INTEGER PROGRAMMING. Integer Programming. Prototype example. BIP model. BIP models

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INTEGER

PROGRAMMING

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 If this is the only deviation from linear programming, it

is called an integer programming (IP)problem. If only somevariables are required to be 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?

João Miguel da Costa Sousa / Alexandra Moutinho 261

Integer

Programming

In integer programming the divisibility assumption

must be dropped.

Another area of application relates to problems

involvingg “yes‐or‐no” decisions, which have binary variables.

These IP problems are called binary integer

programming (BIP)problems.

A small example of a typical BIP problem is given in

the following.

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 the total net present value. 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? x2 $5 million $3 million 3 Build warehouse in Los Angeles? x3 $6 million $5 million 4 Build warehouse in San Francisco? x4 $4 million $2 million

All decision variables have the binary form:

Z= total net present value of these decisions.

BIP

model

1 if decision is yes,

1,2,3,4 0 if decision is no,

j

x j

j

⎧

=⎨ =

⎩

Contingent decisions Mutually exclusive alternatives Maximize Z= 9x1+ 5x2+ 6x3+ 4x4.

Constraints:

6x₁+ 3x₂+ 5x₃+ 2x₄≤10

x₃+ x₄≤1

x₃≤x₁andx₄≤x₂ x_jis binary, for j= 1,2,3,4.

BIP

models

Groups of yes‐or‐no decisions often constitute groups

of mutually exclusive alternatives: only onedecision

in the group can be yes.

Occasionally, decisions of the yes‐or‐no type are contingent decisions: decision that depends upon contingent decisions: decision that depends upon

previous ones.

¾Software options for solving BIP, IP or MIP models:

Excel

MatLab LINGO/LINDO MPL/CPLEX

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BIP

applications

Investment analysis, such as the California Man. Co. Site selection, of factories, warehouses, etc. Designing a production and distribution network, or

more generally the entire global supply‐chain. Dispatching shipments, scheduling routes, vehicles

and time period for departure and arrivals.

Airline applications, as e.g. fleet assignment and crew

scheduling.

Scheduling interrelated activities, asset divestures, etc.

Formulation

examples

Example 1: making choices when decision variables

are continuous. R&D Division of Good ProductsCo.

has developed 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 another

restriction:

Requirement 2:just one of the two plants can be chosen as

the producer of the new products.

Example

1

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

Objectives:choose the products, the plant and the

production rates of the chosen products to maximize

total profit.

Unit profit 5 7 3 (103_$)

Sales

potential 7 5 9

(units per week)

Formulation

of

the

problem

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

Glass Co. if we drop the two restrictions andrequire each

product to use production hours in both plants.

Let x₁, x₂, x₃be the production rates of the respective products:

1 2 3

Maximize Z=5x +7x +3x3

1 2 3

subject to 3 4 2 30

4 6 2 40

7 5 9 , , 0

x x x x x x x x x x x x

+ + ≤

≤ ≤ ≤ ≥

Formulation

of

the

problem

For real problem, restriction 1 adds 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. Restriction 2 requires replacing the first two Restriction 2 requires replacing the first two

functional constraints by:

This again requires an auxiliary binary variable.

+ + ≤

1 2 3

Either 3 4 2 30

or 4 6 2 40 must hold.

x x x

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 an extremely

1 if 0 can hold (can produce product )

0 if 0 must hold (cannot produce product )

j j

j

x j

y

x j

> ⎧ = ⎨ ₌

⎩

This is introduced in the model with the help of an extremely

large positive number M, adding the constraints:

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 variable y4is introduced:

1 2 3

4

1 2 3

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

0 if 3 4 2 30 must hold (choose Plant 1)

x x x y

x x x

+ + ≤ ⎧

= ⎨ ₊ ₊ _≤ ⎩

This adds the constraints:

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

+ + ≤ + + + ≤ + −

Complete

model

(MIP)

1 2 3

1 1

Maximize 5 7 3

subject to 7

5 9 0 0

Z x x x

x x x x My x My = + + ≤ ≤ ≤ − ≤ ≤ 2 2 3 3

1 2 3

1 2 3 4

0 0 2

3 4 2 30

4 6 2 40

and 0, for 1,2,3

is binary, for 1,2,3,4 i

j

x My

y y y

x x x My

x x x My M

x i y j − ≤ − ≤ + + ≤ + + − ≤ + + + ≤ + ≥ = =

Solution

MIP problem with 3 continuous and four binary

variables. Optimal solution:

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

y₁ 1, y₂ 0, y₃ 1, y₄ 1, x₁ 5.5, x₂ 0, x₃ 9. That is, produce products 1 and 3 with production

rates 5.5 units per week and 9 units per week

respectively, and choose Plant 2 for production. Resulting total profit is $54,500 per week.

