od integer programming large 2010

7/24/2019 OD Integer Programming LARGE 2010

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INTEGERPROGRAMMING


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Integer Programming

In many problems the decision variables must have

integer values. Example: assign people, machines, and vehicles to

activities in integer quantities.

If this is the only deviation from linear programming,

it is called an integer programming (IP) problem.

If only somevariables are required to integer, the

model is called a mixed integer programming (MIP)

San Francisco Police Dep. problem is an IP problem. Wyndor Glass Co. problem could be an IP problem;

how?

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Integer Programming

In integer programming, divisibility assumption must

be dropped. Another area of application is in problem involving

yes-or-no decisions, which have binary variables.

These IP problems are called binary integer

programming (BIP) problems.

A very small example of a typical BIP problem is given

in the following.

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Prototype example

California Manufacturing Company is considering expansion

building a factory in Los Angeles, San Francisco or in both cities. One new warehouse can also be considered in a city where a

new factory is being built. Maximum $10 million to invest.

Objective: find feasible combination of alternatives that

maximizes the total net present value.

266

Decision

number

Yes or no question Decision

variable

Net present

value

Capital

required

1 Build factory in Los Angeles? x1 $9 million $6 million

2 Build factory in San Francisco? x

2 $5 million $3 million3 Build warehouse in Los Angeles? x3 $6 million $5 million

4 Build warehouse in San Francisco? x4 $4 million $2 million


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Contingent decisions

Mutually exclusive alternatives

BIP model

All decisions variables have the binary form:

Z= total net present value of these decisions.

Maximize Z= 9x1 + 5x2 + 6x3 + 4x4.

Constraints:

6x1 + 3x2 + 5x3 + 2x4 10

x3 +x4 1

x3 x1 and x4 x2xj is binary, forj = 1,2,3,4.

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1 if decision is yes,1,2,3,4

0 if decision is no,j

jx j

j


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BIP models

Groups of yes-or-no decision often constitute groups

ofmutually exclusive alternatives: only onedecisionin the group can be yes.

Occasionally, decisions of the yes-or-no type are

contingent decisions: decision that depend upon

previous ones.

Software options for solving BIP, IP or MIP models:

Excel

LINGO/LINDO MPL/CPLEX

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BIP applications

Investment analysis, such as the California Man. Co.

Site select ion, of factories, warehouses, etc.

Designing a product ion and dist ribut ion network, or

more generally the entire global supply-chain.

Dispatching shipments, scheduling routes, vehiclesand time period for departure and arrivals.

Airline applicat ions, as e.g. fleet assignment and crew

scheduling.

Scheduling interrelated act ivit ies, asset divestures,etc.

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Formulation examples

Example 1: making choices when decision variables

are continuous. R&D Division ofGood Products Co. hasdeveloped three possible new products.

Requirement 1: from the three, at mosttwo can be

chosen to be produced.

Each product can be produced in either of two plants.However, management has imposed a restriction:

Requirement 2:just one of the two plants can be

chosen as the producer of the new products.

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Example 1

Objectives: choose the products, the plant and the

production rates of the chosen products to maximizetotal profit.

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Production time used for each unit

produced

Production time

available per

week

Product 1 Product 2 Product 3

Plant 1 3 hours 4 hours 2 hours 30 hours

Plant 2 4 hours 6 hours 2 hours 40 hours

Unit profit 5 7 3 (103 euros)

Sale potential 7 5 9 (units per week)


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Formulation of the problem

Similar to a standard product mix problem, such as the Wyndor

Glass Co. It is such a problem, if we drop the two restrictions and by

requiring that a product uses production hours in both plants.

Letx1,x2,x3 be the production rates of the respective products:

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1 2 3Maximize 5 7 3Z x x x

1 2 3

1 2 3

1

2

3 1 2 3

3 4 2 30

4 6 2 40

75

9, , , 0

x x x

x x x

xx

x x x x

subject to


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For the real problem, Requirement 1 add the

constraint:Number of strictly positive variables (x1,x2,x3) must be 2

This must be converted to an IP problem. It needs the

introduction of auxiliary binary variables.

Requirement 2 requires replacing the first twoconstraints to:

must hold. This again requires an auxiliary binary

variable.

