Chapter 1 Integer programming problemspdf

Operations Research(Math3151)
Lecture 1
Integer programming
Teklebirhan A.
Aksum University
January 13, 2025
Teklebirhan A. Aksum University Chapter 1 Integer programming 1 / 124

Outlines
1 Integer programming
2 Deterministic dynamic programming
3 Inventory models
4 Forecasting
5 Decision theory
6 Queuing systems
7 Simulation modeling

Motivation
resources in an organization
Management is the attainment of organizational goals in an
effective and efficient manner through planning, organizing,
leading, and controlling organizational resources.
What are resources in an organization?
Setting an objective requires strategic thinking, experiences,
vision, etc.
"Resource allocation" uses Operations Research (OR).
Names of similar subjects/ideas:
Management science.
Decision science.
Optimization method/algorithm.
Mathematical programming.

Example: job allocation
Two people are going to hold
an event, and they need to
complete some tasks.
One task must be assigned to
exactly one person; one
person can work on one task
at a time.
How to assign the tasks so
that they can complete all
tasks the fastest?
What are the resources? What
is the objective?
ID Task Processing Time (min)
1 Boiling boba 20
2 Brewing milk tea 30
3 Baking cookies 60
4 Designing poster 15
5 Renting handcart 25

Example: Project Management
n workers are going to
complete m jobs in a project.
Some jobs must be
processed with precedence
rules.
Some jobs cannot be done
by certain workers.
Some jobs can be split and
allocated to several
workers.
Some jobs require different
processing time if allocated
to different workers.
How many days does it take to
complete this project?

What is Operations Research?
Operations Research (OR) is:
the methodology to allocate the available resources to the various
activities in a way that is most effective for the organization as a
whole.
It is applied to problems that concern how to conduct and
coordinate the operations (i.e., activities) within an organization.
It aims to support decision making.
Typical tools: intuitions, business senses, and experiences.
By doing OR studies, we generate some suggestions to decision
makers.

The process of conducting an OR study

Example: Knapsack Problem
You are preparing for hiking.
There are some useful items,
but your backpack (knapsack)
can only carry 5 kilograms.
An item cannot be split: Each
item should be either chosen
or discarded.
Which items should you bring
to maximize the total value?
ID Item Weight Value
1 Compass 0.5 6
2 Hatchet 1.5 5
3 Matches 0.4 4
4 Tarpaulin 1.0 4
5 Telescope 1.1 3
6 Cylinder 1.6 4
7 Rilakkuma 0.8 1

A Mathematical Programming Model
Decision variables: What may we determine?
xi =







1 if item i is chosen;
0 otherwise.
Objective function: What do we want?
max6x1 + 5x2 + 4x3 + 4x4 + 3x5 + 4x6 + x7
Constraints: What are the limitations?
0.5x1 + 1.5x2 + 0.4x3 + x4 + 1.1x5 + 1.6x6 + 0.8x7 ≤ 5
and
xi ∈ {0,1},∀i = 1,...,7
Complete Formulation (Model)
max 6x1 + 5x2 + 4x3 + 4x4 + 3x5 + 4x6 + x7
s.t. 0.5x1 + 1.5x2 + 0.4x3 + x4 + 1.1x5 + 1.6x6 + 0.8x7 ≤ 5
xi ∈ {0,1}, ∀i = 1,...,7

A Compact Formulation (Advanced)
Let wi and vi be the weight and value of item i.
Let n be the number of items and B be the maximum allowable
weight.
A compact (and more abstract) formulation is:
max
n
¼
i=1
vi xi
subject to
n
¼
i=1
wi xi ≤ B
xi ∈ {0,1} ∀i = 1,...,n.

Application: Order Selection
Let wi and vi be the processing time and gross profit of order i.
Let n be the number of orders and B be the factory capacity (the
maximum total processing time that a factory has).
The formulation is (still)
max
n
¼
i=1
vi xi
s.t.
n
¼
i=1
wi xi ≤ B
xi ∈ {0,1} ∀i = 1,...,n

Application: Portfolio Optimization
Let wi and vi be the maximum possible investment amount and
expected return of stock i.
Let n be the number of assets and B be the total budget.
The formulation is
max
n
¼
i=1
vi xi
s.t.
n
¼
i=1
wi xi ≤ B
0 ≤ xi ≤ 1 ∀i = 1,...,n
Note that the investment decision is not all-or-nothing.
Here, xi ∈ {0,1} becomes xi ∈ [0,1] or 0 ≤ xi ≤ 1.

