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Explorations in Artificial Intelligence. Prof. Carla P. Gomes gomes@cs.cornell.edu Module 7 Part 3 Integer Programming . Divisibility

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### Explorations in Artificial Intelligence

Prof. Carla P. Gomes

gomes@cs.cornell.edu

Module 7

Part 3

Integer Programming

Decision variables in an LP model are allowed to have any values, including noninteger values, that satisfy the functional and nonnegativity constraints. i.e., activities can be run at fractional levels.

What to do when divisibility assumption violated:

realm of integer programming!!!

### Revisiting the TBA Airlines Problem An Example where Integrality Matters

• TBA Airlines is a small regional company that specializes in short flights in small airplanes.

• The company has been doing well and has decided to expand its operations.

• The basic issue facing management is whether to purchase more small airplanes to add some new short flights, or start moving into the national market by purchasing some large airplanes, or both.

Question: How many airplanes of each type should be purchased to maximize their total net annual profit?

Let S = Number of small airplanes to purchaseL = Number of large airplanes to purchaseMaximize Profit = S + 5L (\$millions)subject to Capital Available: 5S + 50L ≤ 100 (\$millions) Max Small Planes: S ≤ 2and

S ≥ 0, L ≥ 0.

• Divisibility Assumption of Linear Programming: Decision variables in a linear programming model are allowed to have any values, including fractional values, that satisfy the functional and nonnegativity constraints. Thus, these variables are not restricted to just integer values.

• Since the number of airplanes purchased by TBA must have an integer value, the divisibility assumption is violated.

Let S = Number of small airplanes to purchaseL = Number of large airplanes to purchaseMaximize Profit = S + 5L (\$millions)subject to Capital Available: 5S + 50L ≤ 100 (\$millions) Max Small Planes: S ≤ 2and

S ≥ 0, L ≥ 0S, L are integers.

• When an integer programming problem has just two decision variables, its optimal solution can be found by applying the graphical method for linear programming with just one change at the end.

• We begin as usual by graphing the feasible region for the LP relaxation, determining the slope of the objective function lines, and moving a straight edge with this slope through this feasible region in the direction of improving values of the objective function.

• However, rather than stopping at the last instant the straight edge passes through this feasible region, we now stop at the last instant the straight edge passes through an integer point that lies within this feasible region.

• This integer point is the optimal solution.

• Advantages of restricting variables to take on integer values

• More realistic

• More flexibility

• More difficult to model

• Can be much more difficult to solve

• When are “non-integer” solutions okay?

• Solution is naturally divisible

• e.g., \$, pounds, hours

• Solution represents a rate

• e.g., units per week

• Solution only for planning purposes

• When is rounding okay?

• When numbers are large

• e.g., rounding 114.286 to 114 is probably okay.

• When is rounding not okay?

• When numbers are small

• e.g., rounding 2.6 to 2 or 3 may be a problem.

• Binary variables

• yes-or-no decisions

• Pure integer programming problems are those where all the decision variables must be integers.

• Mixed integer programming problems only require some of the variables (the “integer variables”) to have integer values so the divisibility assumption holds for the rest (the “continuous variables”).

• Binary variables are variables whose only possible values are 0 and 1.

• Binary integer programming(BIP) problems are those where all the decision variables restricted to integer values are further restricted to be binary variables.

• Such problems can be further characterized as either pure BIP problems or mixed BIP problems, depending on whether all the decision variables or only some of them are binary variables.

• Making “yes-or-no” type decisions

• Build a factory?

• Manufacture a product?

• Do a project?

• Assign a person to a task?

• Logical constraints

• Alternative constraints

• Conditional constraints

• Representing non-linear functions

• Fixed Charge Problem

• If a product is produced, must incur a fixed setup cost.

• If a warehouse is operated, must incur a fixed cost.

• Piecewise linear representation

• Diseconomies of scale

• Approximation of nonlinear functions

• Set-covering, and set partitioning

• Make a set of assignments that “cover” a set of requirements.

• Partition a set into subsets meeting given requirements

StockCompany ExampleCapital Budgeting Allocation Problem

StockCompany is considering 6 investments. The cash required from each investment as well as the NPV of the investment is given next. The cash available for the investments is \$14,000. Stockco wants to maximize its NPV. What is the optimal strategy?

