od metaheuristics 2010

7/24/2019 OD Metaheuristics 2010

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1

METAHEURISTICS

Introduction

Some problems are so complicated that are not

possible to solve for an optimal solution.

In these problems, it is still important to find a good

feasible solution close to the optimal.

A heuristic method is a procedure to find a very good

feasible solution of a considered problem.

Procedure should be efficient to deal with very large

problems, and is an iterative algorithm .

Heuristic methods usually fit a specific problem rather

than a variety of problems.

409



2

Introduction

For a new problem, OR team would need to start from

scratch to develop a heuristic method.

This changed with the development of metaheuristics.

A metaheuristic is a general solution method that

provides:

a general structure;

guidelines for developing a heuristic method for a

particular type of problem.

410

Nature of metaheuristics

Example: maximize

Function has three local optima. The example is a nonconvex programming problem.

f ( x) is sufficiently complicated to solve analitically.

Simple heuristic method: conduct a local

improvement procedure.

411

5 4 3 2( ) 12 975 28000 345000 1800000 f x x x x x x

subject to 0 31 x



3

Example: objective function

412

Local improvement procedure

Starts with initial trial and uses a hill-climbing

procedure .

Example: gradient search procedure, bisection method,

etc.

Converges to a local optimum. Stops without reaching

global optimum (depends on initialization).

Typical sequence: see figure.

Drawback: procedure converges to local optimum. This

is only a global optimum if search begins in the

neighborhood of this global optimum.

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4

Local improvement procedure

414

Nature of metaheuristics

How to overcome this drawback?

What happens in large problems with many variables?

Metaheuristic: solution method that orchestrates the

interaction between local improvement procedures

and a process to escape from local opt ima in a robust

way.

A trial solution after a local optimum can be inferior to

this local optimum.

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5

Solutions by metaheuristics

416

Metaheuristics

Advantage: deals well with large complicated

problems.

Disadvantage: no guarantee to find optimal solution

or even a nearly optimal solution.

When possible, an algorithm that can guarantee

optimality should be used instead.

Can be applied to nonlinear or integer programming.

Most commonly is applied to combinatorial

optimization .

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6

Most common metaheuristics

Tabu search

Simulated Annealing (SA)

Genetic Algorithms (GA)

Ant Colony Optimization (ACO)

Particle Swarm Optimization (PSO), etc.

418

Common characteristics

Derivat ive freeness : methods rely on evaluations of

objective function; search direction follows heuristic

guidelines.

Intuit ive guidelines : concepts are usually bio-inspired.

Slowness : slower than derivative-based optimization

for continuous optimization problems.

Flexibility : allows any objective function (even

structure of data-fitting model).

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7

Common characteristics

Randomness : stochastic methods – use random

numbers to determine search directions; may be

“global optimizers” given enough computation time

(optimistic view).

Analytic opacity : knowledge based on empirical

studies due to randomness and problem-specific

nature.

Iterative nature : need of stopping criteria to

determine when to terminate the optimization

process.

420

Traveling Salesman Problem

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8


Can be symmetric or asymmetric .

Objective: route that minimizes the distance (cost,

time).

Applications: truck delivering goods, drilling holes

when manufacturing printed circuit boards, etc, etc.

Problem with n cities has (n – 1)!/2 feasible routes.

10 cities: 181 440 feasible solutions

20 cities: 61016

feasible solutions 50 cities: 31062 feasible solutions

422


Some TSP problems can be solved using branch-and-

cut algorithms .

Heuristic methods are more general. A new solution is

obtained by making small adjustments to the current

solution.

Example:

Sub-tour reversal adjusts sequence of visited cities by

reversing order in which subsequence is visited.

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9

Solving example

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Sub-tour reversal 3-4 of

1-2-3-4-5-6-7-1 1-2-4-3-5-6-7-1

Sub-tour reversal algorithm

Initialization: start with any feasible solution.

Iteration: from current solution consider all possible

ways of performing a sub-tour reversal. Select the one

that provides the largest decrease in traveled distance.

Stopping rule: stop when no sub-tour reversal improve

current trial solution.

