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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 is called an integer programming (IP) problem.

� If only some variables 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.

104

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.

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Decision Decision Decision Decision

numbernumbernumbernumber

YesYesYesYes----orororor----no questionno questionno questionno question Decision Decision Decision Decision

variablevariablevariablevariable

Net present Net present Net present Net present

valuevaluevaluevalue

Capital Capital Capital Capital

requiredrequiredrequiredrequired

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

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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 ≤ x2

xj is binary, for j = 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

= =

BIP models

� Groups of yes-or-no decision often constitute groups

of mutually exclusive alternatives: only one decision

in 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 selection, of factories, warehouses, energy power

sources, 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.

108

Formulation examples

� Example 1: making choices when decision variables

are continuous. R&D Division of Good Products Co. has

developed three possible new products.

� Requirement 1: from the three, at most two 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 maximize

total profit.

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Production time used for each unit Production time used for each unit Production time used for each unit Production time used for each unit

producedproducedproducedproduced

Production time Production time Production time Production time

available per available per available per available per

weekweekweekweekProduct 1Product 1Product 1Product 1 Product 2Product 2Product 2Product 2 Product 3Product 3Product 3Product 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)

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.

� Let x1, 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

7

5

9, , , 0

x x x

x x x

x

x

x x x x

+ + ≤

+ + ≤

≤

≤

≤ ≥

subject to

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Formulation of the problem

� 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 two

constraints 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

+ + ≤

+ + ≤

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 large positive number M, 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 variable y4

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

+ + ≤ +

+ + ≤ + −

Complete model (MIP)

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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 7

5

9

0

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

Z x x x

x

x

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 Hillier’s 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 sequence

of flights.

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Data for Southwestern Airways

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Feasible sequence of flightsFeasible sequence of flightsFeasible sequence of flightsFeasible sequence of flights

FlightFlightFlightFlight 1111 2222 3333 4444 5555 6666 7777 8888 9999 10101010 11111111 12121212

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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Formulation of the problem

� Objective: minimize the total cost for the three crews.

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

Should sequence j be 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

=

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

1

1

1

1

1

1

1

3 (assign three crewsj

j

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

x x x

x x x x

x x x x x

x=

+ + + ≥

+ + + ≥

+ + + ≥

+ + + + ≥

+ + + ≥

+ + ≥

+ + + + ≥

+ + + ≥

+ + ≥

+ + + ≥

+ + + + ≥

=∑ )

Formulation of the problem

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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 xj = 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 characteristic i.

� A constraint is included for each characteristic i:

� In set partitioning problems the constraint is

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1j

j

j S

x∈

=∑

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 finite number of feasible solutions.

� However:

� Finite numbers can be astronomically large! With n

variables a BIP problem has 2n solutions, having

exponential growth.

� LP assures that a CFP solution can be optimal,

guaranteeing the remarkable efficiency of the simplex

method. LP problems are much easier to solve than IP

problems.

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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 is the

solution of the IP problem, such as:

� Minimum cost flow problem, including transportation

problem, assignment problem, shortest-path problem

and maximum flow problem.

� Special structures (see examples 2 and 3): mutually

exclusive alternatives, contingent decisions or set-

covering constraints can also simplify the problem.

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Solving IP problems

� Primary determinants of computational complexity:

1. number of integer variables,

2. these variables are binary or general integer 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 round the obtained solution...

Yes??

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Example 1

127

2Maximize Z x=

1 2

1 2

1 2

subject to 0.5

3.5,

and , 0, integers.

x x

x x

x x

− + ≤

+ ≤

≥

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Example 2

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1 2Maximize 5Z x x= +

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 metaheuristics are extremely efficient

for very large problems, but do not guarantee to 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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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-bound with 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 bounding how good the

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

discard it.

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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, for j = 1, 2, 3, 4.

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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 and x1 = 1 for another subset.

� Subproblem 1 (fix x1 = 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, for j = 2, 3, 4.

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Branching

� Subproblem 2 (fix x1 = 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, for j = 2, 3, 4.

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Branching

� Dividing (branching) into suproblems creates a tree

with branches (arcs) for the All node.

� 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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Bounding

� A bound is needed for the best feasible solution of the

subproblem.

� Standard way is to perform a relaxation of the

problem, e.g. by deleting one set of constraints that

makes the problem difficult.

� Most common is to require integer variables, so LP

relaxation is the most widely used.

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Bounding in example

� Example: for the whole problem, (5) is

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

Using simplex:

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

� Thus, Z ≤ 16.5 for all feasible solutions for BIP

problem. Can be rounded to Z ≤ 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

� 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: 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).

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Fathoming

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

bound ≤ 9. Thus, a problem is fathomed when

bound ≤ Z*

� In Subproblem 2 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 solution and can dismissed.