Linear Programming vs. Integer Programming
Linear Programming
This is a linear program:
max
n
¼
i=1
vi xi
s.t.
n
¼
i=1
wi xi ≤ B
0 ≤ xi ≤ 1 ∀i = 1,...,n
Integer Programming
This is an integer program:
max
n
¼
i=1
vi xi
s.t.
n
¼
i=1
wi xi ≤ B
xi ∈ {0,1} ∀i = 1,...,n

Introduction to Linear Programming
What is Linear Programming?
Linear Programming (LP) is a mathematical optimization
technique used to find the best outcome in a mathematical model
with linear constraints.
Linear Programming is the process of formulating and solving
linear programs (also abbreviated as LPs).
A LP is a mathematical program with some special properties.

In a two-dimensional linear programming problem, we represent
the constraints as lines and the feasible region as the
intersection of these lines.

The objective function is represented as a line with the slope
determined by the coefficients of the decision variables.

The objective function is represented as a line with the slope
determined by the coefficients of the decision variables.
The optimal solution is the point in the feasible region where the
objective function line is maximized or minimized.

Basic Elements of a Mathematical Program
In general, any mathematical program may be expressed as:
min f(x1,x2,...,xn) (objective function)
s.t. gi (x1,x2,...,xn) ≤ bi ∀i = 1,...,m (constraints)
xj ∈ ‘ ∀j = 1,...,n (decision variable)
There are m constraints and n variables.
x1,x2,...,xn are real-valued decision variables.
We may write
x =










x1
.
.
.
xn










= (x1,...,xn)
as a vector of decision variables (or a decision vector).
f : ‘n → ‘ and gi : ‘n → ‘ are all real-valued functions.
Mostly we will omit xj ∈ ‘.

Transformation
Transformation of Objective and Constraints
How about a maximization objective function?
maxf(x) ⇔ min−f(x).
How about "=" or "≥" constraints?
gi (x) ≥ bi ⇔ −gi (x) ≤ −bi .
gi (x) = bi ⇔ gi (x) ≤ bi and gi (x) ≥ bi , i.e., −gi (x) ≤ −bi .
Example
max x1 − x2
s.t. − 2x1 + x2 ≥ −3
x1 + 4x2 = 5
⇔
min − x1 + x2
s.t. 2x1 − x2 ≤ 3
x1 + 4x2 ≤ 5
− x1 − 4x2 ≤ −5.

Feasible Solutions
For a mathematical program:
A feasible solution satisfies all the constraints.
An infeasible solution violates at least one constraint.
For example:
min2x1 + x2
s.t x1 ≤ 10
x1 + 2x2 ≤ 12
x1 − 2x2 ≥ −8
x1 ≥ 0,x2 ≥ 0
Feasible Solutions
x1 = (2,3)
x2 = (6,0)
x3 = (6,6)

Feasible region and optimal solutions
Feasible Region
The feasible region (or feasible set) is the set of feasible solutions.
The feasible region may be empty.
Optimal Solutions
An optimal solution is a feasible solution that:
Attains the largest objective value for a maximization problem.
Attains the smallest objective value for a minimization problem.
In short, no feasible solution is better than it.
An optimal solution may not be unique.
There may be multiple optimal solutions.
There may be no optimal solution.

Binding Constraints
Definition
At a solution, a constraint may be binding:
Let g(·) ≤ b be an inequality constraint and x̄ be a solution. g(·) ≤ b
is binding at x̄ if g(x̄) = b.
An inequality is nonbinding at a point if it is strict at that point.
An equality constraint is always binding at any feasible solution.
Some Examples:
x1 + x2 ≤ 10 is binding at (x1,x2) = (2,8).
2x1 + x2 ≥ 6 is nonbinding at (x1,x2) = (2,8).
x1 + 3x2 = 9 is binding at (x1,x2) = (6,1).

Linear Programs
A mathematical program
minf(x) s.t. gi (x) ≤ bi ∀i = 1,...,m,
is a LP if f and gi are all linear functions.
Each of these linear functions may be expressed as
a1x1 + a2x2 + ··· + anxn =
´n
j=1 aj xj ,
where aj ∈ ‘, j = 1,...,n, are the coefficients.
We may write a = (a1,...,an) and f(x) = aT x.
An example:
minimize x1 + x2
subject to
x1 + 2x2 ≤ 6
2x1 + x2 ≤ 6
x1 ≥ 0, x2 ≥ 0.

