Mixing
Optimization model where a certain conditions affect objective rather than being a constraint
$begingroup$
I have a minimization problem related to packing. I have to distribute a given amount of items from different sets to different packs. Each pack must be either empty or have a certain minimum of items. There is a penalty on each pack so i have to minimize the amount of packs
But there is also a penalty (rather than a constraint) for having items from different sets in the same packs so it is
$$min C_1 sum_{i}^{packs}if(X_{i,any} >0 ) + C_2sum_{j}^{packs}if[exists
X_{ij}>0,X_{kj}>0;kneq i
]$$
when $x_{i,j }$ is number of items from $j$ in pack $i$. Also it's a constraint that $ sum X_{i,j} = text{constant}$
What model can I use to solve that problem?
edit:
My problem is that my objective function is not continuous, and as far as I found out is not good for any linear or non-linear programming solution.
is there a way to either replace it with a function that gives a similar solution to the problem or is there another way to solve it?
optimization linear-programming nonlinear-optimization
asked Jan 29 at 21:44
Michael RazgonerMichael Razgoner
32
$endgroup$
add a comment |
$begingroup$
I have a minimization problem related to packing. I have to distribute a given amount of items from different sets to different packs. Each pack must be either empty or have a certain minimum of items. There is a penalty on each pack so i have to minimize the amount of packs
But there is also a penalty (rather than a constraint) for having items from different sets in the same packs so it is
$$min C_1 sum_{i}^{packs}if(X_{i,any} >0 ) + C_2sum_{j}^{packs}if[exists
X_{ij}>0,X_{kj}>0;kneq i
]$$
when $x_{i,j }$ is number of items from $j$ in pack $i$. Also it's a constraint that $ sum X_{i,j} = text{constant}$
What model can I use to solve that problem?
edit:
My problem is that my objective function is not continuous, and as far as I found out is not good for any linear or non-linear programming solution.
is there a way to either replace it with a function that gives a similar solution to the problem or is there another way to solve it?
optimization linear-programming nonlinear-optimization
asked Jan 29 at 21:44
Michael RazgonerMichael Razgoner
32
$endgroup$
add a comment |
$begingroup$
I have a minimization problem related to packing. I have to distribute a given amount of items from different sets to different packs. Each pack must be either empty or have a certain minimum of items. There is a penalty on each pack so i have to minimize the amount of packs
But there is also a penalty (rather than a constraint) for having items from different sets in the same packs so it is
$$min C_1 sum_{i}^{packs}if(X_{i,any} >0 ) + C_2sum_{j}^{packs}if[exists
X_{ij}>0,X_{kj}>0;kneq i
]$$
when $x_{i,j }$ is number of items from $j$ in pack $i$. Also it's a constraint that $ sum X_{i,j} = text{constant}$
What model can I use to solve that problem?
edit:
My problem is that my objective function is not continuous, and as far as I found out is not good for any linear or non-linear programming solution.
is there a way to either replace it with a function that gives a similar solution to the problem or is there another way to solve it?
optimization linear-programming nonlinear-optimization
asked Jan 29 at 21:44
Michael RazgonerMichael Razgoner
32
$endgroup$
I have a minimization problem related to packing. I have to distribute a given amount of items from different sets to different packs. Each pack must be either empty or have a certain minimum of items. There is a penalty on each pack so i have to minimize the amount of packs
But there is also a penalty (rather than a constraint) for having items from different sets in the same packs so it is
$$min C_1 sum_{i}^{packs}if(X_{i,any} >0 ) + C_2sum_{j}^{packs}if[exists
X_{ij}>0,X_{kj}>0;kneq i
]$$
when $x_{i,j }$ is number of items from $j$ in pack $i$. Also it's a constraint that $ sum X_{i,j} = text{constant}$
What model can I use to solve that problem?
edit:
My problem is that my objective function is not continuous, and as far as I found out is not good for any linear or non-linear programming solution.
is there a way to either replace it with a function that gives a similar solution to the problem or is there another way to solve it?
optimization linear-programming nonlinear-optimization
optimization linear-programming nonlinear-optimization
asked Jan 29 at 21:44
Michael RazgonerMichael Razgoner
32
asked Jan 29 at 21:44
Michael RazgonerMichael Razgoner
32
edited Jan 31 at 16:27
Michael Razgoner
asked Jan 29 at 21:44
Michael RazgonerMichael Razgoner
32
asked Jan 29 at 21:44
Michael RazgonerMichael Razgoner
32
asked Jan 29 at 21:44
Michael RazgonerMichael Razgoner
32
32
add a comment |
add a comment |
1 Answer
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$begingroup$
Your math is a little bit difficult to read. I can try to parse the first term. Let's introduce a binary variable $y_i$ indicating if bin $i$ is used. Then we can write:
$$begin{align} min> & C_1 sum_i y_i\
&M_i y_i ge x_{i,j} &forall i,j \
&y_{i} in {0,1}
end{align}$$
This is linear. $M_i$ is a large enough constant, to be chosen with care. I think this is what you want to achieve in the first part.
