Peng Zhang - Shandong Universitycourse.sdu.edu.cn/.../uploadfile/20151224010142463.pdf · 2015. 12....

Introduction to Operations Research

Peng Zhang ∗

June 7, 2015

∗School of Computer Science and Technology, Shandong University, Jinan, 250101,

China. Email: algzhang@sdu.edu.cn.

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1 Introduction

1

2 Overview of the Operations Research

Modeling Approach

2

3 Introduction to Linear Programming

3

4 Solving Linear Programming

Problems: The Simplex Method

4

5 The Theory of the Simplex Method

5

6 Duality Theory and Sensitivity

Analysis

6

7 Other Algorithms for Linear

Programming

7

8 The Transportation and Assignment

Problems

8

9 Network Optimization Models

9

10 Dynamic Programming

10

11 Integer Programming

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Integer programming (IP): every variable is restricted to be an

integer.

Binary integer programming (BIP): every variable is restricted

to have values 0 or 1.

Mixed integer programming (MIP): only some of the variables

are restricted to have integer values.

11.1 PROTOTYPE EXAMPLE

The CALIFORNIA MANUFACTURING COMPANY problem.

The company has 10 million dollars. It wants to build factories and

warehouses in Los Angeles or San Francisco. The number of factories is

at most two, and the number of warehouses is at most one. Only when

a factory is built in some place, a warehouse can be built there.

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The values and costs of factories and warehouses in different places

are given in the following table.

The question is, where it will build factories and warehouses, so that

the total value of the built factories and warehouses is maximized?

This problem can be formulated as the following binary integer

programming (BIP).

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max 9x1 + 5x2 + 6x3 + 4x4

s.t. 6x1 + 3x2 + 5x3 + 2x4 ≤ 10

x3 + x4 ≤ 1

x3 ≤ x1

x4 ≤ x2

xj ∈ {0, 1}, ∀j

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11.2 SOME BIP APPLICATIONS

11.3 INNOVATIVE USES OF BINARYVARIABLES

IN MODEL FORMULATION

11.4 SOME FORMULATION EXAMPLES

The Knapsack problem

Instance: There is a knapsack with capacity W , and n items each of

which has a weight wi > 0 and a value vi > 0.

Task: Put some items into the knapsack such that their total weight

is at most W and their total value is maximized.

Solution.

We use the decision variable xi to indicate whether item i is put into

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the knapsack.

max∑i

vixi

s.t.∑i

wixi ≤W

xi ∈ {0, 1}, ∀i

The Minimum Spanning Tree problem

Instance: We are given an undirected graph G = (V,E) with non-

negative costs {ce} defined on edges.

Task: The goal is to find a minimum cost tree spanning all vertices

in G.

Solution.

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For each edge e ∈ E, there is a decision variable xe. If xe = 1, e is

included in the solution. If xe = 0, it is not.

min∑e

cexe

s.t.∑

e∈δ(S)

xe ≥ 1, ∀∅ ̸= S ⊂ V

xe ∈ {0, 1}, ∀e ∈ E

The Traveling Salesman Problem

Instance: We are given n cities. For every two cities i ̸= j, there is a

length (e.g., distance) cij ≥ 0.

Task: Find a shortest tour that visits every city exactly once and

returns to the original city.

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Solution 1.

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For each ordered pari (i, j), we define a decision variable xij . If

xij = 1, then the edge (i, j) is used in the tour to be found. If xij = 0,

then the edge (i, j) is not used.

min∑i ̸=j

cijxij

s.t.∑j ̸=i

xij = 1, ∀i

∑j ̸=i

xji = 1, ∀i

∑i∈S,j∈S̄

xij ≥ 1, ∀∅ ̸= S ⊂ [n]

xij ∈ {0, 1}, ∀i ̸= j

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Solution 2.

The decision variables {xij} are defined in the same way as above.

For each city i ∈ {1, · · · , n}, there is a variable ui, whose value can

be any real number.

Thus we get a MIP formulation for the TSP problem. This formula-

tion is by Miller, Tucker, and Zemlin [JACM’60].

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min∑i ̸=j

cijxij (1)

s.t.∑j ̸=i

xij = 1, ∀i (2)

∑j ̸=i

xji = 1, ∀i (3)

ui − uj + nxij ≤ n− 1, ∀2 ≤ i ̸= j ≤ n (4)

xij ∈ {0, 1}, ∀i ̸= j

ui ≷ 0, ∀i

Theorem 11.1. The MIP (1) defines TSP.

Proof.

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• We first prove that any solution to (1) is a TSP tour (a circuit visiting

every city exactly once).