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1 2 3

1 2 3

Either 3 4 2 30

or 4 6 2 40

x x x

x x x


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Auxiliary binary variables

For Requirement 1, three auxiliary binary variables (y1,

y2,y3) are introduced:

This is introduced in the model with the help of anextremely large positive numberM, adding the

constraints:

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1 if 0 can hold (can produce product )

0 if 0 must hold (cannot produce product )

j

j

j

x jy

x j

1 1

2 2

3 3

1 2 3 2

is binary, for 1, 2,3.j

x My

x My

x My

y y y

y j


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Auxiliary binary variables

For Requirement 2, another auxiliary binary variabley4

is introduced:

This add the constraints:

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1 2 3

4

1 2 3

1 if 4 6 2 40 must hold (choose Plant 1)

0 if 3 4 2 30 must hold (choose Plant 2)

x x xy

x x x

1 2 3 4

1 2 3 4

4

3 4 2 30

4 6 2 40 (1 )

is binary

x x x My

x x x M y

y


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Complete model (MIP)

276

1 2 3

1

2

3

1 1

2 2

3 3

1 2 3

1 2 3 4

1 2 3 4

Maximize 5 7 3

subject to 75

9

0

00

2

3 4 2 30

4 6 2 40

and 0, for 1, 2,3

is binary, for 1,2,3, 4

i

j

Z x x x

xx

x

x My

x My

x My

y y y

x x x My

x x x My M

x i

y j


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Solution

MIP problem with 3 continuous and 4 binary variables.

Optimal solution:y1 = 1,y2 = 0,y3 = 1,y4 = 1,x1 = 5.5,x2 = 0,x3 = 9.

That is, choose products 1 and 3 to produce, choose

Plant 2 for production, choose production rates of 5.5

units per week for product 1 and 9 units per week for

product 2.

Resulting profit is 54500 per week.

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Example 2

See Hilliers book (page 496). Profit from additional sales

violates proportionality:

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Example: Southwestern Airways

Southwestern Airways needs to assign three crews to

cover all the upcoming flights. Table shows the flights in the first column.

Other 12 columns show the 12 feasible sequences of flights

for a crew.

Numbers in each column indicate the order of the flights.

Exactly three sequences must be chosen (one for each crew).

More than one crew can be assigned to a flight, but it must

be paid as if it was working.

Last column show the cost of assigning a crew to a sequenceof flights.

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Data for Southwestern Airways

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Feasible sequence of flights

Flight 1 2 3 4 5 6 7 8 9 10 11 12

1. San Francisco to Los Angeles 1 1 1 1

2. San Francisco to Denver 1 1 1 1

3. San Francisco to Seattle 1 1 1 1

4. Los Angeles to Chicago 2 2 3 2 3

5. Los Angeles to San Francisco 2 3 5 5

6. Chicago to Denver 3 3 4

7. Chicago to Seattle 3 3 3 3 4

8. Denver to San Francisco 2 4 4 5

9. Denver to Chicago 2 2 2

10. Seattle to San Francisco 2 4 4 5

11. Seattle to Los Angeles 2 2 4 4 2

Cost (1000) 2 3 4 6 7 5 7 8 9 9 8 9


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Objective: minimize the total cost for the three crews.

12 feasible sequence of flights: 12 yes-or-no decisions:

Should sequencejbe assigned to a crew?

The 12 binary variables to represent the decisions are:

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1 if sequence is assigned to a crew

0 otherwisej

jx


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1 2 3 4 5 6 7 8 9

10 11 12

Minimize 2 3 4 6 7 5 7 8 9

9 8 9

Z x x x x x x x x x

x x x

and is binary, for 1,2, ,12jx j

subject to1 4 7 10

2 5 8 11

3 6 9 12

4 7 9 10 12

1 6 10 11

4 5 9

7 8 10 11 12

2 4 5 9

5 8 11

3 7 8 12

6 9 10 11 1212

1

1 (SF to LA)

1

1

1

11

1

1

1

1

1

3 (assign three crewsjj

x x x x

x x x x

x x x x

x x x x x

x x x xx x x

x x x x x

x x x x

x x x

x x x x

x x x x x

x

)


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Solution

One optimal solution is:

x3 = 1 (assign sequence 3 to a crew)



And all otherxj

= 0.

