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Presented By 
Jayatra Majumdar 
SEMESTER III CMSMP3.5 
Roll No: 13G4MMS126 
Seminar On 
1 
9/21/2014
OUTLINE 
 Introduction (Ant Colony Optimization) 
 Traveling Salesman Problem (TSP) 
 An Example of solving TSP Using A...
SECTION I 
Ant Colony optimization 
3 
9/21/2014
WHAT IS ANT COLONY OPTIMIZATION (ACO) ? 
Ant Colony Optimization (ACO) studies artificial systems that take 
inspiration f...
SECTION II 
Travelling Salesman Problem (TSP) 
5 
9/21/2014
TRAVELING SALESMAN PROBLEM (TSP) 
Informally: The problem of finding the shortest tour through 
a set of cities starting a...
A Hamiltonian path (or 
traceable path) is a path in an 
undirected graph which visits 
each vertex exactly once. A 
Hamil...
TECHNIQUES FOR SOLVING TSP 
TSP can be solved through various way. Some of those are, 
 Exact algorithms: The most direct...
SECTION III 
An Example of solving TSP Using Ant 
Colony Optimization (ACO) 
9 
9/21/2014
A simple TSP example 
[] 
1 
[] 
[] 
2 
3 
[] 
4 
[] 
5 
Distances 
Between 
Cities: 
dAB =100; 
dAD = 50; 
dAE = 70; 
dAC...
ITERATION 1 
A 
E 
D 
C 
B 
[A] 
1 
[B] 
[E] 
5 
[C] 
3 
2 
[D] 
4 
And at first iteration , 
each of them are 
assigned t...
HOW TO BUILD NEXT SUB-SOLUTION? 
[A] 
1 
A 
D 
B 
E 
C 
A1 has four different option i.e. 
AB=100,AC=60, AD=50 and 
AE=70 ...
[A] 
τk = ij 
Amount of Pheromone 
ηij 
k= Reciprocal of the distance 
The amount of pheromone on 
edge(A,B),edge(A,C), 
e...
EVAPORATION RATE IN ITERATION 1 
14 
m 
 
ij ij ij t t t 
Pheromone decay:  ( )  (1   ). ( )   
 ( ) 
k 
1 
Iter...
ITERATION 2 
[A] 
1 
[A] 
1 
[A] 
1 
[A,D] 
1 
A 
D 
B 
E 
C 
15 
9/21/2014
HOW TO BUILD NEXT SUB-SOLUTION? 
A 
[A,D] 
D 
B 
E 
C 
Now A1 has three different 
option i.e. 
DB=70,DC=90,DE=150 
Again,...
The amount of pheromone on 
edge(D,B) ,edge(D,C) and 
edge(D,E) are τD,B(t),τD,C(t) 
and τD,E(t) respectively with 
initia...
EVAPORATION RATE IN ITERATION 2 
18 
m 
 
ij ij ij t t t 
Pheromone decay:  ( )  (1   ). ( )   
 ( ) 
k 
1 
Iter...
ITERATION 3 
A 
D 
B 
E 
[A,D,C] 
C 
[A,D] 
1 
[A,D] 
1 
1 
19 
9/21/2014
HOW TO BUILD NEXT SUB-SOLUTION? 
A 
D 
B 
E 
[A,D,C] 
C 
Now A1 has another two option 
i.e. CE=40,CB=60 
Again, A1 will c...
[A,D,C] 
1 
The amount of pheromone on 
Edge(C,B) and edge(C,E) are 
τC,B(t) and τC,E(t) respectively 
with initial value ...
EVAPORATION RATE IN ITERATION 3 
22 
m 
 
ij ij ij t t t 
Pheromone decay:  ( )  (1   ). ( )   
 ( ) 
k 
1 
Iter...
ITERATION 4 
A 
D 
B 
E 
[A,D,C] 
C 
1 
[A,D,C] 
1 
[A,D,C,E] 
1 
23 
9/21/2014
HOW TO BUILD NEXT SUB-SOLUTION? 
A 
D 
B 
E 
C 
Now A1 has only one path i.e. 
EB=60 as A1 has already 
visited city A,C a...
The amount of pheromone on 
Edge(E,B) is τE,B(t) with initial 
value 0.2. 
The pheromone on path CE 
after fourth iteratio...
