0/1 fractional programming study notes
Due to simulate race out of this question and this Konjac fractional plan ignorant, so learn from their mistakes to learn scores planning to introduce the relevant knowledge of planning and 0/1 fractional share my shallow understanding the hearts of scores of planning, if wrong or wrong, please correct me in the comments section chiefs from various quarters.
(This is what I wrote most serious mathematical notes)
1,0 / 1 programming model scores
0/1 fractional programming model means that, given a series of integer \ (A_1 \) , \ (A_2 \) ...... \ (A_N \) and \ (B_1 \) , \ (B_2 \) .. .... \ (B_n \) , find a set of solutions \ (x_i \) ( \ (. 1 \ Leq I \ n-Leq \) , \ (x_i = 0 |. 1 \) ), so that the following equation is maximized.
\ [\ Frac {\ sum_ {
i = 1} ^ {n} {a_i} \ cdot {x_i}} {\ sum_ {i = 1} ^ {n} {b_i} \ cdot {x_i}} \] There a misunderstanding(I existed misunderstanding)Is 0/1 Fractional Programming \ (x_i \) values are constant, that is, for the above equation, there is no molecule \ (x_i \) is 0 and the denominator \ (x_i \) case was 1 . But must at the same time get 0 or 1 at the same time too.
Therefore, we will find that this model can be interpreted this way:
A given length equal to the number of columns of two (i.e. pairs of integers \ (a_i \) , \ (B_i \) ), is selected from a plurality of such pairs of the selected number \ (A \) sum and \ (B \) the sum of the largest providers.
This model is very important ! ! !
2 inference for 0/1 fractional programming model
We consider addressing the 0/1 fractional programming problem when, in fact, looking for a set of solutions { \ (x_1, x_2, x_3, \ cdots x_n \) }, \ (x_i \ in {0, 1} \) , so that the above formula is established.
We guess a random value \ (A \) , such that the following holds:
\ [\ sum_. 1 = {I}} ^ {n-({a_i} -A \ B_i CDOT {}) \ {x_i CDOT} \ geq0 \]
If such a group Solutions
So we have the formula obtained by deforming:
\ [(\ sum_ {I =. 1} ^ {n-} {a_i} \ CDOT {x_i}) - A \ CDOT (\ sum_ {I =. 1} ^ {n-} {B_i} \ cdot {x_i}) \ geq0
\] That is:
\ [\ EXISTS \ {{x_1}, {x_2}, \ cdots {x_n} \} \ Rightarrow \ FRAC {\ sum_ {I =. 1} ^ {n-} { a_i} \ cdot {x_i}}
{\ sum_ {i = 1} ^ {n} {b_i} \ cdot {x_i}} \ geq a \] that is, out of the set value \ (a \) than we seek the maximum small
If we present a set of solutions that any of the following holds
\ [\ sum_ {i = 1
} ^ {n} ({a_i} -A \ cdot {b_i}) \ cdot {x_i} <0 \] So we also can formula obtained by deforming:
\ [(\ sum_ {I =. 1} ^ {n-} {a_i} \ CDOT {x_i}) - A \ CDOT (\ sum_ {I =. 1} ^ {n-} {B_i} \ CDOT { x_i}) <0 \]
Similarly, we obtain:
\ [\ FORALL \ {} {x_1, x_2 {}, \ cdots {x_n} \} \ Rightarrow \ FRAC {\ sum_. 1 = {I}} ^ {n- {a_i} \ cdot {x_i}
} {\ sum_ {i = 1} ^ {n} {b_i} \ cdot {x_i}} <a \] that is, out of the set value \ (a \) than we seek the maximum larger.
Therefore, the above discussion, we find that the maximum we ask, in fact you can use half the answer to achieve, in fact, it is indeed the most common is the 0/1 fraction of planning, the most common implementation.
Achieve 3,0 / 1 Points Plan
Just as we have said, the most common 0/1 Points plan, the most common implementation is half the answer, think of the dichotomy algorithm, we were able to come up with the essence and difficult-half of the answer lies written judgment function check (), then directed to binary fractional programming answer to 0/1, we can find by just derived, we only need to determine "whether there is a set of solutions satisfying \ (\ sum_ {i = 1 } ^ {n} ({a_i} -A \ {} B_i CDOT) \ CDOT x_i} {\ geq0 \) ", and should thus give answers to expand upwardly expanding segment or section downwardly answer.
So, we put into the original problem: given \ (A_1 \) , \ (A_2 \) ...... \ (A_N \) , \ (B_1 \) , \ (B_2 \) ..... . \ (B_n \) and \ (a \) , find a set of solutions: \ (x_1, x_2, X_3, \ cdots x_n \) , so \ (\ sum_ {i = 1 } ^ {n} ({a_i} -A \ cdot {b_i}) \ cdot {x_i} \) is maximized.
Just by derivation, we find that, in this respect, we need only determine whether the maximum value can be non-negative, so we have come half ideas:
In the real domain (possible range of solution) mid-half, we take every mid actually just a problem in \ (A \) , that is to say, we need to calculate the maximum value of this formula to check whether non-negative, if non-negative, then the right range, or to the left.
Solving completed.