TSP Using Branch and Bound

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This lecture provides a comprehensive tutorial on solving the Travelling Salesman Problem (TSP) using the Branch and Bound algorithm with cost matrix reduction. The instructor begins by defining TSP as finding a minimum cost Hamiltonian cycle that visits each city exactly once and returns to the start. Using a 5-city example (A-E), he demonstrates how to represent the problem as a weighted graph and convert it into an adjacency cost matrix. The core of the lecture focuses on the step-by-step execution of Branch and Bound: calculating initial lower bounds via row and column reductions, branching by selecting edges (e.g., 1->2), setting forbidden transitions to infinity to prevent cycles, and recursively computing lower bounds for child nodes. The instructor illustrates how to prune branches where the cost exceeds the current best solution, ultimately identifying the optimal tour with a minimum total cost of 28.

Chapters

1. 0:00 – 2:00 00:00-02:00

   The instructor introduces the Travelling Salesman Problem (TSP) using Branch and Bound, defining it as finding a minimum cost tour that visits each city exactly once and returns to the start. He begins illustrating the problem by drawing a graph with nodes representing cities (A, B, C, D, E) and labeling the edge between A and B with a cost of 10. On-screen text specifies 'Input: n cities, Cost matrix C[i][j]' and 'Output: Minimum cost Hamiltonian Cycle'. The instructor points to specific edges while explaining how distances are stored in a cost matrix, emphasizing the goal of finding a minimum cost Hamiltonian cycle.

2. 2:00 – 5:00 02:00-05:00

   The instructor explains the mapping of a directed graph with weighted edges to an adjacency cost matrix. He points out specific distances between cities (nodes A, B, C, D, E) and fills in the corresponding values in the matrix. The visual evidence shows the progression of filling the first row of the cost matrix based on the graph's edge weights. He writes 'n=5' indicating the number of cities and draws a large bracket for the cost matrix. Text on screen notes 'Diagonal entries = infinity (a city cannot be visited again immediately)' and 'C[i][j] = cost of travelling from city i to city j'.

3. 5:00 – 10:00 05:00-10:00

   The instructor works through a Traveling Salesperson Problem (TSP) example using a 5x5 cost matrix. He begins by pointing to the original matrix displayed on the screen, which contains infinity symbols representing non-existent paths. He then proceeds to manually copy this matrix onto the whiteboard, likely as part of a step-by-step solution process such as row or column reduction. The instructor writes '5x5' to indicate matrix dimensions and starts the initial setup for calculation, preparing to perform row reduction on the copied matrix.

4. 10:00 – 15:00 10:00-15:00

   The instructor demonstrates the solution steps for a Traveling Salesperson Problem (TSP) using a cost matrix. He begins by performing row reduction on the initial 5x5 matrix, subtracting the minimum element of each row from all elements in that row. Subsequently, he proceeds to column reduction on the resulting matrix to further minimize costs. The visual evidence shows 'Step 1 Row reduction' and 'Step 2 Column reduction' written on the board. The final reduced matrix is displayed with zeros in both rows and columns, representing potential assignments with zero additional cost.

5. 15:00 – 20:00 15:00-20:00

   The instructor demonstrates the Branch and Bound algorithm for solving TSP. He performs row reduction on a cost matrix to calculate an initial lower bound, then proceeds with column reduction. The process involves creating nodes in a search tree by fixing specific paths (e.g., 1->2) and recalculating the reduced cost matrix for subsequent steps. Text on screen shows 'Sum of row reduction + Sum of Column reduction = 31 + 4 = 35' and 'Node 2 Path (1->2)'. The instructor explains how to apply row and column reduction again for the new node.

6. 20:00 – 25:00 20:00-25:00

   The instructor demonstrates the Branch and Bound algorithm for solving TSP. The lesson progresses through calculating lower bounds by performing row and column reductions on cost matrices. Specific steps involve generating child nodes, setting selected edges to infinity to prevent cycles, and computing new lower bounds for branching. The instructor writes 'Step 5: Generate child nodes by selecting the next city (Branching)' and 'Set all elements of row i to infinity'. He explains calculating the total cost as sum of reductions and shows 'LB = Sum of Row Reductions + Sum of Column Reductions'.

7. 25:00 – 30:00 25:00-30:00

   The instructor explains the TSP using Branch and Bound with cost matrix reduction. He demonstrates how to calculate the lower bound (cost) for a root node by summing row and column reductions. The process involves creating a reduced cost matrix where infinity symbols represent forbidden self-loops, and then branching to create sub-nodes (like Node 4 and Node 5) by eliminating specific paths. Text on screen shows 'Root Node Cost = Sum of row reduction + Sum of column reduction' and '= 21 + 4 = 25'. He marks 'Path (1->4) elimination' and 'Path (1->5) elimination'.

8. 30:00 – 35:00 30:00-35:00

   The instructor is solving a Traveling Salesperson Problem (TSP) using the Branch and Bound method with Cost Matrix Reduction. He is working on Node 6, where he has established a path starting from node 1 to node 4. He is now extending this path by adding the transition from node 4 to node 2, creating a partial tour (1 -> 4 -> 2). The instructor writes 'Node 6' and '(1 -> 4)' on the board. He marks infinite costs for invalid transitions, such as '(2 -> 1) = infinity', and shows cost matrix reduction steps.

