How to prove that Fractional Knapsack exhibits Greedy Strategy.?

Question

How can i prove that Fractional Knapsack exhibits Greedy Strategy,

I can do it practically but i am not able to find a method to prove it theoretically.?

Please help Thanks in Advance

Answer 1

We need to show that this problem has the greedy choice property. To do this, we need to show that any solution X which does not include the greedy choice a does not have get a worse solution after swapping some choice with a.

For fractional knapsack, this is very easy to show: we take any element of X, say b. If w_a >= w'_b (where w_a is the weight of a, and w'_b is the weight b has in the solution X), we can replace b with as large a fraction of a as possible. Because a is the item with the largest value-density (this is our greedy choice), this will not make the solution worse. If w_a < w'_b we can take all of a, and make w'_b = w'_b - w_a. Again, because a has the largest value-density, this does not make the solution worse.

That's it! We technically also need to show optimal substructure, but that should be fairly straightforward for this problem.

Answer 2

The (primal) fractional knapsack LP is

maximize sum_{i=1}^n v_i x_i
subject to
y: sum_{i=1}^n w_i x_i <= W
z_i: x_i <= 1  (for i=1 to n)
x_i >= 0  (for i=1 to n),

where v_i is the value of item i, and w_i is the weight. The dual LP is

minimize W y + sum_{i=1}^n z_i
subject to
x_i: w_i y + z_i >= v_i
y >= 0
z_i >= 0  (for i=1 to n).

By (weak) LP duality, if the greedy solution to the primal has the same objective as a solution to the dual, then both are optimal. Assume that all weights are positive, that their sum is greater than W, and that the item are ordered so that v_1/w_1 >= v_2/w_2 >= ... >= v_n/w_n. Let j be the pivot item, so that the greedy primal solution is

x_1, x_2, ..., x_{j-1} = 1
x_j = (W - sum_{i=1}^{j-1} w_i) / w_j
x_{j+1}, x_{j+2}, ..., x_n = 0.

By complementary slackness, we can guess that, in the dual, we should have

z_j, z_{j+1}, ..., z_n = 0.

The x_i constraint in the dual is equivalent to

y + z_i/w_i >= v_i/w_i,

so we need to set

y >= v_j/w_j >= v_{j+1}/w_{j+1} >= ... >= v_n/w_n

in order to satisfy the constraints where we have zeroed z_i. On a hunch, we set

y = v_j/w_j,

which intuitively forces the assignments

z_i = (v_i/w_i - v_j/w_j) w_i  (for i=1 to j-1).

Now comes the only part of this argument that needs to be rigorous: verifying that this is a feasible solution to the dual (tedious and thus left as an exercise) and that the objective matches the greedy primal solution. The objective is

W y + sum_{i=1}^{j-1} z_i =
W (v_j/w_j) + sum_{i=1}^{j-1} (v_i/w_i - v_j/w_j) w_i =
sum_{i=1}^{j-1} (v_i/w_i) w_i + (v_j/w_j) (W - sum_{i=1}^{j-1} w_i) =
sum_{i=1}^{j-1} v_i + ((W - sum_{i=1}^{j-1} w_i) / w_j) v_j,

which is indeed the primal objective.