0
$\begingroup$

In the book of Papadimitriou. Combinatorial optimization, there is the instance of Linear Programming (LP) s.t. \begin{equation} \begin{split} &{\rm minimize:\;} c_1x_1 + c_2x_2 + c_3x_3,\\ &{\rm subect\; to:\; }x_1 + x_2 + x_3 = 2 \end{split} \end{equation} The authors argue that the minimum is found at the corners of the 3d triangle (see figure) $v_1,v_2,v_3$.enter image description here

I have a couple of questions:

  1. If the $c_i$'s are all positive, then the minimum of the cost function is simply 2ドルc_k$, where 0ドル\leq c_k\leq c_j \leq c_i$, right?

  2. If for instance two $c_i$'s are negative, then one has to pick the two $x_i$'s that multiply them, s.t. $x_i+x_j = 2$, right? In this case, it cannot be that only $v_1,$v_2$ or $v_3$ are the only solutions to the LP program.

asked Oct 26 at 22:01
$\endgroup$

1 Answer 1

1
$\begingroup$

Presumably, the additional assumption is that all $x_j$ are non-negative, right?

  1. Yes. The minimum is always 2ドル \min_j c_j$. This is always true, even if some $c_i$'s are negative.
  2. If two $c_i$'s are negative, one of them is still smaller and you can pick the corresponding $x_i$ to be equal to 2ドル$ (and all others to 0ドル$).
answered Oct 26 at 22:17
$\endgroup$

Your Answer

Draft saved
Draft discarded

Sign up or log in

Sign up using Google
Sign up using Email and Password

Post as a guest

Required, but never shown

Post as a guest

Required, but never shown

By clicking "Post Your Answer", you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.