Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Solving Integer Optimization Problems
Mustafa Ç. Pınar
Bilkent University
IE 303, Chapter 2: Lecture 1
October 2021
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation
The Branch-and-Bound Method
Example 2
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Introduction
I In IE202, you have studied linear optimization problems and
the simplex method for solving the linear optimization
problems.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Introduction
I In IE202, you have studied linear optimization problems and
the simplex method for solving the linear optimization
problems.
I Unfortunately, the simplex method cannot always solve integer
optimization problems!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Introduction
I In IE202, you have studied linear optimization problems and
the simplex method for solving the linear optimization
problems.
I Unfortunately, the simplex method cannot always solve integer
optimization problems!
I Let us look at the following example:
z IP = max z = 3x + 4y
subject to
5x + 8y ≤ 24
x, y ≥ 0, integer.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Introduction
I In IE202, you have studied linear optimization problems and
the simplex method for solving the linear optimization
problems.
I Unfortunately, the simplex method cannot always solve integer
optimization problems!
I Let us look at the following example:
z IP = max z = 3x + 4y
subject to
5x + 8y ≤ 24
x, y ≥ 0, integer.
I We can solve the LP relaxation (ignoring the integer
requirement) graphically to get the solution x = 24/5, y = 0
with z = 14.4.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Introduction
I In IE202, you have studied linear optimization problems and
the simplex method for solving the linear optimization
problems.
I Unfortunately, the simplex method cannot always solve integer
optimization problems!
I Let us look at the following example:
z IP = max z = 3x + 4y
subject to
5x + 8y ≤ 24
x, y ≥ 0, integer.
I We can solve the LP relaxation (ignoring the integer
requirement) graphically to get the solution x = 24/5, y = 0
with z = 14.4.
I This is obviously notM. something
Ç. Pınar
IE303
Models
Integer Variables
weSolving
want!
x with
is fractional.
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Example LP
Figure: LP relaxation of example.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation
I In general, ignoring the integer value requirement gives us a
linear optimization problem called the LP Relaxation.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation
I In general, ignoring the integer value requirement gives us a
linear optimization problem called the LP Relaxation.
I We have obtained a fractional solution by solving the LP
relaxation!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation
I In general, ignoring the integer value requirement gives us a
linear optimization problem called the LP Relaxation.
I We have obtained a fractional solution by solving the LP
relaxation!
I Can we round it to the nearest integer? (5, 0) is not feasible!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation
I In general, ignoring the integer value requirement gives us a
linear optimization problem called the LP Relaxation.
I We have obtained a fractional solution by solving the LP
relaxation!
I Can we round it to the nearest integer? (5, 0) is not feasible!
I Can we round it down? (4, 0) is feasible with value z = 12. Is
it optimal, i.e. is it the best integer feasible point?
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation
I In general, ignoring the integer value requirement gives us a
linear optimization problem called the LP Relaxation.
I We have obtained a fractional solution by solving the LP
relaxation!
I Can we round it to the nearest integer? (5, 0) is not feasible!
I Can we round it down? (4, 0) is feasible with value z = 12. Is
it optimal, i.e. is it the best integer feasible point?
I The answer is no as we can see by trial and error that (3, 1) is
a better integer feasible point with value z = 13.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation
I In general, ignoring the integer value requirement gives us a
linear optimization problem called the LP Relaxation.
I We have obtained a fractional solution by solving the LP
relaxation!
I Can we round it to the nearest integer? (5, 0) is not feasible!
I Can we round it down? (4, 0) is feasible with value z = 12. Is
it optimal, i.e. is it the best integer feasible point?
I The answer is no as we can see by trial and error that (3, 1) is
a better integer feasible point with value z = 13.
I In fact (3, 1) is the optimal integer solution as we shall prove
later.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation
I In general, ignoring the integer value requirement gives us a
linear optimization problem called the LP Relaxation.
I We have obtained a fractional solution by solving the LP
relaxation!
I Can we round it to the nearest integer? (5, 0) is not feasible!
I Can we round it down? (4, 0) is feasible with value z = 12. Is
it optimal, i.e. is it the best integer feasible point?
I The answer is no as we can see by trial and error that (3, 1) is
a better integer feasible point with value z = 13.
I In fact (3, 1) is the optimal integer solution as we shall prove
later.
I But there is no way to guess (3, 1) from (24/5, 0).
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation: is it useless?
I The answer is absolutely no! It is the most valuable tool. If
the answer is integer we stop!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation: is it useless?
