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Bounds for the travelling salesman problemEdexcel International A Level Maths: Subtopic test

10 questions, 27 marks

Edexcel International A Level Maths

Bounds for the travelling salesman problem

Total 27 marks

Name

Class

Date

  1. 1
    A complete network has five vertices PP, QQ, RR, SS and TT, with lengths PQ=14PQ=14, PR=10PR=10, PS=17PS=17, PT=9PT=9, QR=11QR=11, QS=8QS=8, QT=15QT=15, RS=12RS=12, RT=13RT=13 and ST=16ST=16.
    (a)
    The nearest neighbour algorithm is applied starting at QQ. Which tour is produced?
    [1 mark]
    • AQQ-RR-PP-TT-SS-QQ
    • BQQ-SS-PP-TT-RR-QQ
    • CQQ-PP-TT-RR-SS-QQ
    • DQQ-SS-RR-PP-TT-QQ
    (b)
    The tour from QQ in part (a) has length 5454. Which statement about the travelling salesman problem is correct?
    [1 mark]
    • AThe shortest tour has length exactly 5454
    • BThe shortest tour has length at most 5454
    • CThe shortest tour has length at least 5454
    • DEvery starting vertex gives a tour of length 5454
    (c)
    Apply the nearest neighbour algorithm starting at PP. Write down the tour and its length, and state what this tells you about the shortest tour.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    A complete network has five depots VV, WW, XX, YY and ZZ, with lengths VW=11VW=11, VX=7VX=7, VY=9VY=9, VZ=14VZ=14, WX=6WX=6, WY=13WY=13, WZ=10WZ=10, XY=8XY=8, XZ=12XZ=12 and YZ=15YZ=15. A lower bound for the travelling salesman problem is found by deleting vertex VV.
    (a)
    What is the total length of a minimum spanning tree of the network with VV deleted?
    [1 mark]
    • A2424
    • B2626
    • C3434
    • D4040
    (b)
    Which two arcs at VV are added to the minimum spanning tree to form the lower bound?
    [1 mark]
    • AVZVZ and VWVW
    • BVXVX and VWVW
    • CVXVX and VYVY
    • DVYVY and VZVZ
    (c)
    Find the lower bound for the length of the shortest tour from deleting VV, and explain why it is a lower bound.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    A delivery firm has five depots AA to EE. The only direct roads and their lengths in km are AB=7AB=7, BC=5BC=5, CD=6CD=6, DE=4DE=4 and BE=9BE=9. A driver must visit every depot and return to the start, and may pass through a depot more than once.
    (a)
    Find the shortest distances between AA and DD, between BB and DD, and between CC and EE.
    [3 marks]
    (b)
    The shortest distances are AB=7AB=7, AC=12AC=12, AD=18AD=18, AE=16AE=16, BC=5BC=5, BD=11BD=11, BE=9BE=9, CD=6CD=6, CE=10CE=10 and DE=4DE=4. Apply the nearest neighbour algorithm to this complete network starting at AA, to find an upper bound. Give the route that the driver takes in the original network.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    A complete network has five vertices AA, BB, CC, DD and EE, with lengths AB=12AB=12, AC=9AC=9, AD=15AD=15, AE=7AE=7, BC=8BC=8, BD=11BD=11, BE=14BE=14, CD=10CD=10, CE=13CE=13 and DE=6DE=6.
    (a)
    (i) Use Kruskal's algorithm to find a minimum spanning tree and its weight.
    (ii) Hence find an upper bound by walking around the tree, and use shortcuts to improve this bound.
    [6 marks]
    (b)
    Find a lower bound by deleting BB. Hence show that the shortest tour has length 4141.
    [6 marks]

    Total for question 4: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).