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Summary of fathoming tests

� A subproblem is fathomed (dismissed) if

� Test 1: Its bound ≤ Z*

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

� 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

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

bound by applying its LP relaxation.

� Round down Z for resulting optimal solution.

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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.

Optimality test: Stop when there are no remaining

subproblems. The current incumbent is optimal.

Otherwise, perform another iteration.

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Completing example

� Iteration 2. Remaining subproblems are for x1 = 1.

� Subproblem 3 (fix x1 = 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, for j = 3, 4.

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Example

� Subproblem 4 (fix x1 = 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, for j = 3, 4.

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Example

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

= 3, 4. Optimal solutions are:

� LP relaxation for Subproblem 3:

� (x1, x2, x3, x4) = (1, 1, 0.8, 0), with Z = 13.8

� 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

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Example

� All three fathoming tests fail, so both are unfathomed.

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Iteration 3

� Subproblem 4 has the larger bound, so next branching

is done from (x1, x2) = (1, 1).

� Subproblem 5 (fix x1 = 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.)

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

Maximize Z = 20 + 4x4

subject to

(1) 2x4 ≤ –4

(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.

� Bound for subproblem 5: Z ≤ 16

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Iteration 3 (concl.)

� Subproblem 6 is fathomed by test 2, but not

Subproblem 5.

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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, with Z = 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, with Z* = 14

� Reapplying fathoming test 1 to remaining branch to

remaining Subproblem 3, it is fathomed:

� Bound = 13 ≤ Z* = 14.

150

Solution tree after Iteration 4

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Other options in B&B

� Branching can be done e.g. from the best bound rather

than from the most recently created subproblem.

� 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 optimal solution.

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Branch-and-bound for MIP

� General form of the problem:

153

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 i

j

j

j

Z c x

a x b i m

x j n

x j I I n

=

=

=

≤ =

≥ =

= ≤

∑

∑ …

…

…

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Branch-and-bound for MIP

� Similar to BIP algorithm. Solving LP relaxations are

the basis for bounding and fathoming.

� Four changes are needed:

1. Choice of branching variable. Only integer variables

that have a noninteger value in the optimal solution

for the LP relaxation can be chosen.

154

Branch-and-bound for MIP

2. As integer variables can have a large number of

possible values, create just two new 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 additional constraint.

� Example: xj* = 3.5, then: xj

* ≤ 3 and xj* ≥ 4.

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

branching variable can occur, see figure.

155

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Recurring branching variable

156

Branch-and-bound for MIP

� Changes needed:

3. Bounding step: value of Z was rounded down in BIP

algorithm. Now some variables are not integer-

restricted so bound is value of Z without rounding.

4. Fathoming test 3: optimal solution for the

subproblem’s LP relaxation must only be integer for

integer-restricted variables.

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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:

1. Branching: Among the remaining subproblems, select

the one created most recently.

� From integer variables that have a noninteger value

in the optimal solution for the LP relaxation chose the

first one. Let xj be this variable and xj* its value.

� Branch from this creating two subproblems by adding

the respective constraints: xj* ≤ [xj

*] and xj* ≥ [xj

*] + 1.

158

MIP branch-and-bound algorithm

2. Bounding: For each new subproblem, obtain its

bound by applying its LP relaxation.

� Use Z without rounding 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: Its bound ≤ Z*, where Z* is value of Z for current

incumbent.

Test 2: Its LP relaxation has no feasible solutions.

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MIP branch-and-bound algorithm

3. Fathoming (cont.):

Test 3: Optimal solution for its LP relaxation is integer for

integer-restricted variables. (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 remaining

subproblems. The current incumbent is optimal.

Otherwise, perform another iteration.

� See MIP example in PL#7 and in page 518 of Hillier’s

book.

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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% of nonzero values in the functional

constraints).

� Uses a combination of: automatic problem processing,

generation of cutting planes and B&B techniques.

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Automatic problem processing for BIP

� “Computer inspection” of IP formulation to spot

reformulations that make the problem quicker to

solve:

� Fixing variables: identify variables that can be fixed at 0

or 1, because other value cannot be optimal.

� Eliminating redundant constraints: identify and

eliminate 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.

162

Tightening constraints

163

� LP relaxation including

feasible region.

� LP relaxation after tightening

constraint.

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Generating cutting planes for BIP

� 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 of N variables such that

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

all other variables = 0.

b) It is satisfied if value of any variables 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-1980s.

� Flexibility in stating (nonlinear) constraints:

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.

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Stating constraints

4. Explicit constraints, e.g., x and y 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., if x = 5, y ∈ {6, 7, 8}.

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

can be used.

� Constraint programming applies domain reduction

and constraint propagation.

� The process creates a tree search.

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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}.

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Constraint Programming

� Steps in Constraint Programming:

1. Formulate a compact model of the problem by using

a 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!

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