## Travelling Salesman Problem-

You are given-

• A set of some cities
• Distance between every pair of cities

Travelling Salesman Problem states-

• A salesman has to visit every city exactly once.
• He has to come back to the city from where he starts his journey.
• What is the shortest possible route that the salesman must follow to complete his tour?

### Example-

The following graph shows a set of cities and distance between every pair of cities- If salesman starting city is A, then a TSP tour in the graph is-

A → B → D → C → A

Cost of the tour

= 10 + 25 + 30 + 15

= 80 units

In this article, we will discuss how to solve travelling salesman problem using branch and bound approach with example.

## Problem-

Solve Travelling Salesman Problem using Branch and Bound Algorithm in the following graph- ## Step-01:

Write the initial cost matrix and reduce it- ### Rules

• To reduce a matrix, perform the row reduction and column reduction of the matrix separately.
• A row or a column is said to be reduced if it contains at least one entry ‘0’ in it.

### Row Reduction-

Consider the rows of above matrix one by one.

If the row already contains an entry ‘0’, then-

• There is no need to reduce that row.

If the row does not contains an entry ‘0’, then-

• Reduce that particular row.
• Select the least value element from that row.
• Subtract that element from each element of that row.
• This will create an entry ‘0’ in that row, thus reducing that row.

Following this, we have-

• Reduce the elements of row-1 by 4.
• Reduce the elements of row-2 by 5.
• Reduce the elements of row-3 by 6.
• Reduce the elements of row-4 by 2.

Performing this, we obtain the following row-reduced matrix- ### Column Reduction-

Consider the columns of above row-reduced matrix one by one.

If the column already contains an entry ‘0’, then-

• There is no need to reduce that column.

If the column does not contains an entry ‘0’, then-

• Reduce that particular column.
• Select the least value element from that column.
• Subtract that element from each element of that column.
• This will create an entry ‘0’ in that column, thus reducing that column.

Following this, we have-

• There is no need to reduce column-1.
• There is no need to reduce column-2.
• Reduce the elements of column-3 by 1.
• There is no need to reduce column-4.

Performing this, we obtain the following column-reduced matrix- Finally, the initial distance matrix is completely reduced.

Now, we calculate the cost of node-1 by adding all the reduction elements.

Cost(1)

= Sum of all reduction elements

= 4 + 5 + 6 + 2 + 1

= 18

## Step-02:

• We consider all other vertices one by one.
• We select the best vertex where we can land upon to minimize the tour cost.

### Choosing To Go To Vertex-B: Node-2 (Path A → B)

• From the reduced matrix of step-01, M[A,B] = 0
• Set row-A and column-B to
• Set M[B,A] =

Now, resulting cost matrix is- Now,

• We reduce this matrix.
• Then, we find out the cost of node-02.

### Row Reduction-

• We can not reduce row-1 as all its elements are ∞.
• Reduce all the elements of row-2 by 13.
• There is no need to reduce row-3.
• There is no need to reduce row-4.

Performing this, we obtain the following row-reduced matrix- ### Column Reduction-

• Reduce the elements of column-1 by 5.
• We can not reduce column-2 as all its elements are ∞.
• There is no need to reduce column-3.
• There is no need to reduce column-4.

Performing this, we obtain the following column-reduced matrix- Finally, the matrix is completely reduced.

Now, we calculate the cost of node-2.

Cost(2)

= Cost(1) + Sum of reduction elements + M[A,B]

= 18 + (13 + 5) + 0

= 36

### Choosing To Go To Vertex-C: Node-3 (Path A → C)

• From the reduced matrix of step-01, M[A,C] = 7
• Set row-A and column-C to
• Set M[C,A] =

Now, resulting cost matrix is- Now,

• We reduce this matrix.
• Then, we find out the cost of node-03.

### Row Reduction-

• We can not reduce row-1 as all its elements are ∞.
• There is no need to reduce row-2.
• There is no need to reduce row-3.
• There is no need to reduce row-4.

Thus, the matrix is already row-reduced.

### Column Reduction-

• There is no need to reduce column-1.
• There is no need to reduce column-2.
• We can not reduce column-3 as all its elements are ∞.
• There is no need to reduce column-4.

Thus, the matrix is already column reduced.

Finally, the matrix is completely reduced.

Now, we calculate the cost of node-3.

Cost(3)

= Cost(1) + Sum of reduction elements + M[A,C]

= 18 + 0 + 7

= 25

### Choosing To Go To Vertex-D: Node-4 (Path A → D)

• From the reduced matrix of step-01, M[A,D] = 3
• Set row-A and column-D to
• Set M[D,A] =

Now, resulting cost matrix is- Now,

• We reduce this matrix.
• Then, we find out the cost of node-04.

