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International A LevelMathsEdexcelDecision 1Revision NotesAlgorithms on GraphsThe Travelling Salesman ProblemUpper & Lower Bounds for the Travelling Salesman Problem

Upper & Lower Bounds for the Travelling Salesman Problem (Edexcel International A Level Maths): Revision Note

Naomi C

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Naomi C

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Finding an Upper Bound using a Minimum Connector

Why do I need to find an upper bound?

  • There is no efficient algorithm to find the optimal solution for the travelling salesman problem

    • An upper and lower bound for the solution to the problem can be found

    • The optimal solution (the shortest route) will lie within these bounds

      • Lower boundless or equal than optimal solution less or equal than upper bound

  • The aim is to find a minimum value for the upper bound

    • This will reduce the size of the interval between the upper and lower bounds

How do I find the upper bound?

  • Use a minimum spanning tree to find an upper bound for a network
     

  • STEP 1
    Find a minimum spanning tree for the network using either Prim's or Kruskal's algorithm

    • If you have the table of distances, use Prim's algorithm

    • If you have the graph only, use Prim's or Kruskal's algorithm
       

  • STEP 2
    Double the weight of the minimum spanning tree to find an initial upper bound

    • Each vertex can therefore definitely be visited and the starting vertex returned to
       

  • STEP 3
    Improve the upper bound by finding shortcuts

    • Use some of the edges not included in the minimum spanning tree to reduce the weight

Worked Example

The network below contains six vertices representing villages and the edges (roads) that connect them. The weighting of the edges represents the time, in minutes, that it takes to walk along a particular road between two villages.

3-10-6-ib-ai-hl-bounds-for-travelling-salesman-problem-we-1

a) Find an initial upper bound for the travelling salesman problem.

  • STEP 1
    The question contains the graph but you have not been asked to find the table of least distances
    Apply Kruskal's algorithm to find a minimum spanning tree of the network
    List the edges in order of increasing weight

 

CD (5)
CE (6)
AC (7)
EF (8)
BF (9)
DE (10)
AD (11)
BC (13)
BE (13)
AB (14)
DF (20)
  

Add the edge of least weight (CD), then the next edge of least weight (CE)
Repeat adding the next edge of least weight each time provided it does not create a cycle until all vertices are connected
 

Edges added in the order: CD (5), CE (6), AC (7), EF (8), BF (9) 
  

WdntKlV2_mst

 

  • STEP 2
    Find the weight of the minimum spanning tree

5 + 6 + 7 + 8 + 9 = 35

Double the weight of the minimum spanning tree to find the initial upper bound of the solution

35 x 2 = 70

Initial upper bound = 70 minutes
 

b) Show that the initial upper bound can be reduced to 54 by finding one shortcut.

  • STEP 3
    Identify which edges could be used to replace repeated edges in the minimum spanning tree

Single edge: AB (14)

Repeated pathway: AC + CE + EF + FB (30)

Subtract the weight of the single edge from the repeated pathway to calculate the weight saved by using the short cut

30 - 14 = 16

Subtract the weight saving from the initial upper bound

70 - 16 = 54

Improved upper bound = 54 minutes

This can also be seen as 70 - 30 + 14 = 54
Subtract the 'repeated pathway' then add the 'shortcut'

Finding a Lower Bound using a Minimum Connector

Why do I need to find the lower bound?

  • There is no efficient algorithm to find the optimal solution for the travelling salesman problem

    • An upper and lower bound for the solution to the problem can be found

    • The optimal solution (the shortest route) will lie within these bounds

      • Lower bound less or equal than optimal solution less or equal than upper bound

  • The aim is to find a maximum value for the lower bound 

    • This will reduce the size of the interval between the upper and lower bounds

  • An optimal solution is found if

    • The lower bound gives a Hamiltonian cycle

      • This is only true if the MST is created from a table of least distances

    • The lower bound has the same value as the upper bound

  • If the lower bound is not a Hamiltonian cycle then 

    • lower bound < optimal solution

How do I find the lower bound?