I The answer is absolutely no! It is the most valuable tool. If
the answer is integer we stop!
I Else, the optimal value of the LP relaxation gives us an upper
bound on the integer optimal value z IP .
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation: is it useless?
I The answer is absolutely no! It is the most valuable tool. If
the answer is integer we stop!
I Else, the optimal value of the LP relaxation gives us an upper
bound on the integer optimal value z IP .
I We have obtained 14.4 as the optimal value of the LP
relaxation.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation: is it useless?
I The answer is absolutely no! It is the most valuable tool. If
the answer is integer we stop!
I Else, the optimal value of the LP relaxation gives us an upper
bound on the integer optimal value z IP .
I We have obtained 14.4 as the optimal value of the LP
relaxation.
I This means that in the entire LP feasible region, the best
integer feasible solution cannot have a value bigger than 14.4!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation: is it useless?
I The answer is absolutely no! It is the most valuable tool. If
the answer is integer we stop!
I Else, the optimal value of the LP relaxation gives us an upper
bound on the integer optimal value z IP .
I We have obtained 14.4 as the optimal value of the LP
relaxation.
I This means that in the entire LP feasible region, the best
integer feasible solution cannot have a value bigger than 14.4!
I Since objective function coefficients are integer numbers, we
can conclude that the optimal integer solution cannot have a
value bigger than 14!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation: is it useless?
I The answer is absolutely no! It is the most valuable tool. If
the answer is integer we stop!
I Else, the optimal value of the LP relaxation gives us an upper
bound on the integer optimal value z IP .
I We have obtained 14.4 as the optimal value of the LP
relaxation.
I This means that in the entire LP feasible region, the best
integer feasible solution cannot have a value bigger than 14.4!
I Since objective function coefficients are integer numbers, we
can conclude that the optimal integer solution cannot have a
value bigger than 14!
I Hence, we know that z IP ≤ 14. This is our first upper bound!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Introduction
I We shall now see a method that will give us the integer
optimal solution. Better than trying all feasible integer points!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Introduction
I We shall now see a method that will give us the integer
optimal solution. Better than trying all feasible integer points!
I The method proceeds by dividing recursively the LP relaxation
feasible region into sub-regions, and searching for the best
integer solution in each sub-region. Some sub-regions will
then be eliminated because they may not contain a
“promising” integer point or for other reasons!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Introduction
I We shall now see a method that will give us the integer
optimal solution. Better than trying all feasible integer points!
I The method proceeds by dividing recursively the LP relaxation
feasible region into sub-regions, and searching for the best
integer solution in each sub-region. Some sub-regions will
then be eliminated because they may not contain a
“promising” integer point or for other reasons!
I Therefore, two important steps are used: 1. Branching, i.e.
dividing a problem into two sub-problems and solving the LP
2. Bounding, i.e., updating upper and lower bounds on the
optimal integer value z IP .
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Introduction
I We shall now see a method that will give us the integer
optimal solution. Better than trying all feasible integer points!
I The method proceeds by dividing recursively the LP relaxation
feasible region into sub-regions, and searching for the best
integer solution in each sub-region. Some sub-regions will
then be eliminated because they may not contain a
“promising” integer point or for other reasons!
I Therefore, two important steps are used: 1. Branching, i.e.
dividing a problem into two sub-problems and solving the LP
2. Bounding, i.e., updating upper and lower bounds on the
optimal integer value z IP .
I The method will stop if the lower and upper bounds become
equal or no more sub-regions to search remain. We shall first
see the method with the example above.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Branching
I When a (sub)-problem has a fractional LP solution, if the
method continues from that subproblem then we create two
subproblems by picking a fractional variable, say x = f , and
create two new subproblems by adding x ≤ bf c and x ≥ df e.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Branching
I When a (sub)-problem has a fractional LP solution, if the
method continues from that subproblem then we create two
subproblems by picking a fractional variable, say x = f , and
create two new subproblems by adding x ≤ bf c and x ≥ df e.
I In our example, we first create two subproblems: P1
max z = 3x + 4y
subject to
5x + 8y ≤ 24, x ≥ 5
x, y ≥ 0, integer
which is infeasible, hence discarded.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Branching II
I and P2
max z = 3x + 4y
subject to
5x + 8y ≤ 24, x ≤ 4
x, y ≥ 0, integer
which is feasible.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Branching II
I and P2
max z = 3x + 4y
subject to
5x + 8y ≤ 24, x ≤ 4
x, y ≥ 0, integer
which is feasible.
