BREAKING
Imaginary number feels seen, refuses to be real★Matrices multiplied in wrong order, student loses faith★Proof by induction ends, nobody knows why it worked★
  1. Home
  2. A Level
  3. Further Mathematics
  4. Edexcel
  5. Travelling salesman problem

Edexcel A Level Further Maths

Topic questions Decision mathematics Travelling salesman problem

Travelling salesman problem

5 questions · 54 marks · about 1 hour

Travelling salesman problem questions

Edexcel 9FM0 Travelling salesman problem 5 questions

Paper 3D: Decision Mathematics 1 June 2024, Question 2 13 marks

The table below represents a network of shortest distances, in miles, to travel between nine castles, A, B, C, D, E, F, G, H and J.

ABCDEFGHJ
A–5059265040876359
B50–28617963456448
C5928–335735703645
D266133–2464713733
E50795724–40643031
F4063356440–477071
G874570716447–3467
H63643637307034–33
J5948453331716733–
Reveal mark scheme Mark scheme for question 2
(a)

Use Prim’s algorithm, starting at D, to find the minimum spanning tree for this network. You must clearly state the order in which you select the arcs of your tree.

[3]
Reveal mark scheme Mark scheme for question 2(a)
(b)

State the weight of the minimum spanning tree found in part (a).

[1]
Reveal mark scheme Mark scheme for question 2(b)
(c)

Draw the minimum spanning tree using the vertices given in Diagram 1 in the answer book.

[1]
Reveal mark scheme Mark scheme for question 2(c)
(d)

A historian needs to visit all of the castles, starting and finishing at the same castle, and wishes to minimise the total distance travelled.

Use your answer to part (b) to calculate an initial upper bound for the length of the historian’s route.

[1]
Reveal mark scheme Mark scheme for question 2(d)
(e)(i)

Use the nearest neighbour algorithm, starting at D, to find an upper bound for the length of the historian’s route.

(e)(ii)

Write down the route which gives this upper bound.

[3]
(f)

Using the nearest neighbour algorithm, starting at F, an upper bound of length 352 miles was found.

State the best upper bound that can be obtained by using this information and your answers from parts (d) and (e). Give the reason for your answer.

[1]
Reveal mark scheme Mark scheme for question 2(f)
(g)

By deleting A and all of its arcs, find a lower bound for the length of the historian’s route.

[2]
Reveal mark scheme Mark scheme for question 2(g)
(h)

By deleting J and all of its arcs, a lower bound of length 274 miles was found.

State the best lower bound that can be obtained by using this information and your answer to part (g). Give the reason for your answer.

[1]
Reveal mark scheme Mark scheme for question 2(h)

Paper 3D: Decision Mathematics 1 June 2022, Question 3 10 marks

The initial distance matrix (Table 1) shows the lengths, in metres, of the corridors connecting six classrooms, A, B, C, D, E and F, in a school. For safety reasons, some of the corridors are one‑way only.

Figure for question 3
(a)

By adding the arcs from vertex A, along with their weights, complete the drawing of this network on Diagram 1 in the answer book.

[2]
Reveal mark scheme Mark scheme for question 3(a)
(b)

Floyd’s algorithm is to be used to find the complete network of shortest distances between the six classrooms.

The distance matrix after two iterations of Floyd’s algorithm is shown in Table 2.

Figure for question 3

Perform the next two iterations of Floyd’s algorithm that follow from Table 2. You should show the distance matrix after each iteration.

[4]
Reveal mark scheme Mark scheme for question 3(b)
(c)(i)

The final distance matrix after completion of Floyd’s algorithm is shown in Table 3.

Figure for question 3

Yinka must visit each classroom. He will start and finish at E and wishes to minimise the total distance travelled.

Use the nearest neighbour algorithm, starting at E, to find two Hamiltonian cycles in the completed network of shortest distances.

(c)(ii)

Find the length of each of the two cycles.

(c)(iii)

State, with a reason, which of the two cycles provides the better upper bound for the length of Yinka’s route.

[4]

Paper 3D: Decision Mathematics 1 November 2021, Question 3 8 marks

Figure for question 3

Table 1 shows the shortest distances, in miles, between eight towns, A, B, C, D, E, F, G and H.

(a)

Use Prim’s algorithm, starting at A, to find the minimum spanning tree for this table of distances. You must clearly state the order in which you select the edges of your tree.

[3]
Reveal mark scheme Mark scheme for question 3(a)
(b)

State the weight of the minimum spanning tree.

