Integer Linear Programming

INTEGER PROGRAMMING – AN INTRODUCTION
• An integer programming problem is a mathematical optimization or
feasibility program in which some or all of the variables are restricted
to be integers.
• All the LPP assumptions are inherited except divisibility.
• In 1958, Ralph E. Gomory transformed the field of integer
programming when he published a short paper that described his
cutting-plane algorithm for pure integer programs and announced
that the method could be refined to give a finite algorithm for integer
programming.
• Applications: 1. Knapsack Problem
2. Production Planning
3. Scheduling
4. Telecommunication Networks
5. Cellular Networks

TYPES OF INTEGER PROGRAMMING PROBLEMS
• Pure – Integer Problems
- Require that all decision variables have integer solutions.
• Mixed – Integer Problems
- Require some, but not all, of the decision variables to have
integer values in the final solution, whereas others need not
have integer values.
• 0-1 Integer Problems
- Require integer variables to have value of 0 or 1, such as
situations in which decision variables are of the yes – no type.

METHODS FOR SOLVING ILP PROBLEMS
1. Rounding – off A non-integer solution (Enumeration Method)
• Developed by: A. M. Geoffrion
2. Cutting – Plane Method
• Developed by: Ralph E. Gomory
3. Branch and Bound Method
• Developed by: A. H. Land and A. G. Doing
4. The Additive algorithm for Zero-One integer programming problem
(Balas’ Method)
• Developed by: E. Balas

Maximize Profit = 7X1 + 6X2
Subject to:
2X1 + 3X2 <= 12
6X1 + 5X2 <= 30
X1, X2 >= 0 and Integer
1. ENUMERATION METHOD

• We found that optimal solution to LPP relaxation i.e. X1 =
3.75 and X2 = 1.5 do not have integer values.
• Rounding the values X1 = 4 and X2 = 2.
• However, this is not in feasible solution.
• Rounding X2 down to 1 gives a feasible solution, but it may
not be optimal.
• This could be solved using the enumeration method but it
is generally not possible for large problems.
1. ENUMERATION METHOD - Continue

• The Rounding solution of X1 = 4 and X2
= 1 gives a profit of 34
• Optimal solution to integer
programming problem is X1 = 5 and X2
= 0 with Maximum Z = 35
• The optimal integer solution is less
than the optimal LP solution.
• An integer solution can never be
better than the LP solution.
X1 X2 Z = 7X1 + 6X2
0 0
1
2
3
4
0
6
12
18
24
1 0
1
2
3
7
13
19
25
2 0
1
2
14
20
26
3 0
1
2
21
27
33
4 0
1
28
34
5 0 35*
1. ENUMERATION METHOD - Continue

• An Algorithm for solving pure and mixed integer programming problems
• Developed by Ralph E. Gomory
• Steps to solve:
1. Relax the integer requirements.
2. Solve the resulting LP problem using Simplex Method
3. If all the basic variables have integer values, Optimality of the Integer programming problem
is reached. So, go to step 7; otherwise go to step 4.
4. Examine the constraints corresponding to the current optimal solution. For each basic
variable with non – integer solution in the current optimal table , find the fraction part fi,
Therefore bi = [bi] + fi where [bi] is integer part and fi is fraction part of bi
5. Choose the largest fraction among various fi; i.e. Max (fi). Treat the constraint corresponding
to the maximum fraction as the source row. Based on the source row, develop an additional
constraint (Gomory’s constraint/ fractional cut) as shown:
-Summation[positive fraction for non-basic variables] + Si = -fi
6. Add the fractional cut as the last row in the latest optimal table and proceed further using
dual simplex method, and find the new optimum solution. If the new optimum solution is integer
then go to step 7; otherwise go to step 4.
7. Print the integer solution and optimal value
2. CUTTING – PLANE METHOD

Example: Max Z = 5X1 + 8X2
Subject to:
X1 + 2X2 <= 8
4X1 + X2 <= 10
X1, X2 >= 0 and Integers
Solution: Introducing slack variables
Max Z = 5X1 + 8X2
Subject to:
X1 + 2X2 + S1 = 8
4X1 + X2 + S2 = 10
2. CUTTING – PLANE METHOD - CONTINUE

Cj 5 8 0 0
CB XB X1 X2 S1 S2 RHS Ratio
0 S1 1 2 1 0 8 4
0 S2 4 1 0 1 10 10
Cj - Zj 5 8 0 0 0
Initial table
Iteration 1
Cj 5 8 0 0
CB XB X1 X2 S1 S2 RHS Ratio
8 X2 0.5 1 0.5 0 4 8
0 S2 3.5 0 -0.5 1 6 1.714286
Cj - Zj 1 0 -4 0 32

