An algorithm for solving integer linear programming problems

IJRET: International Journal of Research in Engineering and Technology eISSN: 2319-1163 | pISSN: 2321-7308
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Volume: 02 Issue: 07 | Jul-2013, Available @ http://www.ijret.org 107
AN ALGORITHM FOR SOLVING INTEGER LINEAR PROGRAMMING
PROBLEMS
Shinto K.G1
, C.M. Sushama2
1
Assistant Professor, Department of Mathematics, Christ College Irinjalakkuda, Kerala, India,
2
Associate Professor, Department of Mathematics, National Institute of Technology Calicut, Kerala, India,
shintomaths@gmail.com, sushama@nitc.ac.in
Abstract
The paper describes a method to solve an ILP by describing whether an approximated integer solution to the RLP is an optimal
solution to the ILP. If the approximated solution fails to satisfy the optimality condition, then a search will be conducted on the
optimal hyperplane to obtain an optimal integer solution using a modified form of Branch and Bound Algorithm.
Index Terms: ILP, Linear Diophantine equations, Optimal hyperplane, Branch and Bound algorithm
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1. INTRODUCTION
Integer Linear Programming (ILP) problems belong to a
particular class of Linear programming Problems (LPP). In
LPP, the decision variables are assumed to be continuous, but
in ILP the decision variables are restricted so that it can take
only discrete values. Many practical problems can be
formulated as general Integer Programming Problems and so
ILP problems have an outstanding relevance in many fields.
Various researchers have studied about the applications of
Integer Programming Problems. Brenda Dietrich [7] discusses
its application in IBM, Melvin A. Breuer [6] discuss the
application in Design Automation, and Chui-Yen and Chen [5]
discuss the application in Crew Scheduling Problem. Some
other problems which can be formulated as general Integer
Programming Problems include transportation , generalised
assignment and manpower planning problem. So Integer
Programming technique is so useful in these areas. But
because of the discreteness restriction on variables ILP cannot
consider as an LPP. A general ILP can be defined as
Maximize z = cx
Subject to
Ax ≤ b
x ≥ 0 and integers,
Where c is an n row vector of real entries, A is an m × n
matrix with real entries and b is an m column vector of real
entries.
Associated with each ILP, we can define relaxation RLP as
Maximize z = cx
Subject to
Ax ≤ b
x ≥ 0.
By definition, the feasible space of RLP contains the solution
space of ILP and hence its optimal solution is an upper bound
for ILP optimal solution.
The ILP becomes computationally intractable with increase
problem size and/or variable bounds. In any search for an
optimal solution, early knowledge of a good solution is
important for reasonable computational effort. Most ILP
solution methodology starts with RLP optimal in In [1],
different ILP solution methods are given. In addition to these
methods, the hyperplane search methods are explained in
[2],[3] and [4]. But the most commonly used methods are
Gomory’s Cuttingplane Method and Branch and Bound
Method. To apply any of these methods, the optimum solution
of the RLP is required. If the optimum solution to the RLP is
an integral one, then this solution can be considered as the
optimum solution of the ILP. If the solution is non-integral,
further investigation is necessary to obtain the optimum
integer solution.
In this paper, if the optimum solution of the RLP is non
integral, then some integer approximated optimum feasible
solutions of the ILP will be considered. But it cannot be
ascertained that one of these solution is an optimal solution to
the ILP.
Consider the objective hyperplane
cjxj = z
where each cj ∈ Z, which is a linear Diophantine equation in
integers. Let d = gcd⁡{cj, cj ≠ 0; j = 1,2 … … n}. It has an
integer solution if and only if d|z. Also if a linear Diophantine
equation has an integer solution, then there will be infinitely
many solutions for this equation [8].

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This paper aims to provide a test to examine the optimality of
the approximated solution obtained from RLP to be an optimal
solution of the ILP, using the concept of linear Diophantine
equations. If it does not satisfy the optimality condition, then a
modification of Branch and Bound method is suggested to
arrive at the optimal solution of the ILP.
In the next section some results are derived to suggest a
computational procedure to obtain the optimal solution of the
ILP using RLP. In section 3 an algorithm is developed for this
purpose and in section 4 examples are cited to establish the
effectiveness of the algorithm.
