If-Then Constraint: If $F(X) > 3$, then $Y = 1$
1
$begingroup$
$F(x) = x_1+x_2+x_3+x_4$
Scenario: Amongst binary variables $X_1$, $X_2$, $X_3$, $X_4$, if more than $3$ are chosen, then another binary variable $Y = 1$. Otherwise, $y = 0$. How can I formulate constraints based on this for a linear programming model?
Thanks!
linear-programming integer-programming
edited Dec 16 '18 at 19:45
t.ysn
1397
asked Dec 16 '18 at 19:29
HFCHHFCH
83
$endgroup$
|
show 1 more comment
1
$begingroup$
$F(x) = x_1+x_2+x_3+x_4$
Scenario: Amongst binary variables $X_1$, $X_2$, $X_3$, $X_4$, if more than $3$ are chosen, then another binary variable $Y = 1$. Otherwise, $y = 0$. How can I formulate constraints based on this for a linear programming model?
Thanks!
linear-programming integer-programming
edited Dec 16 '18 at 19:45
t.ysn
1397
asked Dec 16 '18 at 19:29
HFCHHFCH
83
$endgroup$
1
$begingroup$
According to your specification, $Y=1$ if and only if all four $X_i = 1$. Hence $Y = X_1X_2X_3X_4$.
$endgroup$
– Song
Dec 16 '18 at 20:02
$begingroup$
Thank you for your comment. That's what I did at first but I need to run this in a linear programming model. Which is why I need to impose if-then constraint but I'm not sure how..
$endgroup$
– HFCH
Dec 16 '18 at 20:04
$begingroup$
The knowledge of the whole model, especially the objective function, is usually very helpful when you want to formulate such a constraint. Has y to be minimized or maximized?
$endgroup$
– callculus
Dec 16 '18 at 21:58
$begingroup$
Sorry I forgot to mention that y is to be minimised. This is because there will be a penalty if F(x) > 3 (y = 1).
$endgroup$
– HFCH
Dec 16 '18 at 23:42
$begingroup$
In the complete model, there is a value assigned to each x-variable and I need to choose 5 x-variables out of 9 to maximise the total values of the chosen variables, while bearing in mind that if I choose all X1 , X2, X3, X4 I will incur a small penalty. It doesn't mean that I shouldn't choose all these 4 variables because the total values assigned may still be higher to compensate for the penalty. The objective function is (Max. Total values of the 5 chosen variables - 1.2y)
$endgroup$
– HFCH
Dec 16 '18 at 23:50
|
show 1 more comment
1
1
1
$begingroup$
$F(x) = x_1+x_2+x_3+x_4$
Scenario: Amongst binary variables $X_1$, $X_2$, $X_3$, $X_4$, if more than $3$ are chosen, then another binary variable $Y = 1$. Otherwise, $y = 0$. How can I formulate constraints based on this for a linear programming model?
Thanks!
linear-programming integer-programming
edited Dec 16 '18 at 19:45
t.ysn
1397
asked Dec 16 '18 at 19:29
HFCHHFCH
83
$endgroup$
$F(x) = x_1+x_2+x_3+x_4$
Scenario: Amongst binary variables $X_1$, $X_2$, $X_3$, $X_4$, if more than $3$ are chosen, then another binary variable $Y = 1$. Otherwise, $y = 0$. How can I formulate constraints based on this for a linear programming model?
Thanks!
linear-programming integer-programming
linear-programming integer-programming
edited Dec 16 '18 at 19:45
t.ysn
1397
asked Dec 16 '18 at 19:29
HFCHHFCH
83
edited Dec 16 '18 at 19:45
t.ysn
1397
asked Dec 16 '18 at 19:29
HFCHHFCH
83
edited Dec 16 '18 at 19:45
t.ysn
1397
edited Dec 16 '18 at 19:45
t.ysn
1397
edited Dec 16 '18 at 19:45
t.ysn
1397
1397
asked Dec 16 '18 at 19:29
HFCHHFCH
83
asked Dec 16 '18 at 19:29
HFCHHFCH
83
asked Dec 16 '18 at 19:29
HFCHHFCH
83
83
1
$begingroup$
According to your specification, $Y=1$ if and only if all four $X_i = 1$. Hence $Y = X_1X_2X_3X_4$.
$endgroup$
– Song
Dec 16 '18 at 20:02
$begingroup$
Thank you for your comment. That's what I did at first but I need to run this in a linear programming model. Which is why I need to impose if-then constraint but I'm not sure how..