  • Use a minimum spanning tree to find a lower bound for a network

    • Apply the steps of the algorithm below to a table of least distances
       

  • STEP 1
    Remove a vertex and all of its associated edges from the network
     

  • STEP 2
    Find a minimum spanning tree for the remaining network and write down its total weight

    • This spanning tree is known as the residual minimum spanning tree (RMST)
       

  • STEP 3
    Identify the two shortest individual edges that can connect the missing vertex to the RMST
    Find the sum of the total weight of the RMST and the two connecting edges
     

  • STEP 4
    Repeat Steps 1 to 3 for each vertex in the network
     

  • STEP 5
    When the network has been tested for each missing vertex, identify the maximum value

    • This maximum value is the greatest possible lower bound

  • In many problems Steps 4 and 5 will not be required

    • An exam question will often dictate which vertex to delete

    • There will be no need to repeat for each vertex

  • This method may be referred to as the deleted vertex method

Worked Example

The table below contains six vertices representing villages and the roads that connect them. The weighting of the edges represents the time in minutes that it takes to walk along a particular road between two villages.
 

 

A

B

C

D

E

F

A

-

14

7

11

13

21

B

14

-

13

18

13

9

C

7

13

-

5

6

14

D

11

18

5

-

10

18

E

13

13

6

10

-

8

F

21

9

14

18

8

-

 

a) By deleting vertex A and using Prim’s algorithm, find a lower bound for the time taken to start at village A, visit each of the other villages and return to village A
 

  • STEP 1
    Delete vertex A in the table
     

  • STEP 2
    Perform Prim's algorithm on the remaining vertices in the table
      

table-1

 
Edges added in the order: BF (9), EF (8), CE (6), CD (5)

Total weight of the residual minimum spanning tree = 9 + 8 + 6 + 5 = 28
  

  • STEP 3
    Identify the two shortest edges that connect A to the minimum spanning tree

AC (7), AD (11)

Add together the total weight of the minimum spanning tree and the weights of the two shortest lengths connecting it to A

28 + 7 + 11 = 46

Lower bound = 46 minutes

 

b) Show that by deleting vertex E instead, a higher lower bound can be found.

  • STEP 1
    Delete vertex A in the table
     

  • STEP 2
    Perform Prim's algorithm on the remaining vertices in the table

delete-vertex-e

  
Edges added in the order: AC (7), CD (5), BD/DF (18), BF (9) 

Total weight of the residual minimum spanning tree = 7 + 5 + 18 + 9 = 39
  

  • STEP 3
    Identify the two shortest edges that connect E to the minimum spanning tree

CE (6), EF (8)

Add together the total weight of the minimum spanning tree and the weights of the two shortest lengths connecting it to E

39 + 6 + 8 = 53

Lower bound = 53 minutes
This is a higher lower bound than found when vertex A was deleted (46 minutes)

Nearest Neighbour Algorithm

What is the nearest neighbour algorithm?

  • The nearest neighbour algorithm is another method to find an upper bound

  • The algorithm can be performed on a complete network with at least 3 vertices

    • It will generate a low weight Hamiltonian cycle

      • This low weight cycle will not necessarily be the least weight cycle

    • The weight of the cycle can be considered an upper bound

  • The best upper bound is the upper bound with the smallest value

  • The nearest neighbour algorithm can only be used on certain types of graph

    • The graph must be complete and satisfy the triangle inequality

  • If needs be, the table of least distances should be found first

What are the steps of the nearest neighbour algorithm?