2. BASIC RESULTS
Without loss of generality, let the row vector c of the
definition of the ILP has only integer components. If the
components are not integers, then try to make them integers by
multiplying all components cj with a suitable positive integer.
Since the decision variables x are also required to be integers
for an ILP, c1x1 + c2x2 + ⋯ … … … cnxn should be an integer
for all integer values of cj and xj for j = 1,2, … … … . n.
Consider the optimum solution to the RLP say xLP . Let zLP
denotes the optimum value of the objective function. That is
zLP = cxLP . If z∗
denotes the value of the objective function
corresponding to a feasible solution of the ILP. Then z∗
is an
integer less than or equal to zLP .
That is z∗
≤ cxLP .
Since z∗
≤ zLP , let z∗
= zLP , the greatest integer less than or
equal to zLP . Then consider the linear Diophantine equation
c1x1 + c2x2 + ⋯ … … … cnxn = z∗
. Also let d =
gcd c1, c2, … … … … cn /ci ≠ 0 . If d ∤ z∗
, there will not
exist any integer solution to the linear Diophantine equation.
In this case let z∗
= zLP − 1 and the same argument is
applicable for the new z∗
also. Repeat the process until we get
an z∗
so that z∗
is an integer less than or equal to zLP and
d| z∗
.
Proposition1. Let 𝑧𝐼𝑃 denotes the value of the objective
function corresponding to the optimum solution of the ILP.
Then 𝑧𝐼𝑃 ≤ 𝑧∗
where 𝑧∗
is an integer defined as above.
Proof
Let 𝑧 𝐿𝑃 denotes the maximum objective value to the RLP.
Then clearly 𝑧𝐼𝑃 ≤ 𝑧 𝐿𝑃
If possible assume 𝑧𝐼𝑃 > 𝑧∗
But by definition, 𝑧∗
≤ 𝑧 𝐿𝑃.
Then 𝑧∗
< 𝑧𝐼𝑃 ≤ 𝑧 𝐿𝑃
Let 𝒙 𝑰𝑷 denotes the optimum solution of the ILP so that
𝒄𝒙 𝑰𝑷 = 𝑧𝐼𝑃
So the linear Diophantine equation 𝑐1 𝑥1 + 𝑐2 𝑥2 +
⋯ … … … 𝑐 𝑛 𝑥 𝑛 = 𝑧𝐼𝑃 has a solution 𝒙 𝑰𝑷. Hence there will exist
infinitely many integer solutions to this equation and 𝑑 =
𝑔𝑐𝑑 𝑐1, 𝑐2, … … … … 𝑐 𝑛 /𝑐𝑖 ≠ 0 divides 𝑧𝐼𝑃. Thus 𝑧∗
<
𝑧𝐼𝑃 ≤ 𝑧 𝐿𝑃 and 𝑑| 𝑧∗
which is a contradiction to the assumption
of 𝑧∗
and so 𝑧𝐼𝑃 ≤ 𝑧∗
.
Proposition2. If 𝑧∗
< 𝑧 𝐿𝑃, (strict inequality) then there is no
integer solution to the ILP lie between the hyperplanes
𝒄𝒙 = 𝑧 𝐿𝑃 and 𝒄𝒙 = 𝑧∗
.
Proof
Given 𝑧∗
< 𝑧 𝐿𝑃.
If possible assume an integer point (say 𝒙 𝑘 ) is lie between the
given hyperplanes so that 𝒄𝒙 𝒌 < 𝑧 𝐿𝑃 and 𝒄𝒙 𝒌 > 𝑧∗
. Let
𝒄𝒙 𝒌 = 𝑧 𝑘 . Then the linear Diophantine equation 𝑐1 𝑥1 +
𝑐2 𝑥2 + ⋯ … … … + 𝑐 𝑛 𝑥 𝑛 = 𝑧 𝑘 will have infinitely many
integer solutions and 𝑔𝑐𝑑⁡| 𝑧 𝑘 , since 𝒙 𝒌 is an integer solution
of the equation. Thus we have an integer 𝑧 𝑘 so that 𝑧∗
< 𝑧 𝑘 <
𝑧 𝐿𝑃 which is a contradiction and so the assumption is wrong.