$endgroup$
– HFCH
Dec 16 '18 at 20:04
$begingroup$
The knowledge of the whole model, especially the objective function, is usually very helpful when you want to formulate such a constraint. Has y to be minimized or maximized?
$endgroup$
– callculus
Dec 16 '18 at 21:58
$begingroup$
Sorry I forgot to mention that y is to be minimised. This is because there will be a penalty if F(x) > 3 (y = 1).
$endgroup$
– HFCH
Dec 16 '18 at 23:42
$begingroup$
In the complete model, there is a value assigned to each x-variable and I need to choose 5 x-variables out of 9 to maximise the total values of the chosen variables, while bearing in mind that if I choose all X1 , X2, X3, X4 I will incur a small penalty. It doesn't mean that I shouldn't choose all these 4 variables because the total values assigned may still be higher to compensate for the penalty. The objective function is (Max. Total values of the 5 chosen variables - 1.2y)
$endgroup$
– HFCH
Dec 16 '18 at 23:50
|
show 1 more comment
1
$begingroup$
According to your specification, $Y=1$ if and only if all four $X_i = 1$. Hence $Y = X_1X_2X_3X_4$.
$endgroup$
– Song
Dec 16 '18 at 20:02
$begingroup$
Thank you for your comment. That's what I did at first but I need to run this in a linear programming model. Which is why I need to impose if-then constraint but I'm not sure how..
$endgroup$
– HFCH
Dec 16 '18 at 20:04
$begingroup$
The knowledge of the whole model, especially the objective function, is usually very helpful when you want to formulate such a constraint. Has y to be minimized or maximized?
$endgroup$
– callculus
Dec 16 '18 at 21:58
$begingroup$
Sorry I forgot to mention that y is to be minimised. This is because there will be a penalty if F(x) > 3 (y = 1).
$endgroup$
– HFCH
Dec 16 '18 at 23:42
$begingroup$
In the complete model, there is a value assigned to each x-variable and I need to choose 5 x-variables out of 9 to maximise the total values of the chosen variables, while bearing in mind that if I choose all X1 , X2, X3, X4 I will incur a small penalty. It doesn't mean that I shouldn't choose all these 4 variables because the total values assigned may still be higher to compensate for the penalty. The objective function is (Max. Total values of the 5 chosen variables - 1.2y)
$endgroup$
– HFCH
Dec 16 '18 at 23:50
1
1
$begingroup$
According to your specification, $Y=1$ if and only if all four $X_i = 1$. Hence $Y = X_1X_2X_3X_4$.
$endgroup$
– Song
Dec 16 '18 at 20:02
$begingroup$
According to your specification, $Y=1$ if and only if all four $X_i = 1$. Hence $Y = X_1X_2X_3X_4$.
$endgroup$
– Song
Dec 16 '18 at 20:02
$begingroup$
Thank you for your comment. That's what I did at first but I need to run this in a linear programming model. Which is why I need to impose if-then constraint but I'm not sure how..
$endgroup$
– HFCH
Dec 16 '18 at 20:04
$begingroup$
Thank you for your comment. That's what I did at first but I need to run this in a linear programming model. Which is why I need to impose if-then constraint but I'm not sure how..
$endgroup$
– HFCH
Dec 16 '18 at 20:04
$begingroup$
The knowledge of the whole model, especially the objective function, is usually very helpful when you want to formulate such a constraint. Has y to be minimized or maximized?
$endgroup$
– callculus
Dec 16 '18 at 21:58
$begingroup$
The knowledge of the whole model, especially the objective function, is usually very helpful when you want to formulate such a constraint. Has y to be minimized or maximized?
$endgroup$
– callculus
Dec 16 '18 at 21:58
$begingroup$
Sorry I forgot to mention that y is to be minimised. This is because there will be a penalty if F(x) > 3 (y = 1).
$endgroup$
– HFCH
Dec 16 '18 at 23:42
$begingroup$
Sorry I forgot to mention that y is to be minimised. This is because there will be a penalty if F(x) > 3 (y = 1).