An investment can be selected or not. One cannot select a fraction of an investment.

Investment budget = \$14,000

What are the decision variables?

Objective and Constraints?

Max 16x1+ 22x2+ 12x3+ 8x4+ 11x5+ 19x6

5x1+ 7x2+ 4x3+ 3x4+ 4x5+ 6x6  14

xje {0,1} for each j = 1 to 6

Capital Budgeting Allocation Problem (one resource) Knapsack Problem

• Why is a problem with the characteristics of the previous problem called the Knapsack Problem?

• It is an abstraction, considering the simple problem:

A hiker trying to fill her knapsack to maximum total value.

Each item she considers taking with her has a certain value and a

certain weight. An overall weight limitation gives the single

constraint.

Practical applications:

Project selection and capital budgeting allocation problems

Storing a warehouse to maximum value given the indivisibility of goods and space limitations

Sub-problem of other problems e.g., generation of columns for a given model in the course of optimization – cutting stock problem (beyond the scope of this course)

• The previous constraints represent “economic indivisibilities”, either a project is selected, or it is not. There is no selecting of a fraction of a project.

• Similarly, integer variables can model logical requirements (e.g., if stock 2 is selected, then so is stock 1.)

• Exactly 3 stocks are selected.

• If stock 2 is selected, then so is stock 1.

• If stock 1 is selected, then stock 3 is not selected.

• Either stock 4 is selected or stock 5 is selected, but not both.

• Exactly 3 stocks are selected

x1+ x2+ x3+ x4+ x5+ x6=3

A 2-dimensional representation

Stock 2

Stock 1

The integer programming constraint:

If stock 2 is selected then so is stock 1

x1  x2

A 2-dimensional representation

Stock 3

Stock 1

The integer programming constraint:

If stock 1 is selected then stock 3 is not selected

x1 + x3  1

A 2-dimensional representation

stock 5

stock 4

The integer programming constraint:

Either stock 4 is selected or stock 5 is selected, but not both.

x4 + x5 = 1

• The California Manufacturing Company is a diversified company with several factories and warehouses throughout California, but none yet in Los Angeles or San Francisco.

• A basic issue is whether to build a new factory in Los Angeles or San Francisco, or perhaps even both.

• Management is also considering building at most one new warehouse, but will restrict the choice to a city where a new factory is being built.

Question: Should the California Manufacturing Company expand with factories and/or warehouses in Los Angeles and/or San Francisco?

Let x1 = 1 if build a factory in L.A.; 0 otherwisex2 = 1 if build a factory in S.F.; 0 otherwisex3 = 1 if build a warehouse in Los Angeles; 0 otherwisex4 = 1 if build a warehouse in San Francisco; 0 otherwiseMaximize NPV = 8x1 + 5x2 + 6x3 + 4x4 (\$millions)subject to Capital Spent: 6x1 + 3x2 + 5x3 + 2x4 ≤ 10 (\$millions) Max 1 Warehouse: x3 + x4 ≤ 1 Warehouse only if Factory: x3 ≤ x1x4 ≤ x2andx1, x2, x3, x4 are binary variables.

Resource Availability

Mutually exclusive decisions

Contingent decisions

Max 8x1 + 5x2 + 6x3 + 4x4;

subject to

6x1 + 3x2 + 5x3 + 2x4 <= 10;

x3 + x4 <= 1;

x3 <= x1;

x4 <= x2;

BINARY

x1;

x2;

x3;

x4;

If a product is produced, must incur a fixed setup cost.

If a warehouse is operated, must incur a fixed cost.

• The problem is non-linear.

x – quantity of product to be manufactured

x = 0  cost =0;

x > 0  cost = C1x + C2

• How to model it? Using an indicator variable y

y = 1  x is produced; y = 0  x is not produced

Objective function becomes  C1x + C2y

Additional Constraint  x ≤ My

Suppose that two changes are made to the original Wyndor problem:

• For each product, producing any units requires a substantial one-time setup cost for setting up the production facilities.