Linear Programs
In general, an LP may always
be expressed as
min
n
¼
j=1
cj xj
s.t.
n
¼
j=1
Aij xj ≤ bi ∀i = 1,...,m.
Aij s: constraint coefficients
bi s: right-hand-side values
(RHS).
cj s: objective coefficients.
Or by vectors:
mincT
x
s.t. aT
i x ≤ bi ∀i = 1,...,m.
ai ∈ ‘n, bi ∈ ‘, c ∈ ‘n
x ∈ ‘n
Or by matrices:
min cT
x
subject to: Ax ≤ b
A ∈ ‘m×n, b ∈ ‘m

Standard Form of LP
Standard Form:
Maximize cT
x
Subject to Ax ≤ b
x ≥ 0
where:
x is the vector of decision variables.
c is the vector of coefficients of the objective function.
A is the matrix of coefficients of the constraints.
b is the vector of right-hand side constants of the constraints.

Solution Techniques
Graphical Method
It is applicable only for LP problems with two decision variables.
Steps:
1 Graph the Constraints:
Convert each inequality constraint into an equation (replace ≤ with
=).
Plot the equations on a graph. Use the axes for x1 and x2.
Identify the feasible region by testing a point (often the origin) in the
inequalities. The resulting area where all inequalities overlap is the
feasible region.
2 Determine the Corners of the Feasible Region:
Identify the vertices (corner points) of the feasible region. These
vertices are typically where the constraint lines intersect.
3 Calculate the objective function value at each corner point of the
feasible region.
4 Identify the Optimal Solution:
If the LPP is maximization, take the highest value.
If the LPP is minimization, take the lowest value.

Graphical approach
For LPs with only two decision variables, we may solve them with
the graphical approach.
Consider the following example:
maximize 2x1 + x2
subject to x1 ≤ 10
x1 + 2x2 ≤ 12
x1 − 2x2 ≥ −8
x1 ≥ 0
x2 ≥ 0

Graphical Approach
Step 1: Draw the Feasible Region
Draw each constraint one by one, and then find the intersection.
max 2x1 + x2
s.t. x1 ≤ 10
x1 + 2x2 ≤ 12
x1 − 2x2 ≥ −8
x1 ≥ 0
x2 ≥ 0

Graphical Approach
Step 2: Draw Some Isoquant Lines.
A line such that all points on it result in the same objective value.
Also called isoprofit or isocost lines when it is appropriate.
Also called indifference lines (curves) in Economics.
max 2x1 + x2
s.t. x1 ≤ 10
x1 + 2x2 ≤ 12
x1 − 2x2 ≥ −8
x1 ≥ 0
x2 ≥ 0

Graphical Approach
Step 3: Indicate the direction to push the isoquant line.
The direction that decreases/increases the objective value for a
minimization/maximization problem.
max 2x1 + x2
s.t. x1 ≤ 10
x1 + 2x2 ≤ 12
x1 − 2x2 ≥ −8
x1 ≥ 0
x2 ≥ 0

Step 4: Push the isoquant line to the "end" of the feasible region.
Stop when any further step makes all points on the isoquant line
infeasible.

Step 5: Identify the binding constraints at an optimal solution.

Graphical Approach
Step 6: Set the binding constraints to equalities and then solve
the linear system for an optimal solution.
In the example, the binding constraints are: x1 ≤ 10 and
x1 + 2x2 ≤ 12
We may solve the linear system:
x1 = 10
x1 + 2x2 = 12
in any way and obtain an optimal solution (x∗
1,x∗
2) = (10,1).
For example, through Gaussian elimination:
1 0 10
1 2 12
!
→
1 0 10
0 2 2
!
→
1 0 10
0 1 1
!
Step 7: Plug in an optimal solution obtained into the objective
function to get the associated objective value.
In the example, 2x∗
1 + x∗
2 = 21.

Three types of LPs
For any LPs, it must be one of the following:
Infeasible.
Unbounded.
Finitely optimal (having an optimal solution).
A finitely optimal LP may have:
A unique optimal solution.
Multiple optimal solutions.