$endgroup$
– HFCH
Dec 16 '18 at 23:42
$begingroup$
In the complete model, there is a value assigned to each x-variable and I need to choose 5 x-variables out of 9 to maximise the total values of the chosen variables, while bearing in mind that if I choose all X1 , X2, X3, X4 I will incur a small penalty. It doesn't mean that I shouldn't choose all these 4 variables because the total values assigned may still be higher to compensate for the penalty. The objective function is (Max. Total values of the 5 chosen variables - 1.2y)
$endgroup$
– HFCH
Dec 16 '18 at 23:50
$begingroup$
In the complete model, there is a value assigned to each x-variable and I need to choose 5 x-variables out of 9 to maximise the total values of the chosen variables, while bearing in mind that if I choose all X1 , X2, X3, X4 I will incur a small penalty. It doesn't mean that I shouldn't choose all these 4 variables because the total values assigned may still be higher to compensate for the penalty. The objective function is (Max. Total values of the 5 chosen variables - 1.2y)
$endgroup$
– HFCH
Dec 16 '18 at 23:50
|
show 1 more comment
1 Answer
1
active
oldest
votes
1
$begingroup$
Something like this :
$$
mbox{Max } sum_{i=1}^9 C_iX_i - pY
$$
subject to
$$
X_1+X_2+X_3+X_4 le 3 + Y
$$
Variable $Y$ takes value $1$ only if all $4$ $X$ variables take value $1$.
You probably also have the following constraint to ensure exactly $5$ $X$ variables are chosen :
$$
sum_{i=1}^9 X_i = 5
$$
answered Dec 17 '18 at 8:03
KuifjeKuifje
7,2652726
$endgroup$
$begingroup$
Yes! That's the solution I was looking for, much obliged!
$endgroup$
– HFCH
Dec 17 '18 at 11:33
$begingroup$
There is another problem concerning if-then constraint that I would also like to discuss here: If someone selects 2 or more items in category A, then they must also select at least 1 item in category B. X-variables in Category A: X1,X2,X3; X-variables in Category B: X4,X5,X6,X7,X8 I know I need to introduce another binary variable z that takes the value of 1 if 2 or more X variables in Category A are chosen. This will not go into the objective function, but rather act as a constraint.
$endgroup$
– HFCH
Dec 17 '18 at 11:47
$begingroup$
begin{align*} X_1+X_2 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_2+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_2+X_3 le 2+ X_4+X_5+X_6+X_7+X_8 end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 12:40
$begingroup$
Hi Kuifje, thanks for the speedy reply! Is there a way I could simply this by imposing a if-then constraint? I want to make use of the new binary variable z and come up with something like the first solution you posted above :)
$endgroup$
– HFCH
Dec 17 '18 at 13:11
$begingroup$
begin{align*} sum_{i in A} X_i & le 1+|A| delta_A \ delta_A & le sum_{i in B} X_i \ delta_A &in {0,1} end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 13:16
|
show 5 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Something like this :
$$
mbox{Max } sum_{i=1}^9 C_iX_i - pY
$$
subject to
$$
X_1+X_2+X_3+X_4 le 3 + Y
$$
Variable $Y$ takes value $1$ only if all $4$ $X$ variables take value $1$.
You probably also have the following constraint to ensure exactly $5$ $X$ variables are chosen :
$$
sum_{i=1}^9 X_i = 5
$$
answered Dec 17 '18 at 8:03
KuifjeKuifje
7,2652726
$endgroup$
$begingroup$
Yes! That's the solution I was looking for, much obliged!
$endgroup$
– HFCH
Dec 17 '18 at 11:33
$begingroup$
There is another problem concerning if-then constraint that I would also like to discuss here: If someone selects 2 or more items in category A, then they must also select at least 1 item in category B. X-variables in Category A: X1,X2,X3; X-variables in Category B: X4,X5,X6,X7,X8 I know I need to introduce another binary variable z that takes the value of 1 if 2 or more X variables in Category A are chosen. This will not go into the objective function, but rather act as a constraint.
$endgroup$
– HFCH
Dec 17 '18 at 11:47
$begingroup$
begin{align*} X_1+X_2 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_2+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_2+X_3 le 2+ X_4+X_5+X_6+X_7+X_8 end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 12:40
$begingroup$
Hi Kuifje, thanks for the speedy reply! Is there a way I could simply this by imposing a if-then constraint? I want to make use of the new binary variable z and come up with something like the first solution you posted above :)
$endgroup$
– HFCH
Dec 17 '18 at 13:11
$begingroup$
begin{align*} sum_{i in A} X_i & le 1+|A| delta_A \ delta_A & le sum_{i in B} X_i \ delta_A &in {0,1} end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 13:16
|
show 5 more comments
1
$begingroup$
Something like this :
$$
mbox{Max } sum_{i=1}^9 C_iX_i - pY
$$
subject to
$$
X_1+X_2+X_3+X_4 le 3 + Y
$$
Variable $Y$ takes value $1$ only if all $4$ $X$ variables take value $1$.