• The production runs for these products will be ended after one week, so D and W in the original model now represent the total number of doors and windows produced, respectively, rather than production rates. Therefore, these two variables need to be restricted to integer values.

Optimal solution

Let D = Number of doors to produce,W = Number of windows to produce,y1 = 1 if perform setup to produce doors; 0 otherwise,y2 = 1 if perform setup to produce windows; 0 otherwise .Maximize P = 300D + 500W – 700y1 – 1,300y2subject to Original Constraints: Plant 1: D ≤ 4 Plant 2: 2W ≤ 12 Plant 3: 3D + 2W ≤ 18 Produce only if Setup: Doors: D ≤ My1 Windows: W ≤ My2andD ≥ 0, W ≥ 0, y1 and y2 are binary.

Wyndor with Mutually Exclusive Products(Variation 2)

Suppose that now the only change from the original Wyndor problem is:

• The two potential new products (doors and windows) would compete for the same customers. Therefore, management has decided not to produce both of them together.

• At most one can be chosen for production, so either D = 0 or W = 0, or both.

Feasible Solution forWyndor with Mutually Exclusive Products(for non-binary variables)

Let D = Number of doors to produce,W = Number of windows to produce,y1 = 1 if produce doors; 0 otherwise,y2 = 1 if produce windows; 0 otherwise.Maximize P = 300D + 500Wsubject to Original Constraints: Plant 1: D ≤ 4 Plant 2: 2W ≤ 12 Plant 3: 3D + 2W ≤ 18 Auxiliary variables must =1 if produce any: Doors: D ≤ My1 Windows: W ≤ My2 Mutually Exclusive: y1 + y2 ≤ 1 andD ≥ 0, W ≥ 0, y1 and y2 are binary.

Wyndor with Either-Or Constraints(Variation 3)

Suppose that now the only change from the original Wyndor problem is:

• The company has just opened a new plant (plant 4) that is similar to plant 3, so the new plant can perform the same operations as plant 3 to help produce the two new products (doors and windows).

• However, management wants just one of the plants to be chosen to work on these new products. The plant chosen should be the one that provides the most profitable product mix.

Let D = Number of doors to produce,W = Number of windows to produce,y = 1 if plant 4 is used; 0 if plant 3 is usedMaximize P = 300D + 500Wsubject to Plant 1: D ≤ 4 Plant 2: 2W ≤ 12 Plant 3: 3D + 2W ≤ 18 + My Plant 4: 2D + 4W ≤ 28 + M(1 – y) andD ≥ 0, W ≥ 0, y is binary.

• Making “yes-or-no” type decisions

• Build a factory?

• Manufacture a product?

• Do a project?

• Assign a person to a task?

• Fixed costs

• If a product is produced, must incur a fixed setup cost.

• If a warehouse is operated, must incur a fixed cost.

• Either-or constraints

• Production must either be 0 or ≥ 100.

• Subset of constraints

• meet 3 out of 4 constraints.

• Knapsack Problem

• Set Covering Problem

• Set Partitioning Problem

• Set Packing Problem

• The Traveling Salesman Problem

• The Quadratic Assignment Problem

• We are given a set of objects S = {1, 2, 3, …, n}.

• We are also given a set of subsets of S, S. Each subset has a cost associated with it.

• Problem:

• to “cover” all the members of S at the minimum cost using members of S.

• Properties:

• The problem is a minimization and all constraints are >=;

• All RHS coefficients are 1;

• All other matrix coefficients are 0 or 1.

Fire Station ProblemSet Covering Problem

Locate fire stations so that each district has a fire station in it, or next to it.

1

2

3

5

6

7

Minimize the number of fire stations needed.

4

8

9

11

12

13

10

14

15

16

Set

Covers

1

2

3

1

1, 2, 4, 5

2

1, 2, 3, 5, 6

5

6

7

3

2, 3, 6, 7

4

8

9

11

12

13

10

16

13, 15, 16

14

15

16

1

2

3

7

5

6

7

4

8

9

11

11

12

13

15

10

14

15

16

Representation as Integer program

xj = 1 if node j is selectedxj = 0 otherwise

Minimize x1 + x2 + … + x16

s.t. x1 + x2 + x4 + x5  1

x1 + x2 + x3 + x5 + x6  1

x13 + x15 + x16  1

xj {0, 1} for each j.

