Resource Allocation Problem
$begingroup$
Let $I, J, n in mathbb N$. Furthermore, let $mathbf M in mathbb N^{I times J}$. Finally, for $i in {1, dots, I}$ and $j in {1, dots, J}$, let $M(i,j)$ denote the element in the $i$th row and $j$th column of $mathbf M$. I'd like to solve the following optimization problem:
begin{equation}
begin{aligned}
&max_{mathbf x in mathbb N^I} && sum_{i=1}^Isum_{j = 1}^Jmin{M(i,j), x_i} \
&text{ s.t.} && sum_{i=1}^I x_i = n.
end{aligned}
end{equation}
I suspect that the problem can be solved by a "greedy" algorithm with the following form:
- Set $mathbf x equiv mathbf 0$.
- Pick any $mathbf k in arg max sum_{i=1}^Isum_{j = 1}^Jmin{M(i,j), x_i + delta_k}$. In words, find a component of $mathbf x$ which maximally increases the objective value when incremented by one -- allowing for the possibility that there may be multiple components which give maximal increase.
- Set $mathbf x equiv mathbf x + mathbf e_k$, where $mathbf e_k$ is the $k$th coordinate vector.
- If $sum_{i=1}^I x_i = n$, stop; otherwise, return to step 2.
However, I am not sure how to prove that this greedy algorithm leads to an optimal solution.
Can you help me by either providing a proof or counter-example?
optimization convex-optimization integer-programming dynamic-programming mixed-integer-programming
asked Dec 19 '18 at 7:08
GarrettGarrett
5761314
$endgroup$
add a comment |
$begingroup$
Let $I, J, n in mathbb N$. Furthermore, let $mathbf M in mathbb N^{I times J}$. Finally, for $i in {1, dots, I}$ and $j in {1, dots, J}$, let $M(i,j)$ denote the element in the $i$th row and $j$th column of $mathbf M$. I'd like to solve the following optimization problem:
begin{equation}
begin{aligned}
&max_{mathbf x in mathbb N^I} && sum_{i=1}^Isum_{j = 1}^Jmin{M(i,j), x_i} \
&text{ s.t.} && sum_{i=1}^I x_i = n.
end{aligned}
end{equation}
I suspect that the problem can be solved by a "greedy" algorithm with the following form:
- Set $mathbf x equiv mathbf 0$.
- Pick any $mathbf k in arg max sum_{i=1}^Isum_{j = 1}^Jmin{M(i,j), x_i + delta_k}$. In words, find a component of $mathbf x$ which maximally increases the objective value when incremented by one -- allowing for the possibility that there may be multiple components which give maximal increase.
- Set $mathbf x equiv mathbf x + mathbf e_k$, where $mathbf e_k$ is the $k$th coordinate vector.
- If $sum_{i=1}^I x_i = n$, stop; otherwise, return to step 2.
However, I am not sure how to prove that this greedy algorithm leads to an optimal solution.
Can you help me by either providing a proof or counter-example?
optimization convex-optimization integer-programming dynamic-programming mixed-integer-programming
asked Dec 19 '18 at 7:08
GarrettGarrett
5761314
$endgroup$
add a comment |
$begingroup$
Let $I, J, n in mathbb N$. Furthermore, let $mathbf M in mathbb N^{I times J}$. Finally, for $i in {1, dots, I}$ and $j in {1, dots, J}$, let $M(i,j)$ denote the element in the $i$th row and $j$th column of $mathbf M$. I'd like to solve the following optimization problem:
begin{equation}
begin{aligned}
&max_{mathbf x in mathbb N^I} && sum_{i=1}^Isum_{j = 1}^Jmin{M(i,j), x_i} \
&text{ s.t.} && sum_{i=1}^I x_i = n.
end{aligned}
end{equation}
I suspect that the problem can be solved by a "greedy" algorithm with the following form:
- Set $mathbf x equiv mathbf 0$.
- Pick any $mathbf k in arg max sum_{i=1}^Isum_{j = 1}^Jmin{M(i,j), x_i + delta_k}$. In words, find a component of $mathbf x$ which maximally increases the objective value when incremented by one -- allowing for the possibility that there may be multiple components which give maximal increase.
- Set $mathbf x equiv mathbf x + mathbf e_k$, where $mathbf e_k$ is the $k$th coordinate vector.
- If $sum_{i=1}^I x_i = n$, stop; otherwise, return to step 2.
However, I am not sure how to prove that this greedy algorithm leads to an optimal solution.
Can you help me by either providing a proof or counter-example?
optimization convex-optimization integer-programming dynamic-programming mixed-integer-programming
asked Dec 19 '18 at 7:08
GarrettGarrett
5761314
$endgroup$
Let $I, J, n in mathbb N$. Furthermore, let $mathbf M in mathbb N^{I times J}$. Finally, for $i in {1, dots, I}$ and $j in {1, dots, J}$, let $M(i,j)$ denote the element in the $i$th row and $j$th column of $mathbf M$. I'd like to solve the following optimization problem:
begin{equation}
begin{aligned}
&max_{mathbf x in mathbb N^I} && sum_{i=1}^Isum_{j = 1}^Jmin{M(i,j), x_i} \
&text{ s.t.} && sum_{i=1}^I x_i = n.
end{aligned}
end{equation}
I suspect that the problem can be solved by a "greedy" algorithm with the following form:
- Set $mathbf x equiv mathbf 0$.
- Pick any $mathbf k in arg max sum_{i=1}^Isum_{j = 1}^Jmin{M(i,j), x_i + delta_k}$. In words, find a component of $mathbf x$ which maximally increases the objective value when incremented by one -- allowing for the possibility that there may be multiple components which give maximal increase.
- Set $mathbf x equiv mathbf x + mathbf e_k$, where $mathbf e_k$ is the $k$th coordinate vector.
- If $sum_{i=1}^I x_i = n$, stop; otherwise, return to step 2.
However, I am not sure how to prove that this greedy algorithm leads to an optimal solution.
Can you help me by either providing a proof or counter-example?
optimization convex-optimization integer-programming dynamic-programming mixed-integer-programming
optimization convex-optimization integer-programming dynamic-programming mixed-integer-programming
asked Dec 19 '18 at 7:08
GarrettGarrett
5761314
asked Dec 19 '18 at 7:08
GarrettGarrett
5761314
asked Dec 19 '18 at 7:08
GarrettGarrett
5761314
asked Dec 19 '18 at 7:08
GarrettGarrett
5761314
asked Dec 19 '18 at 7:08
GarrettGarrett
5761314
5761314
add a comment |
add a comment |
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