Local improvement algorithm: does not assure

optimal solution!

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Example

Iteration 1: starting with 1-2-3-4-5-6-7-1 (Distance =

69), 4 possible sub-tour reversals that improve

solution are:

Reverse 2-3: 1-3-2-4-5-6-7-1 Distance = 68




426

Example

Iteration 2: continuing with 1-2-4-3-5-6-7-1 only 1 sub-tour

reversal leads to improvement; reverse 3-5-6: 1-2-4-6-5-

3-7-1 with Distance = 64.

Algorit hm stops . Last solution is final solution.

This solut ion is not t he opt imal!

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11

Tabu Search

Fred Glover, 1977

Includes a local search procedure, allowing non-

improvement moves to the best solution.

Referred to as steepest ascent/mildest descent

approach.

To avoid cycle in local optimum, a tabu list is added.

Tabu list records forbidden moves, knows as tabu

moves. Thus, it uses memory to guide the search.

Can include intensification or diversification.

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Basic tabu search algorithm

Initialization: start with a feasible initial solution.

Iteration:

1. Use local search to define feasible moves in

neighborhood.

2. Eliminates moves in tabu list, unless it results in a

better solution.

3. Determine which move provides best solution.

4. Adopt this solution as next trial solution.

5. Update tabu list.

Stopping rule: stop using fixed number of iterations,

fixed number of iterations without improvement, etc.

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Questions in tabu search

1. Which local search procedure should be used?

2. How to define the neighborhood structure ?

3. How to represent tabu moves in the tabu list?

4. Which tabu list should be added to the tabu list in

each iteration?

5. How long should a tabu move remain in the tabu list?

6. Which stopping rule should be used?

430

Ex: minimum spanning tree problem

Problem without constraints: solved using “greedy”

algorithm.

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Added constraints

Constraint 1: link AD can only be included together

with link DE.

Constraint 2: at most one of the three links – AD, CD

and AB – can be included.

Previous solution violates both constraints.

Applying tabu search:

Charge a penalty of 100 if Constraint 1 is violated.

Charge a penalty of 100 if two of the three links inConstraint 2 are included. Increase penalty to 200 if all

three links are included.

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Tabu search implementation

1. Local search procedure: choose best immediate

neighbor not ruled out by tabu status.

2. Neighborhood structure: immediate neighbor is the

one reached by adding a link and deleting one of the

other links in the cycle.

3. Form of tabu moves: list links not to be deleted.

4. Addition of a tabu move: add chosen link to tabu list.

5. Maximum size of tabu list: two (half of total links).

6. Stopping rule: three iterations without improvement.

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Solving example

Initial solution: solution of unconstrained version

Cost = 20 + 10 + 5 + 15 + 200 (why?) = 250

Iteration 1: Options to add a link are BE, CD and DE.

434

Add Delete Cost

BE CE 75 +200 =275

BE AC 70 +200 =270

BE AB 60 +100 =160

CD AD 60 +100 =160

CD AC 65 +300 =365

DE CE 85 +100 =185

DE AC 80 +100 =180

DE AD 75 + 0 = 75 Minimum

Application of tabu search

Add DE to network.

Delete AD from network.

Add DE to tabu list

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Iteration 2

Options to add a link are AD, BE and CD.

436

Add Delete Cost

BE CE 100 +0 =100

BE AC 95 +0 = 95

BE AB 85 + 0 = 85

Minimum

CD DE* 60 +100 =160

CD CE 95 +100 =195

AD DE* (Tabu move)

AD CE 85 +100 =185

AD AC 80 +100 =180*A tabu move; only considered if result in better solutionthan best trial solution found previously.

Iteration 3

Options to add a link are AB, AD and CD.

437

Add Delete Cost

AB BE* (Tabu move)

AB CE 100 +0 =100

AB AC 95 +0 = 95

AD DE* 60 +100 =160

AD CE 95 +0 = 95

AD AC 90 +0 = 90

CD DE

70 + 0 = 70

Minimum

CD CE 105 +0 =105

*A tabu move; only considered if result in better solution thanbest trial solution found previously.