• Southwestern Airways needs to assign crews to cover all its upcoming flights.

• We will focus on assigning 3 crews based in San Francisco (SFO) to 11 flights.

Question: How should the 3 crews be assigned 3 sequences of flights so that every one of the 11 flights is covered?

Let xj = 1 if flight sequence (paring) j is assigned to a crew; 0 otherwise. (j = 1, 2, … , 12).Minimize Cost = 2x1 + 3x2 + 4x3 + 6x4 + 7x5 + 5x6 + 7x7 + 8x8 + 9x9 + 9x10 + 8x11 + 9x12 (in \$thousands)subject to Flight 1 covered: x1 + x4+ x7 + x10 ≥ 1 Flight 2 covered: x2 + x5 + x8 + x11 ≥ 1 : : Flight 11 covered: x6 + x9 + x10 + x11 + x12 ≥ 1 Three Crews: x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12 ≤ 3 andxj are binary (j = 1, 2, … , 12).

pairings

• We are given a set of objects S = {1, 2, 3, …, n}.

• We are also given S, a set of subsets of S. Each subset has a cost associated with it.

• Problem:

• to “cover” all the members of S at the minimum cost using members of S.

• Properties:

• The problem is a minimization and all constraints are >=;

• All RHS coefficients are 1;

• All other matrix coefficients are 0 or 1.

• There are often multiple ways of modeling the same integer program.

• Solvers for integer programs are extremelysensitive to the formulation. (not true for LPs)

• Dramatically improves the modeling capability

• Economic indivisibilities

• Logical constraints

• Modeling nonlinearities (e.g., fixed cost)

• classical problems in capital budgeting and in supply chain management

• Lots of other applications and models

• Not as easy to model

• Not as easy to solve.

• Rounded Solution may not be feasible.

• Rounded solution may not be close to optimal.

• There can be many rounded solutions.

• Example: Consider a problem with 30 variables that are non-integer in the LP-solution. How many possible rounded solutions are there?

• Enumeration Techniques

• Complete Enumeration

• list all “solutions” and choose the best

• Branch and Bound

• Implicitly search all solutions, but cleverly eliminate the vast majority before they are even searched

• Implicit Enumeration

• Branch and Bound applied to binary variables

• Cutting Plane Techniques

• Use LP to solve integer programs by adding constraints to eliminate the fractional solutions.

• It is the starting point for all solution techniques for integer programming

• Lots of research has been carried out over the past 40 years to make it more and more efficient

• But, it is an art form to make it efficient. (We shall get a sense why.)

• Integer programming is intrinsically difficult.

Investment budget = \$14,000

maximize 16x1 + 22x2 + 12x3 + 8x4 +11x5 + 19x6

subject to 5x1 + 7x2 + 4x3 + 3x4 +4x5 + 6x6  14

xj binary for j = 1 to 6

• Systematically considers all possible values of the decision variables.

• If there are n binary variables, there are 2n different ways.

• Usual idea: iteratively break the problem in two. At the first iteration, we consider separately the case that x1 = 0 and x1 = 1.

x2 = 1

x2 = 0

x2 = 1

x2 = 0

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

x3 = 1

An Enumeration Tree

Original problem

x1 = 0

x1 = 1

• Suppose that we could evaluate 1 billion solutions per second.

• Let n = number of binary variables

• Solutions times (approx.)

• n = 30, 1 second

• n = 40, 18 minutes

• n = 50 13 days

• n = 60 31 years

• Suppose that we could evaluate 1 trillion solutions per second, and instantaneously eliminate 99.9999999% of all solutions as not worth considering

• Let n = number of binary variables

• Solutions times

• n = 70, 1 second

• n = 80, 17 minutes

• n = 90 11.6 days

• n = 100 31 years

The essential idea: search the enumeration tree, but at each node

• Solve the linear program at the node

• Eliminate the subtree (fathom it) if

• The solution is integer (there is no need to go further- why?) or

• The best solution in the subtree cannot be as good as the best available solution (the incumbent- how does that happen?) or

• There is no feasible solution

Node 1 is the original LP Relaxation

44 3/7

1

maximize 16x1 + 22x2 + 12x3 + 8x4 +11x5 + 19x6

subject to 5x1 + 7x2 + 4x3 + 3x4 +4x5 + 6x6  14

0  xj  1 for j = 1 to 6

Solution at node 1:

x1 =1 x2 = 3/7 x3 = x4 = x5 = 0 x6 =1 z = 44 3/7

The IP cannot have value higher than 44 3/7.

x1 = 0

2

Branch and Bound

Node 2 is the original LP Relaxation plus the constraint x1 = 0.