You probably also have the following constraint to ensure exactly $5$ $X$ variables are chosen :
$$
sum_{i=1}^9 X_i = 5
$$
answered Dec 17 '18 at 8:03
KuifjeKuifje
7,2652726
$endgroup$
$begingroup$
Yes! That's the solution I was looking for, much obliged!
$endgroup$
– HFCH
Dec 17 '18 at 11:33
$begingroup$
There is another problem concerning if-then constraint that I would also like to discuss here: If someone selects 2 or more items in category A, then they must also select at least 1 item in category B. X-variables in Category A: X1,X2,X3; X-variables in Category B: X4,X5,X6,X7,X8 I know I need to introduce another binary variable z that takes the value of 1 if 2 or more X variables in Category A are chosen. This will not go into the objective function, but rather act as a constraint.
$endgroup$
– HFCH
Dec 17 '18 at 11:47
$begingroup$
begin{align*} X_1+X_2 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_2+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_2+X_3 le 2+ X_4+X_5+X_6+X_7+X_8 end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 12:40
$begingroup$
Hi Kuifje, thanks for the speedy reply! Is there a way I could simply this by imposing a if-then constraint? I want to make use of the new binary variable z and come up with something like the first solution you posted above :)
$endgroup$
– HFCH
Dec 17 '18 at 13:11
$begingroup$
begin{align*} sum_{i in A} X_i & le 1+|A| delta_A \ delta_A & le sum_{i in B} X_i \ delta_A &in {0,1} end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 13:16
|
show 5 more comments
1
1
1
$begingroup$
Something like this :
$$
mbox{Max } sum_{i=1}^9 C_iX_i - pY
$$
subject to
$$
X_1+X_2+X_3+X_4 le 3 + Y
$$
Variable $Y$ takes value $1$ only if all $4$ $X$ variables take value $1$.
You probably also have the following constraint to ensure exactly $5$ $X$ variables are chosen :
$$
sum_{i=1}^9 X_i = 5
$$
answered Dec 17 '18 at 8:03
KuifjeKuifje
7,2652726
$endgroup$
Something like this :
$$
mbox{Max } sum_{i=1}^9 C_iX_i - pY
$$
subject to
$$
X_1+X_2+X_3+X_4 le 3 + Y
$$
Variable $Y$ takes value $1$ only if all $4$ $X$ variables take value $1$.
You probably also have the following constraint to ensure exactly $5$ $X$ variables are chosen :
$$
sum_{i=1}^9 X_i = 5
$$
answered Dec 17 '18 at 8:03
KuifjeKuifje
7,2652726
answered Dec 17 '18 at 8:03
KuifjeKuifje
7,2652726
answered Dec 17 '18 at 8:03
KuifjeKuifje
7,2652726
answered Dec 17 '18 at 8:03
KuifjeKuifje
7,2652726
7,2652726
$begingroup$
Yes! That's the solution I was looking for, much obliged!
$endgroup$
– HFCH
Dec 17 '18 at 11:33
$begingroup$
There is another problem concerning if-then constraint that I would also like to discuss here: If someone selects 2 or more items in category A, then they must also select at least 1 item in category B. X-variables in Category A: X1,X2,X3; X-variables in Category B: X4,X5,X6,X7,X8 I know I need to introduce another binary variable z that takes the value of 1 if 2 or more X variables in Category A are chosen. This will not go into the objective function, but rather act as a constraint.
$endgroup$
– HFCH
Dec 17 '18 at 11:47
$begingroup$
begin{align*} X_1+X_2 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_2+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_2+X_3 le 2+ X_4+X_5+X_6+X_7+X_8 end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 12:40
$begingroup$
Hi Kuifje, thanks for the speedy reply! Is there a way I could simply this by imposing a if-then constraint? I want to make use of the new binary variable z and come up with something like the first solution you posted above :)
$endgroup$
– HFCH
Dec 17 '18 at 13:11
$begingroup$
begin{align*} sum_{i in A} X_i & le 1+|A| delta_A \ delta_A & le sum_{i in B} X_i \ delta_A &in {0,1} end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 13:16
|
show 5 more comments
$begingroup$
Yes! That's the solution I was looking for, much obliged!
$endgroup$
– HFCH
Dec 17 '18 at 11:33
$begingroup$
There is another problem concerning if-then constraint that I would also like to discuss here: If someone selects 2 or more items in category A, then they must also select at least 1 item in category B. X-variables in Category A: X1,X2,X3; X-variables in Category B: X4,X5,X6,X7,X8 I know I need to introduce another binary variable z that takes the value of 1 if 2 or more X variables in Category A are chosen. This will not go into the objective function, but rather act as a constraint.