• Constraint (2) says that every city has exactly one outgoing edge in the

solution. Constraint (3) says that every city has exactly one ingoing edge

in the solution. So, these two constraints guarantee that the solution

must consist of circuits. In the following we will prove that there is only

one circuit in the solution, by constraint (4).

• Actually, we will prove every circuit in the solution passes through city

1. Since city 1 can only be visited exactly once (by constraints (2) and

(3)), this means there is only one circuit in the solution.

• Suppose for the contradiction that in the solution there is a circuit

(i1, i2, · · · , ik, i1) consisting of length k edges which does not visit c-

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ity 1. Since for every edge (i, j) in the circuit we have xij = 1, we have

the following k inequalities for this circuit by constraint (4):

ui1 − ui2 + n ≤ n− 1,

ui2 − ui3 + n ≤ n− 1,

· · · ,

uik − ui1 + n ≤ n− 1.

• Adding these k inequalities, we get kn ≤ k(n− 1), which is absurd.

• Next we prove any TSP tour defines a solution to MIP (1).

• Given a TSP tour, we define xij = 1 if edge (i, j) is used in the tour,

and xij = 0 otherwise. So, constraints (2) and (3) are satisfied.

• The tour must visit every city. Let city 1 be the first city on the tour,

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and we define u1 = 1. In general, if city i is the k-th city (2 ≤ k ≤ n) on

the tour, we define ui = k.

• Now we consider constraint (4). Let (i, j) be any pair such that i ̸= j

and i, j ∈ {2, · · · , n}. Then we have two cases to consider: xij = 0 and

xij = 1.

• If xij = 0, then constraint (4) reduces to

ui − uj ≤ n− 1,

which obviously holds since ui ≤ n and uj ≥ 2.

• If xij = 1, then constraint (4) becomes

ui − uj + n ≤ n− 1,

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which also holds since city j is the successor of city i on the tour and

hence ui − uj = −1.

• Note that since city 1 is not included in the condition of constraint (4),

we need not to consider constraint (4) on the “last” edge on the tour

ingoing to city 1 . (Suppose city i′ is the predecessor of city 1 on the

tour. Then we have xi′1 = 1, and ui′ = n and u1 = 1.)

11.5 SOME PERSPECTIVES ON SOLVING INTE-

GER PROGRAMMING PROBLEMS

IP may be easier than LP. Wrong!

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Since LP can be efficiently solved, and IP has far fewer feasible solu-

tions than LP, it may seem that IP should be relatively easy to solve.

However, the opposite is true: IP is much harder to solve than LP.

There are two fallacies in the above reasoning.

(i) One fallacy is that having a finite number of feasible solutions

ensures the problem is readily solvable. The fact is that, finite

numbers can be astronomically large. For example, if there are

100 0-1 variables in an IP, then there will be 2100 feasible solutions

to be considered.

(ii) The second fallacy is that removing some non-integer solutions

from an LP will make it easier to solve. To the contrary, it is just

because all these feasible solutions are there so that there will be a

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bfs which is also an optimal solution. By contrast, people cannot

prove similar beautiful property for IP.

One may think to use rounding to the closest integers. Does-

n’t work!

Example. Rounding may lead to unfeasible solutions.

max x2

s.t. − x1 + x2 ≤1

2

x1 + x2 ≤ 31

2

x1, x2 ∈ Z+

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The optimal solution to the corresponding LP is (23 , 2). No matter

to round ( 23 , 2) to which solution of (1, 2) or (2, 2), it is not a feasible

solution.

Example. Even leading to a feasible solution, the rounded solution

may be far away from the optimal solution.

max x1 + 5x2

s.t. x1 + 10x2 ≤ 20

x1 ≤ 2

x1, x2 ∈ Z+

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The optimal solution to the corresponding LP will be rounded to

(2, 1). However, the optimal solution to IP is (0, 2).

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11.6 THE BRANCH-AND-BOUND TECHNIQUE

AND ITS APPLICATION TO BINARY IN-

TEGER PROGRAMMING

To illustrate the branch-and-bound method, we begin with an example.

(The California Manufacturing Company problem)

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max 9x1 + 5x2 + 6x3 + 4x4 (IP0)

s.t. 6x1 + 3x2 + 5x3 + 2x4 ≤ 10

x3 + x4 ≤ 1

x3 ≤ x1

x4 ≤ x2

xj ∈ {0, 1}, ∀j

Branching

Since each variable is a 0-1 variable, a natural idea is to branch into

two subproblems on some variable xi. For example, if we branch on x1,

then we get two subproblems.