Example:

Southwestern

Airways

Southwestern Airways needs to assign three crews to

cover all the upcoming flights.

Tableshows the flights in the first column.

Other 12 columns show the 12 feasible sequences of flights

for a crew for a crew.

Numbers in each column indicate the order of the flights. Exactly three sequences must be chosen (one per crew). More than one crew can be assigned to a flight, but it must

be paid as if it was working.

Last row shows the cost of assigning a crew to a particular

sequence of flights.

Data

for

Southwestern

Airways

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

Formulation

of

the

problem

Objective:minimize the total cost for the three crew

assignments that cover all flights.

12 feasible sequences of flights: 12 yes‐or‐no

decisions:

Should sequence jbe assigned to a crew? The 12 binary variables to represent the decisions are:

1 if sequence is assigned to a crew 0 otherwise

j

j x = ⎨⎧

⎩

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Formulation

of

the

problem

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

= + + + + + + + +

+ + +

subject to _{+ + +} _≥

+ + + ≥

+ + + + ≥

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

9 0 2

1 (SF to LA) 1 1 1 x x x x x x x x x x x x x x x x x

and is binary, for 1,2, ,12

j x

j= …

+ + + + ≥ + + + ≥ + + ≥ + + + + ≥ + + + ≥ + + ≥ + + + ≥ + + + + ≥

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 12

1 1 1 1 1 1 1 1 x x x x x

x x x x x x x x x x x x

x x x x x x x x x x x x x x x x

João Miguel da Costa Sousa / Alexandra Moutinho = 278

= ∑12

1

3 (assign three crews)

j j

x

Solution

One optimal solution is:

x3= 1 (assign sequence 3 to a crew) x₄= 1 (assign sequence 4 to a crew) x11= 1 (assign sequence 11 to a crew) And all other x_j= 0.

Total cost is $18,000.

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

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).

Objective: determine the least costly combination of activities

that collectively possess each characteristic at least once. Siis the set of all activities that possess characteristic i.

A constraint is included for each characteristic i:

In set partitioning problemsthe constraint is

1 j j j S x ∈ =

∑

1 i j j S x ∈ ≥

∑

Solving

IP

problems

Are integer problems easy to solve?

Difference to LP is that IP have far fewer solutions. IP problems have a finitenumber of feasible solutions.

However:

¾Fi it b b t i ll l ! With

¾Finite numbers can be astronomically large! With n variables a BIP problem has 2n_solutions,_having exponential growth.

¾LP assuresthat a CPF solution can be optimal,

guaranteeing the remarkable efficiency of the simplex

method. LP problems are much easier to solve than IP LP problems are much easier to solve than IP problems

problems!!

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 problemand

maximum flow problem.

Special structures (see examples 2 and 3): mutually

exclusive alternatives, contingent decisionsor set‐

covering constraintscan also simplify the problem. João Miguel da Costa Sousa / Alexandra Moutinho 282

Solving

IP

problems

Primary determinants of computational complexity: 1. number of integer variables,

2. these variables are binaryor generalinteger variables, 3. any special structurein 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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Example

1

2

1 2

Minimize

subject to 0.5

3.5

and , 0,

Z x

x x

x x

= − + ≤

+ ≤ ≥

1, 2 integers.

x x

Example

2

1 2

Minimize Z x= +5x 1 2 1

1 2

subject to 10 20 2 and , 0, integers.

x x

x x x

+ ≤ ≤ ≥

Solving

IP

problems

Thus, a better approach to deal with IP problems that

are too large to be solved exactly are heuristic

algorithms.

Heuristics and metaheuristicsare extremely efficient

for very large problems, but do not guaranteeto find

an optimal solution.

These algorithms will be discussed later.

Most popular traditional method for solving IP

problems is the branch‐and‐bound technique.

Branch

‐

and

‐

bound

applied

to

BIP

Pure IP problems can consider some type of

enumeration procedure.

This should be done in a clever way such that only a

tiny fraction of the feasible solutions is examined. Branch‐and‐boundwith a divide to conquer

technique 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. João Miguel da Costa Sousa / Alexandra Moutinho 287

Example:

California

Manuf,

Co.

Recall prototype example: Maximize Z= 9x1+ 5x2+ 6x3+ 4x4 subject to

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

(2) x₃+ x₄≤1

(3) –x1+ x3 ≤0

(4) –x₂+ x₄≤0

and

(5) xjis binary, for j= 1, 2, 3, 4.