Linear Programming Problem
Example
Solve the Linear programming problem
Maximize: Z = 4x1 + 7x2
Subject to:
2x1 + 5x2 ≤ 40
x1 + x2 ≤ 11
x2 ≥ 4
x1,x2 ≥ 0

Summary

Simple LP Formulations: Introduction
It is important to learn how to model a practical situation as an LP.
Once you do so, you have "solved" the problem.
This process is typically called LP formulation or modeling.
Here we will see some examples of LP formulation.
Then we formulate large-scale problems with compact
formulations.
Example
A product mix problem
We produce several products to sell.
Each product requires some resources. Resources are limited.
We want to maximize the total sales revenue with available
resources.

Description
Problem description
We produce desks and tables.
Producing a desk requires three units of wood, one hour of labor,
and 50 minutes of machine time.
Producing a table requires five units of wood, two hours of labor,
and 20 minutes of machine time.
We may sell everything we produce.
For each day, we have
Two hundred workers that each works for eight hours.
Fifty machines that each runs for sixteen hours.
A supply of 3600 units of wood.
Desks and tables are sold at $700 and $900 per unit, respectively.

Variables
Define Variables
What do we need to decide?
Let
x1 = number of desks produced in a day
x2 = number of tables produced in a day
With these variables, we now try to express how much we will
earn and how many resources we will consume.
Formulate the Objective Function
We want to maximize the total sales revenue.
Given our variables x1 and x2, the sales revenue is 700x1 + 900x2.
The objective function is thus:
max 700x1 + 900x2.

Formulate Constraints
For each restriction or limitation, we write a constraint:
The supply of wood is limited: 3x1 + 5x2 ≤ 3600
The number of labor hours is limited: x1 + 2x2 ≤ 1600
The amount of machine time is limited: 50x1 + 20x2 ≤ 48000
Note: Use the same unit of measurement!

Complete Formulation
Collectively, our formulation is:
max 700x1 + 900x2 (Objective Function)
s.t. 3x1 + 5x2 ≤ 3600 (wood)
x1 + 2x2 ≤ 1600 (labor)
50x1 + 20x2 ≤ 48000 (machine)
x1 ≥ 0
x2 ≥ 0
In any case:
Clearly define decision variables in front of your formulation.
Write comments after the objective function and constraints.

Solve and Interpret
An optimal solution of this LP is (884.21,189.47).
So, the interpretation is... to produce 884.21 desks and 189.47
tables?
Producing 884.21 desks and 189.47 tables seems weird, but in
fact:
We may produce 884.21 desks and 189.47 tables per day on
average (i.e., roughly 88,420 desks and 18,947 tables per 100
days).
We may suggest to produce, e.g., 884 desks and 189 tables.
It still supports our decision making.
It may not really be optimal, but we spend a very short time to
make a good suggestion.
"All models are wrong, but some are useful."

Personnel scheduling

Constraints
Demand fulfillment:
110 employees are needed on Monday:
x1 + x4 + x5 + x6 + x7 ≥ 110.
80 employees are needed on Tuesday:
x1 + x2 + x5 + x6 + x7 ≥ 80.
120 employees are needed on Sunday:
x3 + x4 + x5 + x6 + x7 ≥ 120.
Nonnegativity constraints:
xi ≥ 0 ∀i = 1,...,7.

Compact

Problems vs. instances

Integer Programming
Introduction to integer programming
We have worked with LPs.
In linear programming problem, the variables are allowed to take
any real or fractional value, however in some cases, variables
must only take integer values.
The subject of formulating and solving models with integer
variables is Integer Programming (IP).
An IP is typically a linear IP (LIP).
If the objective function or any functional constraint is nonlinear, it
is a nonlinear IP (NLIP).
We will focus on linear IP.

Personnel Scheduling (Again)
We know that United Airlines developed an LP to determine the
number of staff in each of their service locations.
The same problem is faced by Taco Bell.
It has more than 6500 restaurants in the US.
It asks how many staff to have at each restaurant in each shift.
Taco Bell developed an Integer Program (i.e., an LP with integer
variables) to solve its workforce scheduling problem.
The number of staff is typically small! Rounding is very inaccurate.
$13 million are saved per year.