I Since P1 is eliminated, we are going to look for the best
integer point in P2.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Branching II
I and P2
max z = 3x + 4y
subject to
5x + 8y ≤ 24, x ≤ 4
x, y ≥ 0, integer
which is feasible.
I Since P1 is eliminated, we are going to look for the best
integer point in P2.
I We do this by solving again the LP relaxation of P2 (hence
recursion!). See the figure on the next slide.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Branching II
I and P2
max z = 3x + 4y
subject to
5x + 8y ≤ 24, x ≤ 4
x, y ≥ 0, integer
which is feasible.
I Since P1 is eliminated, we are going to look for the best
integer point in P2.
I We do this by solving again the LP relaxation of P2 (hence
recursion!). See the figure on the next slide.
I The solution is (4, 1/2) still not integer with value exactly 14!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Subproblem P2
Figure: LP relaxation of P2 solved.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Creating the B & B Tree
I Now, we shall pause for a moment, and start creating the
so-called Branch-and-Bound Tree to help visualise things
better.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Creating the B & B Tree
I Now, we shall pause for a moment, and start creating the
so-called Branch-and-Bound Tree to help visualise things
better.
I The root node of the tree or Node 0 is the LP relaxation of
the original integer problem.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Creating the B & B Tree
I Now, we shall pause for a moment, and start creating the
so-called Branch-and-Bound Tree to help visualise things
better.
I The root node of the tree or Node 0 is the LP relaxation of
the original integer problem.
I Node 1 corresponds to P1, Node 2 corresponds to P2.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Creating the B & B Tree
I Now, we shall pause for a moment, and start creating the
so-called Branch-and-Bound Tree to help visualise things
better.
I The root node of the tree or Node 0 is the LP relaxation of
the original integer problem.
I Node 1 corresponds to P1, Node 2 corresponds to P2.
I We write next to each node the LP relaxation solution and
the objective function value, see next slide.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Creating the B & B Tree
I Now, we shall pause for a moment, and start creating the
so-called Branch-and-Bound Tree to help visualise things
better.
I The root node of the tree or Node 0 is the LP relaxation of
the original integer problem.
I Node 1 corresponds to P1, Node 2 corresponds to P2.
I We write next to each node the LP relaxation solution and
the objective function value, see next slide.
I In our example, Node 1 is “pruned” by infeasibility of P1. The
only node “alive” is Node 2 now.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The B & B Tree so far
Figure: B & B tree at this point.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree
I Still we do not have any integer feasible solution
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree
I Still we do not have any integer feasible solution
I We continue by branching from the only node alive: Node 2.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree
I Still we do not have any integer feasible solution
I We continue by branching from the only node alive: Node 2.
I We create two subproblems by adding to P2 the constraints
y ≤ 0 and y ≥ 1
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree
I Still we do not have any integer feasible solution
I We continue by branching from the only node alive: Node 2.
I We create two subproblems by adding to P2 the constraints
y ≤ 0 and y ≥ 1
I We shall name these two problems P3 and P4.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree
I Still we do not have any integer feasible solution
I We continue by branching from the only node alive: Node 2.
I We create two subproblems by adding to P2 the constraints
y ≤ 0 and y ≥ 1
I We shall name these two problems P3 and P4.
I It is easy to see that for P3 where y ≤ 0 the feasible region is
the segment from (0, 0) to (4, 0) because this new constraint
along with y ≥ 0 fixes y to zero! Therefore, the optimal point
of P3 is (4, 0). But this is an integer feasible point! It gives us
a lower bound on the optimal integer value 12. So we have
now 12 ≤ z IP ≤ 14. This Node 3 is “pruned” by integer
solution now!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree II
I Now let us solve the LP relaxation P4 at Node 4
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree II
I Now let us solve the LP relaxation P4 at Node 4
I The solution is (3.2, 1) with value 13.6! But this is great
because Node 4 is the only node alive where we can continue
the search, and it says that it cannot yield an integer point
with value bigger than 13! So we can update our upper bound
to 13: 12 ≤ z IP ≤ 13
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree II
I Now let us solve the LP relaxation P4 at Node 4
I The solution is (3.2, 1) with value 13.6! But this is great
because Node 4 is the only node alive where we can continue
the search, and it says that it cannot yield an integer point
with value bigger than 13! So we can update our upper bound
to 13: 12 ≤ z IP ≤ 13
I Check out P4 and the B & B tree now in the next two slides.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Subproblem P4
Figure: LP relaxation of P4 solved.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The B & B Tree so far
Figure: B & B tree at this point.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree III
I Now, we branch from Node 4, we add x ≤ 3 and x ≥ 4 to
create sub-problems P5 and P6.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree III
I Now, we branch from Node 4, we add x ≤ 3 and x ≥ 4 to
create sub-problems P5 and P6.
