Mixing
Non linear optimisation with min functions
$begingroup$
I have the following nonlinear optimisation problem under bounds constraints and involving $min$ functions and the euclidean norm in the objective function :
$$underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$$
subject to the constraints
$a in big[underline{a},overline{a}big]$
$b in big[underline{b},overline{b}big]$,
$c in big[underline{c},overline{c}big]$,
$d in big[underline{d},overline{d}big]$
where X is a matrix of size $ntimes4$ and a,b,c,d $in mathbb{R}$ and $X_{:,i}$ stands for the vector corresponding to the $i-th$ colum of X.
I'd like to know if there a way to convert the objective function to standard LP format ? or a way to solve it ? Thank you.
optimization nonlinear-optimization non-convex-optimization
$endgroup$
add a comment |
$begingroup$
I have the following nonlinear optimisation problem under bounds constraints and involving $min$ functions and the euclidean norm in the objective function :
$$underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$$
subject to the constraints
$a in big[underline{a},overline{a}big]$
$b in big[underline{b},overline{b}big]$,
$c in big[underline{c},overline{c}big]$,
$d in big[underline{d},overline{d}big]$
where X is a matrix of size $ntimes4$ and a,b,c,d $in mathbb{R}$ and $X_{:,i}$ stands for the vector corresponding to the $i-th$ colum of X.
I'd like to know if there a way to convert the objective function to standard LP format ? or a way to solve it ? Thank you.
optimization nonlinear-optimization non-convex-optimization
$endgroup$
$begingroup$
What is $X_{i,d}$? As we minimize over $d$ it is necessary to know this.
$endgroup$
– Bertrand
Jan 31 at 12:45
$begingroup$
Since X is a matrix, $X_{i,d}$ corresponds to the data in $X$ at line $i$ and column $d$. There was a typo sorry, i corrected this now.
$endgroup$
– user99905
Jan 31 at 12:47
$begingroup$
So if you found the number minimizing this expression whithout the euclidean norm, what is the difference between this number and its euclidean norm?
$endgroup$
– Bertrand
Jan 31 at 12:51
$begingroup$
Indeed, a better objective function would this one : $underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$, where $X_{:,i}$ is the vector corresponding to the $ith$ column of $X$.
$endgroup$
– user99905
Jan 31 at 13:00
$begingroup$
I have rewritten my problem according to your comments.
$endgroup$
– user99905
Jan 31 at 13:08
add a comment |
$begingroup$
I have the following nonlinear optimisation problem under bounds constraints and involving $min$ functions and the euclidean norm in the objective function :
$$underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$$
subject to the constraints
$a in big[underline{a},overline{a}big]$
$b in big[underline{b},overline{b}big]$,
$c in big[underline{c},overline{c}big]$,
$d in big[underline{d},overline{d}big]$
where X is a matrix of size $ntimes4$ and a,b,c,d $in mathbb{R}$ and $X_{:,i}$ stands for the vector corresponding to the $i-th$ colum of X.
I'd like to know if there a way to convert the objective function to standard LP format ? or a way to solve it ? Thank you.
optimization nonlinear-optimization non-convex-optimization
$endgroup$
I have the following nonlinear optimisation problem under bounds constraints and involving $min$ functions and the euclidean norm in the objective function :
$$underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$$
subject to the constraints
$a in big[underline{a},overline{a}big]$
$b in big[underline{b},overline{b}big]$,
$c in big[underline{c},overline{c}big]$,
$d in big[underline{d},overline{d}big]$
where X is a matrix of size $ntimes4$ and a,b,c,d $in mathbb{R}$ and $X_{:,i}$ stands for the vector corresponding to the $i-th$ colum of X.
I'd like to know if there a way to convert the objective function to standard LP format ? or a way to solve it ? Thank you.
optimization nonlinear-optimization non-convex-optimization
optimization nonlinear-optimization non-convex-optimization
edited Jan 31 at 13:07
asked Jan 31 at 12:34
user99905
$begingroup$
What is $X_{i,d}$? As we minimize over $d$ it is necessary to know this.
