Tsp branch and-bound
Download as PPTX, PDF•16 likes•24,570 views
The document discusses the Travelling Salesman Problem (TSP). TSP aims to find the shortest possible route for a salesman to visit each city in a list only once and return to the origin city. It describes the problem as finding the optimal or least cost Hamiltonian circuit in a graph where cities are nodes and distances between cities are edge costs. The document provides an example problem with 5 cities, calculates possible routes and costs, and illustrates the branch and bound algorithm to solve TSP by systematically eliminating suboptimal routes until the optimal route is found.
Tsp branch and-bound
- 2. Travelling salesman Problem-Definition 3 1 2 4 5 •Let us look at a situation that there are 5 cities, Which are represented as NODES •There is a Person at NODE-1 •This PERSON HAS TO REACH EACH NODES ONE AND ONLY ONCE AND COME BACK TO ORIGINAL (STARTING)POSITION. •This process has to occur with minimum cost or minimum distance travelled. •Note that starting point can start with any Node. For Example: 1-5-2-3-4-1 2-3-4-1-5-2
- 3. Travelling salesman Problem-Definition • If there are ‘n’ nodes there are (n-1)! Feasible solutions • From these (n-1)! Feasible solutions we have to find OPTIMAL SOLUTION. • This can be related to GRAPH THEORY. • Graph is a collection of Nodes and Arcs(Edges).
- 4. Travelling salesman Problem-Definition • Let us say there are Nodes Connected as shown • We can find a Sub graph as 1-3-2-1.Hence this GRAPH IS HAMILTONIAN 1 2 3
- 5. Travelling salesman Problem-Definition • But let us consider this graph • We can go to 1-3-4-3-2-1 But we are reaching 3 again to make a cycle. HENCE THIS GRAPH IS NOT HAMILTONIAN 1 2 3 4
- 6. HAMILTONIAN GRAPHS • The Given Graph is Hamiltonian • If a graph is Hamiltonian, it may have more than one Hamiltonian Circuits. • For eg: 1-4-2-3-1 1-2-3-4-1 etc., 4 1 2 3
- 7. Hamiltonian Graphs And Travelling Salesman Problem • Graphs Which are Completely Connected i.e., if we have Graphs with every vertex connected to every other vertex, then Clearly That graph is HAMILTONIAN. • So Travelling Salesman Problem is nothing but finding out LEAST COST HAMILTONIAN CIRCUIT
- 8. Travelling salesman Problem Example 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - Here Every Node is connected to every other Node. But the cost of reaching the same node from that node is Nil. So only a DASH is put over there. Since Every Node is connected to every other Node various Hamiltonian Circuits are Possible.
- 9. Travelling salesman Problem Example 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - We can have various Feasible Solutions. For Example 1-2-4-5-3-1 2-5-1-4-3-2 Etc… But From these Feasible Solutions We want to find the optimal Solution. We should not have SUBTOURS. It should comprise of TOURS.
- 10. Travelling salesman Problem Example- Formulations • Xij = 1,if person moves IMMEDIATELY from I to j. • Objective Function is to minimize the total distance travelled which is given by ∑∑Cij Xij Where Cij is given by Cost incurred or Distance Travelled For j=1 to n, ∑ Xij=1, ɏ i For i= 1 to n, ∑ Xij=1, ɏj Xij=0 or 1
- 11. Sub Tour Elimination Constraints • We can have Sub tours of length n-1 • We eliminate sub tour of length 1 By making Cost to travel from j to j as infinity. Cjj=∞ • To eliminate Sub tour of Length 2 we have Xij+Xji<=1 • To eliminate Sub tour of Length 3 we have Xij+Xjk+Xki<=2 • If there are n nodes Then we have the following constraints • nc2 for length 2 • nc3 for length 3 • …… • ncn-1 for length n-1