Route Selection
Waste Management Inc. operates an recycling network with 293
landfill sites, 16 waste-to-energy plants, 72 gas-to-energy
facilities, 146 recycling plants, 346 transfer stations, and 435
collection depots.
20,000 routes must be traversed by its vehicles each day.
How to determine a route?
Construct a network with nodes and edges.
Assign a binary variable to each edge: 1 if included, 0 otherwise.
Constraints are required to ensure that selected edges form a valid
route.
A large Integer Programming (IP) model is constructed to save the
company $498 million in operational expenses over a 5-year
period.

Integer programming formulation
The Knapsack Problem
We start our illustration with the classic knapsack problem.
There are four items to select:
Item 1 2 3 4
Value ($) 16 22 12 8
Weight (kg) 5 7 4 3
The knapsack capacity is 10 kg.
We maximize the total value without exceeding the knapsack
capacity.
The Complete Formulation is
max16x1 + 22x2 + 12x3 + 8x4
s.t. 5x1 + 7x2 + 4x3 + 3x4 ≤ 10
xi ∈ {0,1} ∀i = 1,...,4

Types of Integer Programming Problems
Integer programming problems can be classified into three categories:
1 Pure (all) integer programming problems in which all decision
variables are restricted to integer values.
2 Mixed integer programming problems in which some, but not all,
of the decision variables are restricted to integer values.
3 Zero-one integer programming problems in which all decision
variables are restricted to either zero (0) or one (1).

Standard form of Pure Integer Programming Problem
The pure integer programming problem in its standard form can be
written as follows:
Maximize Z = c1x1 + c2x2 + ··· + cnxn,
subject to Constraints:
a11x1 + a12x2 + ··· + a1nxn = b1,
a21x1 + a22x2 + ··· + a2nxn = b2,
.
.
.
am1x1 + am2x2 + ··· + amnxn = bm,
x1,x2,...,xn ≥ 0 and integer.

Solving IP
The branch-and-bound algorithm finds an optimal solution for
any IP.
It decomposes an IP to multiple LPs, solve all the LPs, and
compares those outcomes to reach a conclusion.
Each LP is solved separately (with the simplex method or other
ways).
In general, the process may take a very long time.
As finding an optimal solution for an IP may be too
time-consuming, we often look for a near-optimal feasible
solution instead in practice.
An algorithm that generates a feasible solution in a short time is
called a heuristic algorithm.
Hopefully it is near-optimal.
A good heuristic algorithm does not always work, but it works for
most of the time.

Solving Integer Programs
Suppose we are given an IP,
how may we solve it?
The simplex method does not
work!
The feasible region is not a
region.
It is discrete.
There is no way to move
along edges.
But all we know is how to solve
LPs. How about solving a
linear relaxation first?
Definition
The LP obtained by removing all
the integer constraints from the
given IP is called Linear Relaxation.

Linear Relaxation
What is the linear relaxation of
maxx1 + x2
s.t.x1 + 3x2 ≤ 10
2x1 − x2 ≥ 5
xi ∈ ™+
∀i = 1,2
where ™ is the set of all integers and ™+ is the set of all
non-negative integers.
The linear relaxation is
max x1 + x2
s.t. x1 + 3x2 ≤ 10
2x1 − x2 ≥ 5
xi ≥ 0 ∀i = 1,2

Linear Relaxation
For the Knapsack Problem:
maximize 16x1 + 22x2 + 12x3 + 8x4
s.t. 5x1 + 7x2 + 4x3 + 3x4 ≤ 10
xi ∈ {0,1} ∀i = 1,...,4
The Linear Relaxation is:
maximize 16x1 + 22x2 + 12x3 + 8x4
s.t. 5x1 + 7x2 + 4x3 + 3x4 ≤ 10
xi ∈ [0,1] ∀i = 1,...,4
Note: xi ∈ [0,1] is equivalent to xi ≥ 0 and xi ≤ 1.

Linear Relaxation Provides a Bound
For a minimization IP, its linear relaxation provides a lower bound.
Proposition
Let z∗ and z0 be the objective values associated with optimal solutions
of a minimization IP and its linear relaxation, respectively. Then,
z0 ≤ z∗.
Proof
Both have the same objective function.
The feasible region of the linear relaxation is (weakly) larger than that
of the IP.
For a maximization IP, linear relaxation provides an upper bound.