EVAPORATION RATE IN ITERATION 4 
26 
m 
 
ij ij ij t t t 
Pheromone decay:  ( )  (1   ). ( )   
 ( ) 
k 
1 
Iter...
ITERATION 5 
A 
D 
B 
E 
C 
[A,D,C,E,B] 
[A,D,C,E] 
1 
[A,D,C,E] 
1 
1 
27 
9/21/2014
HOW TO BUILD NEXT SUB-SOLUTION? 
A 
D 
[A,D,C,E,B] 
B 
E 
C 
A1 now has only one path i.e. 
BA=100 as A1 has already 
visi...
The amount of pheromone on 
Edge(B,A) is τBA(t) with initial 
value 0.2. 
The pheromone on path EB 
after fifth iteration ...
EVAPORATION RATE IN ITERATION 5 
30 
m 
 
ij ij ij t t t 
Pheromone decay:  ( )  (1   ). ( )   
 ( ) 
k 
1 
Iter...
ITERATION 6 
[A,D,C,E,B,A] 
A 
D 
B 
E 
[A,D,C,E,B] 
C 
[A,D,C,E,B] 
1 
[A,D,C,E,B] 
1 
1 
1 
Indicated path is the 
optim...
PATH FOR EACH ANT AND PATH LENGTH 
[A,D,C,E,B,A] 
1 
L1=50+90+40+60+100 
L1 =340 
L2 =60+90+50+70+60 
L2 =330 
L3 =60+60+1...
REFERENCES 
 E. Bonabeau, M. Dorigo, G. Theraulaz. Swarm 
Intelligence: From Natural to Artificial Systems, 1999 
 M. Do...
Thank you for yours attention 
9/21/2014 
34
Any 
Question? 
9/21/2014 
35
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A example of Travelling salesman problem solved using ANT COLONY OPTIMIZATION.

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  • DOWNLOAD FULL BOOKS INTO AVAILABLE FORMAT ......................................................................................................................... ......................................................................................................................... 1.DOWNLOAD FULL PDF EBOOK here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL EPUB Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL doc Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL PDF EBOOK here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL EPUB Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL doc Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... .............. Browse by Genre Available eBooks ......................................................................................................................... Art, Biography, Business, Chick Lit, Children's, Christian, Classics, Comics, Contemporary, Cookbooks, Crime, Ebooks, Fantasy, Fiction, Graphic Novels, Historical Fiction, History, Horror, Humor And Comedy, Manga, Memoir, Music, Mystery, Non Fiction, Paranormal, Philosophy, Poetry, Psychology, Religion, Romance, Science, Science Fiction, Self Help, Suspense, Spirituality, Sports, Thriller, Travel, Young Adult,
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  • at slid no 13 behind this calculation what you write i can't understand ,
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  • @jayatra can u plz give me it's ppt formet
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  • Sorry guys, here i have some custom animation for better explanation of the path of ant, but it is not working here...Just final slide appear here. So, I am sorry for this trouble. You guys may send me a message on facebook through www.facebook.com/jayatra.cs
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Show ant-colony-optimization-for-solving-the-traveling-salesman-problem

  1. 1. Presented By Jayatra Majumdar SEMESTER III CMSMP3.5 Roll No: 13G4MMS126 Seminar On 1 9/21/2014
  2. 2. OUTLINE  Introduction (Ant Colony Optimization)  Traveling Salesman Problem (TSP)  An Example of solving TSP Using Ant Colony Optimization (ACO)  Reference 2 9/21/2014