$endgroup$
– Bertrand
Jan 31 at 12:45
$begingroup$
Since X is a matrix, $X_{i,d}$ corresponds to the data in $X$ at line $i$ and column $d$. There was a typo sorry, i corrected this now.
$endgroup$
– user99905
Jan 31 at 12:47
$begingroup$
So if you found the number minimizing this expression whithout the euclidean norm, what is the difference between this number and its euclidean norm?
$endgroup$
– Bertrand
Jan 31 at 12:51
$begingroup$
Indeed, a better objective function would this one : $underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$, where $X_{:,i}$ is the vector corresponding to the $ith$ column of $X$.
$endgroup$
– user99905
Jan 31 at 13:00
$begingroup$
I have rewritten my problem according to your comments.
$endgroup$
– user99905
Jan 31 at 13:08
add a comment |
$begingroup$
What is $X_{i,d}$? As we minimize over $d$ it is necessary to know this.
$endgroup$
– Bertrand
Jan 31 at 12:45
$begingroup$
Since X is a matrix, $X_{i,d}$ corresponds to the data in $X$ at line $i$ and column $d$. There was a typo sorry, i corrected this now.
$endgroup$
– user99905
Jan 31 at 12:47
$begingroup$
So if you found the number minimizing this expression whithout the euclidean norm, what is the difference between this number and its euclidean norm?
$endgroup$
– Bertrand
Jan 31 at 12:51
$begingroup$
Indeed, a better objective function would this one : $underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$, where $X_{:,i}$ is the vector corresponding to the $ith$ column of $X$.
$endgroup$
– user99905
Jan 31 at 13:00
$begingroup$
I have rewritten my problem according to your comments.
$endgroup$
– user99905
Jan 31 at 13:08
$begingroup$
What is $X_{i,d}$? As we minimize over $d$ it is necessary to know this.
$endgroup$
– Bertrand
Jan 31 at 12:45
$begingroup$
What is $X_{i,d}$? As we minimize over $d$ it is necessary to know this.
$endgroup$
– Bertrand
Jan 31 at 12:45
$begingroup$
Since X is a matrix, $X_{i,d}$ corresponds to the data in $X$ at line $i$ and column $d$. There was a typo sorry, i corrected this now.
$endgroup$
– user99905
Jan 31 at 12:47
$begingroup$
Since X is a matrix, $X_{i,d}$ corresponds to the data in $X$ at line $i$ and column $d$. There was a typo sorry, i corrected this now.
$endgroup$
– user99905
Jan 31 at 12:47
$begingroup$
So if you found the number minimizing this expression whithout the euclidean norm, what is the difference between this number and its euclidean norm?
$endgroup$
– Bertrand
Jan 31 at 12:51
$begingroup$
So if you found the number minimizing this expression whithout the euclidean norm, what is the difference between this number and its euclidean norm?
$endgroup$
– Bertrand
Jan 31 at 12:51
$begingroup$
Indeed, a better objective function would this one : $underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$, where $X_{:,i}$ is the vector corresponding to the $ith$ column of $X$.
$endgroup$
– user99905
Jan 31 at 13:00
$begingroup$
Indeed, a better objective function would this one : $underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$, where $X_{:,i}$ is the vector corresponding to the $ith$ column of $X$.
$endgroup$
– user99905
Jan 31 at 13:00
$begingroup$
I have rewritten my problem according to your comments.
$endgroup$
– user99905
Jan 31 at 13:08
$begingroup$
I have rewritten my problem according to your comments.
$endgroup$
– user99905
Jan 31 at 13:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I would approach this in steps.
(1) Linearize $y_{i,1} = min(X_{i,1}-a,X_{i,2}-b)$
(2) Linearize $y_{i,2} = min(X_{i,3}-c,X_{i,4}-d)$
(3) Form $z_i = y_{i,1}-y_{i,2}$
(4) Minimize $sum_i z_i^2$
In general $z=min(x,y)$ can be linearized as:
$$begin{align} & z le x\ &z le y \& z ge x - Mdelta \ & z ge y - M(1-delta) \ &delta in {0,1} end{align}$$ where $delta$ is a binary variable and $M$ is a large enough constant (judiciously chosen). Notes:
- Some solvers have a $min$ function built-in (technically, behind the scenes, they use similar transformations as shown here)
- If there are no good bounds on $M$ we can use a SOS1 approach (some solvers support SOS1 constraints)
The quadratic objective would make this a MIQP problem. If you allow the 2-norm to be approximated by the sum of absolute values, you can make a linear MIP problem out of this. There are several formulations for this. One is to change steps (3) and (4) into:
(3a) Form $-z_i le y_{i,1}-y_{i,2} le z_i$ (you have to split this into two inequalities) with $z_ige 0$ .