Proposition3. Let the optimum solution to the RLP is unique
and non integral. Also let 𝑧 𝐿𝑃 denote the value of z
corresponding to this optimum solution. Thus 𝒙 𝑳𝑷. If the
optimum solution to the ILP exists, then there exist 𝑧 𝑘 < 𝑧 𝐿𝑃
such that the optimum solution to the ILP lies on the
hyperplane 𝒄𝒙 = 𝑧 𝑘.
Proof
Let 𝒙 𝑳𝑷 denote the optimum solution of the RLP. Let 𝑆1
denote the set of feasible solutions to the RLP. Then 𝑆1 ∩
𝒙 ∕ 𝒄𝒙 = 𝑧 𝐿𝑃 is the singleton set 𝒙 𝑳𝑷. It is also given that
𝒙 𝑳𝑷 is non integral. Let 𝑆 denotes the feasible solutions to the
ILP which is non empty.
Now
𝑆 ⊆ 𝑆1; 𝒙 𝑳𝑷 ∈ 𝑆1 and 𝒙 𝑳𝑷 ∉ 𝑆.
Also 𝑆1 ∩ 𝒙 ∕ 𝒄𝒙 = 𝑧 𝐿𝑃 = 𝒙 𝑳𝑷.
So 𝑆 ∩ 𝒙 ∕ 𝒄𝒙 = 𝑧 𝐿𝑃 = 𝜙.
Also 𝑧 𝐿𝑃 gives an upper bound for ILP. So the value of the
objective function for the ILP, 𝑧𝐼𝑃 should be strictly less than
𝑧 𝐿𝑃 and the optimum solution to the ILP will lie on the
hyperplane 𝒄𝒙 = 𝑧𝐼𝑃.
Remark: If 𝑧 𝐿𝑃 is an integer, the solution 𝒙 𝑳𝑷 is non integral
and if there exist alternate optimum solution for the ILP (that
is 𝒙 𝑳𝑷 is not unique), then the optimum solution to the ILP
can lie on the optimum hyperplane of the RLP. That is the
optimum solution to the ILP can lie on the hyperplane
𝒄𝒙 = 𝑧 𝐿𝑃. An example in the next section illustrates this fact.
Theorem1. Let 𝑧∗
be an integer defined as in section 2. Let 𝑆
denote the set of all feasible solutions to the ILP. If 𝑆 ∩
𝒙 ∕ 𝒄𝒙 = z∗
is non empty, then the optimum solution to the
ILP will lie on the hyperplane 𝒄𝒙 = z∗
.
Proof

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Let 𝑧 𝐿𝑃 denote the value of the objective function
corresponding to the optimal solution of the RLP. Then the
hyperplane 𝒄𝒙 = zLP is the optimum hyperplane for the RLP.
Also by definition of 𝑧∗
, 𝑧∗
≤ 𝑧 𝐿𝑃.
Assume z∗
= zLP .
It is also given that S ∩ x ∕ cx = z∗
≠ ϕ . So S ∩
x ∕ cx = zLP ≠ ϕ. Hence there exist an integer point in the
optimum hyperplane of the RLP which is a feasible solution of
the ILP. Also the objective function value of RLP for its
optimal solution is an upper bound for the ILP optimal
solution. So the integral point must be optimal.
Now assume z∗
< zLP . Then by proposition2, there is no
feasible solution to the ILP lying between the hyperplanes
cx = zLP and cx = z∗
. So no hyperplane containing integer
solution to the ILP exists between these two hyperplanes.
Now since S ∩ x ∕ cx = z∗
≠ ϕ , it is possible to find a
feasible integer point on the hyperplane which gives value z∗
for the objective function. From proposition1 the optimum
solution to the ILP can give objective function value less than
or equal to z∗
. So clearly the integer point on the hyperplane
cx = z∗
is the optimum solution to the ILP.
Thus in both cases, the optimum solution to the ILP lie on the
hyperplane cx = z∗
.
