How to linearize a constraint including product of two binary variables in summation with different indexes?
I am trying to linearize the following two expressions:
$sum_{k=1}^K sum_{t=1}^Tsum_{h=1}^W x_{ijkt} a_{hjt} =sum_{k=1}^K sum_{t=1}^T x_{ijkt} k , iin N, j in M$
$sum_{k=1}^K sum_{t=p_{ijk}}^Tsum_{l=t-P_{ijk}+1}^t x_{ijkt} a_{hjl} =sum_{k=1}^K sum_{t=p_{ijk}}^T x_{ijkt} a_{hjt} p_{ijk} , iin N, j in M, h in W$
$x_{ijkt}$: binary variable
$a_{hjt}$: binary variable
$p_{ijk}$: parameter
K: parameter
I already know product of two binary variables can be linearized as follows:
ab=z
$a le z$
$b le z$
$zge a+b-1$
Accordingly I did as follows to linearized the first constraint:
$x_{ijkt} a_{hjt}=z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, x_{ijkt} le z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, a_{hjt} le z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, x_{ijkt} + a_{hjt}-1 ge z_{ijkth}$
And finally converted the constraint as follows:
$sum_{k=1}^K sum_{t=1}^Tsum_{h=1}^W z_{ijkth} =sum_{k=1}^K sum_{t=1}^T x_{ijkt} k , iin N, j in M$
However it makes the model infeasible!
optimization linear-programming linearization
asked Dec 13 '17 at 2:19
araz nasirian
11
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
I am trying to linearize the following two expressions:
$sum_{k=1}^K sum_{t=1}^Tsum_{h=1}^W x_{ijkt} a_{hjt} =sum_{k=1}^K sum_{t=1}^T x_{ijkt} k , iin N, j in M$
$sum_{k=1}^K sum_{t=p_{ijk}}^Tsum_{l=t-P_{ijk}+1}^t x_{ijkt} a_{hjl} =sum_{k=1}^K sum_{t=p_{ijk}}^T x_{ijkt} a_{hjt} p_{ijk} , iin N, j in M, h in W$
$x_{ijkt}$: binary variable
$a_{hjt}$: binary variable
$p_{ijk}$: parameter
K: parameter
I already know product of two binary variables can be linearized as follows:
ab=z
$a le z$
$b le z$
$zge a+b-1$
Accordingly I did as follows to linearized the first constraint:
$x_{ijkt} a_{hjt}=z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, x_{ijkt} le z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, a_{hjt} le z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, x_{ijkt} + a_{hjt}-1 ge z_{ijkth}$
And finally converted the constraint as follows:
$sum_{k=1}^K sum_{t=1}^Tsum_{h=1}^W z_{ijkth} =sum_{k=1}^K sum_{t=1}^T x_{ijkt} k , iin N, j in M$
However it makes the model infeasible!
optimization linear-programming linearization
asked Dec 13 '17 at 2:19
araz nasirian
11
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Your binary multiplication $z=ab$ is not correct. The constraints $ale z$ and $b le z$ should read $z le a $ and $z le b$.
– Erwin Kalvelagen
Dec 13 '17 at 2:38
add a comment |
I am trying to linearize the following two expressions:
$sum_{k=1}^K sum_{t=1}^Tsum_{h=1}^W x_{ijkt} a_{hjt} =sum_{k=1}^K sum_{t=1}^T x_{ijkt} k , iin N, j in M$
$sum_{k=1}^K sum_{t=p_{ijk}}^Tsum_{l=t-P_{ijk}+1}^t x_{ijkt} a_{hjl} =sum_{k=1}^K sum_{t=p_{ijk}}^T x_{ijkt} a_{hjt} p_{ijk} , iin N, j in M, h in W$
$x_{ijkt}$: binary variable
$a_{hjt}$: binary variable
$p_{ijk}$: parameter
K: parameter
I already know product of two binary variables can be linearized as follows:
ab=z
$a le z$
$b le z$
$zge a+b-1$
Accordingly I did as follows to linearized the first constraint:
$x_{ijkt} a_{hjt}=z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, x_{ijkt} le z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, a_{hjt} le z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, x_{ijkt} + a_{hjt}-1 ge z_{ijkth}$
And finally converted the constraint as follows:
$sum_{k=1}^K sum_{t=1}^Tsum_{h=1}^W z_{ijkth} =sum_{k=1}^K sum_{t=1}^T x_{ijkt} k , iin N, j in M$
However it makes the model infeasible!
