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-1

$begingroup$

I have the following problem:

Given $p =[p_1;p_2;cdots;p_n]^Tin R^n$, $t =[t_1;t_2;cdots;t_n]^Tin R^n$, $q in R$, $l^{[1]}, l^{[2]}, ..., l^{[M]}$, where $l^{[k]} = [l_1;l_2;cdots;l_n]^Tin R^n, forall_{k in {1, dots, n} } $, I have to find b where $b =[b_1;b_2;cdots;b_M]^Tin R^M$, and $b_k in {0,1}$ and $x^{[1]}, x^{[2]}, ..., x^{[M]}$, where $x^{[k]} in R^n, forall_{k in {1, dots, n} } $

$$text{maximize}, sum_{k=1}^{M} b_{k}(x^{[k]T}p - t_{k}) \ text{subject to}, sum_{k=1}^{N} b_{k} x^{[k]} leq q,\ x^{[k]} leq l^{[k]}, b in {0,1} $$

In other words, I have $M$ vectors $l^{[k]} in R^n, k in {1, dots, M} $ and I have to create $Z$ vectors $x^{[k]}$, $Z leq M$ with respect to those 2 constraints formulated above. Each vector has a "penalty" constant ($t_{k}$) and I have to adjust their elements so that $ sum_k (x^{[k]T}p - t_k)$ is maximum. As you can see, in order to formulate the idea of finding the best Z vectors, $Z leq M$, and have a LP problem, I artificially created binary vector $b$. So, $b$ would have $M-Z$ values of $0$ and $Z$ values of $1$. However, this binary vector complicates my problem, which is not LP anymore and I do not know how to solve.

So, I have two questions:

1. Is there any way to better formulate the problem of choosing the best Z vectors that can maximize that sum? Can I get rid of $b$? Should I get rid of $p$ somehow? Should I merge p and $t_k$ so that $x^{[k]} = [x_1;x_2;cdots;x_n; -1] in R^{n+1}$ and $p_k = [p_1;p_2;cdots;p_n; t_k] in R^{n+1}$?




2. How should I approach solving this problem? For all $Z leq M$, find all the posssible combinations of $Z$ vectors, and solving the LP problem (maximize the sum where $sum x^{[k]} < q$ and $x^{[k]} leq l^{[k]}$) for each of these problems would take an enormous ammount of time. I am interested in any software that can solve in reasonable time my problem.

optimization linear-programming

share|cite|improve this question

asked Mar 13 at 23:43

S'en douS'en dou

11

$endgroup$

add a comment | 

-1

$begingroup$

I have the following problem:

Given $p =[p_1;p_2;cdots;p_n]^Tin R^n$, $t =[t_1;t_2;cdots;t_n]^Tin R^n$, $q in R$, $l^{[1]}, l^{[2]}, ..., l^{[M]}$, where $l^{[k]} = [l_1;l_2;cdots;l_n]^Tin R^n, forall_{k in {1, dots, n} } $, I have to find b where $b =[b_1;b_2;cdots;b_M]^Tin R^M$, and $b_k in {0,1}$ and $x^{[1]}, x^{[2]}, ..., x^{[M]}$, where $x^{[k]} in R^n, forall_{k in {1, dots, n} } $

$$text{maximize}, sum_{k=1}^{M} b_{k}(x^{[k]T}p - t_{k}) \ text{subject to}, sum_{k=1}^{N} b_{k} x^{[k]} leq q,\ x^{[k]} leq l^{[k]}, b in {0,1} $$

In other words, I have $M$ vectors $l^{[k]} in R^n, k in {1, dots, M} $ and I have to create $Z$ vectors $x^{[k]}$, $Z leq M$ with respect to those 2 constraints formulated above. Each vector has a "penalty" constant ($t_{k}$) and I have to adjust their elements so that $ sum_k (x^{[k]T}p - t_k)$ is maximum. As you can see, in order to formulate the idea of finding the best Z vectors, $Z leq M$, and have a LP problem, I artificially created binary vector $b$. So, $b$ would have $M-Z$ values of $0$ and $Z$ values of $1$. However, this binary vector complicates my problem, which is not LP anymore and I do not know how to solve.