Branching

Most straightforward way to divide the problem:

fix the value of a variable:

e.g. x₁= 0 for one subset and x₁= 1 for another subset. ¾Subproblem 1(fixx1= 0): ¾Subproblem 2(fixx = 1):

p 1

MaximizeZ= 5x2+ 6x3+ 4x4 subject to

(1) 3x2+ 5x3+ 2x4≤10 (2) x3+ x4≤1 (3) x3 ≤0 (4) –x2+ x4≤0 (5) xjis binary, forj= 2, 3, 4.

¾Subproblem 2(fixx1 1): MaximizeZ= 9 + 5x2+ 6x3+ 4x4 subject to

(1) 3x2+ 5x3+ 2x4≤4 (2) x3+ x4≤1 (3) x3 ≤1 (4) –x2+ x4≤0 (5) xjis binary, forj= 2, 3, 4.

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Branching

Dividing (branching) into suproblems creates a tree

with branches(arcs) for the Allnode. This is the solution treeor enumeration tree. Branching variableis the one used for branching. Branching variableis the one used for branching. The branching continues or not after evaluating the

subproblem.

Other IP problems usually creates as many branches

as needed.

Bounding

A boundis needed for the best feasible solution of

each of the subproblems.

Standard way is to perform a relaxationof the

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

makes the problem difficult to solve.

Most common is to require integer variables, so LP

relaxationis the most widely used.

Bounding

in

example

Example: for the whole problem, (5) is

replaced by x_j≤1 and x_j≥0 for j=1,2,3,4. Using

simplex:

(x₁,x₂,x₃,x₄) = (5/6, 1, 0, 1), with Z = 16.5 Thus,, Z ≤16.55 for all feasible solutions for BIP

problem. Can be rounded to Z≤16 (why?) LP relaxation for subproblem 1(x₁=0):

(x₁,x₂,x₃,x₄) = (0, 1, 0, 1), with Z = 9 LP relaxation for subproblem 2(x1=1):

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

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.

E l f ( ) ( ) i i

Example: for x₁=0, (x₁,x₂,x₃,x₄) = (0, 1, 0, 1), is integer. It must be stored as first incumbent(best feasible

solution found so far) for the whole problem, along with

value of Z:

Z*₌_value_of_Z_for_first_incumbent In the example Z*₌_9.

Subproblem 1 is solved, so it is fathomed(dismissed). João Miguel da Costa Sousa / Alexandra Moutinho 293

Fathoming

2. As Z*₌_9,_we_should_not_consider_{subproblems with} bound≤9. Thus, a problem is fathomed when

Bound ≤Z*

In Subproblem 2p that does not occur,, the bound of 16 is

larger than 9. However, it can occur for descendants. As new incumbents with larger values of Z*_are_found,_it

becomes easier to fathom in this way.

3. If the simplex method finds that a subproblem’s LP

relaxation has no feasible solution, the subproblem has no feasible solutionand can be dismissed. João Miguel da Costa Sousa / Alexandra Moutinho 294

Summary

of

fathoming

tests

A subproblem is fathomed (dismissed) if ¾ Test 1:Its bound ≤Z*

or

¾ Test 2:Its LP relaxation has no feasible solutions ¾ Test 2:Its LP relaxation has no feasible solutions or

¾ 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

Result of applying the three tests is in figure below. Subproblem 1is fathomed by test 3.

BIP

branch

‐

and

‐

bound

algorithm

Initialization:Set Z*₌_–_∞_._Apply_bounding,_fathoming

and optimization steps described below to the

whole problem. If not fathomed, perform iteration. Steps for each iteration:

1. Branching:Among the remaining subproblems,

select the one created most recently. Branch from

this node by fixing the next variable as either 0 or 1. 2. Bounding:For each new subproblem, obtain its

boundby applying its LP relaxation. Round down Zfor resulting optimal solution.

BIP

branch

‐

and

‐

bound

algorithm

3. Fathoming:For each new subproblem, apply the

three fathoming tests, and discard subproblems that are fathomed by the tests.

Optimalityy test:Stop when there are no remainingg

subproblems. The current incumbent is optimal.

Otherwise, perform another iteration.

Completing

example

¾Subproblem 3 (fix x₁= 1, x₂= 0):

¾Subproblem 4 (fix x₁= 1, x₂= 1): Iteration 2.Remaining

subproblems are for x₁= 1.

( ₁ , ₂ ) Maximize Z= 9 + 6x3+ 4x4 subject to

(1) 5x3+ 2x4≤4 (2) x3+ x4≤1 (3) x3 ≤1 (4) x4≤0 (5) xjis binary, for j= 3, 4.