3. SOLUTION METHODOLOGY
Using the results derived from the last section, it is possible to
check whether an integer approximated solution to the RLP
gives the optimum solution for the ILP. If it fails to give the
optimum solution, a modified form of Branch and Bound
algorithm is suggested to obtain the optimum solution. For the
development of the algorithm, it is necessary to explain the
procedure to obtain
1. Integer approximated solution to the RLP sub
problems
2. An objective hyperplane for the ILP.
3.1 Integer Approximated Solutions to an RLP Sub
problem
Given RLP is an LPP. To obtain the integer approximated
solution, first of all solve this LPP and find the solution. Let
the solution as x i
and the maximum objective value as zi. If
x i
satisfies the integer requirement, then consider x∗
i
= x i
as the integer approximated solution with objective value
zi
∗
= zi, while if x i
is infeasible, there is no integer
approximated solution. So consider x i
as feasible and non
integral, consider the integer approximated solution as
follows,
For each component xj of x i
, consider xj and xj and then
all combinations of these components to obtain the integer
approximated solutions. (That is there are two choices x1
and x1 for x1 and two choices x2 and x2 for x2 and so on.
If some particular component xk is an integer, then there is
only one choice for this component. So there will be atmost 2n
approximated solutions). Then consider the feasibility of each
of these points. If a particular point is feasible, then put this
point in the set Yi, consider the set Yi
∗
= Yi Yi ∩ Y where Y
is the set of integer approximated solutions. Initially Y = ∅. If
Yi
∗
= ∅, then set Zi
∗
= −∞ and x∗
i
= ∅. Otherwise find
Zi
∗
= max cx/x ∈ Yi
∗
and x∗
i
= x ∈ Yi
∗
/cx = Zi
∗
. Then set
Y ≔ Y ∪ Yi
∗
.
Then x∗
i
is the integer approximated solution to the ith sub
problem with the objective function valueZi
∗
.
3.2 Search for the Optimal Objective value and the
Optimal Hyperplane
Let xLP be the optimum solution of the RLP with objective
value zLP (That is cxLP = zLP ). Also let d denote the gcd of
the cost coefficients. Let zIP denote the maximum value of the
objective function for the ILP. Then zIP will be an integer such
that zIP ≤ zLP and d|zIP .hence the integer z can assume so
that z = zLP . If d ∤ z then reduce the value zby unity and
check whether it is divisible by d. If not reduce the value of z
by unity again and so on. Hence the maximum possible value
of z for the ILP is given by z and cx = z is a hyperplane
having infinitely many integer points on it. If xLP is unique
then z < zLP . If any of the integer feasible point of the ILP lie
on this hyperplane, then cx = z should be the optimal
hyperplane and so zIP = z. If none of the feasible integer oint
lie on the hyperplane, then we have to reduce the value z to
obtain the optimal hyperplane and zIP . In the algorithm cx = z
will be considered as the optimal hyperplane and on which an
integer point is searched. Hence this procedure explains how
to obtain the maximum possible value z when zLP and gcd of
the cost coefficients of the LPP are known.
3.3 Modification to Branch and Bound Method
Now it is possible to suggest a modification to the Branch and
Bound algorithm. The problem to be solved is an all integer
linear programming problem PIP with maximization objective.
Its linear programming relaxation is called PLP . It is assumed
that a specific node and branch selection strategies have been
chosen. The different strategies are given in [1]. The algorithm
is initialized with node n1. Take i = 1 (That is we are
considering the first LPP), and Y = ∅ (there are no integer
approximated solutions are known). Obtain optimal solution
and integer approximated solution of the RLP at node 1 (Use
the procedure in section 3.1 with i = 1). Let x 1
denote its
optimum solution with objective function value z1. Also let

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x∗
1
with objective value z1
∗
is the integer approximated
solution. If x 1
has only integer components, then it is the
optimal solution of the ILP, procedure stops here (then the
optimum solution of the ILP will be xIP = x 1
with zIP = z1).
Now if x 1
is not integral, then set z∗
= z1
∗
and x∗
= x∗
1
.
Also let d = gcd{ c1, c2 … … … … cn /ci ≠ 0}. Then obtain
the maximum possible objective function value z using z1and
gcd d (using the procedure in section 3.2). now if z = z∗
, then
the integer approximated solution x∗
1
is the optimum solution
of the ILP with objective value z∗
and cx = z∗
is the optimal
hyperplane and procedure terminates. If z∗
< z, then by
setting S = L = n1 , zLP = z1 and xLP = x1 proceed with the
Branch and Bound algorithm.