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Iteration 3 and optimal solution

438

Traveling salesman problem example

1. Local search procedure: choose best immediateneighbor not ruled out by tabu status.

2. Neighborhood structure: immediate neighbor is theone reached by making a sub-tour reversal (add andremove two links of current solution).

3. Form of tabu moves: list links such that a sub-tourreversal would be tabu if both links are in the list.

4. Addition of a tabu move: add two chosen links totabu list.

5. Maximum size of tabu list: four links.

6. Stopping rule: three iterations without improvement.

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Solving problem

Init ial t rial solution: 1-2-3-4-5-6-7-1 Distance = 69

Iterat ion 1: choose to reverse 3-4.

Deleted links: 2-3 and 4-5

Added links (tabu list): 2-4 and 3-5

New trial solution: 1-2-4-3-5-6-7-1 Distance = 65

Iterat ion 2: choose to reverse 3-5-6.

Deleted links: 4-3 and 6-7 (OK since not in tabu list)

Added links: 4-6 and 3-7 Tabu list: 2-4, 3-5, 4-6 and 3-7


440

Solving problem

Only two immediate neighbors:

Reverse 6-5-3: 1-2-4-3-5-6-7-1 Distance = 65. (This

would delete links 4-6 and 3-7 that are in the tabu list.)


Iterat ion 3: choose to reverse 3-7.


Added links: 5-7 and 3-1

Tabu list: 4-6, 3-7, 5-7 and 3-1


441



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Sub-tour reversal of 3-7 (Iteration 3)

442

Iteration 4

Four immediate neighbors:

Reverse 2-4-6-5-7: 1-7-5-6-4-2-3-1 Distance = 65



Reverse 7-3: 1-2-4-5-6-3-7-1 Distance = 64 Iterat ion 4: choose to reverse 5-7.


Added links: 6-7 and 5-3

Tabu list: 5-7, 3-1, 6-7 and 5-3

New trial (final) solution: 1-2-4-6-7-5-3-1 Distance = 63

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Sub-tour reversal of 5-7 (Iteration 4)

444

Simulated Annealing

Kirkpatrick, Gelatt, Vecchi, 1983

Suitable for continuous and discrete optimization

problems.

Effective in finding near optimal solutions for large-

scale combinatorial problems such as TSP.

Enables search process to escape from local minima.

Instead of steepest ascent/mildest descent approach

as in tabu search, it tries to search for the tallest hill.

Early iterations take steps in random directions.

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Simulated Annealing

Principle analogous to metals behavior when cooled at

a controlled rate.

Value of objective function analogous to energy in

thermodynamic systems:

At high temperatures, it is likely to accept a new point

with higher energy.

At low temperatures, likelihood of accepting a new

point with higher energy is much lower.

Annealing or cooling schedule : specifies how rapidly

the temperature is lowered from high to low values.

446

Simulated Annealing

Let

Z c = objective function value for current trial solution.

Z n = objective function value for candidate of next trial

solution.

T = tendency to accept Z n if not better than currentsolution.

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Simulated Annealing

Move selection rule: among all immediate neighbors

of current solution, select one randomly. To maximize ,

accept or reject candidate as follows:

If Z n Z c, always accept candidate,

If Z n Z c, accept candidate with following probability:

(For minimization problems reverse Z n

and Z c.)

If candidate is rejected, repeat for another immediate

neighbor.

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Prob{acceptance} , where x n c

Z Z e x

T

Probability of accepting solutions

Larger T : probability of accepting is higher.

Simulated annealing starts with large T and gradually decreasesit as the iterations proceed.

A temperature scheduling should be chosen.

Implementation of move selection rule: compare randomnumber between 0 and 1 to the probability of acceptance.

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Prob{acceptance}=

–0.01 0.990

-0.25 0.779

-1 0.368

-3 0.050

-5 0.007

n c

Z

x

T



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Simulated annealing algorithm

Initialization: start with a feasible initial trial solution.

Iteration: Use the move selection rule to select next

trial.

If none of immediate neighbors of current solution are

accepted, the algorithm is terminated.

