Branch-and-Bound Technique for Solving Integer Programs

1
Branch-and-Bound Technique for Solving Integer
Programs

• In this handout
• Basic concepts
• Introducing the technique based on an example

2
Basic Concepts

• The basic concept underlying the branch-and-bound
  technique
• is to divide and conquer.
• Since the original large problem is hard to
  solve directly,
• it is divided into smaller and smaller
  subproblems
• until these subproblems can be conquered.
• The dividing (branching) is done by partitioning
• the entire set of feasible solutions into
  smaller and smaller subsets.
• The conquering (fathoming) is done partially by
• (i) giving a bound for the best solution in the
  subset
• (ii) discarding the subset if the bound
  indicates that
• it cant contain an optimal solution.
• These three basic steps branching, bounding,
  and fathoming
• are illustrated on the following example.

3
Example of Branch-and-Bound
6

• Max Z 5x1 8x2
• s.t. x1 x2 ? 6
• 5x1 9x2 ? 45
• x1 , x2 0 integer

x1 x2 6
5
4
(2.25, 3.75)
3
Z41.25
2
5x1 9x2 45
1
Z20
0
7
8
1
2
3
4
5
6
4
Utilizing the information about the optimal
solution of the LP-relaxation

• Fact If LP-relaxation has integral optimal
  solution x, then x is optimal for IP too.
• In our case, (x1, x2) (2.25, 3.75) is the
  optimal solution of the LP-relaxation. But,
  unfortunately, it is not integral.
• The optimal value is 41.25 .
• Fact OPT(LP-relaxation) OPT(IP)
• (for maximization problems)
• That is, the optimal value of the LP-relaxation
• is an upper bound
• for the optimal value of the integer program.
• In our case, 41.25 is an upper bound for OPT(IP).

5
Branching step

• In an attempt to find out more
• about the location of the IPs optimal
  solution,
• partition the feasible region of the
  LP-relaxation.
• Choose a variable that is fractional in the
  optimal solution to the LP-relaxation say, x2 .
  Observe that every feasible IP point must have
  either x2 ? 3 or x2 4 .
• With this in mind, branch on the variable x2 to
  create the following two subproblems
• Subproblem 1 Subproblem 2
• Max Z 5x1 8x2 Max Z 5x1 8x2
• s.t. x1 x2 ? 6 s.t. x1 x2 ? 6
• 5x1 9x2 ? 45 5x1 9x2 ?
  45
• x2 ? 3 x2 4
• x1 , x2 0 x1 , x2 0
• Solve both subproblems
• (note that the original optimal solution
  (2.25, 3.75) cant recur)

6
Branching step (graphically)
Z41

• Subproblem 1 Opt. solution (3,3) with value
  39
• Subproblem 2 Opt. solution (1.8,4) with
  value 41

5
Subproblem 2
4
(1.8, 4)
3
(3, 3)
2
Subproblem 1
1
Z39
Z20
0
7
8
1
2
3
4
5
6
7
Solution tree
S1 x2 ? 3 (3, 3) Z39

• For each subproblem, we record
• the restriction that creates the subproblem
• the optimal LP solution
• the LP optimum value
• The optimal solution for Subproblem 1 is
  integral (3, 3).
• If further branching on a subproblem will yield
  no useful information, then we can fathom
  (dismiss) the subproblem.
• In our case, we can fathom Subproblem 1
  because its solution is integral.
• The best integer solution found so far is stored
  as incumbent. The value of the incumbent is
  denoted by Z.
• In our case, the first incumbent is (3, 3), and
  Z39.
• Z is a lower bound for OPT(IP) OPT(IP) Z .
• In our case, OPT(IP) 39. The upper bound is
  41 OPT(IP) ? 41.

int.
All (2.25, 3.75) Z41.25
S2 x2 4 (1.8, 4) Z41
8
Next branching step (graphically)

• - Fathom Subproblem 1.
• - Branch Subproblem 2 on x1
• Subproblem 3 New restriction is x1 ? 1.
• Opt. solution (1, 4.44) with value
  40.55
• Subproblem 4 New restriction is x1 2.
• The subproblem is infeasible

5
(1, 4.44)
Subpr. 4
Subpr. 3
4
3
Z40.55
2
1
0
7
8
1
2
3
4
5
6
9
Solution tree (cont.)
S1 x2 ? 3 (3, 3) Z39
int.

• If a subproblem is infeasible, then it is
  fathomed.
• In our case, Subproblem 4 is infeasible fathom
  it.
• The upper bound for OPT(IP) is updated 39 ?
  OPT(IP) ? 40.55 .
• Next branch Subproblem 3 on x2 .
• (Note that the branching variable might recur).

All (2.25, 3.75) Z41.25
S3 x1 ? 1 (1, 4.44) Z40.55
S2 x2 4 (1.8, 4) Z41
S4 x1 2 infeasible
10
Next branching step (graphically)

• Branch Subproblem 3 on x2
• Subproblem 5 New restriction is x2 ? 4.
• Feasible region
• the segment joining (0,4) and
  (1,4)
• Opt. solution (1, 4) with value 37
• Subproblem 6 New restriction is x2 5.
• Feasible region is just one point (0,
  5)
• Opt. solution (0, 5) with value 40

(0, 5)
5
4
(1, 4)
3
2
1
0
7
8
1
2
3
4
5
6
11
Solution tree (final)
S1 x2 ? 3 (3, 3) Z39
int.
S5 x2 ? 4 (1, 4) Z37

• If the optimal value of a subproblem is ? Z,
  then it is fathomed.
• In our case, Subproblem 5 is fathomed because 37
  ? 39 Z.
• If a subproblem has integral optimal solution x,
  
• and its value gt Z, then x replaces the
  current incumbent.
• In our case, Subproblem 5 has integral optimal
  solution, and its value 40gt39Z. Thus,
  (0,5) is the new incumbent, and new Z40.
• If there are no unfathomed subproblems left, then
  the current incumbent is an optimal solution for
  (IP).
• In our case, (0, 5) is an optimal solution with
  optimal value 40.

int.
All (2.25, 3.75) Z41.25
S3 x1 ? 1 (1, 4.44) Z40.55
S2 x2 4 (1.8, 4) Z41
S6 x2 5 (0, 5) Z40
int.
S4 x1 2 infeasible
12
In the next handout

• Summary of branch-and-bound for integer programs
• Which variable to branch on?
• Which open subproblem to solve first?
• Updating the bounds (lower and upper) on OPT(IP)
• Summary of fathoming criteria
• Optimality test
• Branch-and-bound applied to
• binary integer programs
• mixed integer programs