Total cost is 18000.

Another optimal solution is: x1 =x5 =x12 = 1.

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Discussion

This example belongs to a class called set covering

problems, with a number of potential activities(e.g.flight sequences) and characteristics(e.g. flights).

Si is the set of all activities that possess characteristici.

A constraint is included for each characteristici:

In set partitioning problems the constraint is

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1j

j

j Sx

1i

j

j S

x

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Solving IP problems

Integer problems are easy to solve?

Difference to LP is that IP have far fewer solutions.

IP problems have a finitenumber of feasible solutions.

However:

Finite numbers can be astronomically large! With n

variables a BIP problem has 2n solutions, having

exponential growth.

LP assuresthat a CFP solution can be optimal,

guaranteeing the remarkable efficiency of the simplex

method. LP problems are much easier t o solve than IP

problems.

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l i bl


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Solving IP problems

Consequently, most IP algorithms incorporate the

simplex method. This is called the LP relaxation. Sometimes, the solution of the LP problem isthe

solution of the IP problem, such as:

Minimum cost flow problem, including transportation

problem, assignment problem, shortest -path problem

and maximum f low problem.

Special structures (see examples 2 and 3): mutually

exclusive alternat ives, cont ingent decisionsor set-

covering const raintscan also simplify the problem.

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S l i I bl


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Solving IP problems

Primary determinants of computational complexity:

1. number of integer variables,2. these variables are binaryor generalinteger variables,

3. any special structure in the problem.

This is in contrast to LP, where number of constraints

is much more important than the number of

variables.

As IP problems are much more difficult than LP, we

could apply LP and roundthe obtained solution...Yes??

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E l 1


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Example 1

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2Minimize Z x

1 2

1 2

1 2

subject to 0.5

3.5,

and , 0, integers.

x x

x x

x x

E l 2


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Example 2

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1 2Minimize 5Z x x

1 2

1

1 2

subject to 10 20

2

and , 0, integers.

x x

x

x x

S l i IP bl


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Solving IP problems

Thus, a better approach to deal with IP problems that

are too large to be solved exactly are heuristicalgorithms.

Heuristics and metaheuristics are extremely efficient

for very large problems, but do not guaranteeto find

an optimal solution. These algorithms will be discussed further later.

Most popular traditional method for solving IP

problems is the branch-and-bound technique.

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B h d b d li d t BIP


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Branch-and-bound applied to BIP

Pure IP problems can consider some type of

enumerat ion procedure. This should be done in a clever way such that only a

tiny fraction of the feasible solutions is examined.

Branch-and-bound with a divide to conquertechnique

can be used.

dividing (branching) the problem into smaller and

smaller subproblems until it can be conquered

conquering (fathoming) by boundinghow good the

best solution can be. If no optimal solution in subset:

discard it.

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E l C lif i M f C


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Example: California Manuf, Co.

Recall prototype example:

Maximize Z= 9x1 + 5x2 + 6x3 + 4x4

subject to

(1) 6x1 + 3x2 + 5x3 + 2x4 10

(2) x3 + x4 1(3) x1 + x3 0

(4) x2 + x4 0

and

(5) xj is binary, forj = 1, 2, 3, 4.

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B hi


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Branching

Most straightforward way to divide the problem: fix

the value of a variable at one set. e.g.x1 = 0 for one subset andx1 = 1 for another subset.

Subproblem 1(fixx1 = 0):

Maximize Z= 5x2 + 6x3 + 4x4

subject to(1) 3x2 + 5x3 + 2x4 10

(2) x3 + x4 1

(3) x3 0

(4) x2 + x4 0(5) xj is binary, forj = 2, 3, 4.

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B hi


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Branching

Subproblem 2(fixx1 = 1):

Maximize Z= 9 + 5x2 + 6x3 + 4x4

subject to

(1) 3x2 + 5x3 + 2x4 4

(2) x3

+ x4

1

(3) x3 1

(4) x2 + x4 0

(5) xj is binary, forj = 2, 3, 4.

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B hi


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Branching

Dividing (branching) into suproblems creates a tree

with branches(arcs) for theAllnode. This is the solution tree or enumeration tree.