  3. 3. SECTION I Ant Colony optimization 3 9/21/2014
  4. 4. WHAT IS ANT COLONY OPTIMIZATION (ACO) ? Ant Colony Optimization (ACO) studies artificial systems that take inspiration from the behavior of real ant colonies and which are used to solve discrete optimization problems. In 1991, the Ant Colony Optimization metaheuristic was defined by Dorigo, Di Caro and Gambardella. 4 9/21/2014
  5. 5. SECTION II Travelling Salesman Problem (TSP) 5 9/21/2014
  6. 6. TRAVELING SALESMAN PROBLEM (TSP) Informally: The problem of finding the shortest tour through a set of cities starting at some city and going through all other cities once and only once, returning to the starting city Formally: Finding the shortest Hamiltonian path in a fully-connected graph TSP belongs to the class of NP-complete problems. Thus, it is assumed that there is no efficient algorithm for solving TSPs. 6 9/21/2014
  7. 7. A Hamiltonian path (or traceable path) is a path in an undirected graph which visits each vertex exactly once. A Hamiltonian cycle (or Hamiltonian circuit) is a cycle in an undirected graph which visits each vertex exactly once and also returns to the starting vertex. Determining whether such paths and cycles exist in graphs is the Hamiltonian path problem which is NP-complete. A graph representing the vertices, edges, and faces of a dodecahedron, with a Hamiltonian cycle shown by red color edges. NP problem can not be solved in polynomial time. 7 9/21/2014
  8. 8. TECHNIQUES FOR SOLVING TSP TSP can be solved through various way. Some of those are,  Exact algorithms: The most direct solution would be to try all permutations (ordered combinations) and see which one is cheapest (using brute force search).  Computational complexity  Complexity of approximation  Heuristic and approximation algorithms  4.3.1 Constructive heuristics  4.3.2 Iterative improvement  4.3.3 Randomized improvement • 4.3.3.1 Ant colony optimization  Special cases  4.4.1 Metric TSP  4.4.2 Euclidean TSP  4.4.3 Asymmetric TSP  4.4.3.1 Solving by conversion to symmetric TSP, etc. 8 9/21/2014
  9. 9. SECTION III An Example of solving TSP Using Ant Colony Optimization (ACO) 9 9/21/2014
  10. 10. A simple TSP example [] 1 [] [] 2 3 [] 4 [] 5 Distances Between Cities: dAB =100; dAD = 50; dAE = 70; dAC = 60; dBE = 60; dBC = 60; dCD = 90; dCE = 40; dDE = 150; A E D C B Here, m=n Where, m= number of ants; n= number of cities 9/21/2014 10
  11. 11. ITERATION 1 A E D C B [A] 1 [B] [E] 5 [C] 3 2 [D] 4 And at first iteration , each of them are assigned to different city. 9/21/2014 11
  12. 12. HOW TO BUILD NEXT SUB-SOLUTION? [A] 1 A D B E C A1 has four different option i.e. AB=100,AC=60, AD=50 and AE=70 A1 will choose path by a probabilistic value   ( t )   .          k  ij ij ij k Ji t    ( ) .  il il p t  ( ) Depending on a random number. 12 9/21/2014
  13. 13. [A] τk = ij Amount of Pheromone ηij k= Reciprocal of the distance The amount of pheromone on edge(A,B),edge(A,C), edge(A,D) and edge(A,E) are τA,B(t), τA,C(t), τA,D(t) and τA,E(t) respectively with same initial value 0.2. The value of η(A,B), η(A,C), η(A,D) and η(A,E) are 1/100, 1/60, 1/50 and 1/70 respectively. Value of α=0.2 and β=0.6 Now the probabilities , , ,and are as shown here respectively. HOW TO BUILD NEXT SUB-SOLUTION? A D B E C Depending on a random number A1 choose AD as path. Assuming random no.=0.25 1 k (t) AB p k (t) AE (t) p k AD (t) p k AC p                     (t) 0.3056 k AE p 0.6 . 1/70 0.2 0.2 0.6 . 1/50 0.2 0.2 0.6 . 1/60 0.2 0.2 0.6 . 1/100 0.2 0.2 0.6 * 1/70 0.2 0.2 (t) k AE p          0.1791 AB 0.6 * 1/100 0.2 0.2 AB                     (t) 0.2432 k AC p 0.6 . 1/70 0.2 0.2 0.6 . 1/50 0.2 0.2 0.6 . 1/60 0.2 0.2 0.6 . 1/100 0.2 0.2 0.6 * 1/60 0.2 0.2 (t) k AC p      0.2719 AD 50 AD 13 9/21/2014