Linear Relaxation and Integer Programming
Linear relaxation may solve the Integer Program (IP):
If we are lucky, the linear relaxation may be infeasible or
unbounded.
The IP is then infeasible or unbounded.
If we are lucky, an optimal solution to the linear relaxation may be
feasible to the original IP. When this happens, the IP is solved.
Proposition
Let x0 be an optimal solution to the linear relaxation of an IP. If x0 is
feasible to the IP, it is optimal to the IP.
Proof
Suppose x0 is not optimal to the IP; there must be another feasible
solution x00 that is better. However, as x00 is feasible to the IP, it is also
feasible to the linear relaxation, which implies that x0 cannot be
optimal to the linear relaxation.
What if we are unlucky?

Rounding a Fractional Solution
Suppose we solve a linear relaxation with an LR-optimal solution
x0.
"LR-optimal" means x0 is optimal to the linear relaxation.
x0, however, has at least one variable violating the integer
constraint in the original Integer Program (IP).
We may choose to round the variable.
Round up or down?
Is the resulting solution always feasible?
Will the resulting solution be close to an IP-optimal solution x∗?

Rounding

Rounding a Fractional Solution
Rounding a fractional solution involves the following steps:
When we solve the linear relaxation and find any variable
violating an integer constraint, we will branch this problem into
two new problems, one with an additional constraint.
The two new programs are still linear programs.
Once we have solved them:
If their LR-optimal solutions are both IP-feasible, compare them and
choose the better one.
If any of them results in a variable violating the integer constraint,
branch on that variable recursively.
Eventually compare all the IP-feasible solutions we obtain.

Illustration of the Branch-and-Bound Algorithm
Example
Let’s illustrate the
branch-and-bound algorithm
with the following example:
maximize 3x1 + x2
(P0) s.t. 4x1 + 2x2 ≤ 11
x1,x2 ∈ ™+

Sub-problem 1
We first solve the linear
relaxation:
max 3x1 + x2
(P1) Subject to:
4x1 + 2x2 ≤ 11
xi ≥ 0 ∀i = 1,2
The optimal solution is:
x1
=

11
4
,0

So we need to branch on x1.

Branching tree
The branch and bound algorithm produces a branching tree.
Each node represents a sub-problem (which is an LP).
Each time we branch on a variable, we create two child nodes.

Sub-problem 2
When we add x1 ≤ 2:
max 3x1 + x2
(P2) Subject to:
4x1 + 2x2 ≤ 11
x1 ≤ 2
xi ≥ 0 ∀i = 1,2
An (P2)-optimal solution is
x2 = (2, 3
2).
So later we need to branch
on x2.
Before that, let’s solve (P3).

Sub-problem 3
When we add x1 ≤ 3:
max 3x1 + x2
Subject to:
(P3) 4x1 + 2x2 ≤ 11
x1 ≥ 3
xi ≥ 0 ∀i = 1,2
The problem is infeasible!
This node is dead and does
not produce any candidate
solution

Branching tree
The current progress can be summarized in the branching tree
Note that z2 = 7.5 8.25 = z1.
In general, when we branch to the next level, the objective value
associated with a sub-problem-optimal solution will always be
weakly lower (for a maximization problem). Why?

Branching tree
As x2 = 3
2 in x2, we will branch sub-problem 2 on x2.

Subproblem-4

Subproblem-5

Summary
In running the branch-and-bound algorithm, we maintain a tree.
If a subproblem-optimal solution is IP-feasible, set it to the
candidate solution if it is currently the best among all IP-feasible
solutions. Stop branching this node.
If a subproblem is infeasible, stop branching this node.
If a subproblem-optimal solution is not IP-feasible:
If it is better than the current candidate solution, branch.
Otherwise, stop branching.

The Knapsack Problem
We start our illustration with the classic knapsack problem.
There are five items to select:
Item 1 2 3 4 5
Value ($) 2 3 4 1 3
Weight (kg) 4 5 3 1 4
The knapsack capacity is 11 kg.
We maximize the total value without exceeding the knapsack
capacity.
The Complete Formulation
max 2x1 + 3x2 + 4x3 + x4 + 3x5
s.t. 4x1 + 5x2 + 3x3 + x4 + 4x5 ≤ 11
xi ∈ {0,1} ∀i = 1,...,5
Let’s solve the knapsack problem with the branch-and-bound
algorithm.