Iteration 2
Cj 5 8 0 0
CB XB X1 X2 S1 S2 RHS
8 X2 0 1 0.571429 -0.14286 3.142857
5 X1 1 0 -0.14286 0.285714 1.714286
Cj - Zj 0 0 -3.85714 -0.28571 33.71429
Optimal solution to relaxation problem is achieved as Cj – Zj <=0 for all j but both the variables have non –
integer values so,
XB bi [bi] fi
X2 3.142857 3 0.142857
X1 1.714286 1 0.714286
Maximum fraction is 0.714286 for X1
the we will make Gomory’s
fractional cut constraint
corresponding to second row
(1-0.14286)S1 + 0.285714S2 >= 0.714286
-0.85714S1 - 0.285714S2 <= -0.714286
-0.85714S1 - 0.285714S2 + S3 = -0.714286
Adding the constraint into the problem and then

Cj 5 8 0 0
CB XB X1 X2 S1 S2 S3 RHS
8 X2 0 1 0.571429 -0.14286 0 3.142857
5 X1 1 0 -0.14286 0.285714 0 1.714286
0 S3 0 0 -0.85714 -0.28571 1 -0.71429
Cj - Zj 0 0 -3.85714 -0.28571 0 33.7143
(Cj - Zj)/yrj {yrj<0} - - 4.5 1 - 47.2
Cj 5 8 0 0
CB XB X1 X2 S1 S2 S3 RHS
8 X2 0 1 1 0 -0.5 3.5
5 X1 1 0 -1 0 1 1
0 S2 0 0 3 1 -3.5 2.5
Cj - Zj 0 0 -3 0 -1 33
Iteration 3
Iteration 3

Now maximum fraction is 0.5 for X2 so we will use first row to make fractional cut constraint then new
constraint is -0.5S3 + S4 = -0.5
Cj 5 8 0 0 0 0
CB XB X1 X2 S1 S2 S3 S4 RHS
8 X2 0 1 1 0 -0.5 0 3.5
5 X1 1 0 -1 0 1 0 1
0 S2 0 0 3 1 -3.5 0 2.5
0 S4 0 0 0 0 -0.5 1 -0.5
Cj - Zj 0 0 -3 0 -1 0 33
(Cj - Zj)/yrj {yrj<0} 2
Cj 5 8 0 0 0 0
CB XB X1 X2 S1 S2 S3 S4 RHS
8 X2 0 1 1 0 0 -1 4
5 X1 1 0 -1 0 0 2 0
0 S2 0 0 3 1 0 -7 6
0 S4 0 0 0 0 1 -2 1
Cj - Zj 0 0 -3 0 0 -2 32
Iteration 5
Iteration 4

Optimal solution to integer programming is
achieved as
X1 = 0 and X2 = 6 and
Max Z = 32

• Creates and solves a sequence of sub – problems to the original problem that are
increasingly more restrictive until an optimal solution is found.
BRANCHING
• Selection of an integer value of a decision variable to examine for a possible integer
solution to a problem.
• If the solution to the LPP contains non – integer values for some or all decision
variables, then the solution space is reduced by introducing constraints with respect
to anyone of those decision variables. If the value of the decision variable X1 is 3.5,
then two more problems will be created by using each of the following constraints.
X1 <= 3 and X1>=4.
BOUND
• An upper or lower limit on the value of the objective function at a given stage of the
analysis of an integer programming problem.
• Lower Bound: Value of the objective function corresponding to the integer parts of
the decision variables in a node.
• Upper Bound: Value of the objective function corresponding to the LPP solution in
that node.
3. BRANCH-AND-BOUND METHOD

• FATHOMED NODE: A sub-problem is said to be fathomed if any of
the following three conditions is true:
1. The values of the decision variables of the problem are integer.
2. The upper bound of the problem which has non-integer values
for its decision variables is not greater than the current best
lower bound.
3. The problem has infeasible solution.
This means that further branching is not necessary.
• CURRENT BEST LOWER BOUND: This is the best lower bound among
the lower bounds of all the fathomed nodes. Initially, it is assumed
as infinity for the root node.
3. BRANCH-AND-BOUND METHOD (continue)

Branch and bound algorithm applied to maximization problem
1. Solve the given linear programming problem. Set, the current best lower bound ZB as ∞
2. Check, whether the problem has integer solution. If yes, print the current solution as the
optimal solution and stop; Otherwise go to step 3.
3. Identify the variable xk which has the maximum fractional part as the branching variable.
(In case of tie, select the variable which has the highest objective function coefficient.)
4. Create two more problems by including each of the following constraints to the current
problem and solve them.
a) Xk <= [Xk]
b) Xk >= [Xk] + 1
5. If any one of the new sub – problems has infeasible solution or fully integer values for the
decision variables, the corresponding node is fathomed. If a new node has integer values
for the decision variables, update the current best lower bound as the lower bound of
that node if its lower bound is greater than the previous current best lower bound.
6. Are all terminal nodes fathomed? If yes, go to step – 7; otherwise, identify the node with
the highest lower bound and go to step – 3.
7. Select the solution of the problem with respect to the fathomed node whose lower
bound is equal to the current best lower bound as the optimal solution.

