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Wednesday, May 28, 2025
Integer Linear Programming (ILP) Solution Using Branch and Bound Method
Problem Statement
We need to solve the following Integer Linear Programming (ILP) problem:
Maximize:
Z=3x1+4x2Subject to:
7x1+16x2≤52
3x1−2x2≤9
x1,x2≥0
x1,x2 must be integers.
Solution Using Branch and Bound Method
Step 1: Solve the Linear Relaxation (Ignore Integer Constraints)
First, we solve the problem as a standard Linear Program (LP) without integer constraints.
Graphical Approach
Constraint 1: 7x1+16x2=52
When x1=0, x2=3.25
When x2=0, x1≈7.43
Constraint 2: 3x1−2x2=9
When x1=0, x2=−4.5 (Not feasible since x2≥0)
When x2=0, x1=3
Feasible Corner Points:
(0,0) → Z=0
(0,3.25) → Z=13
Intersection of Constraints:
Solving:
7x1+16x2=523x1−2x2=9Solution: x1=4, x2=1.5 → Z=18
(3,0) → Z=9
Optimal LP Solution:
(x1,x2)=(4,1.5),Z=18Since x2=1.5 is not integer, we proceed with Branch and Bound.
Step 2: Branching on x2
We branch on the non-integer variable x2=1.5, creating two subproblems:
Subproblem 1: x2≤1
Subproblem 2: x2≥2
Subproblem 1 (x2≤1)
New Constraints:
7x1+16x2≤52
3x1−2x2≤9
x2≤1
Solution:
Intersection of 3x1−2x2=9 and x2=1:
3x1−2(1)=9⟹x1=311≈3.666Optimal LP Solution: (3.666,1), Z=15.666
Since x1 is still non-integer, we branch further on x1:
Branch x1≤3
New Solution: (3,1), Z=13
Feasible Integer Solution!
Current Best: Z=13
Branch x1≥4
New Solution: (4,0.75) → Z=15
Still non-integer, but Z=15>13.
Further branching needed, but we prune since it cannot exceed Z=18.
Subproblem 2 (x2≥2)
New Constraints:
7x1+16x2≤52
3x1−2x2≤9
x2≥2
Solution:
Intersection of 7x1+16x2=52 and x2=2:
7x1+16(2)=52⟹x1=720≈2.857Optimal LP Solution: (2.857,2), Z=16.571
Since x1 is non-integer, we branch further on x1:
Branch x1≤2
New Solution: (2,2.375), Z=15.5
Still non-integer, but Z=15.5>13.
Further branching possible, but we prune since it cannot exceed Z=18.
Branch x1≥3
New Solution:
(3,1.9375), Z=15.75
Still non-integer, but Z=15.75>13.
Prune since it cannot exceed Z=18.
Step 3: Compare All Feasible Integer Solutions
From the branching steps, the feasible integer solutions found are:
(3,1) → Z=13
(2,2) → Z=14 (Check feasibility: 7(2)+16(2)=46≤52, 3(2)−2(2)=2≤9)
Best Integer Solution:
(x1,x2)=(2,2),Z=14Final Answer
The optimal integer solution is:
(x1,x2)=(2,2) with Z=14