Mixing
How to use non-binary variable in a conditional statement MILP?
$begingroup$
I have a conditional statement I want to implement in a MILP.
$A$ is a non-binary variable that has known upper and lower bounds. $B$ is a known parameter. And $C$ is a binary variable.
How do I formulate the condition?
If $A < B$ then $C$ is $1$, otherwise $C$ is $0$ or $1$.
Is this possible? I am new to optimization using GAMS and i'm still figuring out how this works.
optimization linear-programming constraints mixed-integer-programming
edited Jan 9 at 13:19
Klangen
1,72811334
asked Jan 9 at 13:01
Charles uuuCharles uuu
32
$endgroup$
add a comment |
$begingroup$
I have a conditional statement I want to implement in a MILP.
$A$ is a non-binary variable that has known upper and lower bounds. $B$ is a known parameter. And $C$ is a binary variable.
How do I formulate the condition?
If $A < B$ then $C$ is $1$, otherwise $C$ is $0$ or $1$.
Is this possible? I am new to optimization using GAMS and i'm still figuring out how this works.
optimization linear-programming constraints mixed-integer-programming
edited Jan 9 at 13:19
Klangen
1,72811334
asked Jan 9 at 13:01
Charles uuuCharles uuu
32
$endgroup$
add a comment |
$begingroup$
I have a conditional statement I want to implement in a MILP.
$A$ is a non-binary variable that has known upper and lower bounds. $B$ is a known parameter. And $C$ is a binary variable.
How do I formulate the condition?
If $A < B$ then $C$ is $1$, otherwise $C$ is $0$ or $1$.
Is this possible? I am new to optimization using GAMS and i'm still figuring out how this works.
optimization linear-programming constraints mixed-integer-programming
edited Jan 9 at 13:19
Klangen
1,72811334
asked Jan 9 at 13:01
Charles uuuCharles uuu
32
$endgroup$
I have a conditional statement I want to implement in a MILP.
$A$ is a non-binary variable that has known upper and lower bounds. $B$ is a known parameter. And $C$ is a binary variable.
How do I formulate the condition?
If $A < B$ then $C$ is $1$, otherwise $C$ is $0$ or $1$.
Is this possible? I am new to optimization using GAMS and i'm still figuring out how this works.
optimization linear-programming constraints mixed-integer-programming
optimization linear-programming constraints mixed-integer-programming
edited Jan 9 at 13:19
Klangen
1,72811334
asked Jan 9 at 13:01
Charles uuuCharles uuu
32
edited Jan 9 at 13:19
Klangen
1,72811334
asked Jan 9 at 13:01
Charles uuuCharles uuu
32
edited Jan 9 at 13:19
Klangen
1,72811334
edited Jan 9 at 13:19
Klangen
1,72811334
edited Jan 9 at 13:19
Klangen
1,72811334
1,72811334
asked Jan 9 at 13:01
Charles uuuCharles uuu
32
asked Jan 9 at 13:01
Charles uuuCharles uuu
32
asked Jan 9 at 13:01
Charles uuuCharles uuu
32
32
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We know the bounds of $A$, let say $m le A le M$.
if $m ge B$, then we clearly, we have $A ge B$.
Hence we assume that $m < B$, then we know that
$$frac{B-A}{B-m} le 1$$
we consider the constraint $$frac{B-A}{B-m}le C$$
Hence if $A < B$, then we have the left hand side of the inequality being positive and it is at most $1$, forcing $C$ to take value $1$.
If $A ge B$, then the left hand side of the inequality is non-positive, hence not imposing constraint on $C$.
answered Jan 9 at 13:19
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Thanks Siong !!
$endgroup$
– Charles uuu
Jan 9 at 13:52
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
$begingroup$
We know the bounds of $A$, let say $m le A le M$.
if $m ge B$, then we clearly, we have $A ge B$.
Hence we assume that $m < B$, then we know that
$$frac{B-A}{B-m} le 1$$
we consider the constraint $$frac{B-A}{B-m}le C$$
Hence if $A < B$, then we have the left hand side of the inequality being positive and it is at most $1$, forcing $C$ to take value $1$.
If $A ge B$, then the left hand side of the inequality is non-positive, hence not imposing constraint on $C$.
answered Jan 9 at 13:19
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Thanks Siong !!
$endgroup$
– Charles uuu
Jan 9 at 13:52
add a comment |
$begingroup$
We know the bounds of $A$, let say $m le A le M$.
if $m ge B$, then we clearly, we have $A ge B$.
Hence we assume that $m < B$, then we know that
$$frac{B-A}{B-m} le 1$$
we consider the constraint $$frac{B-A}{B-m}le C$$
Hence if $A < B$, then we have the left hand side of the inequality being positive and it is at most $1$, forcing $C$ to take value $1$.
If $A ge B$, then the left hand side of the inequality is non-positive, hence not imposing constraint on $C$.
answered Jan 9 at 13:19
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
$begingroup$
Thanks Siong !!
$endgroup$
– Charles uuu
Jan 9 at 13:52
add a comment |
$begingroup$
We know the bounds of $A$, let say $m le A le M$.
if $m ge B$, then we clearly, we have $A ge B$.
Hence we assume that $m < B$, then we know that
$$frac{B-A}{B-m} le 1$$
we consider the constraint $$frac{B-A}{B-m}le C$$
Hence if $A < B$, then we have the left hand side of the inequality being positive and it is at most $1$, forcing $C$ to take value $1$.
If $A ge B$, then the left hand side of the inequality is non-positive, hence not imposing constraint on $C$.
answered Jan 9 at 13:19
Siong Thye GohSiong Thye Goh
101k1466117
$endgroup$
We know the bounds of $A$, let say $m le A le M$.
if $m ge B$, then we clearly, we have $A ge B$.
Hence we assume that $m < B$, then we know that
$$frac{B-A}{B-m} le 1$$
we consider the constraint $$frac{B-A}{B-m}le C$$
Hence if $A < B$, then we have the left hand side of the inequality being positive and it is at most $1$, forcing $C$ to take value $1$.
If $A ge B$, then the left hand side of the inequality is non-positive, hence not imposing constraint on $C$.
answered Jan 9 at 13:19
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 9 at 13:19
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 9 at 13:19
Siong Thye GohSiong Thye Goh
101k1466117
answered Jan 9 at 13:19
Siong Thye GohSiong Thye Goh
101k1466117
101k1466117
$begingroup$
Thanks Siong !!
$endgroup$
– Charles uuu
Jan 9 at 13:52
add a comment |
$begingroup$
Thanks Siong !!
$endgroup$
– Charles uuu
Jan 9 at 13:52
$begingroup$
Thanks Siong !!
$endgroup$
– Charles uuu
Jan 9 at 13:52
$begingroup$
Thanks Siong !!
$endgroup$
– Charles uuu
Jan 9 at 13:52
add a comment |
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