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Subproblem 1 (x1 = 0):

max 5x2 + 6x3 + 4x4 (IP1)

s.t. 3x2 + 5x3 + 2x4 ≤ 10

x3 + x4 ≤ 1

x3 ≤ 0

x4 ≤ x2

x2, x3, x4 ∈ {0, 1}

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Subproblem 2 (x1 = 1):

max 9 + 5x2 + 6x3 + 4x4 (IP2)

s.t. 3x2 + 5x3 + 2x4 ≤ 4

x3 + x4 ≤ 1

x3 ≤ 1

x4 ≤ x2

x2, x3, x4 ∈ {0, 1}

Bounding

For each of these subproblems, we can get an upper bound on its

optimal value by solving the corresponding LP-relaxation of it.

For example, let (LP0) be the LP-relaxation of (IP0). The optimal

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solution to (LP0) is ( 56 , 1, 0, 1) with value Z∗LP0 = 16 1

2 . Since Z∗IP0 ≤

Z∗LP0 and Z∗

IP0 is an integer, we conclude that Z∗IP0 ≤ 16.

Similarly, we get that the optimal solution to (LP1) is (0, 1, 0, 1) with

value Z∗LP1 = 9, so Z∗

IP1 ≤ 9. The optimal solution to (LP2) is (1, 45 , 0,

45 )

with value Z∗LP2 = 16 1

5 , so Z∗IP2 ≤ 16.

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In the procedure of the branch-and-bound method, we remember the

current best feasible solution (called incumbent) and its value Z∗.

So far, the current best feasible solution is (0, 1, 0, 1) with Z∗ = 9.

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Fathoming1

A subproblem can be conquered (fathomed), and thereby dismissed

from further consideration, in the three ways described below.

(i) The subproblem’s upper bound is ≤ Z∗.

(Since the upper bound is ≤ Z∗, the optimal value of the subprob-

lem is obviously ≤ Z∗, too. Since there is already the incumbent

whose value is Z∗, this subproblem can be dismissed.)

(ii) The subproblem’s LP-relaxation has no feasible solution.

(iii) The subproblem’s LP-relaxation has integer (0-1) optimal solution.

If this solution is better than the incumbent, it becomes the new

1The original meaning of fathom is to understand what something means after

thinking about it carefully. It means here to conquer, not to consider further.

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incumbent, and the fathoming rule 1 is reapplied to all unfathomed

subproblems with the new larger Z∗.

The Branch-and-Bound algorithm for BIP

1 The set of unfathomed subproblems LIST ← ∅. The incumbent is

assigned by empty. The current best value Z∗ ← −∞.

2 Compute the upper bound of the current whole problem. Apply

the three fathoming rules to the current problem. If it is not

fathomed, then insert the problem to LIST .

3 while LIST ̸= ∅ do

4 Take the subproblem that was created most recently. (Break

ties according to which has the larger bound.) Remove it from

LIST .

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5 (Branching) Create two new subproblems by fixing the next

variable (the branching variable) at either 0 or 1.

6 (Bounding) For each of the two new subproblems, compute its

upper bound.

7 (Fathoming) For each of the two new subproblems, apply the

three fathoming rules on it. (If a problem in LIST is fathomed,

then remove it. Recall the fathoming rule 3.)

8 Insert the not fathomed new subproblems to LIST .

9 endwhile

10 return the incumbent.

Completing the Example

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Branching on (IP2) with either x2 = 0 or x2 = 1, we get two sub-

problems (IP3) and (IP4).

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Subproblem 3 (x1 = 1, x2 = 0):

max 9 + 6x3 + 4x4 (IP3)

s.t. 3x2 + 5x3 + 2x4 ≤ 4

x3 + x4 ≤ 1

x3 ≤ 1

x4 ≤ 0

x3, x4 ∈ {0, 1}

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Subproblem 4 (x1 = 1, x2 = 1):

max 14 + 6x3 + 4x4 (IP4)

s.t. 5x3 + 2x4 ≤ 1

x3 + x4 ≤ 1

x3 ≤ 1

x4 ≤ 1

x3, x4 ∈ {0, 1}

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Both (IP3) and (IP4) cannot be fathomed. (Recall the three fathom-

ing rules.)

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Branching on (IP4) with either x3 = 0 or x3 = 1, we get two sub-

problems (IP5) and (IP6).

Subproblem 5 (x1 = 1, x2 = 1, x3 = 0):

max 14 + 4x4 (IP5)

s.t. 2x4 ≤ 1

x4 ≤ 1

x4 ∈ {0, 1}

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Subproblem 6 (x1 = 1, x2 = 1, x3 = 1):

max 20 + 4x4 (IP6)

s.t. 2x4 ≤ −4

x4 ≤ 0

x4 ≤ 1

x4 ∈ {0, 1}

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Since (IP6)’s relaxation (LP6) has no feasible solution, (IP6) is fath-

omed.