44 3/7

1

44

maximize 16x1 + 22x2 + 12x3 + 8x4 +11x5 + 19x6

subject to 5x1 + 7x2 + 4x3 + 3x4 +4x5 + 6x6  14

0  xj  1 for j = 1 to 6, x1 = 0

Solution at node 2: x1 = 0 x2 = 1 x3 = 1/4 x4 = x5 = 0 x6 = 1 z = 44

x1 = 1

x1 = 0

3

2

Branch and Bound

Node 3 is the original LP Relaxation plus the constraint x1 = 1.

44 3/7

1

44 3/7

44

The solution at node 1 was

x1 =1 x2 = 3/7 x3 = x4 = x5 = 0 x6 =1 z = 44 3/7

Note: it was the best solution with no constraint on x1. So, it is also the solution for node 3. (If you add a constraint, and the old optimal solution is feasible, then it is still optimal.)

x2 = 0

4

Branch and Bound

Node 4 is the original LP Relaxation plus the constraints x1 = 0, x2 = 0.

44 3/7

1

1

x1 = 0

x1 = 1

2

3

44

44 3/7

4

42

Solution at node 4: 0 0 1 0 1 1 z = 42

Our first incumbent solution!

No solution in the subtree can have a value better than 42.

No further searching from node 4 because there cannot be a better integer solution.

x2 = 1

x2 = 0

x2 = 1

x2 = 0

4

5

6

7

Branch and Bound

The incumbent solution has value 42

The incumbent is the best solution on hand.

44 3/7

1

1

x1 = 0

x1 = 1

2

3

44

44 3/7

4

42

44

44

44 1/3

We next solved the LP’s associated with nodes 5, 6, and 7

No new integer solutions were found.

We would eliminate (fatham) a subtree if we were guaranteed that no solution in the subtree were better than the incumbent.

x2 = 1

x2 = 0

x2 = 1

x2 = 0

4

5

6

7

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

x3 = 1

8

9

10

11

12

13

Branch and Bound

The incumbent solution has value 42

44 3/7

1

x1 = 0

x1 = 1

2

3

44

44 3/7

4

42

44

44

44 1/3

13

43.75

43.5

43.25

43.8

44.3

-

We next solved the LP’s associated with nodes 8 -13

• We have solved 13 different linear programs so far.

• One integer solution found

• One subtree fathomed (pruned) because the solution was integer (node 4)

• One subtree fathomed because the solution was infeasible (node 13)

• No subtrees fathomed because of the bound

x2 = 1

x2 = 0

x2 = 1

x2 = 0

5

6

7

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

8

9

10

11

12

Branch and Bound

The incumbent solution has value 42

44 3/7

1

x1 = 0

x1 = 1

2

3

44

44 3/7

4

42

44

44

44 1/3

x3 = 1

43.75

43.5

43.25

43.8

44.3

-

13

43.75

42.66

We next solved the LP’s associated with the next nodes.

We can fathom the node with z = 42.66. Why?

• The bound at each node is obtained by solving an LP.

• But all costs are integer, and so the objective value of each integer solution is integer. So, the best integer solution has an integer objective value.

• If the best integer valued solution for a node is at most 42.66, then we know the best bound is at most 42.

• Other bounds can also be rounded down.

x2 = 1

x2 = 0

x2 = 1

x2 = 0

5

6

7

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

8

9

10

11

12

Branch and Bound

The incumbent solution has value 42

44 3/7

1

x1 = 0

x1 = 1

2

3

44

44 3/7

4

42

44

44

44 1/3

x3 = 1

43.75

43.5

43.25

43.8

44.3

-.