Branching variable is the one used for branching.

The branching continues or not after evaluating the

subproblem.

Other IP problems usually creates as many branches as

needed.

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B di


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Bounding

A boundis needed for the best feasible solution of the

subproblem. Standard way is to perform a relaxation of the

problem, e.g. by deletingone set of constraints that

makes the problem difficult.

Most common is to require integer variables, so LP

relaxationis the most widely used.

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Bounding in example


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Bounding in example

Example: for the whole problem, (5) is

replaced byxj 1 andxj 0 forj=1, 2, 3, 4.Using simplex:

(x1,x2,x3,x4) = (5/6, 1, 0, 1), with Z = 16.5

Thus, Z 16.5 for all feasible solutions for BIPproblem. Can be rounded toZ 16 (why?)

LP relaxation for subproblem 1:

(x1,x2,x3,x4) = (0, 1, 0, 1), with Z = 9

LP relaxation for subproblem 2:

(x1,x2,x3,x4) = (1, 4/5, 0, 4/5), with Z = 16.5

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Fathoming


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Fathoming

A subproblem can be conquered (fathomed, i.e.

search tree is pruned) in three ways:1. When the optimal solution for the LP relaxation of a

subproblem is integer, it must be optimal.

Example: forx1=0, (x1,x2,x3,x4) = (0, 1, 0, 1), is integer.

It must be stored as first incumbent (best feasible solutionfound so far) for the whole problem, along with value ofZ:

Z* = value ofZ for first incumbent

In the exampleZ* = 9.

Subproblem 1 is solved, so it is fathomed (dismissed).

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Fathoming


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Fathoming

2. AsZ* = 9, we should not consider subproblems with

bound 9. Thus, a problem is fathomed whenboundZ*

In Subproblem 2 that does not occur, the bound of16 is

larger than 9. However, it can occur for descendants.

As new incumbents with larger values ofZ* are found, itbecomes easier to fathom in this way.

3. If the simplex method finds that a subproblems LP

relaxation has no feasible solut ion, the subproblem

has nofeasible solution and can dismissed.

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Summary of fathoming tests


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Summary of fathoming tests

A subproblem is fathomed (dismissed) if

Test 1: ItsboundZ*

or

Test 2: Its LP relaxation has no feasible solutions

or Test 3: Optimal solution for its LP relaxation is

integer.

If better, this solution becomes new incumbent, and

Test 1 is reapplied for all unfathomed subproblems.

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Fathoming in example


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Fathoming in example

Result of applying the three tests is in figure below.

Subproblem 1 is fathomed by test 3.

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BIP branch and bound algorithm


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BIP branch-and-bound algorithm

Initialization:SetZ* = . Apply bounding, fathoming

and optimization steps described below to the wholeproblem. If not fathomed, perform iteration.

Steps for each it erat ion:

1. Branching: Among the remaining subproblems, select

the one created most recently. Branch from this nodeby fixing the next variable as either 0 or 1.

2. Bounding: For each new subproblem, obtain its

boundby applying its LP relaxation.

Round downZfor resulting optimal solution.

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BIP branch and bound algorithm


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BIP branch-and-bound algorithm

3. Fathoming: For each new subproblem, apply the

three fathoming tests, and discard subproblems thatare fathomed by the tests.

Optimalit y test :Stop when there are no remaining

subproblems. The current incumbent is optimal.

Otherwise, perform another iteration.

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Completing example


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Completing example

Iteration 2. Remaining subproblems are forx1 = 1.

Subproblem 3(fixx1 = 1,x2 = 0):Maximize Z= 9 + 6x3 + 4x4

subject to

(1) 5x3 + 2x4 4(2) x3 + x4 1

(3) x3 1

(4) x4 0

(5) xj is binary, forj = 3, 4.

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Example


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Example

Subproblem 4(fixx1 = 1,x2 = 1):

Maximize Z= 14 + 6x3 + 4x4

subject to

(1) 5x3 + 2x4 1

(2) x3 + x4 1(3) x3 1

(4) x4 1

(5) xj is binary, forj = 3, 4.