### Row Reduction-

• We can not reduce row-1 as all its elements are ∞.
• There is no need to reduce row-2.
• Reduce all the elements of row-3 by 5.
• There is no need to reduce row-4.

Performing this, we obtain the following row-reduced matrix- ### Column Reduction-

• There is no need to reduce column-1.
• There is no need to reduce column-2.
• There is no need to reduce column-3.
• We can not reduce column-4 as all its elements are ∞.

Thus, the matrix is already column-reduced.

Finally, the matrix is completely reduced.

Now, we calculate the cost of node-4.

Cost(4)

= Cost(1) + Sum of reduction elements + M[A,D]

= 18 + 5 + 3

= 26

Thus, we have-

• Cost(2) = 36 (for Path A → B)
• Cost(3) = 25 (for Path A → C)
• Cost(4) = 26 (for Path A → D)

We choose the node with the lowest cost.

Since cost for node-3 is lowest, so we prefer to visit node-3.

Thus, we choose node-3 i.e. path A → C.

### Step-03:

We explore the vertices B and D from node-3.

We now start from the cost matrix at node-3 which is- Cost(3) = 25

### Choosing To Go To Vertex-B: Node-5 (Path A → C → B)

• From the reduced matrix of step-02, M[C,B] =
• Set row-C and column-B to
• Set M[B,A] =

Now, resulting cost matrix is- Now,

• We reduce this matrix.
• Then, we find out the cost of node-5.

### Row Reduction-

• We can not reduce row-1 as all its elements are ∞.
• Reduce all the elements of row-2 by 13.
• We can not reduce row-3 as all its elements are ∞.
• Reduce all the elements of row-4 by 8.

Performing this, we obtain the following row-reduced matrix- ### Column Reduction-

• There is no need to reduce column-1.
• We can not reduce column-2 as all its elements are ∞.
• We can not reduce column-3 as all its elements are ∞.
• There is no need to reduce column-4.

Thus, the matrix is already column reduced.

Finally, the matrix is completely reduced.

Now, we calculate the cost of node-5.

Cost(5)

= cost(3) + Sum of reduction elements + M[C,B]

= 25 + (13 + 8) +

=

### Choosing To Go To Vertex-D: Node-6 (Path A → C → D)

• From the reduced matrix of step-02, M[C,D] =
• Set row-C and column-D to
• Set M[D,A] =

Now, resulting cost matrix is- Now,

• We reduce this matrix.
• Then, we find out the cost of node-6.

### Row Reduction-

• We can not reduce row-1 as all its elements are ∞.
• There is no need to reduce row-2.
• We can not reduce row-3 as all its elements are ∞.
• We can not reduce row-4 as all its elements are ∞.

Thus, the matrix is already row reduced.

### Column Reduction-

• There is no need to reduce column-1.
• We can not reduce column-2 as all its elements are ∞.
• We can not reduce column-3 as all its elements are ∞.
• We can not reduce column-4 as all its elements are ∞.

Thus, the matrix is already column reduced.

Finally, the matrix is completely reduced.

Now, we calculate the cost of node-6.

Cost(6)

= cost(3) + Sum of reduction elements + M[C,D]

= 25 + 0 + 0

= 25

Thus, we have-

• Cost(5) = (for Path A → C → B)
• Cost(6) = 25 (for Path A → C → D)

We choose the node with the lowest cost.

Since cost for node-6 is lowest, so we prefer to visit node-6.

Thus, we choose node-6 i.e. path C → D.

## Step-04:

We explore vertex B from node-6. Cost(6) = 25

### Choosing To Go To Vertex-B: Node-7 (Path A → C → D → B)

• From the reduced matrix of step-03, M[D,B] = 0
• Set row-D and column-B to
• Set M[B,A] =

Now, resulting cost matrix is- Now,

• We reduce this matrix.
• Then, we find out the cost of node-7.

### Row Reduction-

• We can not reduce row-1 as all its elements are ∞.
• We can not reduce row-2 as all its elements are ∞.
• We can not reduce row-3 as all its elements are ∞.
• We can not reduce row-4 as all its elements are ∞.

### Column Reduction-

• We can not reduce column-1 as all its elements are ∞.
• We can not reduce column-2 as all its elements are ∞.
• We can not reduce column-3 as all its elements are ∞.
• We can not reduce column-4 as all its elements are ∞.

Thus, the matrix is already column reduced.

Finally, the matrix is completely reduced.

All the entries have become ∞.

Now, we calculate the cost of node-7.

Cost(7)

= cost(6) + Sum of reduction elements + M[D,B]

= 25 + 0 + 0

= 25

Thus,

• Optimal path is: A → C → D → B → A
• Cost of Optimal path = 25 units

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