[1]
Reveal mark scheme Mark scheme for question 3(b)
(c)
Figure for question 3

Table 2 shows the distances, in miles, between town J and towns A, B, C, D, E, F, G and H.

Pranav needs to visit all of the towns, starting and finishing at J, and wishes to minimise the total distance he travels.

Starting at J, use the nearest neighbour algorithm to obtain an upper bound for the length of Pranav’s route. You must state your route and its length.

[2]
Reveal mark scheme Mark scheme for question 3(c)
(d)

Starting by deleting J, and all of its edges, find a lower bound for the length of Pranav’s route.

[2]
Reveal mark scheme Mark scheme for question 3(d)

Paper 3D: Decision Mathematics 1 October 2020, Question 3 9 marks

Figure for question 3

Direct roads between five villages, A, B, C, D and E, are shown in Figure 2. The weight on each arc is the time, in minutes, it takes to travel along the corresponding road. The road from D to C is one-way as indicated by the arrow on the corresponding arc.

Floyd’s algorithm is to be used to find the complete network of shortest times between the five villages.

(a)

Set up initial time and route matrices.

The matrices after two iterations of Floyd’s algorithm are shown below.

Time matrix
Route matrix

ABCDEABCDE
A–84718AABCDB
B8–31510BABCAE
C43–116CABCAE
D7151–1DAACDE
E181061–EBBCDE
[2]
Reveal mark scheme Mark scheme for question 3(a)
(b)

Perform the next two iterations of Floyd’s algorithm that follow from the tables above. You should show the time and route matrices after each iteration.

The final time matrix after completion of Floyd’s algorithm is shown below.

Final time matrix

ABCDE
A–7478
B7–3109
C43–76
D541–1
E6521–
[4]
Reveal mark scheme Mark scheme for question 3(b)
(c)(i)

Use the nearest neighbour algorithm, starting at A, to find a Hamiltonian cycle in the complete network of shortest times.

(c)(ii)

Find the time taken for this cycle.

(c)(iii)

Interpret the cycle in terms of the actual villages visited.

[3]
Reveal mark scheme Mark scheme for question 3(c)(iii)

Paper 3D: Decision Mathematics 1 June 2019, Question 3 14 marks

Figure for question 3

The network in Figure 2 shows the direct roads linking five villages, A, B, C, D and E. The number on each arc represents the length, in miles, of the corresponding road. The roads from A to E and from C to B are one-way, as indicated by the arrows.

(a)

Complete the initial distance and route tables for the network provided in the answer book.

[2]
Reveal mark scheme Mark scheme for question 3(a)
(b)

Perform the first three iterations of Floyd’s algorithm. You should show the distance table and the route table after each of the three iterations.

Figure for question 3
[5]
Reveal mark scheme Mark scheme for question 3(b)
(c)(i)

Explain how the partially completed final route table can be used to find the shortest route from E to A.

Reveal mark scheme Mark scheme for question 3(c)(i)
(c)(ii)

State this route.

[3]
Reveal mark scheme Mark scheme for question 3(c)(ii)
(d)(i)

Mabintou decides to use the distance table to try to find the shortest cycle that passes through each vertex. Starting at D, she applies the nearest neighbour algorithm to the final distance table.

State the cycle obtained using the nearest neighbour algorithm.

(d)(ii)

State the length of this cycle.

Reveal mark scheme Mark scheme for question 3(d)(ii)
(d)(iii)

Interpret the cycle in terms of the actual villages visited.

(d)(iv)

Prove that Mabintou’s cycle is not optimal.

[4]
Reveal mark scheme Mark scheme for question 3(d)(iv)

Report a problem with a question on this page
What is wrong?

Goes to whoever maintains the question split. No account needed and nothing is published on this page.

Memes

Certified Further Maths memes. Peer-reviewed by nobody.

Laugh now. Cry in the exam hall later. Then come back and do a past paper.

Prove it by induction
Write ‘clearly’ and move on
Me on the final ‘show that’
Using the formula booklet
Proving it by induction
Writing ‘by symmetry’ to skip half the work
Writing ‘trivially’ and daring the examiner
Further enlightenment, ranked

Mum 💖

how's further maths going?? 🥰

my answers are all complex rn

keep it simple love 💖

📱 Screen time this week · Spotify4h 59m

The answer was i. Nobody knows why

Further Maths starter pack

Calculator that does matricesProof that’s 3 pages longArgand diagram doodlesThis website
Collect all four. Prove it.