Sorry, I don't understand what you try to achieve in the second part.
In your text you indicate you want to model: "Each pack must be either empty or have a certain minimum of items." In other words: if $y_{i} = 1$ then $sum_j x_{i,j}ge L$. This can be modeled as:
$$sum_j x_{i,j} ge y_i L
$$
answered Jan 31 at 11:13
Erwin KalvelagenErwin Kalvelagen
3,2542512
$endgroup$
$begingroup$
Thank you for your help so far, you are correct and it helped me, what i am trying to do in the second part is some kind of variable that is true when more than one cell is not zero in a row (with the same j), a pack of items that come from mix of sets.
$endgroup$
– Michael Razgoner
Jan 31 at 14:33
$begingroup$
But X(i,j) variables are not binary, how can I create that step function to indicate when X(i,j) is non zero?
$endgroup$
– Michael Razgoner
Jan 31 at 15:13
$begingroup$
Indeed: see the $M_i$ in the first constraint. It is an upper bound on the number of items $j$ in bin $i$.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 15:33
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
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votes
$begingroup$
Your math is a little bit difficult to read. I can try to parse the first term. Let's introduce a binary variable $y_i$ indicating if bin $i$ is used. Then we can write:
$$begin{align} min> & C_1 sum_i y_i\
&M_i y_i ge x_{i,j} &forall i,j \
&y_{i} in {0,1}
end{align}$$
This is linear. $M_i$ is a large enough constant, to be chosen with care. I think this is what you want to achieve in the first part.
Sorry, I don't understand what you try to achieve in the second part.
In your text you indicate you want to model: "Each pack must be either empty or have a certain minimum of items." In other words: if $y_{i} = 1$ then $sum_j x_{i,j}ge L$. This can be modeled as:
$$sum_j x_{i,j} ge y_i L
$$
answered Jan 31 at 11:13
Erwin KalvelagenErwin Kalvelagen
3,2542512
$endgroup$
$begingroup$
Thank you for your help so far, you are correct and it helped me, what i am trying to do in the second part is some kind of variable that is true when more than one cell is not zero in a row (with the same j), a pack of items that come from mix of sets.
$endgroup$
– Michael Razgoner
Jan 31 at 14:33
$begingroup$
But X(i,j) variables are not binary, how can I create that step function to indicate when X(i,j) is non zero?
$endgroup$
– Michael Razgoner
Jan 31 at 15:13
$begingroup$
Indeed: see the $M_i$ in the first constraint. It is an upper bound on the number of items $j$ in bin $i$.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 15:33
add a comment |
$begingroup$
Your math is a little bit difficult to read. I can try to parse the first term. Let's introduce a binary variable $y_i$ indicating if bin $i$ is used. Then we can write:
$$begin{align} min> & C_1 sum_i y_i\
&M_i y_i ge x_{i,j} &forall i,j \
&y_{i} in {0,1}
end{align}$$
This is linear. $M_i$ is a large enough constant, to be chosen with care. I think this is what you want to achieve in the first part.
Sorry, I don't understand what you try to achieve in the second part.
In your text you indicate you want to model: "Each pack must be either empty or have a certain minimum of items." In other words: if $y_{i} = 1$ then $sum_j x_{i,j}ge L$. This can be modeled as:
$$sum_j x_{i,j} ge y_i L
$$
answered Jan 31 at 11:13
Erwin KalvelagenErwin Kalvelagen
3,2542512
$endgroup$
$begingroup$
Thank you for your help so far, you are correct and it helped me, what i am trying to do in the second part is some kind of variable that is true when more than one cell is not zero in a row (with the same j), a pack of items that come from mix of sets.
$endgroup$
– Michael Razgoner
Jan 31 at 14:33
$begingroup$
But X(i,j) variables are not binary, how can I create that step function to indicate when X(i,j) is non zero?