  14. 14. EVAPORATION RATE IN ITERATION 1 14 m  ij ij ij t t t Pheromone decay:  ( )  (1   ). ( )    ( ) k 1 Iteration Path Decay Rate 1 AB 0.08 9/21/2014 AC 0.08 AD 0.08 AE 0.08 BC 0.08 BD 0.08 BE 0.08 CD 0.08 CE 0.08 DE 0.08 Assume that ρ =0.6
  15. 15. ITERATION 2 [A] 1 [A] 1 [A] 1 [A,D] 1 A D B E C 15 9/21/2014
  16. 16. HOW TO BUILD NEXT SUB-SOLUTION? A [A,D] D B E C Now A1 has three different option i.e. DB=70,DC=90,DE=150 Again, A1 will choose path by the probabilistic value   ( t )   .          k  ij ij ij k Ji t    ( ) .  il il p t  ( ) Depending on a random number. 1 16 9/21/2014
  17. 17. The amount of pheromone on edge(D,B) ,edge(D,C) and edge(D,E) are τD,B(t),τD,C(t) and τD,E(t) respectively with initial value 0.2. The pheromone on path AD after first iteration will be updated by. HOW TO BUILD NEXT SUB-SOLUTION? A [A,D] D B E C    / ( ) ( , ) ( ) 0. 100/ 50 2  SO , A   AD A    AD Assuming a random number = 0.32, A1 choose DC as path. t else k t if i j T k Q L k ij  Where, Q=100 and LK =50 1 Value of α=0.2 and β=0.6 ( ) A p t DC And the probabilities , and p A ( t ) DE are as shown here respectively. ( ) A p t DB 0.4009 DE DB 70 DB                 (t) 0.3461 k DC p 0.6 . 1/70 0.2 0.2 0.6 . 1/150 0.2 0.2 0.6 . 1/90 0.2 0.2 0.6 * 1/90 0.2 0.2 (t) k DC p             (t) 0.2528 . 1/70 0.2 0.2 0.6 . 1/50 0.2 0.2 0.6 150 (t) DE    17 9/21/2014
  18. 18. EVAPORATION RATE IN ITERATION 2 18 m  ij ij ij t t t Pheromone decay:  ( )  (1   ). ( )    ( ) k 1 Iteration Path Decay Rate 2 AB 0.032 9/21/2014 AC 0.032 AD 0.832 AE 0.032 BC 0.032 BD 0.032 BE 0.032 CD 0.032 CE 0.032 DE 0.032 Assume that ρ =0.6
  19. 19. ITERATION 3 A D B E [A,D,C] C [A,D] 1 [A,D] 1 1 19 9/21/2014
  20. 20. HOW TO BUILD NEXT SUB-SOLUTION? A D B E [A,D,C] C Now A1 has another two option i.e. CE=40,CB=60 Again, A1 will choose path by the probabilistic value   ( t )   .          k  ij ij ij k Ji t    ( ) .  il il p t  ( ) Depending on a random number. 1 20 9/21/2014
  21. 21. [A,D,C] 1 The amount of pheromone on Edge(C,B) and edge(C,E) are τC,B(t) and τC,E(t) respectively with initial value 0.2. The pheromone on path DC after second iteration will be updated by. HOW TO BUILD NEXT SUB-SOLUTION? A D B E C    / ( ) ( , ) ( ) 0. 100/ 140 0.7142  SO , A    DC A   DC ( ) A p t CB Assuming a random number =0.5, A1 choose CE as path. t else k t if i j T k Q L k ij  Where, Q=100 and LK =50+90 =140 Value of α=0.2 and β=0.6 And the probabilities and p A ( t ) CE are as shown here respectively.             (t) 0.4386 CB k p 0.6 . 1/40 0.2 0.2 0.6 . 1/60 0.2 0.2 0.6 * 1/60 0.2 0.2 (t) k p CB                0.5613 0.6 . 1/40 0.2 0.2 0.6 . 1/60 0.2 0.2 0.6 * 1/40 0.2 0.2 (t) CE CE   21 9/21/2014
  22. 22. EVAPORATION RATE IN ITERATION 3 22 m  ij ij ij t t t Pheromone decay:  ( )  (1   ). ( )    ( ) k 1 Iteration Path Decay Rate 3 AB 0.0128 9/21/2014 AC 0.0128 AD 0.3328 AE 0.0128 BC 0.0128 BD 0.0128 BE 0.0128 CD 0.2984 CE 0.0128 DE 0.0128 Assume that ρ =0.6
  23. 23. ITERATION 4 A D B E [A,D,C] C 1 [A,D,C] 1 [A,D,C,E] 1 23 9/21/2014
