Integer Linear Programming
A linear program with the added restriction that the decision variables must have integer variables is called an integer linear program (ILP) or simply an integer program (IP).
One approach to solving integer programs is to ignore or relax the integer restriction and solve the resulting LP. For some types of problems, for example minimum cost network flow problems, there will always be an optimal LP solution that is also an integer-valued solutions. Such a solution must also be the optimal IP solution. This will not work for ``most'' integer programs, however, because the optimal LP solution may contain variables with non-integer values. When this happens, it might be possible to round the values of LP solution up or down to get an integer-valued solution. For the reasons outlined below, this technique does work very often.
The optimal LP solution is 
. Since each variable could be rounded either up or down, there are eight possible ways of rounding the optimal LP solution to get an integer-valued solution.
Unfortunately, all eight ways lead to infeasible solutions.
An LP-Based Branch-and-Bound Algorithm for Integer Programming
Consider the following binary integer program (BIP). A binary variable is one that is constrained to be either 1 or 0.
In this case, we create the LP relaxation by replacing the binary constraints with constraints of the form 
. The LP relaxation and optimal solution (obtained with CPLEX) are shown below.
Maximize obj: 75 x1 + 6 x2 + 3 x3 + 33 x4 Subject To c1: 774 x1 + 76 x2 + 22 x3 + 42 x4 <= 875 c2: 67 x1 + 27 x2 + 794 x3 + 53 x4 <= 875 Bounds 0 <= x1 <= 1 0 <= x2 <= 1 0 <= x3 <= 1 0 <= x4 <= 1 Primal - Optimal: Objective = 1.1383727197e+02 Solution time = 0.00 sec. Iterations = 4 (0) Variable Name Solution Value x1 1.000000 x2 0.506042 x3 0.933674 x4 1.000000
Although solving the LP relaxation did not give us an integer solution, it does give us some information. For example, we know that the best integer solution, if there is one, has an objective function value of at most 113. This is because any integer solution is also a solution to the LP relaxation and therefore cannot have an objective function of more than 113.84. Since all the objective function coefficients are integer-valued, we know that an integer solution could have an objective function of at most 113.
In the optimal LP
relaxation, there are two variables that do not have integer
variables: 
and 
. Since two must be either 0 or 1
in a feasible solution to the original BIP, we will create and solve
two new problems, called subproblems by adding constraints to
the LP relaxation.
For the case where 
, we create subproblem 1 by adding the
constraint and solving the resulting problem with CPLEX.
Maximize obj: 75 x1 + 6 x2 + 3 x3 + 33 x4 Subject To c1: 774 x1 + 76 x2 + 22 x3 + 42 x4 <= 875 c2: 67 x1 + 27 x2 + 794 x3 + 53 x4 <= 875 c3: x2 = 0 Bounds 0 <= x1 <= 1 0 <= x2 <= 1 0 <= x3 <= 1 0 <= x4 <= 1 Primal - Optimal: Objective = 1.1085264484e+02 Variable Name Solution Value x1 1.000000 x3 0.950882 x4 1.000000 All other variables in the range 1-4 are zero.
For the case where 
, we create subproblem 2 by adding the
constraint and solving the resulting problem with CPLEX.
Maximize obj: 75 x1 + 6 x2 + 3 x3 + 33 x4 Subject To c1: 774 x1 + 76 x2 + 22 x3 + 42 x4 <= 875 c2: 67 x1 + 27 x2 + 794 x3 + 53 x4 <= 875 c3: x2 = 1 Bounds 0 <= x1 <= 1 0 <= x2 <= 1 0 <= x3 <= 1 0 <= x4 <= 1 Primal - Optimal: Objective = 1.1315228795e+02 Variable Name Solution Value x1 0.951860 x2 1.000000 x3 0.920939 x4 1.000000
We still haven't found a feasible BIP solution, but we do know that if , the best possible objective function value is 110 and that if , the best possible objective function value is 113.
Going back to subproblem 1, we create subproblem 3 by adding the constraint 
and subproblem 4 by adding the constraint 
.
Here are the results for subproblem 3.
Maximize obj: 75 x1 + 6 x2 + 3 x3 + 33 x4 Subject To c1: 774 x1 + 76 x2 + 22 x3 + 42 x4 <= 875 c2: 67 x1 + 27 x2 + 794 x3 + 53 x4 <= 875 c3: x2 = 0 c4: x3 = 0 Bounds 0 <= x1 <= 1 0 <= x2 <= 1 0 <= x3 <= 1 0 <= x4 <= 1 Primal - Optimal: Objective = 1.0800000000e+02 Variable Name Solution Value x1 1.000000 x4 1.000000 All other variables in the range 1-4 are zero.
We have just found our first feasible BIP solution. Let 
be the objective function value of the best BIP solution, called the incumbent solution, found so far. In this case, 
and the incumbent solution is 
, , and 
.