13

43.75

42.66

x2 = 1

x2 = 0

x2 = 1

x2 = 0

5

6

7

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

8

9

10

11

12

Branch and Bound

The incumbent solution has value 42

44

1

x1 = 0

x1 = 1

2

3

44

44

4

42

44

44

44

x3 = 1

43

43

43

43

44

-

13

43

42

43

42

43

43

We found a new incumbent solution!

x1 = 1, x2 = x3 = 0, x4 = 1, x5 = 0, x6 = 1 z = 43

x2 = 1

x2 = 0

x2 = 1

x2 = 0

5

6

7

Branch and Bound

The new incumbent solution has value 43

44

1

x1 = 0

x1 = 1

2

3

44

44

4

42

44

44

44

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

x3 = 1

8

9

10

11

43

43

43

43

44

8

9

10

11

12

-

13

43

42

43

42

43

43

We found a new incumbent solution!

x1 = 1, x2 = x3 = 0, x4 = 1, x5 = 0, x6 = 1 z = 43

x2 = 1

x2 = 0

x2 = 1

x2 = 0

5

6

7

Branch and Bound

The new incumbent solution has value 43

44

1

x1 = 0

x1 = 1

2

3

44

44

4

42

44

44

44

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

x3 = 1

8

9

10

11

43

43

43

43

44

8

9

10

11

12

-

13

If we had found this incumbent earlier, we could have saved some searching.

x2 = 1

x2 = 0

x2 = 1

x2 = 0

5

6

7

44

14

15

-

44

16

17

-

38

-

18

19

Finishing Up

The new incumbent solution has value 43

44

1

x1 = 0

x1 = 1

2

3

44

44

4

42

44

44

44

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

x3 = 1

8

9

10

11

43

43

43

43

44

8

9

10

11

12

-

13

• Branch and Bound can speed up the search

• Only 25 nodes (linear programs) were evaluated

• Other nodes were fathomed

• Obtaining a good incumbent earlier can be valuable

• only 19 nodes would have been evaluated.

• Solve linear programs faster, because we start with an excellent or optimal solution

• uses a technique called the dual simplex method

• Obtaining better bounds can be valuable.

• We sometimes use properties that are obvious to us, such as the fact that integer solutions have integer solution values

Notation:

• z* = optimal integer solution value

• Subdivision: a node of the B&B Tree

• Incumbent: the best solution on hand

• zI: value of the incumbent

• zLP: value of the LP relaxation of the current node

• Children of a node: the two problems created for a node, e.g., by saying xj = 1 or xj = 0.

• LIST: the collection of active (not fathomed) nodes, with no active children.

NOTE: zIz*

1

x1 = 0

x1 = 1

2

3

44

44 3/7

x2 = 1

x2 = 0

x2 = 1

x2 = 0

4

4

5

6

7

42

44

44

44 1/3

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

x3 = 1

13

43.75

8

43.5

9

43.25

10

43.8

11

44.3

12

-

13

Illustrating the definitions

The incumbent solution has value 42

z* = 43 = optimal integer solution value. (We found it later in the search)

Incumbent is 0 0 1 0 1 1 zI = 42. It is the optimal solution for the subdivision 4.

The zLP values for each subdivision are next to the nodes.

1

x1 = 0

x1 = 1

2

3

44

44 3/7

x2 = 1

x2 = 0

x2 = 1

x2 = 0

4

4

5

6

7

42

44

44

44 1/3

x3 = 0

x3 = 1

x3 = 0

x3 = 1

x3 = 0

x3 = 1

13

43.75

8

43.5

9

43.25

10

43.8

11

44.3

12

-

13

Illustrating the definitions

The incumbent solution has value 42

The children of node (subdivision) 1 are nodes 2 and 3.The children of node 3 are nodes 6 and 7.

LIST = { 8, 9, 10, 11, 12 } = unfathomed nodes with no active children

44 3/7

-

13

Branch and Bound Algorithm

INITIALIZE LIST = {original problem}Incumbent: = zI = -

SELECT:

If LIST = , then the Incumbent is optimal if it exists, and the problem is infeasible if no incumbent exists;

else, let S be a node (subdivision) from LIST.