Check t he temperature scheduling: decrease T if a

certain number of iterations have been performed.

Stopping rule: stop after a predetermined number of iterations (or if there are no accepted solutions). Best

trial solution at any iteration is the final solution.

450

Questions in simulated annealing

1. How should the initial solution be selected?

2. What is the neighborhood structure that specifies

which solutions are immediate neighbors?

3. What device should be used in the move selection

rule to randomly select one of the immediate

neighbors?

4. What is an appropriate temperature scheduling?

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1. Initial trial solution: any feasible solution. Can be

chosen randomly or it can be 1-2-3-4-5-6-7-1.

2. Neighborhood structure: is one reached by a sub-tour

reversal , as described previously.

3. Random selection of an immediate neighbor: has only

some restrictions.

4. Temperature schedule: five iterations for each of the

five values T 1 = 0.2 Z c

( Z c

is objective function for initial

solution), T i+1 = 0.5T i , i = 1, 2, 3, 4.

452

Solving problem: iteration 1

Initial trial solution: 1-2-3-4-5-6-7-1, Z c= 69, T 1= 13.8.

Sub-tour to reverse can begin between the second slot(city 2) and the sixth slot (city 6).

Can have equal probabilities to start in any of the fiveslots.

After choosing beginning slot, choose end slot withequal probability.

Suppose that 3-4 was chosen to reverse, Z n= 65.

Solutions are not always feasible, if algorithm choosesto reverse 2-3-4-5, solution 1-5-4-3-2-6-7-1 is notfeasible.

As Z n= 65 < Z c= 69, this is the next trial solution.

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Solving problem

Suppose that Iteration 2 results in reversing 3-5-6, to

obtain 1-2-4-6-5-3-7-1, with Z n= 64.

Suppose now that Iteration 3 results in reversing 3-7,

to obtain 1-2-4-6-5-7-3-1, with Z n= 66. As Z n > Z c:

One application of SA gave best solution 1-3-5-7-6-4-2-1

at iterations 13 and 15 (out of 25) with distance = 63.

454

2/13.8

Prob{acceptance}

0.865

n c Z Z

T e

e

Nonlinear programming application

Problems of the type:

1. Initial trial solution: any feasible solution. Can be

x j = L j + (U j – L j)/2.

2. Neighborhood structure: any feasible solution, see

random selection of immediate neighbor.

455

1Maximize ( , , ) n f x x

subject to

, for 1, ,

j j j L x U j n



25

Nonlinear programming application

3. Random selection of an immediate neighbor: Set

where N (0, j) is a normal distribution.

4. Temperature schedule: five iterations for each of the

five values T 1 = 0.2 Z c, T i+1 = 0.5T i , i = 1, 2, 3, 4.

456

6

j j

j

U L

reset (0, ), for 1, , j j j x x N j n

Example

Initial solution: x = 15.5, Z c = f (15.5) = 3 741 121, and

T 1 = 748 224. = (31 - 0)/6 = 5.167.

Iteration 1: x = 15.5 + N (0, 5.167) = 15.5 – 7.5 = 8.

Z n = f (8) = 3 055 616. As

After 25 iterations, gives optimal value of x = 20.031

(optimal is x = 20).

457

3055616 3714121 0.916748224 n c Z Z

T

0.916Prob{acceptance} 0.400 e



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GENETICALGORITHMS

Genetic Algorithms

Motivation

What evolution brings us?

Vision

Hearing

Smelling Taste

Touch

Learning and reasoning

Can we emulate the evolutionary process with today’s

fast computers?

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Genetic Algorithms

Introduced by John Holland in 1975.

Randomized search algorithms based on mechanics of

natural selection and genetics.

Principle of natural selection through ‘survival of the

fittest’ with randomized search.

Search efficiently in large spaces.

Robust with respect to the complexity of the search

problem. Use a population of solutions instead of searching only

one solution at a time: easily parallelized algorithms.

460

Basic elements

Candidate solut ion is encoded as a string of characters

in binary or real. Bit string is called a chromosome.

Solution represented by a chromosome is the

individual . A number of individuals form a population.