Here are the results for subproblem 4.
Maximize obj: 75 x1 + 6 x2 + 3 x3 + 33 x4 Subject To c1: 774 x1 + 76 x2 + 22 x3 + 42 x4 <= 875 c2: 67 x1 + 27 x2 + 794 x3 + 53 x4 <= 875 c3: x2 = 0 c4: x3 = 1 Bounds 0 <= x1 <= 1 0 <= x2 <= 1 0 <= x3 <= 1 0 <= x4 <= 1 Primal - Optimal: Objective = 8.6716981132e+01 Variable Name Solution Value x1 1.000000 x3 1.000000 x4 0.264151 All other variables in the range 1-4 are zero.
Notice that the objective function value of subproblem 4 is less than
. This means that we do not need to consider creating new
subproblem by adding constraints to subproblem 4 because any BIP
solutions they could generated would have worse objective function
value than our incumbent solution. Also, there is no point in
creating subproblems by adding constraints to subproblem 3 because
its solution is already integer-valued and adding additional
constraints could make the solution worse. So, we know that of all
the solutions where BIP solutions where (half of the 16
possible solutions), the best possible objective function value is 108.
Now, we go back to subproblem 2 and create subproblems 5 and 6 by
adding the constraint that 
or . Notice that we could
also have created subproblems by constraining to be either 0 or
1.
Here are the results for subproblem 5.
Maximize obj: 75 x1 + 6 x2 + 3 x3 + 33 x4 Subject To c1: 774 x1 + 76 x2 + 22 x3 + 42 x4 <= 875 c2: 67 x1 + 27 x2 + 794 x3 + 53 x4 <= 875 c3: x2 = 1 c4: x1 = 0 Bounds 0 <= x1 <= 1 0 <= x2 <= 1 0 <= x3 <= 1 0 <= x4 <= 1 Primal - Optimal: Objective = 4.2000000000e+01 Variable Name Solution Value x2 1.000000 x3 1.000000 x4 1.000000 All other variables in the range 1-4 are zero.
This gives us an integer-valued solution, but it is not as good as the incumbent solution.
Here are the results for subproblem 6.
Maximize obj: 75 x1 + 6 x2 + 3 x3 + 33 x4 Subject To c1: 774 x1 + 76 x2 + 22 x3 + 42 x4 <= 875 c2: 67 x1 + 27 x2 + 794 x3 + 53 x4 <= 875 c3: x2 = 1 c4: x1 = 1 Bounds 0 <= x1 <= 1 0 <= x2 <= 1 0 <= x3 <= 1 0 <= x4 <= 1 Primal - Optimal: Objective = 1.0064285714e+02 Variable Name Solution Value x1 1.000000 x2 1.000000 x4 0.595238 All other variables in the range 1-4 are zero.
Notice that the objective function value of subproblem 6 is 100.64
which is less than . This means that we do not need to
consider creating new subproblems by adding constraints to subproblem 6
because any BIP solutions they could generated would have worse
objective function values than our incumbent solution. Also, there is
no point in creating subproblems by adding constraints to subproblem 5
because its solution is already integer-valued and adding additional
constraints could make the solution worse. So, we know that none of
the solutions where BIP solutions where have an objective
function value of more than 108. Thus our incumbent solution is an
optimal BIP solution.
The process followed above, know as the branch-and-bound algorithm can be presented graphically as shown below.
Summary of the Generic Branch-and-Bound Technique for Maximization IP's
to 
. Throughout the algorithm, is equal to the objective function value of the best feasible integer solution found so far, the incumbent solution. If no feasible integer solution is known,
.If the optimal solution to the LP relaxation is integer-valued, then it is also an optimal IP solution and we can stop. Otherwise, the LP relaxation becomes the first unfathomed node of the branch-and-bound tree. Let Z, the optimal objective function of the LP relaxation, be the bound on the the first unfathomed node
be the value of 
. Choose a variable such
that is not an integer.
be integer value found by rounding down.
For example, if 
, 
.
to the current subproblem. Add a new node to the branch-and-bound tree that is associated with this subproblem.
to the current subproblem.
Add a new node to the branch-and-bound tree that is associated with this subproblem.
Find the bound for each of the two new nodes of the branch-and-bound tree by solving the associated subproblems (LP's) and letting the Z values (the optimal objective function values) be the bounds.
Let Z be the bound for a new node. The node becomes fathomed (discarded) if any of the following conditions are true.
.
, then this is the new incumbent solution.
If there are no more unfathomed nodes in the branch-and-bound tree and there is an incumbent solution, then the incumbent solution is optimal. If there are no more unfathomed nodes in the branch-and-bound tree and there is no incumbent solution, then the problem has no feasible solution. Otherwise, return to the Branching Step.