  • STEP 1
    Choose a starting vertex
     

  • STEP 2
    Select the edge of least weight from the current vertex to an adjacent unvisited vertex

    • If there is more than one edge of least weight pick one at random

    • Move to this vertex (column in matrix form)

    • If this is the final vertex to be visited go to STEP 3

    • Otherwise disregard (cross out in matrix form) any edges that go to vertices already visited and repeat STEP 2
       

  • STEP 3
    Add a final edge to return to the starting vertex
     

  • STEP 4
    Find the weight of the resulting Hamiltonian cycle
    This is an upper bound for the travelling salesman problem
     

  • STEP 5
    Repeat STEP 1 to STEP 4 starting with a different vertex in the network
    When the network has been tested for each vertex as the starting vertex, identify the minimum value
    This minimum value is the least possible upper bound that can be found by the nearest neighbour algorithm

  • In many problems STEP 5 will not be required

    • An exam question will dictate which vertex to start at

    • There will be no requirement to repeat with all vertices as the starting vertex

Examiner Tips and Tricks

  • Be careful not to confuse the nearest neighbour algorithm with Prim's algorithm 

    • This is a common mistake because they are both performed on tables!

  • Questions may ask you to explain the difference between the two algorithms

    • The nearest neighbour algorithm selects the nearest vertex adjacent to the current vertex

    • Prim's algorithm selects the nearest vertex adjacent to any vertex already visited

Worked Example

The table below contains six vertices representing villages and the roads that connect them. The weighting of the edges represents the time in minutes that it takes to walk along a particular road between two villages.
 

 

A

B

C

D

E

F

A

-

14

7

11

13

21

B

14

-

13

18

13

9

C

7

13

-

5

6

14

D

11

18

5

-

10

18

E

13

13

6

10

-

8

F

21

9

14

18

8

-

Starting at village A, use the nearest neighbour algorithm to find an upper bound for the time it would take to visit each village and return to village A.
 

  • STEP 1
    Start at vertex A (column A)

  • STEP 2
    Select the edge of least weight in column A (AC) and write it down

~8028F-4_table-1

 AC (7)

Move to vertex/column C
C is not the final vertex to be visited
Disregard (cross out) the edge (value) in column C that would connect C to A as vertex A has already been visited
Repeat Step 2

  • STEP 2
    Select the edge of least weight (from those remaining) in column C (CD) and write it down

0IaJRg_O_table-2

AC (7), CD (5)

Move to column D
D is not the final vertex
Cross out the values for the vertices already visited (A and C)
Repeat Step 2

  • STEP 2
    Select the edge of least weight (from the remaining) in column D (DE) and write it down

table-3

 AC (7), CD (5), DE (10)

Move to column E
E is not the final vertex
Cross out the values for the vertices already visited (A, C and D)
Repeat Step 2

  • STEP 2
    Select the edge of least weight (from the remaining values) in column E (EF) and write it down

table-4

AC (7), CD (5), DE (10), EF (8)

Move to vertex F
F is not the final vertex
Cross out the values for the vertices already visited (A, C, D and E)
Repeat Step 2

  • STEP 2
    Select the edge of least weight (from the remaining values) in column F (BF) and write it down

table-5

AC (7), CD (5), DE (10), EF (8), BF (9)

Move to vertex B
B is the final vertex to be visited
Go to Step 3

  • STEP 3
    Choose the final edge (value) to get back from the last vertex (B) to the starting vertex (A)

table-6

AC (7), CD (5), DE (10), EF (8), BF (9), AB (14)

Hamiltonian cycle: ACDEFBA

  • STEP 4
    Find the weight of the Hamiltonian cycle by adding together the weights of the edges forming it 

Total weight = 7 + 5 + 10 + 8 + 9 + 14 = 53

This is an upper bound for the travelling salesman problem

Upper bound = 53 minutes

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    Naomi C

    Author: Naomi C

    Expertise: Maths

    Naomi graduated from Durham University in 2007 with a Masters degree in Civil Engineering. She has taught Mathematics in the UK, Malaysia and Switzerland covering GCSE, IGCSE, A-Level and IB. She particularly enjoys applying Mathematics to real life and endeavours to bring creativity to the content she creates.