(4a) Minimize $sum_i z_i$
answered Jan 31 at 14:11
Erwin KalvelagenErwin Kalvelagen
3,2542512
$endgroup$
$begingroup$
+1, your answer contains what I was about to answer
$endgroup$
– LinAlg
Jan 31 at 18:51
$begingroup$
Sorry, I have no clue what you are saying here. You should probably talk to your teacher as there are some conceptual problems here.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 19:31
$begingroup$
Thank you for your answer ! And how do write your solution considering the 2-norm to be approximated by the sum of absolute values ? thank you.
$endgroup$
– user99905
Feb 2 at 19:34
$begingroup$
Updated with absolute value formulation.
$endgroup$
– Erwin Kalvelagen
Feb 2 at 19:47
$begingroup$
great ! thanks !
$endgroup$
– user99905
Feb 2 at 20:04
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
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$begingroup$
I would approach this in steps.
(1) Linearize $y_{i,1} = min(X_{i,1}-a,X_{i,2}-b)$
(2) Linearize $y_{i,2} = min(X_{i,3}-c,X_{i,4}-d)$
(3) Form $z_i = y_{i,1}-y_{i,2}$
(4) Minimize $sum_i z_i^2$
In general $z=min(x,y)$ can be linearized as:
$$begin{align} & z le x\ &z le y \& z ge x - Mdelta \ & z ge y - M(1-delta) \ &delta in {0,1} end{align}$$ where $delta$ is a binary variable and $M$ is a large enough constant (judiciously chosen). Notes:
- Some solvers have a $min$ function built-in (technically, behind the scenes, they use similar transformations as shown here)
- If there are no good bounds on $M$ we can use a SOS1 approach (some solvers support SOS1 constraints)
The quadratic objective would make this a MIQP problem. If you allow the 2-norm to be approximated by the sum of absolute values, you can make a linear MIP problem out of this. There are several formulations for this. One is to change steps (3) and (4) into:
(3a) Form $-z_i le y_{i,1}-y_{i,2} le z_i$ (you have to split this into two inequalities) with $z_ige 0$ .
(4a) Minimize $sum_i z_i$
answered Jan 31 at 14:11
Erwin KalvelagenErwin Kalvelagen
3,2542512
$endgroup$
$begingroup$
+1, your answer contains what I was about to answer
$endgroup$
– LinAlg
Jan 31 at 18:51
$begingroup$
Sorry, I have no clue what you are saying here. You should probably talk to your teacher as there are some conceptual problems here.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 19:31
$begingroup$
Thank you for your answer ! And how do write your solution considering the 2-norm to be approximated by the sum of absolute values ? thank you.
$endgroup$
– user99905
Feb 2 at 19:34
$begingroup$
Updated with absolute value formulation.
$endgroup$
– Erwin Kalvelagen
Feb 2 at 19:47
$begingroup$
great ! thanks !
$endgroup$
– user99905
Feb 2 at 20:04
add a comment |
$begingroup$
I would approach this in steps.
(1) Linearize $y_{i,1} = min(X_{i,1}-a,X_{i,2}-b)$
(2) Linearize $y_{i,2} = min(X_{i,3}-c,X_{i,4}-d)$
(3) Form $z_i = y_{i,1}-y_{i,2}$
(4) Minimize $sum_i z_i^2$
In general $z=min(x,y)$ can be linearized as:
$$begin{align} & z le x\ &z le y \& z ge x - Mdelta \ & z ge y - M(1-delta) \ &delta in {0,1} end{align}$$ where $delta$ is a binary variable and $M$ is a large enough constant (judiciously chosen). Notes:
- Some solvers have a $min$ function built-in (technically, behind the scenes, they use similar transformations as shown here)
- If there are no good bounds on $M$ we can use a SOS1 approach (some solvers support SOS1 constraints)
The quadratic objective would make this a MIQP problem. If you allow the 2-norm to be approximated by the sum of absolute values, you can make a linear MIP problem out of this. There are several formulations for this. One is to change steps (3) and (4) into:
(3a) Form $-z_i le y_{i,1}-y_{i,2} le z_i$ (you have to split this into two inequalities) with $z_ige 0$ .