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Example


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Example

LP relaxation is obtained by replacing (5) by 0xj 1 j

= 3, 4. Optimal solutions are: LP relaxation for Subproblem 3:

(x1,x2,x3,x4) = (1, 1, 0.8, 0), withZ= 13.8

LP relaxation for Subproblem 4:

(x1,x2,x3,x4) = (1, 1, 0, 0.5), withZ = 16

Resulting bounds:

Bound for subproblem 3: Z 13


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Example


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Example

All three fathoming tests fail, so both are unfathomed.

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Iteration 3


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Iteration 3

Subproblem 4 has the larger bound, so next branching

is done from (x1,x2) = (1, 1). Subproblem 5(fixx1 = 1,x2 = 1,x3 = 0):

Maximize Z= 14 + 4x4

subject to

(1) 5x3 + 2x4 1

(2), (4) x4 1

(5) x4 is binary

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Iteration 3 (cont )


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Iteration 3 (cont.)

Subproblem 6(fixx1 = 1,x2 = 1,x3 = 1):

Maximize Z= 20 + 4x4

subject to

(1) 2x44

(2) x4 0

(4) x4 1(5) x4 is binary

LP relaxation: replace (5) by 0 x4 1. Optimal solutions are:

LP relaxation for subproblem 5: (x1,x2,x3,x4) = (1, 1, 0, 0.5),Z= 16

LP relaxation for subproblem 6: No feasible solutions.


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Iteration 3 (concl )


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Iteration 3 (concl.)

Subproblem 6 is fathomed by test 2, but not

Subproblem 5.

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Iteration 4


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Iteration 4

Node created most recently is selected for branching:

x4 = 0: (x1,x2,x3,x4) = (1,1,0,0) is feasible, withZ= 14

x4 = 1: (x1,x2,x3,x4) = (1,1,0,1) is infeasible.

First solution passes test 3 and second passes test 2

for fathoming.

First solution is better than incumbent, so it becomes

new incumbent, withZ* = 14

Reapplying fathoming test 1 to remaining branch to

remaining Subproblem 3, it is fathomed: Bound = 13 Z* = 14.

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Solution tree after Iteration 4


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Solution tree after Iteration 4

312

Other options in B&B


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Other options in B&B

Branching can be done e.g. from the best boundrather

than from the most recent ly createdsubproblem. Bounding is done by solving a relaxation. Another

possible one is e.g. the Lagrangian relaxation.

Fathoming criteria can be generally stated as:

Criterion 1: feasible solutions of subproblem must have

Z Z*

Criterion 2: the subproblem has no feasible solutions, or

Criterion 3: an optimal solution of subproblem has been

found.

Branch-and-bound can find a nearly opt imal solut ion.

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Branch-and-bound for MIP


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Branch and bound for MIP

General form of the problem:

314

1

1

Maximize

subject to , for 1,2, , ,

0, for 1,2, ,and

is integer, for 1, 2, , ; .

n

j j

j

n

ij j ij

j

j

Z c x

a x b i m

x j n

x j I I n



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Similar to BIP algorithm. Solving LP relaxations are

the basis for boundingand fathoming.

Four changes are needed:

1. Choice ofbranching variable. Only integer variables

that have a noninteger valuein the optimal solution

for the LP relaxation can be chosen.

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2. As integer variables can have a large number of

possible values, create just twonew subproblems: xj

*: noninteger value of optimal solution for LP

relaxation.

[xj*] = greatest integer xj

*.

Range of variables for two new subproblems:

xj* [xj

*] and xj* [xj

*] + 1.

Each inequality becomes an addit ional constraint.

Example:xj

* = 3.5, then:xj

* 3 and xj

* 4.

When changes 1. and 2. are combined, a recurring

branching variablecan occur, see figure.

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Recurring branching variable


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Recurring branching variable

317



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Changes needed:

3. Bounding step: value ofZwas rounded down in BIPalgorithm. Now some variables are not integer-

restricted so bound is value ofZwithoutrounding.

4. Fathoming test 3: optimal solution for the

subproblems LP relaxation must only be integerforinteger-restrictedvariables.

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MIP branch-and-bound algorithm


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MIP branch and bound algorithm

Initialization:SetZ* =. Apply bounding, fathoming

and optimization steps described below to the wholeproblem. If not fathomed, perform iteration.