So, I have two questions:

1. Is there any way to better formulate the problem of choosing the best Z vectors that can maximize that sum? Can I get rid of $b$? Should I get rid of $p$ somehow? Should I merge p and $t_k$ so that $x^{[k]} = [x_1;x_2;cdots;x_n; -1] in R^{n+1}$ and $p_k = [p_1;p_2;cdots;p_n; t_k] in R^{n+1}$?




2. How should I approach solving this problem? For all $Z leq M$, find all the posssible combinations of $Z$ vectors, and solving the LP problem (maximize the sum where $sum x^{[k]} < q$ and $x^{[k]} leq l^{[k]}$) for each of these problems would take an enormous ammount of time. I am interested in any software that can solve in reasonable time my problem.

optimization linear-programming

share|cite|improve this question

asked Mar 13 at 23:43

S'en douS'en dou

11

$endgroup$

add a comment | 

$begingroup$

I have the following problem:

Given $p =[p_1;p_2;cdots;p_n]^Tin R^n$, $t =[t_1;t_2;cdots;t_n]^Tin R^n$, $q in R$, $l^{[1]}, l^{[2]}, ..., l^{[M]}$, where $l^{[k]} = [l_1;l_2;cdots;l_n]^Tin R^n, forall_{k in {1, dots, n} } $, I have to find b where $b =[b_1;b_2;cdots;b_M]^Tin R^M$, and $b_k in {0,1}$ and $x^{[1]}, x^{[2]}, ..., x^{[M]}$, where $x^{[k]} in R^n, forall_{k in {1, dots, n} } $

$$text{maximize}, sum_{k=1}^{M} b_{k}(x^{[k]T}p - t_{k}) \ text{subject to}, sum_{k=1}^{N} b_{k} x^{[k]} leq q,\ x^{[k]} leq l^{[k]}, b in {0,1} $$

In other words, I have $M$ vectors $l^{[k]} in R^n, k in {1, dots, M} $ and I have to create $Z$ vectors $x^{[k]}$, $Z leq M$ with respect to those 2 constraints formulated above. Each vector has a "penalty" constant ($t_{k}$) and I have to adjust their elements so that $ sum_k (x^{[k]T}p - t_k)$ is maximum. As you can see, in order to formulate the idea of finding the best Z vectors, $Z leq M$, and have a LP problem, I artificially created binary vector $b$. So, $b$ would have $M-Z$ values of $0$ and $Z$ values of $1$. However, this binary vector complicates my problem, which is not LP anymore and I do not know how to solve.

So, I have two questions:

1. Is there any way to better formulate the problem of choosing the best Z vectors that can maximize that sum? Can I get rid of $b$? Should I get rid of $p$ somehow? Should I merge p and $t_k$ so that $x^{[k]} = [x_1;x_2;cdots;x_n; -1] in R^{n+1}$ and $p_k = [p_1;p_2;cdots;p_n; t_k] in R^{n+1}$?




2. How should I approach solving this problem? For all $Z leq M$, find all the posssible combinations of $Z$ vectors, and solving the LP problem (maximize the sum where $sum x^{[k]} < q$ and $x^{[k]} leq l^{[k]}$) for each of these problems would take an enormous ammount of time. I am interested in any software that can solve in reasonable time my problem.

optimization linear-programming

share|cite|improve this question

asked Mar 13 at 23:43

S'en douS'en dou

11

$endgroup$

share|cite|improve this question

asked Mar 13 at 23:43

S'en douS'en dou

11

share|cite|improve this question

asked Mar 13 at 23:43

S'en douS'en dou

11

asked Mar 13 at 23:43

S'en douS'en dou

11

asked Mar 13 at 23:43

S'en douS'en dou

11