Population is updated iteratively; each iteration is

called a generation.

Objective function is called the fitness function.

Fitness value is maximized.

Multiple solutions are evaluated in parallel.

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Definitions

Population: a collection of solutions for the studied

(optimization) problem.

Individual: a single solution in a GA.

Chromosome: (bit string) representation of a single

solution.

Gene: part of a chromosome, usually representing a

variable characterizing part of the solution.

462

Definitions

Encoding: conversion of a solution to its equivalent bit

string representation (chromosome).

Decoding: conversion of a chromosome to its

equivalent solution.

Fitness: scalar value denoting the suitability of a

solution.

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GA terminology

464

1 0 0 0

0 0 1 1

0 1 1 0

0 1 0 1

0 1 0 1

p o p u l a t i o n

x y

gene

chromosome

individual

solution fitness

(2,0)

(1,1)

(0,3)

(1,2)

(1,1)

4

2

3

3

2

Generation t

Basic genetic algorithm

Initialization : Start initial population of solutions, e.g.

randomly. Evaluate the fitness for each individual.

Iteration:

1. Select some members of population to become parents.

2. Cross genetic material of parents in a crossover

operation. Mutation can occur in some genes.

3. Take care of infeasible solutions, by making themfeasible.

4. Evaluate fitness of new members, including the clones.

Stopping rule: stop using fixed number of iterations,fixed number of iterations without improvement, etc.

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Genetic algorithm

466

Genetic AlgorithmGenetic Algorithm

Fitness Function

Assess Fitness

Selection

Crossover

Mutation

Increment Generation

Define Initial Population

Parents

Best IndividualsChildren

Termination criteria

Number of generations

(restart GA if best solution is not satisfactory).

Fitness of best individual.

Average fitness of population.

Difference of best fitness (across generations). Difference of average fitness (across generations).

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Reproduction

Three steps:

Selection

Crossover

Mutation

In GAs, the populat ion size is often kept constant .

User is free to choose which methods to use for all

three steps.

468

Roulette-wheel selection

469

0110010001

11010001111100010110

0110010001

10001110001011010110

3448

23154150

selection

fitnessindividuals

Sum = 211

p = 0.16

p = 0.23

p = 0.11

p = 0.07

p = 0.19

p = 0.24

Cumulative probability: 0.16, 0.39, 0.50, 0.57, 0.76, 1.00



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Roulette-wheel selection

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1

16%

2

23%

3

11%4

7%

5

19%

6

24%

1

2

3

4

5

6

Tournament selection

Select pairs randomly

Fitter individual wins

deterministic

probabilistic

constant probability of winning

probability of winning depends on fitness

It is also possible to combine tournament selection

with roulette-wheel.

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GA iteration

476

10010110

01100010

10100100

10011001

01111101

. . .

. . .

. . .

10010110

01100010

10100100

10011101

01111001

. . .

. . .

. . .

Selection Crossover Mutation

Current

generation

Next

generation

Elitism

reproduction

Spaces in GA iteration

477

0110010001110100011111000

10110

010111011111001000111101001010

1217267

24

22

3448231541

50

genetic-operators

(de)coding fitness functiongeneration N

generation N+1

gene-space problem-space fitness-space

fitness operators



36

Encoding and decoding

Chromosomes represent solutions for a problem in a

selected encoding.

Solutions must be encoded into their chromosome

representation and chromosomes must be decoded to

evaluate their fitness.

Success of a GA can depend on the coding used.

May change the nature of the problem.

Common coding methods: simple binary coding

gray coding (binary)

real valued coding (requires special genetic operators)

478

Handling constraints

Explicit: in the fitness function

penalty function

barrier function

setting fitness of unfeasible solutions to zero

(search may be very inefficient due to unfeasible

solutions)

Implicit: (preferred method) with special encoding

GA searches always for feasible solutions

smaller search space

adhoc method, may be difficult to find

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37

Questions in genetic algorithms

1. What is the encoding scheme?

2. What should the population size be?

3. How should the individuals of the current population

be selected to become parents?

4. How should the genes of the children be derived

from the genes of the parents?