Example: Max Z = 10X1 + 20X2
Subject to:
6X1 + 8X2 <= 48
X1 + 3X2 <= 12
Max Z = 96 at X1 = 4.8 and X2 = 2.4
Zupper = 96
Zlower = 80
Maximum fraction is 0.8 for X1
Sub-Problem 1: X1 <= 4
Sub-Problem 2: X1 >= 5
Extreme Points Z = 10X1 + 20X2
A (0, 4) 80
B(4.8, 2.4) 96
C(8, 0) 80

Zupper = 95 at X1 = 5 and X2 = 2.25
X2 is not integer then
Zlower = 80
A (5, 0) 50
B(5, 2.25) 95
C(8, 0) 80

Zupper = 93.333 at X1 = 4 and X2 = 2.667
Zlower = 80
Zlower is same for both problems but
Zupper is highest in sub-problem 1 So,
Sub-Problem 3: X2 <= 2 and
A (0, 4) 80
B(4, 2.667) 93.333
C(4, 0) 40

No feasible solution exists. So, sub
– problem 4 is infeasible problem
and this node is fathomed
Now, we will make branches of
Sub – problem 2.
Sub-Problem 5: X2 <= 2 and

Zupper = 93.333 at X1 = 5.333 and X2
= 2
Zlower = 90
A (5, 0) 50
B(5, 2) 90
C(5.333, 2) 93.333
D(8,0) 80

Zupper = 60<80 at X1 = 4 and X2 = 2
Zlower = 60 So, it is fathomed.
A (0, 2) 20
B(4, 2) 60
C(4, 0) 40

Zupper = Zlower = 90 at X1 = 4 and X2 = 2
Both variables have integer values.
So, it is feasible solution. And this
node is terminated.
Sub – Problem 3 will be sub divided.
A (0, 4) 60
B(3, 3) 90
C(0, 3) 60

Max Z = 10X1 + 20X2
Subject to:
6X1 + 8X2 <= 48
X1 + 3X2 <= 12
X1, X2 >= 0
Max Z = 95
X1 = 4.8,
X2 = 2.4
Zlower = 80
Max Z = 10X1 + 20X2
S.t. 6X1 + 8X2 <= 48
X1 + 3X2 <= 12
X1<=4
X1, X2 >= 0
Max Z = 10X1 + 20X2
S.t. 6X1 + 8X2 <= 48
X1 + 3X2 <= 12
X1>=5
X1, X2 >= 0
X1>=5 X1<=4
Sub – Problem 1 Sub – Problem 2
Zupper = 95
X1 = 5,
X2 = 2.25
Zlower = 80
Zupper = 93.333
X1 = 4 and
X2 = 2.667
Zlower = 80
Max Z = 10X1 + 20X2
S.t. 6X1 + 8X2 <= 48
X1 + 3X2 <= 12
X1>=5
X2 >=3
X1, X2 >= 0
Max Z = 10X1 + 20X2
S.t. 6X1 + 8X2 <= 48
X1 + 3X2 <= 12
X1>=5
X2<=2
X1, X2 >= 0
X2>=3
Max Z = 10X1 + 20X2
S.t. 6X1 + 8X2 <= 48
X1 + 3X2 <= 12
X1<=4
X2>=3
X1, X2 >= 0
Max Z = 10X1 + 20X2
S.t. 6X1 + 8X2 <= 48
X1 + 3X2 <= 12
X1<=4
X2<=2
X1, X2 >= 0
X2<=2 X2<=2
X2>=3
Zupper = 93.333
X1 = 5.333
X2 = 2
Zlower = 90
Infeasible Problem
Fathomed
Zupper =
90
X1 = 3
X2 = 3
Zlower =
90
Zupper =
60
X1 = 4
X2 = 2
Zlower =
60
Feasible
Solution
Feasible
Solution
It will be terminate
at X1 = 5 and X2 = 2
with Z = 90

Integer Linear Programming

More Related Content

What's hot (20)

Similar to Integer Linear Programming (20)

Recently uploaded (20)

Integer Linear Programming