- 12. Travelling salesman Problem Example Sub tour elimination 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - We can eliminate Sub tours by a formidable method as Ui-Uj+nXij<=n-1 For i=1 to n-1 And j=2 to n
- 13. TSP - SOLUTIONS • Branch and Bound Algorithm • Heuristic Techniques
- 14. Travelling salesman Problem Example Row Minimum 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - Total Minimum Distance =Sum of Row Minima Here Total Minimum Distance =31 Lower Bound=31 that a person should surely travel. Our cost of optimal Solution should be surely greater than or equal to 31
- 15. Travelling salesman Problem Example Column Minimum 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - Total Minimum Distance =Sum of Column Minima Here Total Minimum Distance also =31 Hence the Problem Matrices is Symmetric. TSP USUALLY SATISFIES 1.SQUARE 2.SYMMETRIC 3.TRIANGLE INEQUALITY dij+djk>=dik
- 16. Branch and Bound-Step 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 X12 X13 X14 X15 For X12 10+5+8+6+6=35 1 3 4 5 2 - 10 5 6 3 8 - 8 9 4 9 8 - 6 5 7 9 6 -
- 17. Branch and Bound-Step 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 X12 X13 X14 X15 For X13 8+5+8+5+6=32 1 2 4 5 2 10 - 5 6 3 - 10 8 9 4 9 5 - 6 5 7 6 6 -
- 18. Branch and Bound-Step 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 X12 X13 X14 X15 For X14 9+6+8+5+6=34 1 2 3 5 2 10 - 10 6 3 8 10 - 9 4 - 5 8 6 5 7 6 9 -
- 19. Branch and Bound-Step 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 For X15 7+5+8+5+6=31 1 2 3 4 2 10 - 10 5 3 8 10 - 8 4 9 5 8 - 5 - 6 9 6
- 20. Branch and Bound-Step 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 X21 X23 X24 2 3 4 3 10 - 8 4 5 8 - 5 - 9 6 For X15 And X21 7+10+8+5+6=36 36
- 21. Branch and Bound-Step 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 X21 X23 X24 For X15 And X23 7+10+8+5+6=36 36 1 2 4 3 - 10 8 4 9 5 - 5 7 6 6 36
- 22. Branch and Bound-Step 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 X21 X23 X24 For X15 And X24 7+5+8+8+6=34 36 1 2 3 3 8 10 - 4 9 - 8 5 7 6 9 36 34
- 23. Branch and Bound-Step 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 X21 X23 X24 36 36 34 X21 X24 X25 37 2 4 5 3 - 8 9 4 5 - 6 5 6 6 - For X13 And X21 8+10+8+5+6=37
- 24. Branch and Bound-Step 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 X21 X23 X24 36 36 34 X21 X24 X25 37 For X13 And X24 8+5+8+6+6=33 1 2 5 3 8 10 9 4 9 - 6 5 7 6 - 33
- 25. Branch and Bound-Steps 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 X21 X23 X24 36 36 34 X21 X24 X25 37 For X13 And X25 8+6+8+5+6=33 33 1 2 4 3 8 10 8 4 9 5 - 5 7 - 6 33
- 26. Branch and Bound-Steps 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 X21 X23 X24 36 36 34 X21 X24 X25 37 33 33 X32 X35 1-3-2-4-5-1 Distance=8+10+ 5+6+7=36 1-3-5-2-4-1 Distance=8+9+6 +5+9=37
- 27. Branch and Bound-Steps 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 X21 X23 X24 36 36 34 X21 X24 X25 37 33 33 X32 X35 1-3-2-4-5-1 Distance=8+10+ 5+6+7=36 1-3-5-2-4-1 Distance=8+9+6 +5+9=37 X32 X34 1-3-2-5-4-1 Distance=8+10+ 6+6+9=39 1-3-4-2-5-1 Distance=8+8+5+6+ 7=34
- 28. Branch and Bound-Steps 31 1 2 3 4 5 1 - 10 8 9 7 2 10 - 10 5 6 3 8 10 - 8 9 4 9 5 8 - 6 5 7 6 9 6 - 35 32 34 31 X12 X13 X14 X15 X21 X23 X24 36 36 34 X21 X24 X25 37 33 33 X32 X35 1-3-2-4-5-1 Distance=8+10+ 5+6+7=36 1-3-5-2-4-1 Distance=8+9+6 +5+9=37 X32 X34 1-3-2-5-4-1 Distance=8+10+ 6+6+9=39 1-3-4-2-5-1 Distance=8+8+5+6+7=34 This is the Optimal Solution. This is same as 1-5-2-4-3-1