9. 35:00 – 40:00 35:00-40:00

   The instructor is solving a Traveling Salesperson Problem (TSP) using the Branch and Bound method. He calculates the cost of a specific path, specifically focusing on Node 6 where the path is (1->4->3). He computes a lower bound by summing elements from the reduced cost matrix, resulting in a value of 28. The process involves reducing rows and columns to find the minimum cost path. Text on screen shows 'Node (6)' and '(1->4->3)', with a calculation '= 25 + 3 + 0 = 28'. He demonstrates cost calculation formula and uses infinity for non-existent edges.

10. 40:00 – 45:00 40:00-45:00

    The instructor is solving a Traveling Salesperson Problem (TSP) using the Branch and Bound method. He calculates lower bounds for different nodes by summing row reduction costs, column reduction costs, and the cost of the selected edge. The process involves creating a reduced cost matrix for each node to determine the next best path. He calculates lower bound for Node 4 and draws the reduced cost matrix for Node 5. Text on screen shows '= 25 + 12 + 13' and '= 50'. He traces paths like '(1->4->5)' and identifies infinity values in the matrix.

11. 45:00 – 50:00 45:00-50:00

    The instructor is solving a Traveling Salesperson Problem (TSP) using the Branch and Bound method with cost matrix reduction. He calculates a total cost of 52 by summing reduction values (28 + 11 + 13). The lesson progresses to defining a complete path (1 -> 4 -> 2 -> 5) and setting up the next node in the branch and bound tree, specifically handling the constraint that city 5 cannot go to itself (infinity cost). Text on screen shows '= 28 + 11 + 13 = 52' and 'Node 10 Complete Path (1->4->2->5)'. He marks '(5,5) = infinity'.

12. 50:00 – 55:00 50:00-55:00

    The instructor is solving a Traveling Salesperson Problem (TSP) using the Branch and Bound method. He calculates the cost of a specific path (1->4->2->5->3) by summing edge weights from the cost matrix, resulting in a total cost of 28. He then updates the current best cost and discusses pruning branches where the path cost exceeds this limit. Text on screen shows '1->4->2->5->3' and '= 28 + 0 + 0'. He marks invalid edges with infinity, such as '(5,3) = infinity' and '(5,4) = infinity', while updating 'Current Best Cost = 28'.

13. 55:00 – 60:00 55:00-60:00

    The instructor continues to refine the solution by checking validity of edges in the matrix. He calculates total cost for path 1->4->2->5->3, confirming the result of 28. He updates current best cost to 28 and discusses pruning branches where the path cost exceeds this limit. The instructor marks invalid edges with infinity, such as '(5,3) = infinity' and '(5,4) = infinity', while updating 'Current Best Cost = 28'. He ensures all constraints are met for the optimal tour.

14. 60:00 – 65:00 60:00-65:00

    The instructor continues to refine the solution by checking validity of edges in the matrix. He calculates total cost for path 1->4->2->5->3, confirming the result of 28. He updates current best cost to 28 and discusses pruning branches where the path cost exceeds this limit. The instructor marks invalid edges with infinity, such as '(5,3) = infinity' and '(5,4) = infinity', while updating 'Current Best Cost = 28'. He ensures all constraints are met for the optimal tour.

15. 65:00 – 70:00 65:00-70:00

    The instructor continues to refine the solution by checking validity of edges in the matrix. He calculates total cost for path 1->4->2->5->3, confirming the result of 28. He updates current best cost to 28 and discusses pruning branches where the path cost exceeds this limit. The instructor marks invalid edges with infinity, such as '(5,3) = infinity' and '(5,4) = infinity', while updating 'Current Best Cost = 28'. He ensures all constraints are met for the optimal tour.

16. 70:00 – 72:40 70:00-72:40

    The instructor concludes the lecture by summarizing the Branch and Bound method for TSP. He reviews the final optimal path (1->4->2->5->3) with a total cost of 28. He emphasizes the importance of pruning branches where costs exceed the current best solution to optimize efficiency. The instructor reiterates how setting infinity values prevents cycles and ensures a valid Hamiltonian cycle is found. He wraps up by confirming the minimum cost tour for the given 5-city problem.

The lecture systematically guides students through the Branch and Bound algorithm for solving the Travelling Salesman Problem (TSP). It begins with foundational concepts, defining TSP as finding a minimum cost Hamiltonian cycle and representing it via a weighted graph and adjacency cost matrix. The instructor then transitions to the algorithmic solution, starting with row and column reductions to establish an initial lower bound. The core of the lesson involves branching: selecting edges, setting forbidden transitions to infinity to prevent cycles, and recursively computing lower bounds for child nodes. Key techniques include calculating the sum of reductions plus path costs to determine node costs, and pruning branches where costs exceed the current best solution. The example progresses through multiple nodes (Node 1 to Node 10), demonstrating how partial tours are extended and validated. The final result identifies the optimal path (1->4->2->5->3) with a minimum cost of 28, illustrating the effectiveness of Branch and Bound in finding optimal solutions for combinatorial optimization problems.