}$)), your reasoning is correct.
– LinAlg
Dec 4 at 17:53
So, if I understand correctly, since it is a maximization problem under a budget constraint, the original constraint (involving $vec{y}$) should be binding at optimality. Further, since the objective involves only the $vec{x}$ I would set $vec{x}$ "as large as possible" and $vec{y}$ to $0$, but this is prevented by the second set of constraints where $vec{y}$ is greater than $vec{x}$. So, the best solution is to set $vec{x} = vec{y}$. Is this a correct reasoning? Still, I miss the absolute value.
– Libra
Dec 4 at 17:44
So, if I understand correctly, since it is a maximization problem under a budget constraint, the original constraint (involving $vec{y}$) should be binding at optimality. Further, since the objective involves only the $vec{x}$ I would set $vec{x}$ "as large as possible" and $vec{y}$ to $0$, but this is prevented by the second set of constraints where $vec{y}$ is greater than $vec{x}$. So, the best solution is to set $vec{x} = vec{y}$. Is this a correct reasoning? Still, I miss the absolute value.
– Libra
Dec 4 at 17:44
@Libra maximization vs minimization is irrelevant (as you can always multiply $c$ with $-1$). If the objective causes the constraints to be binding (which can be with an "as small as possible" (negative) $x$ as well (depending on the sign of $a_{ij}$)), your reasoning is correct.
– LinAlg
Dec 4 at 17:53
@Libra maximization vs minimization is irrelevant (as you can always multiply $c$ with $-1$). If the objective causes the constraints to be binding (which can be with an "as small as possible" (negative) $x$ as well (depending on the sign of $a_{ij}$)), your reasoning is correct.
– LinAlg
Dec 4 at 17:53
add a comment |
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