Homework 6 - Implicit Enumeration - Integer Programming | ORI 391Q, Assignments of Operational Research

Material Type: Assignment; Professor: Bard; Class: 4-INTEGER PROGRAMMING; Subject: Operations Rsch; University: University of Texas - Austin; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/30/2009

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Integer Programming HW# 6
ORI 391Q
Implicit Enumeration
1. Solve the following 0-1 ILP using the additive (Balas) algorithm:
min z = 5x1 + 6x2 + 7x3 + 8x4 + 9x5
subject to 3x1 x2 + x3 + x4 2x5≥ 2
x1 + 3x2 x3 2x4 + x5≥ 0
x1 x2 + 3x3 + x4 + x5≥ 1
xj binary, j = 1,...,5
2. Find an upper bound for the integer variables in the following ILP and then convert it to a 0-1
problem. Solve the resultant 0-1 ILP using the additive algorithm.
max z = 3x1 + 4x2 + 2x3 + x4 + 2x5
subject to 2x1 x2 + x3 + x4 + x5≤ 3
x1 + 3x2 + x3 x4 2x5≤ 2
2x1 + x2 x3 + x4 + 3x5≤ 1
xj ≥ 0, j = 1,...,5
xj binary, j = 1,...,3
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Integer Programming HW# 6 ORI 391Q Implicit Enumeration

  1. Solve the following 0-1 ILP using the additive (Balas) algorithm: min z = 5 x 1 + 6 x 2 + 7 x 3 + 8 x 4 + 9 x 5 subject to 3 x 1 – x 2 + x 3 + x 4 – 2 x 5 ≥ 2 x 1 + 3 x 2 – x 3 – 2 x 4 + x 5 ≥ 0
  • x 1 – x 2 + 3 x 3 + x 4 + x 5 ≥ 1 x j binary, j = 1,...,
  1. Find an upper bound for the integer variables in the following ILP and then convert it to a 0- problem. Solve the resultant 0-1 ILP using the additive algorithm. max z = 3 x 1 + 4 x 2 + 2 x 3 + x 4 + 2 x 5 subject to 2 x 1 – x 2 + x 3 + x 4 + x 5 ≤ 3 - x 1 + 3 x 2 + x 3 – x 4 – 2 x 5 ≤ 2 2 x 1 + x 2 – x 3 + x 4 + 3 x 5 ≤ 1 xj ≥ 0, j = 1,..., xj binary, j = 1,..., 1
  1. For the 0-1 ILP max z = cx subject to Axb xB n where cj ≤ 0, j = 1,..., n , assume that the additive algorithm is being used to find a solution. At the k th iteration we have: max zk = cj x^ j j ∈ Fk

∑ +^ cj

j∈ Sk^ +

subject to aij x (^) j j ∈ Fk

∑ ≤^ bi −^ aij

j∈ Sk^ +

∑ = si , i = 1,..., m

xj  {0,1}, j  Fk where Fk is the set of free variables and (^) S k

  • (^) is the set of variables assigned a value of 1. Now let Qk = { i : si < 0} (set of infeasible rows) Rk = { j : jFk and aij < 0 for some iQk } (set of free variables that can potentially remove an infeasibility) Show that the condition: x (^) j ≥ 1 j∈ Rk

is satisfied by any feasible completion of Wk in the additive algorithm. 2