3.4 Algorithm
Step 1: Is 𝐿 = ∅?
If YES: Is 𝑧∗
> −∞, then the solution 𝒙 𝑰𝑷 = 𝒙∗
with objective
value 𝑧𝐼𝑃 = 𝑧∗
is optimal and 𝒄𝒙 = 𝑧∗
is the optimal
hyperplane.
Otherwise the integer programming problem has no feasible
solution.
If NO: Go to step 2.
Step 2: Choose some node 𝑛𝑖 ∈ 𝐿 (according to some
prescribed criterion) with solution 𝒙 𝒊
and objective value 𝑧𝑖.
Step 3: Branch on some variable 𝑥𝑗 (according to some
prescribed criterion) to the nodes 𝑛𝑡+1 with 𝑥𝑗 ≤ 𝑥𝑗 and 𝑛𝑡+2
with 𝑥𝑗 ≥ 𝑥𝑗 . Set 𝐿 ≔ 𝐿 𝑛𝑖 and 𝑆 ≔ 𝑆 ∪ 𝑛𝑡+1, 𝑛𝑡+2
𝑛𝑖 . Solve the LPP at 𝑛𝑡+1 and 𝑛𝑡+2. The results are solutions
𝒙 𝒕+𝟏
and 𝒙 𝒕+𝟐
with objective values 𝑧𝑡+1 and 𝑧𝑡+2 and the
improved integer approximated solutions (if it exist) 𝒙∗
𝒕+𝟏
and 𝒙∗
𝒕+𝟐
with objective values 𝑧𝑡+1
∗
and 𝑧𝑡+2
∗
respectively.
Step 4: Is 𝒙 𝒕+𝟏
feasible?
If YES: Go to step 5.
Step 5: Is 𝑧𝑡+1
∗
= 𝑧?
If YES: Set 𝒙∗
= 𝒙∗
𝒕+𝟏
; 𝑧∗
= 𝑧𝑡+1
∗
; 𝐿 = ∅ and go to
step 1.
Step 6: Does 𝒙 𝒕+𝟏
satisfy the integrality requirement?
If YES: Go to step 7
If NO: Is 𝑧𝑡+1 > 𝑧∗
?
If YES: Set 𝐿 ≔ 𝐿 ∪ 𝑛𝑡+1 and go to step
8.
If NO: Go to step 8
Step 7: Is 𝑧𝑡+1 > 𝑧∗
?
If YES: Set 𝒙∗
= 𝒙 𝒕+𝟏
, 𝑧∗
= 𝑧𝑡+1 and go to step 8.
Step 8: Is 𝑡 = 𝑎𝑛 𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟?
If YES: Set 𝑡 = 𝑡 + 1 and go to step 4.
If NO: Consider 𝑧 𝐿𝑃
∗
= max 𝑧𝑖/𝑛𝑖 ∈ 𝐿 and go to
step 9.
Step 9: Is 𝑧 𝐿𝑃
∗
= 𝑧 𝐿𝑃?
If YES: Set 𝑡 = 𝑡 + 1 and go to step 1.
If NO: Set 𝑧 𝐿𝑃 = 𝑧 𝐿𝑃
∗
, obtain the new objective
function value 𝑧 for this 𝑧 𝐿𝑃, go to step 10.
Step 10: Is 𝑧∗
= 𝑧?
If YES: Set 𝐿 = ∅ and go to step 1.
If NO: Set 𝑡 = 𝑡 + 1, go to step 1.
The key of the algorithm is the updating procedure of the sets
S and L. In each step when the procedure branches from some
node ni to two nodes nt+1 and nt+2, the set S of end nodes is
updated to include the new nodes nt+1 and nt+2 and to
exclude ni. From the set of live nodes L, the node from which
the branching takes place is deleted in step 3 and the node
nt+1 or nt+2 is added to the set only if the solution is feasible
(step 4) but not integral with objective value is greater than the
known integer solution (step 6). In addition to these, the
algorithm will be terminated by emptying the set L if the
integer approximated solution lies in the optimal hyperplane
of the ILP (steps 5 to 8). Algorithm is illustrated using the
following examples.