optimization linear-programming linearization
asked Dec 13 '17 at 2:19
araz nasirian
11
I am trying to linearize the following two expressions:
$sum_{k=1}^K sum_{t=1}^Tsum_{h=1}^W x_{ijkt} a_{hjt} =sum_{k=1}^K sum_{t=1}^T x_{ijkt} k , iin N, j in M$
$sum_{k=1}^K sum_{t=p_{ijk}}^Tsum_{l=t-P_{ijk}+1}^t x_{ijkt} a_{hjl} =sum_{k=1}^K sum_{t=p_{ijk}}^T x_{ijkt} a_{hjt} p_{ijk} , iin N, j in M, h in W$
$x_{ijkt}$: binary variable
$a_{hjt}$: binary variable
$p_{ijk}$: parameter
K: parameter
I already know product of two binary variables can be linearized as follows:
ab=z
$a le z$
$b le z$
$zge a+b-1$
Accordingly I did as follows to linearized the first constraint:
$x_{ijkt} a_{hjt}=z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, x_{ijkt} le z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, a_{hjt} le z_{ijkth}$
$i in N, j in M, k in K, t in T, h in w, x_{ijkt} + a_{hjt}-1 ge z_{ijkth}$
And finally converted the constraint as follows:
$sum_{k=1}^K sum_{t=1}^Tsum_{h=1}^W z_{ijkth} =sum_{k=1}^K sum_{t=1}^T x_{ijkt} k , iin N, j in M$
However it makes the model infeasible!
optimization linear-programming linearization
optimization linear-programming linearization
asked Dec 13 '17 at 2:19
araz nasirian
11
asked Dec 13 '17 at 2:19
araz nasirian
11
asked Dec 13 '17 at 2:19
araz nasirian
11
asked Dec 13 '17 at 2:19
araz nasirian
11
asked Dec 13 '17 at 2:19
araz nasirian
11
11
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Your binary multiplication $z=ab$ is not correct. The constraints $ale z$ and $b le z$ should read $z le a $ and $z le b$.
– Erwin Kalvelagen
Dec 13 '17 at 2:38
add a comment |
Your binary multiplication $z=ab$ is not correct. The constraints $ale z$ and $b le z$ should read $z le a $ and $z le b$.
– Erwin Kalvelagen
Dec 13 '17 at 2:38
Your binary multiplication $z=ab$ is not correct. The constraints $ale z$ and $b le z$ should read $z le a $ and $z le b$.
– Erwin Kalvelagen
Dec 13 '17 at 2:38
Your binary multiplication $z=ab$ is not correct. The constraints $ale z$ and $b le z$ should read $z le a $ and $z le b$.
– Erwin Kalvelagen
Dec 13 '17 at 2:38
add a comment |
1 Answer
1
active
oldest
votes
If you want to handle the sum of products
$$ sum_{i,j} x_i y_j $$
with $x_i, y_i in {0,1}$ you need to introduce a new variable $z_{i,j}in {0,1}$ and write:
$$
begin{align}
&sum_{i,j} z_{i,j}\
&z_{i,j} le x_i&forall i,j\
&z_{i,j} le y_j&forall i,j\
&z_{i,j} ge x_i+y_j-1&forall i,j\
& 0 leq z_{i,j} leq 1&forall i,j
end{align}
$$
Note that we can relax $z$ to be continuous (i.e., $zin [0,1]$) as $z$ assumes integer values automatically. This formulation extends naturally to more indices.
edited Sep 27 '18 at 8:38
YukiJ
2,1112928
answered Dec 13 '17 at 10:21
Erwin Kalvelagen
3,0892511
In the last constraint the direction should be $ge$.
– YukiJ
Sep 27 '18 at 7:50
@YukiJ Yes, thanks.
– Erwin Kalvelagen
Sep 27 '18 at 8:40
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you want to handle the sum of products
$$ sum_{i,j} x_i y_j $$
with $x_i, y_i in {0,1}$ you need to introduce a new variable $z_{i,j}in {0,1}$ and write:
$$
begin{align}
&sum_{i,j} z_{i,j}\
&z_{i,j} le x_i&forall i,j\
&z_{i,j} le y_j&forall i,j\
&z_{i,j} ge x_i+y_j-1&forall i,j\
& 0 leq z_{i,j} leq 1&forall i,j
end{align}
$$
Note that we can relax $z$ to be continuous (i.e., $zin [0,1]$) as $z$ assumes integer values automatically. This formulation extends naturally to more indices.
edited Sep 27 '18 at 8:38
YukiJ
2,1112928
answered Dec 13 '17 at 10:21
Erwin Kalvelagen
3,0892511
In the last constraint the direction should be $ge$.