We branch on (IP5) with either x4 = 0 or x4 = 1, getting subproblem

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7 and subproblem 8.

Subproblem 7 (x1 = 1, x2 = 1, x3 = 0, x4 = 0):

max 14 (IP7)

Subproblem 8 (x1 = 1, x2 = 1, x3 = 0, x4 = 1):

max 18 (IP8)

s.t. 2 ≤ 1

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Since (LP8) is not feasible, (IP8) is fathomed.

(IP7) is fathomed too, since it has the integer optimal solution (1, 1, 0, 0)

with Z∗IP8 = 14.

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Since Z∗IP8 is better than the value Z∗ of the incumbent, (1, 1, 0, 0)

becomes the new incumbent.

Since Z∗ = 14 now, (IP3) is fathomed by the fathoming rule 1.

Now LIST is empty (there is no unfathomed problem). The algo-

rithm terminates, and the optimal solution to the original problem is

(1, 1, 0, 0), with optimal value 14.

11.7 A BRANCH-AND-BOUNDALGORITHM FOR

MIXED INTEGER PROGRAMMING

Compared to the Branch-and-Bound algorithm for BIP, there are four

changes for the algorithm for MIP.

(i) Only the integer variables that have a non-integer value in the

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optimal solution to the LP-relaxation of the current subproblem

are considered as branching variables.

(ii) When branching, just create two new subproblems by specifying

two ranges of values for the variable.

For example, if xj is the branching variable, then we add the fol-

lowing constraints to the two new subproblems,

xj ≤ ⌊x∗j⌋ and xj ≥ ⌈x∗

j⌉,

respectively.

(iii) In BIP, the optimal value Z∗LP is rounded down to be used as an

upper bound of the current subproblem. Now, in MIP, since there

may be non-integer variables, the upper bound is Z∗LP without

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rounding down.

(iv) The fathoming rule 3 is now as follows.

If the optimal solution to the LP-relaxation of the current subprob-

lem has integer values for the integer restricted variables, then this

subproblem can be fathomed.

And, if this solution is better than the incumbent, it becomes the

new incumbent and the fathoming rule 1 is reapplied to all the

unfathomed subproblems with the new larger Z∗.

Remarks. In execution of the Branch-and-Bound algorithm for MIP,

a branching variable may recur again as the branching variable in the

future. See the example shown below.

The Branch-and-Bound algorithm for MIP

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1 The set of unfathomed subproblems LIST ← ∅. The incumbent is

assigned by empty. The current best value Z∗ ← −∞.

2 Compute the upper bound of the current whole problem. Apply

the three fathoming rules to the current problem. If it is not

fathomed, then insert the problem to LIST .

3 while LIST ̸= ∅ do

4 Take the subproblem that was created most recently. (Break

ties according to which has the larger bound.) Remove it from

LIST .

5 (Branching) let xj be the branching variable and x∗j be its value

in the optimal solution to the LP-relaxation of the subprob-

lem. Create two new subproblems by adding the respective

constraints xj ≤ ⌊x∗j⌋ and xj ≥ ⌈x∗

j⌉.

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6 (Bounding) For each of the two new subproblems, compute its

upper bound.

7 (Fathoming) For each of the two new subproblems, apply the

three fathoming rules on it. (If a problem in LIST is fathomed,

then remove it. Recall the fathoming rule 3.)

8 Insert the not fathomed new subproblems to LIST .

9 endwhile

10 return the incumbent.

An MIP Example. Solve the following MIP by Branch-and-Bound.

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max 4x1 − 2x2 + 7x3 − x4 (MP0)

s.t. x1 + 5x3 ≤ 10

x1 + x2 − x3 ≤ 1

6x1 − 5x2 ≤ 0

− x1 + 2x3 − 2x4 ≤ 3

x1, x2, x3 ∈ Z+

x4 ≥ 0

Solution.

Solving LP0, we get the following.

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Branch into two subproblems (MP1) and (MP2) on variable x1.