4. EXAMPLES
4.1 Example in which Optimum Solution of the ILP
lie on the Optimal Hyperplane if the RLP
Maximize 𝑧 = 𝑥1 + 𝑥2
Subject to
𝑥1 + 𝑥2 ≤ 7
2𝑥1 ≤ 11
2𝑥2 ≤ 7
𝑥1, 𝑥2 ≥ 0 and integers.
The optimum solution to the RLP using simplex method is
given by x1 =
11
2
, x2 =
3
2
and maximum z = 7. This is a non
integral solution. But alternate solutions exist for this. The
integer approximations of this RLP solutions are
5,1 , 5,2 , 6,1 , 6,2 . Here the set of feasible integer
approximations Y1
∗
= 5,1 , 5,2 . So the integer
approximated solution is given by x∗
= 5,2 with z∗
= 7.
Also the maximum possible objective function value z = 7.
Therefore z∗
= z. So the integer approximated solution is the
optimal solution of the ILP and hence no need to consider the
Branch and Bound algorithm. Also this solution is a point on
the optimal hyperplane of the RLP.

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Hence in this case the optimal solution of the ILP lies on the
optimal hyperplane of the RLP x1 + x2 = 7.
4.2 Example in which gcd of Cost Coefficients Unity
Maximize 𝑧 = 𝑥1 + 2𝑥2
Subject to
𝑥1 + 𝑥2 ≤ 7
2𝑥1 ≤ 11
2𝑥2 ≤ 7
𝑥1, 𝑥2 ≥ 0 and integers.
The unique solution to the RLP at node 1 is given by x 1
=
7
2
,
7
2
with maximum z =
21
2
. Clearly it is not an integer
solution. The integer approximations of this RLP solutions are
{ 4,4 , 4,3 , 3,4 , 3,3 }. Here the set of feasible integer
approximations Y1
∗
= 4,3 , 3,3 . So the integer
approximated solution is given by x∗
= 4,3 with z∗
= 10.
Also the maximum possible objective function value z = 10.
So the integer approximated solution is the optimal solution of
the ILP.
Therefore the optimum solution is given by x1 = 4, x2 = 3
with maximum z = 10 and so the optimum hyperplane is
x1 + 2x2 = 10.
4.3 Example in which gcd of Cost Coefficients
Greater than Unity
Maximize z = 200x1 + 400x2 + 300x3
Subject to
30x1 + 40x2 + 20x3 ≤ 600
20x1 + 10x2 + 20x3 ≤ 400
10x1 + 30x2 + 20x3 ≤ 800
x1, x2, x3 ≥ 0 and integers.
The unique optimum solution of the RLP at node 1 is given by
x 1
= 0,
20
3
,
50
3
with z =
23000
3
which is clearly non integral.
The integer approximations are
0,6,16 , 0,7,16 , 0,6,17 , 0,7,17 . The set of feasible
integer approximations Y1
∗
= 0,6,16 , 0,7,16 , 0,6,17 .
Therefore the integer approximated solution is given by
x∗
= 0,7,16 with z = 7600.
Now d = gcd 200,400,300 = 100.
Hence the optimal objective value using RLP at node 1 is
given by z =
2300
3
= 7666.
But 100 ∤ 7666.
Hence reduce the value to 7600 so that 100 | 7600. So finally
z = 7600.
Here also z = z∗
. Hence the integer approximated solution at
node 1 is the optimal solution to the ILP and hence no need to
consider the Branch and Bound Algorithm.
Therefore the optimum solution is given by x1 = 0, x2 =
7, x3 = 16 with maximum z = 7600 and so the optimum
hyperplane of the problem is 200x1 + 400x2 + 300x3 =
7600.
4.4 Example in which gcd of Cost Coefficients is
Unity and Branch and Bound Algorithm Required to
Obtain Solution
Maximize z = 3x1 + 4x2
Subject to
2x1 + 4x2 ≤ 13
−2x1 + x2 ≤ 2
2x1 + 2x2 ≥ 1
6x1 − 4x2 ≤ 15
x1, x2 ≥ 0 and integers.