( ₁ , ₂ )

Maximize Z= 14 + 6x3+ 4x4 subject to

(1) 5x3+ 2x4≤1 (2) x3+ x4≤1 (3) x3 ≤1 (4) x4≤1 (5) xjis binary, for j= 3, 4. João Miguel da Costa Sousa / Alexandra Moutinho 299

Example

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

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

((xx₁₁,,xx₂₂,,xx₃₃,,xx₄₄)) = (1,(1, 0,0, 0.8,0.8, 0),0), withwith ZZ= 13.813.8 LP relaxation for subproblem 4:

(x1,x2,x3,x4) = (1, 1, 0, 0.5), with Z= 16 Resulting bounds:

Bound for subproblem 3: Z≤13 Bound for subproblem 4: Z≤16

Example

All three fathoming tests fail, so both are unfathomed.

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Iteration

3

¾Subproblem 5 ¾Subproblem 6

Subproblem 4 has the larger

bound, so next branching is

done from (x1, x2) = (1, 1). ¾Subproblem 5

(fix x₁= 1, x₂= 1, x₃= 0): Maximize Z= 14 + 4x4 subject to

(1) 5x3+ 2x4≤1 (2), (4) x4≤1 (5) x4is binary

¾Subproblem 6 (fix x₁= 1, x₂= 1, x₃= 1): Maximize Z= 20 + 4x4 subject to

(1) 2x4≤–4 (2) x4≤ 0 (4) x4≤ 1 (5) x4is binary

Iteration

3 (cont.)

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 LP relaxation for subproblem 6: No feasible solutions. Bound for subproblem 5: Z≤16

Subproblem 6 is fathomed by test 2, but not

subproblem 5.

Iteration

3 (concl.)

Iteration

4

Node created most recently is selected for branching: x₄= 0: (x₁,x₂,x₃,x₄) = (1, 1, 0, 0) is feasible, with Z= 14, x₄= 1: (x₁,x₂,x₃,x₄) = (1, 1, 0, 1) is infeasible. First solution passes test 3 (integer solution) and

second passes test 2 (infeasible) for fathoming. First solution is better than incumbent, so it becomes

new incumbent, with Z*₌₁₄

Reapplying fathoming test 1 (bound) to remaining

branch of Subproblem 3: Bound = 13 ≤Z*₌₁₄_(fathomed).

Solution

tree

after

Iteration

4 Other

options

in

Branch

‐

and

‐

Bound

Branchingcan be done e.g. from the best bound

rather than from the most recently created

subproblem.

Boundingis done by solving a relaxation. Another

possible one is e.g. the Lagrangian relaxation.

p g g g

Fathomingcriteria can be generally stated as: Crit. 1:feasible solutions of subproblem must have Z≤Z*_, Crit. 2:the subproblem has no feasible solutions, or Crit. 3:an optimal solution of subproblem has been found. Some adjustments necessary for Branch‐and‐bound

to find multiple optimal solutions.

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Branch

‐

and

‐

bound

for

MIP

General form of the problem:

1

Maximize

n j j j

Z c x

=

∑

1

subject to , for 1,2, , ,

0, for 1,2, ,

and

is integer, for 1,

2, , ; .

n ij j i j

j j

a x b i m

x j n

x j I I n

=

≤ =

≥ =

= ≤

∑

…

… …

Branch

‐

and

‐

bound

for

MIP

Similar to BIP algorithm. Solving LP relaxationsare

the basis for boundingand fathoming.

4 changes are needed:

1. Choice of branching variable. Only integer variables

that have a noninteger valuein the optimal solution

for the LP relaxation can be chosen.

Branch

‐

and

‐

bound

for

MIP

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.

[x_j*_]₌_greatest_integer_≤_x

j*.

f bl f b bl

Range of variables for two new subproblems:

x_j*_≤_[_x

j*] and xj*≥[xj*] + 1.

Each inequality becomes an additional constraint. Example: x_j*₌_3.5,_then:_x

j*≤3 and xj*≥4.

¾ When changes 1.and 2.are combined a recurring

branching variablecan occur, see figure.

Recurring

branching

variable

Branch

‐

and

‐

bound

for

MIP

Changes needed:

3. Bounding step: value of Zwas rounded downin BIP

algorithm. Now some variables are not integer‐

restricted so bound is value of Zwithoutrounding.g

4. Fathoming test 3: optimal solution for the

subproblem’s LP relaxation must only be integerfor integer‐restrictedvariables.