(4a) Minimize $sum_i z_i$
answered Jan 31 at 14:11
Erwin KalvelagenErwin Kalvelagen
3,2542512
$endgroup$
$begingroup$
+1, your answer contains what I was about to answer
$endgroup$
– LinAlg
Jan 31 at 18:51
$begingroup$
Sorry, I have no clue what you are saying here. You should probably talk to your teacher as there are some conceptual problems here.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 19:31
$begingroup$
Thank you for your answer ! And how do write your solution considering the 2-norm to be approximated by the sum of absolute values ? thank you.
$endgroup$
– user99905
Feb 2 at 19:34
$begingroup$
Updated with absolute value formulation.
$endgroup$
– Erwin Kalvelagen
Feb 2 at 19:47
$begingroup$
great ! thanks !
$endgroup$
– user99905
Feb 2 at 20:04
add a comment |
$begingroup$
I would approach this in steps.
(1) Linearize $y_{i,1} = min(X_{i,1}-a,X_{i,2}-b)$
(2) Linearize $y_{i,2} = min(X_{i,3}-c,X_{i,4}-d)$
(3) Form $z_i = y_{i,1}-y_{i,2}$
(4) Minimize $sum_i z_i^2$
In general $z=min(x,y)$ can be linearized as:
$$begin{align} & z le x\ &z le y \& z ge x - Mdelta \ & z ge y - M(1-delta) \ &delta in {0,1} end{align}$$ where $delta$ is a binary variable and $M$ is a large enough constant (judiciously chosen). Notes:
- Some solvers have a $min$ function built-in (technically, behind the scenes, they use similar transformations as shown here)
- If there are no good bounds on $M$ we can use a SOS1 approach (some solvers support SOS1 constraints)
The quadratic objective would make this a MIQP problem. If you allow the 2-norm to be approximated by the sum of absolute values, you can make a linear MIP problem out of this. There are several formulations for this. One is to change steps (3) and (4) into:
(3a) Form $-z_i le y_{i,1}-y_{i,2} le z_i$ (you have to split this into two inequalities) with $z_ige 0$ .
(4a) Minimize $sum_i z_i$
answered Jan 31 at 14:11
Erwin KalvelagenErwin Kalvelagen
3,2542512
$endgroup$
I would approach this in steps.
(1) Linearize $y_{i,1} = min(X_{i,1}-a,X_{i,2}-b)$
(2) Linearize $y_{i,2} = min(X_{i,3}-c,X_{i,4}-d)$
(3) Form $z_i = y_{i,1}-y_{i,2}$
(4) Minimize $sum_i z_i^2$
In general $z=min(x,y)$ can be linearized as:
$$begin{align} & z le x\ &z le y \& z ge x - Mdelta \ & z ge y - M(1-delta) \ &delta in {0,1} end{align}$$ where $delta$ is a binary variable and $M$ is a large enough constant (judiciously chosen). Notes:
- Some solvers have a $min$ function built-in (technically, behind the scenes, they use similar transformations as shown here)
- If there are no good bounds on $M$ we can use a SOS1 approach (some solvers support SOS1 constraints)
The quadratic objective would make this a MIQP problem. If you allow the 2-norm to be approximated by the sum of absolute values, you can make a linear MIP problem out of this. There are several formulations for this. One is to change steps (3) and (4) into:
(3a) Form $-z_i le y_{i,1}-y_{i,2} le z_i$ (you have to split this into two inequalities) with $z_ige 0$ .