5. How should mutations occur in the genes of the

children?6. Which stopping rule should be used?

480

Example: “peaks” function

Maximization of the “peaks” function using GA

481

2 2 2 2 2 22 ( 1) 3 5 ( 1)

( , )

13(1 ) 10

5 3

x y x y x y

z f x y

x x e x y e e



38


Derivatives of the “peaks” function:

482

…

Example: settings

Search domain: [-3,3][-3,3]

8-bit binary coding

Search space size = 2828 = 65536

Each generation with 20 individuals

Fitness value = value of “peaks” function – minimumfunction value across population

One-point crossover scheme: 1.0 crossover rate

Uniform mutation: 0.01 mutation rate

Elitism of the best 2 individuals across generations

30 generations

483



39


484

Initial population 5th generation 10th generation

GA process:

Example: performance profile

485

0 5 10 15 20 25 30-5

0

5

10

Generations

F i t n e s

s

Best

Average

Poorest



40

Nonlinear programming example

1. Encoding scheme: integers from 0 to 31. Five binary

genes are needed.

Example: x = 21 is 10101 in base 2.

Note that no infeasible solutions are possible.

2. Population size: 10 (problem is simple).

3. Selection of parents: select randomly 4 from the 5

most fit, and 4 from the 5 least fitted.

Select 3 pairs of parents, that will produce 6 children. Elitism: four best solutions are clones in next

generation.

486

Nonlinear programming example

4. Crossover operator: When a bit is difference in the

two parents, select 0 or 1 randomly (uniform dist.)

Parents are 0x01x. Children can be 01011 and 00010.

5. Mutation operator: mutation rate is 0.1 for each

gene.

6. Stopping criteria: stop after 5 consecutive

generations without improvement.

Fit ness value is just the objective function value in

this example.

6 generations were enough in this example.

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41


1. Encoding scheme: exactly as before.

Example: 1-2-3-4-5-6-7-1.

Initial population: generated randomly, using possible

links between cities.

2. Population size: 10 (problem is simple).

3. Selection of parents: select randomly 4 from the 5

most fit, and 4 from the 5 least fitted.

Select 3 pairs of parents, that will produce 6 children. Elitism: four best solutions are clones in next

generation.

488


4. and 5. operators: follows an algorithm:

1. Options for next link: links from current city not in children’s

tour that are used by parents.

2. Select ion of next link: randomly with a uniform distribution.

3. Mutation: if a mutation occurs replace last link with any

other possible one, unless it is impossible.4. Repetition: when there is still more than one link to include

in child, go to step 1.

5. Completion: add last city and complete tour in links exist. If

not, solution is infeasible and all process should be

repeated.

6. Stopping criteria: 5 iterations without improvement.

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42

Traveling salesman problem

Example: P1: 1-2-3-4-5-6-7-1

P2: 1-2-4-6-5-7-3-1

Child: From P1 can be 1-2 or 1-7. From P2 can be 1-2 or

1-3. 1-2 has 50% probability to be chosen. Why?

Suppose 1-2 was chosen. At next step links 2-3 (P1) and

2-4 (P2) can be chosen. Suppose it is 2-4.

Child is 1-2-4. And so on!

Note that problem is too simple, and solutions

(individuals) can be repeated in the population.

See this example in Hillier’s book.

490

ANT COLONYOPTIMIZATION



43

Ant Colony Optimization

Marco Dorigo, 1992

Ant Colony Optimiziation is the most used method of

the Artif icial Life algorithms (Wasp, Bees, Swarm).

Applications: Traveling Salesman Problem, Vehicle

Routing, Quadratic Assignment Problem, Internet

Routing and Logistic Scheduling.

There are also some applications of ACO in clustering

and data mining problems.

492

Ant Colony Optimization

Ants can perform complex tasks:

nest building, food storage.

garbage collection, war.

foraging (to wander in search of food).

There is no management in an ant colony

collective intelligence.

They communicate using:

pheromones (chemical substances), sound, touch.

Curiosities:

Ant colonies exist for more than 100 million years.

Myrmecologists estimate that there are around 20,000 species of ants.