MIP

branch

‐

and

‐

bound

algorithm

Initialization:Set Z*₌_–_∞_._Apply_bounding,_fathoming

and optimization steps described below to the

whole problem. If not fathomed, perform iteration. Steps for each iteration:

h h b bl

1. Branching: Among the remaining subproblems,

select the one created most recently.

From integer variablesthat have a noninteger valuein the

optimal solution for the LP relaxation choose the first one.

Let x_jbe this variable and x_j* _its_value.

Branch from this creating two subproblems by adding the

respective constraints: x_j*_≤_[_x

j*] and xj*≥[xj*] + 1.

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MIP

branch

‐

and

‐

bound

algorithm

2. Bounding:For each new subproblem, obtain its

boundby applying its LP relaxation.

Use Zwithoutrounding for resulting optimal solution.

3. Fathoming:For each new subproblem, apply the

3 g p pp y

three fathoming tests, and discard subproblems

that are fathomed by the tests.

Test 1:Its bound ≤Z*_,_where_Z*_is_value_of_Z_for_current incumbent.

Test 2:Its LP relaxation has no feasible solutions.

MIP

branch

‐

and

‐

bound

algorithm

3. Fathoming (cont.):

Test 3:Optimal solution for its LP relaxation has integer values

for integer‐restrictedvariables. (If this solution is better it

becomes new incumbent, and test 1 is reapplied for all

unfathomed subproblems).p )

Optimality test:Stop when there are no remaining

subproblems. The current incumbent is optimal.

Otherwise, perform another iteration. See MIP examples in PL#7 and in page 518 of

Hillier’s book.

Branch

‐

and

‐

cut

approach

to

BIP

Branch‐and‐bound was develop and refined in the

60’s and early 70’s.

Can solve problems up to 100 variables.

Branch‐and‐cut approachpp was introduced in the mid

80’s, and can solve problems with thousands of

variables.

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

even 1% of nonzerovalues in functional constraints). Uses a combination of automatic problem processing,

generation of cutting planesand B&B techniques. João Miguel da Costa Sousa / Alexandra Moutinho 316

Automatic

problem

processing

for

BIP

“Computer inspection” of IP formulation to spot

reformulationsthat make the problem quicker to

solve:

¾Fixing variables:identify variables that can be fixed at 0 or

1,, because other value cannot lead to feasible and optimalp

solution.

¾Eliminating redundant constraints:identify and eliminate

constraints that are automatically satisfied by solutions

that satisfy all other constraints.

¾Tightening constraints:tighten constraints in a way that

reduces feasible region of LP relaxation without

eliminating any feasible solutions for the BIP problem. João Miguel da Costa Sousa / Alexandra Moutinho 317

Tightening

constraints

LP relaxation including

feasible region.

LP relaxation after tightening

constraint.

Generating

cutting

planes

for

BIP

Cutting plane(or cut) is a new functional constraint that

reduces feasible region for LP relaxation without

eliminating any feasible solutions of IP problem. Procedure for generating cutting planes:

1. Consider functional constraint in ≤ form with only

nonnegative coefficients.

2. Find a group of Nvariables such that

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

other variables = 0.

b) It is satisfied if value of anyvariables changes from 1 to 0. 3. Resulting cutting plane:

sum of variables in group ≤ N– 1.

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Constraint

Programming

Combination of artificial intelligence with computer

programming languages in the mid‐80’s.

Flexibilityin stating (nonlinear) constraints:

1 Mathematical constraints e g x+ y< z

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

2. Disjunctive constraints, e.g., times of certain tasks

cannot overlap.

3. Relational constraints, e.g., at least three tasks

should be assigned to a certain machine.

Stating

constraints

4. Explicit constraints, e.g., xand yhave same domain

{1,2,3,4,5}, but (x, y) must be (1, 1), (2, 3) or (4, 5).

5. Unary constraints, e.g. zis integer between 5 and

10.

6 L i l t i t if th [6 8]

6. Logical constraints, e.g., if x= 5, then y∈[6, 8]. ¾ Allows use of standard logical functions such as IF,

AND, OR, NOT.

¾ Constraint programming applies domain reduction

and constraint propagation.

¾ The process creates a tree search.

Example

Consider:

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

1. All variables 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}.

Constraint

Programming

Steps in Constraint Programming:

1. Formulate a compact model for the problem by

using a variety of constraint types (most not of IP

type).

ff f f f

2. Efficiently find feasible solutions that satisfy all

these constraints.

3. Search among feasible solutions for an optimal one. ¾ Strength of constraint programming is in first two

steps, whereas the main strength of IP is in step 3. ¾ Current research: integrate CP and IP!