  24. 24. HOW TO BUILD NEXT SUB-SOLUTION? A D B E C Now A1 has only one path i.e. EB=60 as A1 has already visited city A,C and D. Again, A1 will choose path by the probabilistic value   ( t )   .          k  ij ij ij k Ji t    ( ) .  il il p t  ( ) Depending on a random number. [A,D,C,E] 1 24 9/21/2014
  25. 25. The amount of pheromone on Edge(E,B) is τE,B(t) with initial value 0.2. The pheromone on path CE after fourth iteration will be updated by. HOW TO BUILD NEXT SUB-SOLUTION? A D B E C    / ( ) ( , ) ( ) 0. 100/ 180 0.5555  SO , A    CE A   CE ( ) A p t EB Assuming a random number =0.5, A1 choose EB as path. t else k t if i j T k Q L k ij  Where, Q=100 and LK =50+90+40 =180 Value of α=0.2 and β=0.6 And the probabilities is as shown here, [A,D,C,E] 1          (t) 1 k p 0.6 . 1/60 0.2 0.2 0.6 * 1/60 0.2 0.2 (t) k EB p EB  25 9/21/2014
  26. 26. EVAPORATION RATE IN ITERATION 4 26 m  ij ij ij t t t Pheromone decay:  ( )  (1   ). ( )    ( ) k 1 Iteration Path Decay Rate 4 AB 0.00512 9/21/2014 AC 0.00512 AD 0.13312 AE 0.00512 BC 0.00512 BD 0.00512 BE 0.00512 CD 0.11936 CE 0.22732 DE 0.00512 Assume that ρ =0.6
  27. 27. ITERATION 5 A D B E C [A,D,C,E,B] [A,D,C,E] 1 [A,D,C,E] 1 1 27 9/21/2014
  28. 28. HOW TO BUILD NEXT SUB-SOLUTION? A D [A,D,C,E,B] B E C A1 now has only one path i.e. BA=100 as A1 has already visited all other city and return to home city. Again, A1 will choose path by the probabilistic value   ( t )   .          k  ij ij ij k Ji t    ( ) .  il il p t  ( ) Depending on a random number. 1 28 9/21/2014
  29. 29. The amount of pheromone on Edge(B,A) is τBA(t) with initial value 0.2. The pheromone on path EB after fifth iteration will be updated by. HOW TO BUILD NEXT SUB-SOLUTION? A D [A,D,C,E,B] B E C    / ( ) ( , ) ( ) 0. 100/ 240 0.4166  SO , A    EB A   E ( ) A p t BA Assuming a random number=0.8, A1 choose BA as path. t else k t if i j T k Q L k ij  B Where, Q=100 and LK= 50+90+40+60 =240 Value of α=0.2 and β=0.6 And the probabilities is as shown here, 1          (t) 1 k p 0.6 . 1/100 0.2 0.2 0.6 * 1/100 0.2 0.2 (t) k BA p BA  29 9/21/2014
  30. 30. EVAPORATION RATE IN ITERATION 5 30 m  ij ij ij t t t Pheromone decay:  ( )  (1   ). ( )    ( ) k 1 Iteration Path Decay Rate 4 AB 0.002048 9/21/2014 AC 0.002048 AD 0.053248 AE 0.002048 BC 0.002048 BD 0.002048 BE 0.16868 CD 0.047744 CE 0.090928 DE 0.002048 Assume that ρ =0.6
  31. 31. ITERATION 6 [A,D,C,E,B,A] A D B E [A,D,C,E,B] C [A,D,C,E,B] 1 [A,D,C,E,B] 1 1 1 Indicated path is the optimal path for ant A1 . 31 9/21/2014
  32. 32. PATH FOR EACH ANT AND PATH LENGTH [A,D,C,E,B,A] 1 L1=50+90+40+60+100 L1 =340 L2 =60+90+50+70+60 L2 =330 L3 =60+60+150+50+60 L3 =380 L4 =150+70+100+60+90 L4 =470 L5 =70+100+60+90+150 L5 = 470 [B,C,D,A,E.B] 2 [C,B,E,D,A,C] 3 [D,E,A,B,C,D] 4 [E,A,B,C,D,E] 5 Similarly, path for other ants are presented here with optimal path length. 9/21/2014 32
  33. 33. REFERENCES  E. Bonabeau, M. Dorigo, G. Theraulaz. Swarm Intelligence: From Natural to Artificial Systems, 1999  M. Dorigo and L. Gambardella. Ant colony system: A cooperative learning approach to the traveling salesman problem, 1997.  M. Dorigo and T. Stützle. Ant Colony Optimization, MIT Press, 2004.  J. Kennedy and R. Eberhart. Swarm Intelligence, 2001 9/21/2014 33
  34. 34. Thank you for yours attention 9/21/2014 34
  35. 35. Any Question? 9/21/2014 35

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