I We notice immediately that P6 is infeasible since x at least 4
and y at least 1 will not do! So Node 6 is discarded (“pruned
by infeasibility”)
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree III
I Now, we branch from Node 4, we add x ≤ 3 and x ≥ 4 to
create sub-problems P5 and P6.
I We notice immediately that P6 is infeasible since x at least 4
and y at least 1 will not do! So Node 6 is discarded (“pruned
by infeasibility”)
I Now, the only node alive is Node 5 with problem P5.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree III
I Now, we branch from Node 4, we add x ≤ 3 and x ≥ 4 to
create sub-problems P5 and P6.
I We notice immediately that P6 is infeasible since x at least 4
and y at least 1 will not do! So Node 6 is discarded (“pruned
by infeasibility”)
I Now, the only node alive is Node 5 with problem P5.
I Its optimal solution is x = 3, y = 9/8 with z = 13.5. (verify
this graphically!) The upper bound does not change since we
still hope for 13.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Subproblem P5
Figure: LP relaxation of P5 solved.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The B & B Tree so far
Figure: B & B tree at this point with 7 nodes.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree IV
I Let us take stock of what we have achieved so far.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree IV
I Let us take stock of what we have achieved so far.
I Only node alive is Node 5 which did not give an integer
solution, and it can still give us a 13 perhaps!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree IV
I Let us take stock of what we have achieved so far.
I Only node alive is Node 5 which did not give an integer
solution, and it can still give us a 13 perhaps!
I Our bounds are 12 ≤ z IP ≤ 13.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree IV
I Let us take stock of what we have achieved so far.
I Only node alive is Node 5 which did not give an integer
solution, and it can still give us a 13 perhaps!
I Our bounds are 12 ≤ z IP ≤ 13.
I We branch from node 5 by adding he constraints y ≤ 1 and
y ≥ 2. Let us name the subproblems P7 and P8, respectively.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree IV
I Let us take stock of what we have achieved so far.
I Only node alive is Node 5 which did not give an integer
solution, and it can still give us a 13 perhaps!
I Our bounds are 12 ≤ z IP ≤ 13.
I We branch from node 5 by adding he constraints y ≤ 1 and
y ≥ 2. Let us name the subproblems P7 and P8, respectively.
I Solving the LP at P7, we find the optimal point (3, 1) which
is integer with value 13. So lower bound becomes 13, and
now upper and lower bounds are equal. We can stop.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The Branch-and-Bound Method: Continuing the Search on
the B & B Tree IV
I Let us take stock of what we have achieved so far.
I Only node alive is Node 5 which did not give an integer
solution, and it can still give us a 13 perhaps!
I Our bounds are 12 ≤ z IP ≤ 13.
I We branch from node 5 by adding he constraints y ≤ 1 and
y ≥ 2. Let us name the subproblems P7 and P8, respectively.
I Solving the LP at P7, we find the optimal point (3, 1) which
is integer with value 13. So lower bound becomes 13, and
now upper and lower bounds are equal. We can stop.
I We have reached our upper bound with an integer feasible
point. This is provably optimal!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The B & B Tree Final
Figure: Final B & B tree.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2
I Consider the problem
z IP = max z = −x1 + 4x2
subject to
−10x1 + 20x2 ≤ 22
5x1 + 10x2 ≤ 49
x1 , x2 ≥ 0, integer.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2
I Consider the problem
z IP = max z = −x1 + 4x2
subject to
−10x1 + 20x2 ≤ 22
5x1 + 10x2 ≤ 49
x1 , x2 ≥ 0, integer.
I Let us solve the LP relaxation graphically (see figure in the
next slide)
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2
I Consider the problem
z IP = max z = −x1 + 4x2
subject to
−10x1 + 20x2 ≤ 22
5x1 + 10x2 ≤ 49
x1 , x2 ≥ 0, integer.
I Let us solve the LP relaxation graphically (see figure in the
next slide)
I The LP solution is fractional: x1 = 19/5, x2 = 3. The optimal
value is z = 41/5. Therefore, the upper bound on optimal
integer value is 8!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2
I Consider the problem
z IP = max z = −x1 + 4x2
subject to
−10x1 + 20x2 ≤ 22
5x1 + 10x2 ≤ 49
x1 , x2 ≥ 0, integer.