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44

The foraging behaviour of ants

How can almost blind animals manage to learn the shortest

route paths from their nests to the food source and back?

494

a) - Ants follow path between theNest and the Food Source

b) - Ants go around the obstacle following oneof two different paths with equal probability

c) - On the shorter path, morepheromones are laid down

Fotos: http://iridia.ulb.ac.be/~mdorigo/ACO/RealAnts.html

d) – At the end, all ants follow the shortestpath.

Artificial ants

Artificial ants move in graphs

nodes / arcs

environment is discrete

As the real ants, they:

choose paths based on pheromone

deposit pheromones on the travelled paths

The environment updates the pheromones

Artificial ants have more abilities:

they can “see” (heuristic ) [former visibility]

they have memory ( feasible neighbourhood N) [former tabu list ]

495

Nest

Source

Food Source

Destination

http://iridia.ulb.ac.be/~mdorigo/ACO/RealAnts.html

http://iridia.ulb.ac.be/~mdorigo/ACO/RealAnts.html



45

Mathematical framework of real ants

Choose trail

Deposit pheromone

Environment (time)

updates pheromones

496

j = 3

j = 2

Ant 1, t=0

12

i = 113

j = 2

12

i = 113

Ant 1, t=1

Ant 2, t=2

j = 2

12

i = 1

j = 3

13

( 1) ( ) (1 )ij

k

ij ijt t

( )k

ij ij p f

Mathematical framework

Choose node

Update Feasible Neighbourhood

Pheromone update

497

, if ( , )

0, otherwise

k

ij

Q f i j S

\ N N j

k

ijll )1()()1(

,

0, otherwise

ij ij

k ij ij

ij j

if j p

InitializationSet ij = 0

For l =1: Nmax

Build a complete tourFor i = 1 to n

For k = 1 to mChoose node

Update NApply Local Heurist ic

endendAnalyze solutionsFor k = 1 to m

Compute f kendUpdate pheromones

end



46

Notation for ants in TSP

cij – cost for transversal from city i to city j;

ij – pheromone in edge (i, j);

ij – amount of pheromone deposited in edge (i, j);

ij = 1/cij – local heuristic;

pij – probability that ant k in city i visits city j;

N – set of cities still to be visited by ant k in city i;

– evaporation coefficient;

– parameters that determine relative importance

of pheromone versus heuristic;

498


n cities (5)

Complexity: (n –1)! / 2

499

1

2

2

5

8 17

2

5 4

3



47

Solution using the nearest cit y heuristic

500

Step #1 Step #2

Step #3 Step #4

2 2

2

2

2

2

17

2

2

Solution using the nearest cit y heuristic

The final solution is

obviously non-optimal

This heuristic can give the

optimal solution if it is

given a proper initial node

501

Step #5

2

2

2

17

17

2

2

2

5

5



48

ACO in Travelling Salesman Problem

m ants

n cities

= 1 d

502

vs

ACO balances the heuristic information with the

experience (pheromone) information

Iteration l = 1, ant m = 1

All paths have the same pheromone intensity 0=0.5.

Pheromone trail and heuristic information have the same

weight = 1, = 1, = 0.1.

An ant is randomly placed.

The probability to choose is, in this case, based only on

heuristic information: p12=30%

p13=14%

p14=26%

p15=30%

Ant m = 1 chooses node 5.

503

Step #1



49


504

Step #2

46%

32%22%

Step #3

47%

53%

Step #4

100%

Step #5

36.121752221 f


All paths have the same pheromone intensity 0=0.5.


weight = 1, = 1, = 0.1.


The probability to choose is, in this case, based only on

heuristic information: p12=30%

p13=14%

p14=26%

p15=30%


505

Step #1



50


506

47.10225522 f

Step #2

39%

27%34%

Step #3

65%

35%

Step #4

100%

Step #5

Iteration l = 1, pheromone update

The final solution of ant m=1 is

D=12.36. The reinforcement

produced by this ant m=1 is 0,08.

The final solution of ant m=2 is

D=10,47. The reinforcement

produced by ant m=2 is 0,095!