Solving the linear relaxation
The first step is always to solve the linear relaxation:
max 2x1 + 3x2 + 4x3 + x4 + 3x5
s.t. 4x1 + 5x2 + 3x3 + x4 + 4x5 ≤ 11
xi ∈ [0,1] ∀i = 1,...,5.
A very intuitive way works:
Sort all items in the descending order of vi
wi
, where vi and wi are the
value and weight of item i, respectively.
Select items one by one according to the order until the knapsack
is full. Note that the last item may be partially selected.
The five ratios are 2
4, 3
5, 4
3, 1
1, and 3
4. The order is item 3, item 4,
item 5, item 2, and lastly item 1.
An optimal solution to the linear relaxation is (0, 3
5 ,1,1,1).

Subproblems

Cont...

Remarks
The major process is still branch and bound.
The only unique thing is that we have a special way to solve the
linear relaxation of the knapsack problem.
In other words, we may customize the branch-and-bound
algorithm for each specific type of IP.

Procedure of Branch and Bound Algorithm
Step: 1 (Initialization)
Consider the following all integer programming problem:
Max Z = c1x1 + c2x2+... + cnxn
Subject to the constraints:
a11x1 + a12x2 + ... + a1nxn = b1
a21x1 + a22x2 + ... + a2nxn = b2
...
am1x1 + am2x2 + ... + amnxn = bm
and xj ≥ 0 and non-negative integer.

Obtain the optimal solution of the given problem
ignoring integer restriction on the variables
1 If the solution to this LP problem (say LP-A) is infeasible or
unbounded, the solution to the all integer programming problem
is also infeasible or unbounded, as the case may be.
2 If the solution satisfies the integer restriction, the optimal integer
solution has been obtained. If one or more basic variables do not
satisfy the integer requirement, then go to step 2. Let the optimal
value of the objective function of LP-A be Z1. This value provides
an initial upper bound on the objective function value and it is
denoted by ZU .
3 Find the feasible value by rounding off each variable value. The
value of the objective function so obtained is the least upper
bound and it is denoted by ZL .

Step 2: Branching Step
1 Let xk be one basic variable which does not have an integer value
and also has the largest fractional value.
2 Branch (or partition) the LP-A into two new LP sub-problems (also
called nodes) based on integer values of xk that are immediately
above and below its non-integer value. That is, it is partitioned by
adding two mutually exclusive constraints: xk ≤ ⌊xk ⌋ and
xk ≥ ⌊xk ⌋ + 1 to the original LP problem.
3 Here ⌊xk ⌋ is the integer portion of the current non-integer value of
the variable xk . This is obviously done to exclude the non-integer
value of the variable xk . The two new LP sub-problems are as
follows:

LP Sub-problem B
Max Z =
n
¼
j=1
cj xj
subject to
n
¼
j=1
aij xj = bi
xk ≤ ⌊xk ⌋
and xj ≥ 0
LP Sub-problem C
Max Z =
n
¼
j=1
cj xj
subject to
n
¼
j=1
aij xj = bi
xk ≥ ⌊xk ⌋ + 1
and xj ≥ 0

Step 3: Bound Step
Obtain the optimal solution of LP Subproblem B. Let the optimal
value of the objective function of LP-B be Z2.
Obtain the optimal solution of LP Subproblem C. Let the optimal
value of the objective function of LP-C be Z3.
The best integer solution value becomes the lower bound on the
integer LP problem objective function value (initially this is the
rounded off value). Let the lower bound be denoted by ZL .

Step 04: (Fathoming Step) Examining the solution of
both LP-B and LP-C:
1 If the sub-problem yields an infeasible solution, then terminate
the branch.
2 If the sub-problems yield a feasible solution, but not an integer
solution, then go to Step 02.
3 If the sub-problem yields an integer feasible solution, then
examine the value of the objective function.
If the value is equal to the upper bound, then the optimal solution
has been obtained.
If it is not equal to the upper bound but exceeds the lower bound,
then this value is considered as the new upper bound and return to
step 2. Finally, If it is less than the lower bound terminate the
branch.

Step 05 (Termination)
Procedure of branching and bounded continues, until no sub
problem remains to be examined.
At this stage, the integer solution corresponding to the current
lower bound is optimal all integer programming problem.

Branch and bound method example
Example
Solve the following integer programming problem using branch and
bound method:
Maximize Z = 2x1 + 3x2
Subject to 6x1 + 5x2 ≤ 25,
x1 + 3x2 ≤ 10,
x1,x2 ≥ 0, x1,x2 are integers.