(4a) Minimize $sum_i z_i$
answered Jan 31 at 14:11
Erwin KalvelagenErwin Kalvelagen
3,2542512
edited Feb 2 at 19:46
answered Jan 31 at 14:11
Erwin KalvelagenErwin Kalvelagen
3,2542512
answered Jan 31 at 14:11
Erwin KalvelagenErwin Kalvelagen
3,2542512
answered Jan 31 at 14:11
Erwin KalvelagenErwin Kalvelagen
3,2542512
3,2542512
$begingroup$
+1, your answer contains what I was about to answer
$endgroup$
– LinAlg
Jan 31 at 18:51
$begingroup$
Sorry, I have no clue what you are saying here. You should probably talk to your teacher as there are some conceptual problems here.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 19:31
$begingroup$
Thank you for your answer ! And how do write your solution considering the 2-norm to be approximated by the sum of absolute values ? thank you.
$endgroup$
– user99905
Feb 2 at 19:34
$begingroup$
Updated with absolute value formulation.
$endgroup$
– Erwin Kalvelagen
Feb 2 at 19:47
$begingroup$
great ! thanks !
$endgroup$
– user99905
Feb 2 at 20:04
add a comment |
$begingroup$
+1, your answer contains what I was about to answer
$endgroup$
– LinAlg
Jan 31 at 18:51
$begingroup$
Sorry, I have no clue what you are saying here. You should probably talk to your teacher as there are some conceptual problems here.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 19:31
$begingroup$
Thank you for your answer ! And how do write your solution considering the 2-norm to be approximated by the sum of absolute values ? thank you.
$endgroup$
– user99905
Feb 2 at 19:34
$begingroup$
Updated with absolute value formulation.
$endgroup$
– Erwin Kalvelagen
Feb 2 at 19:47
$begingroup$
great ! thanks !
$endgroup$
– user99905
Feb 2 at 20:04
$begingroup$
+1, your answer contains what I was about to answer
$endgroup$
– LinAlg
Jan 31 at 18:51
$begingroup$
+1, your answer contains what I was about to answer
$endgroup$
– LinAlg
Jan 31 at 18:51
$begingroup$
Sorry, I have no clue what you are saying here. You should probably talk to your teacher as there are some conceptual problems here.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 19:31
$begingroup$
Sorry, I have no clue what you are saying here. You should probably talk to your teacher as there are some conceptual problems here.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 19:31
$begingroup$
Thank you for your answer ! And how do write your solution considering the 2-norm to be approximated by the sum of absolute values ? thank you.
$endgroup$
– user99905
Feb 2 at 19:34
$begingroup$
Thank you for your answer ! And how do write your solution considering the 2-norm to be approximated by the sum of absolute values ? thank you.
$endgroup$
– user99905
Feb 2 at 19:34
$begingroup$
Updated with absolute value formulation.
$endgroup$
– Erwin Kalvelagen
Feb 2 at 19:47
$begingroup$
Updated with absolute value formulation.
$endgroup$
– Erwin Kalvelagen
Feb 2 at 19:47
$begingroup$
great ! thanks !
$endgroup$
– user99905
Feb 2 at 20:04
$begingroup$
great ! thanks !
$endgroup$
– user99905
Feb 2 at 20:04
add a comment |
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$begingroup$
What is $X_{i,d}$? As we minimize over $d$ it is necessary to know this.
$endgroup$
– Bertrand
Jan 31 at 12:45
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Since X is a matrix, $X_{i,d}$ corresponds to the data in $X$ at line $i$ and column $d$. There was a typo sorry, i corrected this now.
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– user99905
Jan 31 at 12:47
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So if you found the number minimizing this expression whithout the euclidean norm, what is the difference between this number and its euclidean norm?
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– Bertrand
Jan 31 at 12:51
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Indeed, a better objective function would this one : $underset{a,b,c,d}{min} BigVert min(X_{:,1}-a,X_{:,2}-b) - min(X_{:,3}-c,X_{:,4}-d)BigVert_2$, where $X_{:,i}$ is the vector corresponding to the $ith$ column of $X$.
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– user99905
Jan 31 at 13:00
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I have rewritten my problem according to your comments.
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– user99905
Jan 31 at 13:08