507



51

Updating pheromone matrix

The pheromone update can be done following different approaches:

Considering the pheromone dropped by every ants

Considering the pheromone dropped by the best ant of the present iteration

Considering the pheromone dropped by the best ant in all iterations

(after iteration N = 1, this is the same as the previous approach)

508

0000095.0

95.000000

0095.0000

00095.000

000095.00

008.0000

00008.00

000008.0

0008.000

08.00000

1

5.05.05.05.05.0

5.05.05.05.05.0

5.05.05.05.05.0

5.05.05.05.05.0

5.05.05.05.05.0

1 l

0000095.0

95.000000

0095.0000

00095.000

000095.00

1

5.05.05.05.05.0

5.05.05.05.05.0

5.05.05.05.05.0

5.05.05.05.05.0

5.05.05.05.05.0

1 l


The pheromone trails have different intensities.


weight = 1, = 1, = 0.1.


The probability to choose is:

p41=19% p42=26%

p43=23%

p45=32%


509

Step #1



52


510

46%

32%22%

Step #3

29%

71%

Step #4

100%

Step #5

47.10552221 f

Step #2


The pheromone trails have different intensities.


weight = 1, = 1, = 0.1.


The probability to choose is:

p21=26% p23=29%

p24=26%

p25=19%


511

Step #1



53


512

Step #2

52%

24%

24%

Step #3

63%

37%

Step #3 Step #4

47.10222552 f

100%

Iteration l = 2, pheromone update

The final solution of antm=1 and m=2 is D=10,47.

The reinforcement

produced by each ant is

0,095.

513



54

Updating pheromone matrix

Considering the pheromone dropped by every ants

Considering the pheromone dropped by the best ant of thepresent iteration

514

0000095.0

95.000000

0095.0000

00095.000

000095.00

0000095.0

95.000000

0095.0000

00095.000

000095.00

1

45.045.045.045.055.0

55.045.045.045.045.0

45.055.045.045.045.0

45.045.055.045.045.0

45.045.045.055.045.0

1 l

0000095.0

95.000000

0095.0000

00095.000

000095.00

1

45.045.045.045.055.0

55.045.045.045.045.0

45.055.045.045.045.0

45.045.055.045.045.0

45.045.045.055.045.0

1 l

ACO toolbox for Matlab

Place ant at the first node randomly

j = start_ants (number of nodes)

Choose next node using pij k

j = next_node(i, feasible neighborhood, , , , )

Pheromone update

= update_phero(solution, f(solution), , )

515



55

Fujitsu-Siemens logistics

516

Optimization of a logistic system

517

Order #1

Order #2

Order #3Order #1

Order #2

Order #3

Request the components

Suppliers

(Delay)

Component stock

Orders

Scheduling

decision process

Order #1

Poisson

Exponential

A

B

C

B

A

B

C

A

B

C

E

A

D

A

B

C

D

E

B

A

C

E

Order #1

Order #2

Order #3Order #1

Order #2

Order #3

Order #1

Order #2

Order #3

Request the components

Suppliers

(Delay)

Component stock

Scheduling

decision process

Scheduling

decision process

Order #1

PoissonPoisson

ExponentialExponential

A

B

C

B

A

B

C

A

B

C

A

B

C

E

A

D

A

B

C

D

E

B

A

C

E



56

Optimization results

518

Method f L Orders at correct

date

Orders delay max (T) (T) T[s]

GA 25.12 12 23 4 2.61 32

ACO 25.12 23 23 4 2.61 954

Best optimization result for a specific day

Method (f L) Orders at correct date T[s]

GA 25.93 1.06 11.64 0.24 32.1 0.95

ACO 25.35 0.16 11.96 0.40 934.6 20.30

Average optimization results for a specific day

Comparing results

ACO takes less iterations to find the optimum

GA is much faster than ACO

519



Comparing results

GA and ACO present similar performances

ACO has less variation between runs

GA are faster than ACO

Is there any advantage to use ACO?

ACO records information!

How to use this information?

Dynamic environments

Distributed optimization

520

od metaheuristics 2010

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