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max 4x1 − 2x2 + 7x3 − x4 (MP1)

s.t. x1 + 5x3 ≤ 10

x1 + x2 − x3 ≤ 1

6x1 − 5x2 ≤ 0

− x1 + 2x3 − 2x4 ≤ 3

x1 ≤ 1 (*)

x1, x2, x3 ∈ Z+

x4 ≥ 0

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max 4x1 − 2x2 + 7x3 − x4 (MP2)

s.t. x1 + 5x3 ≤ 10

x1 + x2 − x3 ≤ 1

6x1 − 5x2 ≤ 0

− x1 + 2x3 − 2x4 ≤ 3

x1 ≥ 2 (*)

x1, x2, x3 ∈ Z+

x4 ≥ 0

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(MP2) is fathomed since (LP2) has no feasible solution.

Branch into two subproblems (MP3) and (MP4) on variable x2.

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max 4x1 − 2x2 + 7x3 − x4 (MP3)

s.t. x1 + 5x3 ≤ 10

x1 + x2 − x3 ≤ 1

6x1 − 5x2 ≤ 0

− x1 + 2x3 − 2x4 ≤ 3

x1 ≤ 1 (*)

x2 ≤ 1 (*)

x1, x2, x3 ∈ Z+

x4 ≥ 0

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max 4x1 − 2x2 + 7x3 − x4 (MP4)

s.t. x1 + 5x3 ≤ 10

x1 + x2 − x3 ≤ 1

6x1 − 5x2 ≤ 0

− x1 + 2x3 − 2x4 ≤ 3

x1 ≤ 1 (*)

x2 ≥ 2 (*)

x1, x2, x3 ∈ Z+

x4 ≥ 0

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Branch into two subproblems (MP5) and (MP6) on variable x1 (re-

curring).

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max 4x1 − 2x2 + 7x3 − x4 (MP5)

s.t. x1 + 5x3 ≤ 10

x1 + x2 − x3 ≤ 1

6x1 − 5x2 ≤ 0

− x1 + 2x3 − 2x4 ≤ 3

x1 ≤ 1 (*)

x2 ≤ 1 (*)

x1 ≤ 0 (*)

x1, x2, x3 ∈ Z+

x4 ≥ 0

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max 4x1 − 2x2 + 7x3 − x4 (MP6)

s.t. x1 + 5x3 ≤ 10

x1 + x2 − x3 ≤ 1

6x1 − 5x2 ≤ 0

− x1 + 2x3 − 2x4 ≤ 3

x1 ≤ 1 (*)

x2 ≤ 1 (*)

x1 ≥ 1 (*)

x1, x2, x3 ∈ Z+

x4 ≥ 0

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(MP5) is fathomed since we find the new incumbent (0, 0, 2, 12 ) with

Z∗ = 13 12 .

(MP4) is fathomed since its upper bound is ≤ Z∗.

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(MP6) is fathomed since (LP6) has no feasible solution.

LIST is empty now. The optimal solution to (MP0) is (0, 0, 2, 12 ),

with the optimal value Z∗ = 13 12 .

11.8 THE BRANCH-AND-CUT APPROACH TO

SOLVING BIP PROBLEMS

A cutting plane for any IP problem is a new functional constraint that

reduces the feasible solution for the LP-relaxation without eliminating

any feasible solutions for the IP problem.

Example.

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max 3x1 + 2x2

s.t. 2x1 + 3x2 ≤ 4

x1, x2 ∈ {0, 1}

The feasible region for the LP-relaxation of the above IP is as follows.

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If we add the constraint x1+x2 ≤ 1 to the LP-relaxation, the feasible

region will be sharpened. However, no feasible solution to the original

IP problem is harmed.

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In this case, the constraint x1 + x2 ≤ 1 is a cutting plane for the

original IP problem.

Generating Cutting Planes for BIP

There are many ways to generate cutting planes for BIP. Here we

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just give one such method.

1. Consider any functional constraint in≤ form with only nonnegative

coefficients.

For example, 2x1 + 3x2 ≤ 4.

2. Find a group of variables (called a minimal cover of the constraint)

such that

(a) The constraint is violated if every variable in the group equals

to 1 and all other variables equal 0.

(b) But the constraint becomes satisfied if the value of any one of

these variables is changed from 1 to 0.

In our example, the group is {x1, x2}.

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3. By letting N denote the number of variables in the group, the

resulting cutting plane is

Sum of variables in group ≤ N − 1.

So, the cutting plane for our example is x1 + x2 ≤ 1.

Remarks. For a given functional constraint, its cutting plane may

not be unique.

Example.

Consider the constraint

6x1 + 3x2 + 5x3 + 2x4 ≤ 10.

Then

x1 + x2 + x4 ≤ 2

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is a cutting plane, since x1, x2 and x4 cannot be 1 simultaneously, but

every two of them can be 1 at the same time.

It is interesting to notice that this constraint has another cutting

plane

x1 + x3 ≤ 1.

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