Let xLP be the optimal solution to S

Let zLP = its objective value

e.g., S = {1}

e.g., S = {13}

13

CASE 1. zLP = - (the LP is infeasible)

Remove S from LIST (fathom it)

8

8

42

43

14

Branch and Bound Algorithm

INITIALIZE

SELECT:

If LIST = , then the Incumbent is optimal (if it exists), and the problem is infeasible if no incumbent exists;

else, let S be a node from LIST.

Let xLP be the optimal solution to S

Let zLP = its objective value

CASE 2. - < zLP  zI.

That is, the LP is dominated by the incumbent.

Then remove S from LIST (fathom it)

e.g., the incumbent has value 43, and node 14 is selected. zLP = 43.

14

42

Branch and Bound Algorithm

INITIALIZE

SELECT:

If LIST = , then the Incumbent is optimal (if it exists), and the problem is infeasible if no incumbent exists;

else, let S be a subdivision from LIST.

Let xLP be the optimal solution to S

Let zLP = its objective value

CASE 3. zI < zLP and xLP is integral.

That is, the LP solution is integral and dominates the incumbent.

Then Incumbent := xLP;zI := zLP

Remove S from LIST (fathomed by integrality)

e.g., node 4 was selected, and the solution to the LP was integer-valued.

4

x2 = 1

7

x1 = 1

3

x2 = 0

6

Branch and Bound Algorithm

INITIALIZE

SELECT:

If LIST = , then the Incumbent is optimal (if it exists), and the problem is infeasible if no incumbent exists;

else, let S be a subdivision from LIST.

Let xLP be the optimal solution to S

Let zLP = its objective value

CASE 4. zI < zLP and xLP is not integral.

There is not enough information to fathom S

Remove S from LIST

Add the children of S to LIST

e.g., select node 3.

44 3/7

List := List – 3 + {6,7}

• Rule of Thumb 1:Don’t let LIST get too big (the solutions must be stored). So, prefer nodes that are further down in the tree.

• Rule of Thumb 2:Pick a node of LIST that is likely to lead to an improved incumbent. Sometimes special heuristics are used to come up with a good incumbent.

= 0

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One does not have to have the B&B tree be symmetric, and one does not select subtrees by considering variables in order.

Choosing how to branch so as to reduce running time is largely “art” and based on experience.

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• Branching: determining children for a node. There are many choices.

• Rule of thumb 1: if it appears clear that xj = 1 in an optimal solution, it is often good to branch on xj = 0 vs xj = 1.

• The hope is that a subdivision with xj = 0 can be pruned.

• Rule of thumb 2: branching on important variables is worthwhile

• We use the bound obtained by dropping the integrality constraints (LP relaxation). There are other choices.

• Key tradeoff for bounds: time to obtain a bound vs quality of the bound.

• If one can obtain a bound much quicker, sometimes we would be willing to get a bound that is worse

• It usually is worthwhile to get a bound that is better, so long as it doesn’t take too long.

• One can choose children as follows:

• child 1: x1 3 (or xj k)

• child 2 x1 4 (or xj k+1)

• How would one choose the variable j and the value k

• A common choice would be to take a fractional value from xLP. e.g., if x7 = 5.62, then we may branch on x7 5 and x7 6.

• Other choices are also possible.

• Branch and Bound is the standard way of solving IPs to optimality.

• There is art to making it work well in practice.

• Much of the art is built into state-of-the-art solvers such as CPLEX.

New exciting area

• combining LP based branch-and-bound based techniques withconstraint programming techniques - forward checking; arc-consistency and other forms of consistency checking and propagation!

A* - well-known AI algorithm that is a generalization of branch-and-bound with LP relaxations – notion of admissible heuristic that overestimates (underestimates) the objective function for a maximization (minimization) problem (analogously to what a relaxation does)

Used in e.g. MapQuest and Darpa Challenge

It was fun to teach INFO 372! branch-and-bound with LP relaxations – notion of admissible heuristic that overestimates (underestimates) the objective function for a maximization (minimization) problem (analogously to what a relaxation does)

Hope you had fun too!

THE END

!!!