$endgroup$
– Michael Razgoner
Jan 31 at 15:13
$begingroup$
Indeed: see the $M_i$ in the first constraint. It is an upper bound on the number of items $j$ in bin $i$.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 15:33
add a comment |
$begingroup$
Your math is a little bit difficult to read. I can try to parse the first term. Let's introduce a binary variable $y_i$ indicating if bin $i$ is used. Then we can write:
$$begin{align} min> & C_1 sum_i y_i\
&M_i y_i ge x_{i,j} &forall i,j \
&y_{i} in {0,1}
end{align}$$
This is linear. $M_i$ is a large enough constant, to be chosen with care. I think this is what you want to achieve in the first part.
Sorry, I don't understand what you try to achieve in the second part.
In your text you indicate you want to model: "Each pack must be either empty or have a certain minimum of items." In other words: if $y_{i} = 1$ then $sum_j x_{i,j}ge L$. This can be modeled as:
$$sum_j x_{i,j} ge y_i L
$$
answered Jan 31 at 11:13
Erwin KalvelagenErwin Kalvelagen
3,2542512
$endgroup$
Your math is a little bit difficult to read. I can try to parse the first term. Let's introduce a binary variable $y_i$ indicating if bin $i$ is used. Then we can write:
$$begin{align} min> & C_1 sum_i y_i\
&M_i y_i ge x_{i,j} &forall i,j \
&y_{i} in {0,1}
end{align}$$
This is linear. $M_i$ is a large enough constant, to be chosen with care. I think this is what you want to achieve in the first part.
Sorry, I don't understand what you try to achieve in the second part.
In your text you indicate you want to model: "Each pack must be either empty or have a certain minimum of items." In other words: if $y_{i} = 1$ then $sum_j x_{i,j}ge L$. This can be modeled as:
$$sum_j x_{i,j} ge y_i L
$$
answered Jan 31 at 11:13
Erwin KalvelagenErwin Kalvelagen
3,2542512
edited Jan 31 at 15:12
answered Jan 31 at 11:13
Erwin KalvelagenErwin Kalvelagen
3,2542512
answered Jan 31 at 11:13
Erwin KalvelagenErwin Kalvelagen
3,2542512
answered Jan 31 at 11:13
Erwin KalvelagenErwin Kalvelagen
3,2542512
3,2542512
$begingroup$
Thank you for your help so far, you are correct and it helped me, what i am trying to do in the second part is some kind of variable that is true when more than one cell is not zero in a row (with the same j), a pack of items that come from mix of sets.
$endgroup$
– Michael Razgoner
Jan 31 at 14:33
$begingroup$
But X(i,j) variables are not binary, how can I create that step function to indicate when X(i,j) is non zero?
$endgroup$
– Michael Razgoner
Jan 31 at 15:13
$begingroup$
Indeed: see the $M_i$ in the first constraint. It is an upper bound on the number of items $j$ in bin $i$.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 15:33
add a comment |
$begingroup$
Thank you for your help so far, you are correct and it helped me, what i am trying to do in the second part is some kind of variable that is true when more than one cell is not zero in a row (with the same j), a pack of items that come from mix of sets.
$endgroup$
– Michael Razgoner
Jan 31 at 14:33
$begingroup$
But X(i,j) variables are not binary, how can I create that step function to indicate when X(i,j) is non zero?
$endgroup$
– Michael Razgoner
Jan 31 at 15:13
$begingroup$
Indeed: see the $M_i$ in the first constraint. It is an upper bound on the number of items $j$ in bin $i$.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 15:33
$begingroup$
Thank you for your help so far, you are correct and it helped me, what i am trying to do in the second part is some kind of variable that is true when more than one cell is not zero in a row (with the same j), a pack of items that come from mix of sets.
$endgroup$
– Michael Razgoner
Jan 31 at 14:33
$begingroup$
Thank you for your help so far, you are correct and it helped me, what i am trying to do in the second part is some kind of variable that is true when more than one cell is not zero in a row (with the same j), a pack of items that come from mix of sets.
$endgroup$
– Michael Razgoner
Jan 31 at 14:33
$begingroup$
But X(i,j) variables are not binary, how can I create that step function to indicate when X(i,j) is non zero?
$endgroup$
– Michael Razgoner
Jan 31 at 15:13
$begingroup$
But X(i,j) variables are not binary, how can I create that step function to indicate when X(i,j) is non zero?
$endgroup$
– Michael Razgoner
Jan 31 at 15:13
$begingroup$
Indeed: see the $M_i$ in the first constraint. It is an upper bound on the number of items $j$ in bin $i$.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 15:33
$begingroup$
Indeed: see the $M_i$ in the first constraint. It is an upper bound on the number of items $j$ in bin $i$.
$endgroup$
– Erwin Kalvelagen
Jan 31 at 15:33
add a comment |
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