I Let us solve the LP relaxation graphically (see figure in the
next slide)
I The LP solution is fractional: x1 = 19/5, x2 = 3. The optimal
value is z = 41/5. Therefore, the upper bound on optimal
integer value is 8!
I We shall solve by Branch-and-Bound.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The LP Relaxation of Example 2
Figure: LP relaxation of Example 2. LP feasible region in black.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound
I Let us create two subproblems by branching on x1 : we add
x1 ≤ 3 and x1 ≥ 4. The initial LP relaxation becomes Node 0
in the search tree as usual.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound
I Let us create two subproblems by branching on x1 : we add
x1 ≤ 3 and x1 ≥ 4. The initial LP relaxation becomes Node 0
in the search tree as usual.
I The subproblem with x1 ≤ 3 becomes Node 1, the other one
Node 2
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound
I Let us create two subproblems by branching on x1 : we add
x1 ≤ 3 and x1 ≥ 4. The initial LP relaxation becomes Node 0
in the search tree as usual.
I The subproblem with x1 ≤ 3 becomes Node 1, the other one
Node 2
I See Figure in the next slide.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound
I Let us create two subproblems by branching on x1 : we add
x1 ≤ 3 and x1 ≥ 4. The initial LP relaxation becomes Node 0
in the search tree as usual.
I The subproblem with x1 ≤ 3 becomes Node 1, the other one
Node 2
I See Figure in the next slide.
I The black region on the left corresponds to the feasible set of
Subproblem 1; the one on the left to Subproblem 2
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound
I Let us create two subproblems by branching on x1 : we add
x1 ≤ 3 and x1 ≥ 4. The initial LP relaxation becomes Node 0
in the search tree as usual.
I The subproblem with x1 ≤ 3 becomes Node 1, the other one
Node 2
I See Figure in the next slide.
I The black region on the left corresponds to the feasible set of
Subproblem 1; the one on the left to Subproblem 2
I Notice that the blank slab in the middle has been discarded
since it cannot contain an integer point!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound
I Let us create two subproblems by branching on x1 : we add
x1 ≤ 3 and x1 ≥ 4. The initial LP relaxation becomes Node 0
in the search tree as usual.
I The subproblem with x1 ≤ 3 becomes Node 1, the other one
Node 2
I See Figure in the next slide.
I The black region on the left corresponds to the feasible set of
Subproblem 1; the one on the left to Subproblem 2
I Notice that the blank slab in the middle has been discarded
since it cannot contain an integer point!
I Subproblem 1 has optimal solution (3, 2.6); Subproblem 2
(4, 2.9), none of them integer!
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 after the first branching
Figure: LP feasible regions of the two subproblems after the first
branching.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The B & B Tree after the first branching
Figure: B & B tree after the first branching.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound II
I Now, let us take stock of the situation.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound II
I Now, let us take stock of the situation.
I We have Node 1 and Node 2 that gave an optimal value of
7+ both. This means that in both regions there can be an
integer point with value 7 or less.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound II
I Now, let us take stock of the situation.
I We have Node 1 and Node 2 that gave an optimal value of
7+ both. This means that in both regions there can be an
integer point with value 7 or less.
I Now the previous upper bound was 8. But after the first
branching there is no possibility of finding an integer point
with value 8. So the upper bound must be lowered to 7.
z IP ≤ 7.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound II
I Now, let us take stock of the situation.
I We have Node 1 and Node 2 that gave an optimal value of
7+ both. This means that in both regions there can be an
integer point with value 7 or less.
I Now the previous upper bound was 8. But after the first
branching there is no possibility of finding an integer point
with value 8. So the upper bound must be lowered to 7.
z IP ≤ 7.
I Both nodes are alive since both may give 7. Therefore, we
should choose the node to continue branching.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound II
I Now, let us take stock of the situation.
I We have Node 1 and Node 2 that gave an optimal value of
7+ both. This means that in both regions there can be an
integer point with value 7 or less.
I Now the previous upper bound was 8. But after the first
branching there is no possibility of finding an integer point
with value 8. So the upper bound must be lowered to 7.
z IP ≤ 7.
I Both nodes are alive since both may give 7. Therefore, we
should choose the node to continue branching.