$endgroup$
– HFCH
Dec 17 '18 at 11:47
$begingroup$
begin{align*} X_1+X_2 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_2+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_2+X_3 le 2+ X_4+X_5+X_6+X_7+X_8 end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 12:40
$begingroup$
Hi Kuifje, thanks for the speedy reply! Is there a way I could simply this by imposing a if-then constraint? I want to make use of the new binary variable z and come up with something like the first solution you posted above :)
$endgroup$
– HFCH
Dec 17 '18 at 13:11
$begingroup$
begin{align*} sum_{i in A} X_i & le 1+|A| delta_A \ delta_A & le sum_{i in B} X_i \ delta_A &in {0,1} end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 13:16
$begingroup$
Yes! That's the solution I was looking for, much obliged!
$endgroup$
– HFCH
Dec 17 '18 at 11:33
$begingroup$
Yes! That's the solution I was looking for, much obliged!
$endgroup$
– HFCH
Dec 17 '18 at 11:33
$begingroup$
There is another problem concerning if-then constraint that I would also like to discuss here: If someone selects 2 or more items in category A, then they must also select at least 1 item in category B. X-variables in Category A: X1,X2,X3; X-variables in Category B: X4,X5,X6,X7,X8 I know I need to introduce another binary variable z that takes the value of 1 if 2 or more X variables in Category A are chosen. This will not go into the objective function, but rather act as a constraint.
$endgroup$
– HFCH
Dec 17 '18 at 11:47
$begingroup$
There is another problem concerning if-then constraint that I would also like to discuss here: If someone selects 2 or more items in category A, then they must also select at least 1 item in category B. X-variables in Category A: X1,X2,X3; X-variables in Category B: X4,X5,X6,X7,X8 I know I need to introduce another binary variable z that takes the value of 1 if 2 or more X variables in Category A are chosen. This will not go into the objective function, but rather act as a constraint.
$endgroup$
– HFCH
Dec 17 '18 at 11:47
$begingroup$
begin{align*} X_1+X_2 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_2+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_2+X_3 le 2+ X_4+X_5+X_6+X_7+X_8 end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 12:40
$begingroup$
begin{align*} X_1+X_2 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_2+X_3 le 1+ X_4+X_5+X_6+X_7+X_8 \ X_1+X_2+X_3 le 2+ X_4+X_5+X_6+X_7+X_8 end{align*}
$endgroup$
– Kuifje
Dec 17 '18 at 12:40
$begingroup$
Hi Kuifje, thanks for the speedy reply! Is there a way I could simply this by imposing a if-then constraint? I want to make use of the new binary variable z and come up with something like the first solution you posted above :)
$endgroup$
– HFCH
Dec 17 '18 at 13:11
$begingroup$
Hi Kuifje, thanks for the speedy reply! Is there a way I could simply this by imposing a if-then constraint? I want to make use of the new binary variable z and come up with something like the first solution you posted above :)
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– HFCH
Dec 17 '18 at 13:11
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begin{align*} sum_{i in A} X_i & le 1+|A| delta_A \ delta_A & le sum_{i in B} X_i \ delta_A &in {0,1} end{align*}
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– Kuifje
Dec 17 '18 at 13:16
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begin{align*} sum_{i in A} X_i & le 1+|A| delta_A \ delta_A & le sum_{i in B} X_i \ delta_A &in {0,1} end{align*}
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– Kuifje
Dec 17 '18 at 13:16
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According to your specification, $Y=1$ if and only if all four $X_i = 1$. Hence $Y = X_1X_2X_3X_4$.
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– Song
Dec 16 '18 at 20:02
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Thank you for your comment. That's what I did at first but I need to run this in a linear programming model. Which is why I need to impose if-then constraint but I'm not sure how..
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– HFCH
Dec 16 '18 at 20:04
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The knowledge of the whole model, especially the objective function, is usually very helpful when you want to formulate such a constraint. Has y to be minimized or maximized?
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– callculus
Dec 16 '18 at 21:58
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Sorry I forgot to mention that y is to be minimised. This is because there will be a penalty if F(x) > 3 (y = 1).
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– HFCH
Dec 16 '18 at 23:42
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In the complete model, there is a value assigned to each x-variable and I need to choose 5 x-variables out of 9 to maximise the total values of the chosen variables, while bearing in mind that if I choose all X1 , X2, X3, X4 I will incur a small penalty. It doesn't mean that I shouldn't choose all these 4 variables because the total values assigned may still be higher to compensate for the penalty. The objective function is (Max. Total values of the 5 chosen variables - 1.2y)
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– HFCH
Dec 16 '18 at 23:50