Solution: Step 01
Relaxing the integer condition, Let us find the optimal solution to the
given LP problem by graphical method.
Graphical Method
Let us draw the line 6x1 + 4x2 = 25
Let x1 = 0, 6(0) + 4x2 = 25 =⇒ x2 = 25/4 i.e. (0,5)
Let x2 = 0, 6x1 + 4(0) = 25 =⇒ x1 = 25/6 i.e. (4.16,0)
Let us draw the line x1 + 3x2 = 10
Let x1 = 0, 0 + 3x2 = 10 =⇒ x2 = 10/3 i.e. (0,3.33)
Let x2 = 0, x1 + 3(0) = 10 =⇒ x1 = 10 i.e. (10,0)

Step 1 Cont.
To find the intersection point:
(1) ⇒ 6x1 + 5x2 = 25
(2) ⇒ 6x1 + 18x2 = 60
Subtracting the two equations:
−13x2 = −35 =⇒ x2 = 35
13
Substitute x2 value in equation (1) to get x1 = 35
13 i.e. the
intersection point is (1.92,2.69)
At (4.16,0), Max Z = 2(4.16) + 3(0) = 8.32
At (0,3.33), Max Z = 2(0) + 3(3.33) = 9.99
At (1.92,2.69), Max Z = 2(1.92) + 3(2.69) = 11.91

Cont.
The optimum solution is x1 = 1.92, x2 = 2.69 and Max Z1 = 11.91.
Let ZU = 11.91 be the initial upper bound.
The feasible solution by rounding off the solution gives the initial
lower bound ZL = 11 (rounding off x1 = 1, x2 = 3).
The optimal solution lies between these two bounds.

Step 02 (Branching step)
Here both x1 and x2 are not integers, therefore we can select the
variable for branching with the highest fractional value, here x2
has a higher fractional value compared to x1.
Solving the variable x2 for branching, divide the given problem
into two subproblems A and B to eliminate the fractional part of
x2 = 2.69.
The new constraints to be added are x2 ≤ 2 and x2 ≥ 3.

Cont’d

Cont’d
The corresponding graph is given by

Cont’d

Step 03 (Bound step)
Rule: The best integer solution gives the lower bound.
Here the best integer solution is x1 = 1, x2 = 3 and Max Z3 = 11.
The value ZL = 11 is the lower bound for the given LP problem.
Step 04 (Fathoming step)
Both the sub-problems B and C give the same Z value.
In the sub-problem C, both the variables x1 and x2 are integers, so
there is no need to branch the sub-problem C further.
Since ZL ≤ Z3 ZU (11 ≤ 11 11.91), therefore the new upper
bound is ZU = 11.
In the sub-problem B, x1 is not an integer and since ZL ≤ Z2
(11 ≤ 11), therefore the sub-problem B can be branched further.

Stet 05 (Branch step)
Let us divide the sub problem B into two sub problems, D and E, by
taking variable x1 = 2.5.
The new constraints to be added are x1 ≤ 2 and x1 ≥ 3.

Cont’d

Cont’d
Rule: The best integer solution gives the lower bound.
Here the best integer solution is x1 = 2, x2 = 2, and MaxZ4 = 10.
The solution to the LP sub problem D is integer feasible, but Z4 ZL
(10 11), and since the value of the lower bound remains unchanged
(ZL = 11), therefore the sub problem D is not considered for further
branching.

Cont’d
Since the solution of sub problem E is non-integer (x2 is not an
integer), it can be further branched with variable x2, but Z5 ZL
(10.2 11). Therefore, sub problem E is not considered for further
branching.
Thus, the integer solution corresponding to the current lower
bound is the optimal solution.
Hence the integer optimal solution is corresponding to sub
problem C and is given by x1 = 1, x2 = 3, and MaxZ = 11.
The entire branch and bound procedure can be represented by the
following enumeration tree. Each node is a sub problem.

Cont’d

Note
If both sub-problems give non-integer solutions, then the
sub-problem with the maximum objective value should be further
branched.
If one of the sub-problems has no feasible solution and the other
has a feasible solution, but it is not integer, and any other
sub-problem is the same. kind exist, always select the sub
problem which has maximum objective function value for further
branching.

Chapter 1 Integer programming problemspdf

Recommended

More Related Content

Similar to Chapter 1 Integer programming problemspdf (20)

Recently uploaded (20)

Chapter 1 Integer programming problemspdf