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CS149 Homework 5: Combinatorial Optimization - Simplex Algorithm and Log Hauling Schedule, Assignments of Computer Science

The fifth homework assignment for the cs149: introduction to combinatorial optimization course. It includes three optimization problems to be solved using the simplex algorithm and one log hauling schedule problem to be formulated and solved as a linear program. The first part of the assignment deals with the simplex algorithm, where students are asked to perform the algorithm twice using different tie-breaking rules and observe the results. The second part of the assignment does not rely on simplex or duality proofs and is intended to motivate these ideas. The third part is the log hauling schedule problem, where students are required to formulate the problem as a linear program and solve it using the 2-phase simplex algorithm.

Typology: Assignments

2009/2010

Uploaded on 02/25/2010

koofers-user-1ns
koofers-user-1ns 🇺🇸

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Download CS149 Homework 5: Combinatorial Optimization - Simplex Algorithm and Log Hauling Schedule and more Assignments Computer Science in PDF only on Docsity! CS149 Introduction to Combinatorial Optimization Homework 5 Due: 4:00 pm, Wed, Oct. 21st Problem 1 Here we will see an example where Bland’s rule is needed to avoid cycles. We are given the tableau: −3/4 20 −1/2 6 0 0 0 3 1/4 −8 −1 9 1 0 0 0 1/2 −12 −1/2 3 0 1 0 0 0 0 1 0 0 0 1 1 Perform the Simplex algorithm twice, each time using a different tie-breaking rule: a) Always select the non-basic variable with the most negative cj to enter the basis. In case of a tie, select the basic variable with the smallest subscript to leave the basis. Notice that we arrive on a cycle. b) Apply Bland’s anti-cycling tie-breaking rule. Notice that we manage to avoid the cycle. Problem 2 For this problem, you may not use any proofs relating to Simplex or Duality (Strong or Weak). This problem is intended to motivate these ideas and so should not depend on them. It is not intended to be difficult, and a short 2-3 sentence answer will suffice for each part. a) min 2x1 +3x2 −5x3 +x4 s.t. 2x1 +2x2 −6x3 + 1 2 x4 ≥ 5 x ≥ 0. Show that 5 is a lower bound for our objective function.
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