1. Branch-and-Bound Algorithm for Integer Program
Solution set is non-convex. Difficult to solve.
Idea : divide and conquer.
Divide feasible solutions set into many disjoint sets. Find the best solution for each subset. Then choose the best one. To divide the feasible solutions set, we add linear inequalities to the problem. (e.g. x1  1, and x1  0 )
But the divided problem is still integer programming problem. So we need a means to find the best integer solution for each integer programming problem.
Linear programming relaxation :
max 𝑐 ′ 𝑥 max 𝑐 ′ 𝑥
(IP) 𝐴𝑥≤𝑏 (LPR) 𝐴𝑥≤𝑏
𝑥≥0 and integer 𝑥≥0
(𝑥∈ 𝐵 𝑛 , or 0≤𝑥≤1 and integer) (0≤𝑥≤1) OR

2. If 𝑥 ∗ is a feasible solution to (IP)  𝑥 ∗ feasible solution to (LPR)
Let z be the optimal value of (IP) and zLP be the optimal value of (LPR)
If 𝑥 ∗ is a feasible solution to (IP)  𝑥 ∗ feasible solution to (LPR)
Hence {set of feasible solutions to (IP) } ⊆ { set of feasible solutions to (LPR) } and objective functions are same.
If 𝑥 ∗ is an optimal solution to (IP) (i.e. 𝑧=𝑐′ 𝑥 ∗ ), 𝑥 ∗ may not be an optimal solution to (LPR), but it is a feasible solution to (LPR). Hence
 𝑧≤ 𝑧 𝐿𝑃 ( 𝑧 𝐿𝑃 provides an upper bound on 𝑧)
Also a feasible solution to (IP) provides a lower bound on 𝑧.
( If 𝑥 0 is a feasible solution to (IP), then 𝑐′ 𝑥 0 ≤𝑧)
Hence if we solve (LPR) and obtain optimal solution which happens to be integral, then the solution provides a lower bound which is the same as the upper bound.
It implies that the solution is optimal to (IP).
If the obtained solution is not integral, we only know an upper bound on optimal value to (IP), but don’t know an optimal solution. Then we divide the feasible solution set and repeat the procedure again for each divided problem. OR

3. Results of solving LP relaxation.
unbounded  integer program unbounded
infeasible  integer program infeasible
optimal solution which is integer  it is optimal to integer program
optimal solution not integer  only obtain upper bound Need to branch
How to divide the solution set in case of 4.
Suppose 𝑥 ∗ is an optimal solution to LP relaxation and
𝑥 𝑗 ∗ =𝑘+𝛼, 𝑘≥0 integer, 0<𝛼<1 fractional.
Then consider 2 sets
(a) 𝐴𝑥≤𝑏, 𝑥≥0, 𝑥 𝑗 ≤𝑘, 𝑥 integer
(b) 𝐴𝑥≤𝑏, 𝑥≥0, 𝑥 𝑗 ≥𝑘+1, 𝑥 integer
Any feasible solution to integer program is contained in one of (a), (b). So we do not miss any feasible solution. Then we solve LP relaxation of (a), (b) again. (Search procedure with tree structure). Note that the optimal solution 𝑥 ∗ to the initial LP relaxation is not a feasible solution to the LP relaxation of (a) and (b). LP relaxations of (a) and (b) can be solved using (dual) simplex method for general LP and reoptimization OR

4. Example max 17 𝑥 1 +12 𝑥 2 s.t. 10 𝑥 1 +7 𝑥 2 ≤2 𝑥 1 + 𝑥 2 ≤5
𝑥 𝑥 2 ≤5
𝑥 1 , 𝑥 2 ≥0
𝑥 1 , 𝑥 2 integer OR

5. 5 4 3 2 1 1 2 3 4 OR

6. OR

7. 5 4 3 2 1 1 2 3 4 OR

8. 5 4 3 2 1 1 2 3 4 OR

9. Useful tool to facilitate the search:
Suppose 𝑥 ∗ is the best solution currently known and its objective value is 𝑧 ∗ .
If the LP relaxation of a subproblem gives optimal value 𝑧′ such that
𝑧′≤ 𝑧 ∗ , we know that there does not exist an integer solution in this subproblem which gives objective value greater than 𝑧 ∗ .
Hence we do not need to explore the subproblem any further and prune the subproblem (node).
Obtaining a good (small) upper bound in each subproblem is important to reduce the number of subproblems examined. Modern approach adds more valid inequalities, (which are satisfied by feasible integer points but not satisfied by the current optimal solution to the LP relaxation), to tighten the feasible region (cutting plane approach). Note that we can use reoptimization techniques after adding a cutting plane to the problem. Recall adding a constraint in Chap. 10 Sensitivity analysis.
Effectiveness of B-and-B alg. largely depends on our ability to prune many nodes in the B-and-B tree in early stages of the B-and-B algorithm, hence cutting plane approach improves the performance of the alg. on many cases, and it is employed in modern commercial S/W. OR