– YukiJ
Sep 27 '18 at 7:50
@YukiJ Yes, thanks.
– Erwin Kalvelagen
Sep 27 '18 at 8:40
add a comment |
If you want to handle the sum of products
$$ sum_{i,j} x_i y_j $$
with $x_i, y_i in {0,1}$ you need to introduce a new variable $z_{i,j}in {0,1}$ and write:
$$
begin{align}
&sum_{i,j} z_{i,j}\
&z_{i,j} le x_i&forall i,j\
&z_{i,j} le y_j&forall i,j\
&z_{i,j} ge x_i+y_j-1&forall i,j\
& 0 leq z_{i,j} leq 1&forall i,j
end{align}
$$
Note that we can relax $z$ to be continuous (i.e., $zin [0,1]$) as $z$ assumes integer values automatically. This formulation extends naturally to more indices.
edited Sep 27 '18 at 8:38
YukiJ
2,1112928
answered Dec 13 '17 at 10:21
Erwin Kalvelagen
3,0892511
In the last constraint the direction should be $ge$.
– YukiJ
Sep 27 '18 at 7:50
@YukiJ Yes, thanks.
– Erwin Kalvelagen
Sep 27 '18 at 8:40
add a comment |
If you want to handle the sum of products
$$ sum_{i,j} x_i y_j $$
with $x_i, y_i in {0,1}$ you need to introduce a new variable $z_{i,j}in {0,1}$ and write:
$$
begin{align}
&sum_{i,j} z_{i,j}\
&z_{i,j} le x_i&forall i,j\
&z_{i,j} le y_j&forall i,j\
&z_{i,j} ge x_i+y_j-1&forall i,j\
& 0 leq z_{i,j} leq 1&forall i,j
end{align}
$$
Note that we can relax $z$ to be continuous (i.e., $zin [0,1]$) as $z$ assumes integer values automatically. This formulation extends naturally to more indices.
edited Sep 27 '18 at 8:38
YukiJ
2,1112928
answered Dec 13 '17 at 10:21
Erwin Kalvelagen
3,0892511
If you want to handle the sum of products
$$ sum_{i,j} x_i y_j $$
with $x_i, y_i in {0,1}$ you need to introduce a new variable $z_{i,j}in {0,1}$ and write:
$$
begin{align}
&sum_{i,j} z_{i,j}\
&z_{i,j} le x_i&forall i,j\
&z_{i,j} le y_j&forall i,j\
&z_{i,j} ge x_i+y_j-1&forall i,j\
& 0 leq z_{i,j} leq 1&forall i,j
end{align}
$$
Note that we can relax $z$ to be continuous (i.e., $zin [0,1]$) as $z$ assumes integer values automatically. This formulation extends naturally to more indices.
edited Sep 27 '18 at 8:38
YukiJ
2,1112928
answered Dec 13 '17 at 10:21
Erwin Kalvelagen
3,0892511
edited Sep 27 '18 at 8:38
YukiJ
2,1112928
edited Sep 27 '18 at 8:38
YukiJ
2,1112928
edited Sep 27 '18 at 8:38
YukiJ
2,1112928
2,1112928
answered Dec 13 '17 at 10:21
Erwin Kalvelagen
3,0892511
answered Dec 13 '17 at 10:21
Erwin Kalvelagen
3,0892511
answered Dec 13 '17 at 10:21
Erwin Kalvelagen
3,0892511
3,0892511
In the last constraint the direction should be $ge$.
– YukiJ
Sep 27 '18 at 7:50
@YukiJ Yes, thanks.
– Erwin Kalvelagen
Sep 27 '18 at 8:40
add a comment |
In the last constraint the direction should be $ge$.
– YukiJ
Sep 27 '18 at 7:50
@YukiJ Yes, thanks.
– Erwin Kalvelagen
Sep 27 '18 at 8:40
In the last constraint the direction should be $ge$.
– YukiJ
Sep 27 '18 at 7:50
In the last constraint the direction should be $ge$.
– YukiJ
Sep 27 '18 at 7:50
@YukiJ Yes, thanks.
– Erwin Kalvelagen
Sep 27 '18 at 8:40
@YukiJ Yes, thanks.
– Erwin Kalvelagen
Sep 27 '18 at 8:40
add a comment |
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Your binary multiplication $z=ab$ is not correct. The constraints $ale z$ and $b le z$ should read $z le a $ and $z le b$.
– Erwin Kalvelagen
Dec 13 '17 at 2:38