The unique optimum of the RLP at node 1 is given by
x 1
=
7
2
,
3
2
with maximum z =
33
2
, which is clearly non
integral. The integer approximations are given by
4,1 , 4,2 , 3,1 , 3,2 . But the set of feasible solutions
Y1
∗
= 3,1 . Therefore x∗
= x∗
1
= 3,1 with z∗
= z1
∗
= 13.
Now d = gcd 3,4 = 1.
Hence the optimal objective value using RLP at node 1 is
given by z =
33
2
= 16 and z∗
< z. So we have to proceed
with Branch and Bound Algorithm.
Now by selecting node n1 and branching variable x1, we can
obtain subproblems at nodes n2 and n3 and on solving,
x 2
= 3,
7
4
with z2 = 16 and x 3
is infeasible. Also
x∗
2
= ∅ with z2
∗
= −∞. Therefore the new z = 15. Now by
selecting node n2 and branching variable x2 we can obtain
subproblems at nodes n4 and n5 and on solving x 4
= 3,1
=
5
2
, 2 with z5 =
31
2
. So z4
∗
= 13 with
x∗
4
= 3,1 and z5
∗
= 14 with x∗
5
= 2,2 . Therefore the
new z = 15, z∗
= 14 and x∗
= 2,2 . Finally by selecting
node n5 and branching variable x1, we can obtain
subproblems at nodes n6 and n7 and on solving x 6
= 2,
9
4
is infeasible. Also z6
∗
= −∞ with
x∗
6
= ∅ and the new value z = 14. Hence z = z∗
and so the
integer approximated solution in hand is the optimal solution
of the ILP.
Therefore the optimum solution is given by x1 = 2, x2 = 2
with maximum z = 14 and so the optimal hyperplane of the
problem is 3x1 + 4x2 = 14

__________________________________________________________________________________________
CONCLUSIONS
In this paper, a method to solve pure integer linear
programming problems using integer approximated solution
method is discussed. This method is very effective for
problems whose gcd of cost coefficients is greater than unity.
It can also be used to check whether a feasible point is an
optimal solution of the ILP. Here the optimal hyperplane is
identified using the idea of the linear Diophantine equations.
To obtain integer approximated solution the procedure
requires the solution of the RLP at each iteration. But any
method which can give the integer approximations may reduce
the computational efforts. This is a task worth consideration of
future research and if it is possible, then this will be more
effective in integer programming.
REFERENCES:
[1] H. A. Eiselt and C. L. Sandblom, “Integer
Programming and Network Models”; Springer (2009).
[2] L. Cooper, “Hyperplane search algorithm for the
solution of integer programming problems”, IEEE
Transactions on systems, Man, and Cybernetics 3/3
(1973) 234-240.
[3] A. Joseph, S. I. Gass and N. Bryson, “An objective
hyperplane search procedure for solving the general all-
integer linear programming (ILP) problem”, European
Journal of Operational Research 104 (1998) 601-614.
[4] A. Joseph, S.I. Gass and N.A. Bryson, “A
computational study of an objective hyperplane search
heuristic for the general integer linear programming
problem, Mathl. Comput. Modelling Vol. 25, No.
10,(1997) 63-76.
[5] Chui-Yen and Chen, “Applicaton of integer
programming for solving crew scheduling problem in
Taipei Rapid Transist Corporation”, IEEE International
Conference on Systems, Man and Cybernetics, (2008)
2548 – 2553.
[6] M.A. Breuer, “The application of integer programming
in design automation”, Annual ACM IEEE Design
Automation Conference,( 1966) 2.1-2.9.
[7] Brenda Dietrich, “Some of my favorite integer
programming applications at IBM”, Annals of
Operations Research 149,(2007) 75-80.
[8] Alexander Schrijver,“Theory of Linear and Integer
Programming”, John Wiley & Sons Ltd (1986).
[9] Richard Bronson and Govindasami Naadimuthu,
“Schaums Outline of Theory and Problems of
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Laurence A. Wolsey,“Cutting planes for integer
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An algorithm for solving integer linear programming problems

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