Steps for each it erat ion:

1. Branching: Among the remaining subproblems, select

the one created most recently.

From integer variablesthat have a noninteger value

in the optimal solution for the LP relaxation chose the

first one. Letxj be this variable andxj* its value.

Branch from this creating two subproblems by adding

the respective constraints:xj* [xj

*] andxj* [xj

*] + 1.

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2. Bounding: For each new subproblem, obtain its

boundby applying its LP relaxation. UseZwithoutrounding for resulting optimal solution.

3. Fathoming: For each new subproblem, apply the

three fathoming tests, and discard subproblems that

are fathomed by the tests.Test 1: Itsbound Z*, whereZ* is value ofZfor current

incumbent.

Test 2: Its LP relaxation has no feasible solutions.

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3. Fathoming (cont.):

Test 3: Optimal solution for its LP relaxation is integerforinteger-restrictedvariables. (if this solution is better it

becomes new incumbent, and test 1 is reapplied for all

unfathomed subproblems).

Optimality test: Stop when there are no remainingsubproblems. The current incumbent is optimal.

Otherwise, perform another iteration.

See MIP example in PL#7 and in page 518 of Hilliersbook.

321

Branch-and-cut approach to BIP


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Branch and cut approach to BIP

Branch-and-bound was develop and refined in the

1960s and early 1970s. Can solve problems up to 100 variables.

Branch-and-cut approach was introduced in the mid

1980s, which can solve problems with thousands

variables. Only solve large problems if they are sparse (less than 5

or even 1% ofnonzerovalues in the functional

constraints).

Uses a combination of: automatic problem processing,generation of cutting planes and B&B techniques.

322

Automat ic problem processing for BIP


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Automat ic problem processingfor BIP

Computer inspection of IP formulation to spot

reformulationsthat make the problem quicker tosolve:

Fixing variables: identify variables that can be fixed at 0

or 1, because other value cannot be optimal.

Eliminating redundant constraints: identify andeliminate constraints that are automatically satisfied by

solutions that satisfy all other constraints.

Tightening constraints: tighten constraints in a way that

reduces feasible region of LP relaxation.

323

Tightening constraints


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Tightening constraints

324

LP relaxation including

feasible region.

LP relaxation after tightening

constraint.

Generating cutting planes for BIP


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g g p

Cutting plane (or cut) is a new functional constraint

that reduced feasible region for LP relaxation.Procedure for generating cutting planes:

1. Consider functional constraint in form.

2. Find a group ofNvariables such that

a) Constraint is violated if every variable in group = 1 and

all other variables = 0.

b) It is satisfied if value ofanyvariables changes from 1

to 0.

3. Resulting cutting plane:

sum of variables in group N 1.

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Constraint Programming


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g g

Combination of artificial intelligence with computer

programming languages in the mid-1980s. Flexibilityin stating (nonlinear) constraints:

1. Mathematical constraints, e.g.,x +y


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g

4. Explicit constraints, e.g.,x andy have same domain

{1,2,3,4,5}, but (x,y) must be (1, 1), (2, 3) or (4, 5).5. Unary constraints, e.g.z is integer between 5 and 10.

6. Logical constraints, e.g., ifx = 5,y {6, 7, 8}.

Standard logical functions such as IF, AND, OR, NOT

can be used.

Constraint programming applies domain reduct ion

and constraint propagat ion.

The process creates a t ree search.

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Example


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p

Consider:

x1{1,2},x2{1,2},x3{1,2,3},x4{1,2,3,4,5} Constraints:

1. Allvariables must have different values;

2. x1 +x3 = 4.

Apply domain reduction and constraint propagation

to obtain feasible solutions:

x1{1}, x2{2}, x3{3}, x4{4,5}.

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Constraint Programming


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g g

Steps in Constraint Programming:

1. Formulate a compact model of the problem by usinga variety of constraint types (most not of IP type).

2. Efficiently find feasible solutions that satisfy all these

constraints.

3. Search among feasible solutions for an optimal one.

Strength of constraint programming is in first two

steps, whereas the main strength of IP is in step 3.

Current research: integrate CP and IP!

od integer programming large 2010

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