I Let us choose the node 2 on the right. Arbitrarily.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound III
I We branch from node 2 by adding x2 ≤ 2 (Subproblem 3 )and
x2 ≥ 3 (Subproblem 4), resp.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound III
I We branch from node 2 by adding x2 ≤ 2 (Subproblem 3 )and
x2 ≥ 3 (Subproblem 4), resp.
I We notice immediately that Subproblem 4 is infeasible.
Therefore Node 4 is discarded (pruned by infeasibility)
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound III
I We branch from node 2 by adding x2 ≤ 2 (Subproblem 3 )and
x2 ≥ 3 (Subproblem 4), resp.
I We notice immediately that Subproblem 4 is infeasible.
Therefore Node 4 is discarded (pruned by infeasibility)
I Node 3 gives an LP optimal point which is integer: (4, 2)!
This provides a lower bound on the optimal value equal to 4.
Hence, we have now 4 ≤ z IP ≤ 7. Node 3 is also pruned by
integer solution.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound III
I We branch from node 2 by adding x2 ≤ 2 (Subproblem 3 )and
x2 ≥ 3 (Subproblem 4), resp.
I We notice immediately that Subproblem 4 is infeasible.
Therefore Node 4 is discarded (pruned by infeasibility)
I Node 3 gives an LP optimal point which is integer: (4, 2)!
This provides a lower bound on the optimal value equal to 4.
Hence, we have now 4 ≤ z IP ≤ 7. Node 3 is also pruned by
integer solution.
I Node 1 is still alive and it has value 7.4. Therefore, we must
back up to Node 1 continue branching from there.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 after the second branching
Figure: LP feasible regions of the two subproblems after the second
branching.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
The B & B Tree after the second branching
Figure: B & B tree after the 2nd branching.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound IV
I We branch from node 1 by adding x2 ≤ 2 (Subproblem 5 and
Node 5 )and x2 ≥ 3 (Subproblem 6 and Node 6), resp.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound IV
I We branch from node 1 by adding x2 ≤ 2 (Subproblem 5 and
Node 5 )and x2 ≥ 3 (Subproblem 6 and Node 6), resp.
I Subproblem 6 is infeasible, node 6 is pruned.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound IV
I We branch from node 1 by adding x2 ≤ 2 (Subproblem 5 and
Node 5 )and x2 ≥ 3 (Subproblem 6 and Node 6), resp.
I Subproblem 6 is infeasible, node 6 is pruned.
I Subproblem 5 gives an LP optimal point (1.8, 2) with optimal
value equal 6.2. Since Node 5 is the only node alive, the
upper bound becomes 6: 4 ≤ z IP ≤ 6.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound IV
I We branch from node 1 by adding x2 ≤ 2 (Subproblem 5 and
Node 5 )and x2 ≥ 3 (Subproblem 6 and Node 6), resp.
I Subproblem 6 is infeasible, node 6 is pruned.
I Subproblem 5 gives an LP optimal point (1.8, 2) with optimal
value equal 6.2. Since Node 5 is the only node alive, the
upper bound becomes 6: 4 ≤ z IP ≤ 6.
I Node 5 is still alive, upper and lower bounds still differ. We
continue by branching from Node 5: add x1 ≤ 1 (Subproblem
7 and Node 7) and x1 ≥ 2 (Subproblem 8 and Node 8).
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 by Branch-and-Bound IV
I We branch from node 1 by adding x2 ≤ 2 (Subproblem 5 and
Node 5 )and x2 ≥ 3 (Subproblem 6 and Node 6), resp.
I Subproblem 6 is infeasible, node 6 is pruned.
I Subproblem 5 gives an LP optimal point (1.8, 2) with optimal
value equal 6.2. Since Node 5 is the only node alive, the
upper bound becomes 6: 4 ≤ z IP ≤ 6.
I Node 5 is still alive, upper and lower bounds still differ. We
continue by branching from Node 5: add x1 ≤ 1 (Subproblem
7 and Node 7) and x1 ≥ 2 (Subproblem 8 and Node 8).
I But Node 8 Lp soultion is integer (2, 2) with optimal value 6,
which is a new lower bound equal to the upper bound. We
stop.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Example 2 after backing up to Node 1
Figure: After backing up to Node 1.
M. Ç. Pınar
IE303 Solving Models with Integer Variables
Outline
The LP Relaxation
The Branch-and-Bound Method
Example 2
Final B & B Tree
Figure: Final B & B tree.
M. Ç. Pınar
IE303 Solving Models with Integer Variables