EPFO APFC 2023 General Studies · Question 83 of 120

Correct — D, (d) 15. In a mesh topology every device has a dedicated point-to-point link to every other device; that is the definition of the topology and it is the whole of the calculation. Count it in the way that makes the reasoning visible. Each of the six computers must reach the five others, so each carries five links. Six computers with five links apiece gives thirty, but every link has been counted twice — once from each of the two machines it joins — so the number of distinct links is thirty divided by two, that is 15. The same result written as a formula is n(n − 1)/2, which for n = 6 is (6 × 5)/2 = 15. It is also the number of ways of choosing two devices out of six to be joined, since a link is exactly an unordered pair of devices. Two related quantities are worth fixing at the same time, because papers alternate between them. The number of input-output ports each device needs is n − 1, here five, and that is a per-device figure, not a total. The number of links is a whole-network figure, here 15. Confusing the two is the commonest way this item is failed. There is also a quick check on any answer offered. Because the count is n(n − 1)/2, it must be a triangular number: for two, three, four, five, six and seven devices the totals run 1, 3, 6, 10, 15 and 21. Only one of the four options in this set is a triangular number of that series, which settles the item even before the arithmetic is done. Notice too that the growth is quadratic — doubling the devices roughly quadruples the cabling — and that this is precisely why a full mesh is used for small, critical networks and almost never for large ones.

• (a)6 — This is the number of computers, not the number of links, and it is placed first to catch the candidate who reads the figure out of the stem instead of computing anything. It has a second and more interesting claim on attention: 6 is a genuine mesh link count, but for four devices, not six, since (4 × 3)/2 = 6. A candidate who half-remembers the triangular series and picks the familiar-looking member of it lands here. Six links would also be the right answer for a different topology entirely — a ring of six devices, where each machine is joined only to its two neighbours and the ring closes, needs exactly six links. That is the useful contrast to carry away: a ring of n devices needs n links, a star needs n links plus a central hub, and a mesh needs n(n − 1)/2, which is why the mesh becomes expensive so quickly.
• (b)12 — This is twice the number of devices, and it is where two different mistakes both land. The first is counting each device's links as two rather than five — the ring picture rather than the mesh picture — and multiplying by six. The second is starting correctly, finding that each device needs five links, and then failing to halve the thirty, but arriving at twelve by arithmetic slip rather than at thirty. Either way the answer can be rejected without redoing the work, because the number of links in a full mesh is always n(n − 1)/2 and therefore always a triangular number: 1, 3, 6, 10, 15, 21 for two to seven devices. Twelve is not a member of that series, so no full mesh of any size has twelve links, whatever the arithmetic used to reach it.
• (c)13 — No mesh of any size has thirteen links. The count n(n − 1)/2 produces the triangular numbers 1, 3, 6, 10, 15, 21 and so on, and thirteen appears nowhere in that series, so this option can be eliminated before any calculation is attempted. Its function in the option set is to sit next to the correct answer and absorb a candidate who has computed 6 × 5 = 30, halved it approximately, or who has counted links by hand on a sketch and dropped or duplicated a couple of lines. Hand-counting is exactly what the formula exists to replace: with six vertices there are fifteen lines to draw and the drawing becomes crowded near the centre, which is where the miscount happens. Multiply, halve, and check the result against the triangular series.

Network topology describes how the devices on a network are physically or logically interconnected, and the choice of topology fixes the cost of cabling, the difficulty of adding a device and the way the network behaves when something fails. In a mesh topology every device has a dedicated point-to-point link to every other device, so each pair of devices has a private channel. That gives a mesh its distinctive properties: traffic between two machines is not shared with anybody else, a fault on one link isolates only that link and never brings the network down, and privacy and fault identification are both straightforward. The price is the wiring. With n devices the number of links is n(n − 1)/2 and each device needs n − 1 ports, so both cabling and hardware grow with the square of the number of devices, and installation and reconfiguration become impractical at any scale. The alternatives trade differently. A star connects every device to a central hub or switch, needing n links and one port per device; it is cheap and easy to extend, but the hub is a single point of failure. A bus runs every device off one backbone cable through drop lines and taps, using the least cable of all, but a break in the backbone disables the whole segment. A ring joins each device to two neighbours and needs n links, with data travelling in one direction from node to node. A tree is a hierarchy of stars, and most real installations are hybrids that combine several of these shapes.

This is a one-formula item, and the paper sets it precisely because the formula is short enough that the failure mode is not ignorance but haste. Three of the four options are numbers a candidate can produce in a hurry from the figures printed in the stem — the six from the stem itself, twice the six, and a number sitting beside the right answer — which means the item distinguishes candidates who computed from candidates who recognised. The notes record that the options here are laid out two to a line rather than one below the other, a small piece of typography that encourages exactly the skim the question punishes. What makes the topic worth more than the thirty seconds the arithmetic takes is the reason the formula matters. The quadratic growth of a mesh is the standard illustration of why network designers do not connect everything to everything: six machines need fifteen links, but twenty machines would need a hundred and ninety, and each machine would need nineteen ports. That is the argument for the star and for switching, and it is the argument a governance paper is really testing when it asks about topologies. Learn the four counts together — mesh n(n − 1)/2 links and n − 1 ports per device, star n links and a hub, ring n links, bus one backbone — and any variant of this question is answerable in a single line.

• In a mesh topology every device has a dedicated point-to-point link to every other device, so the number of links for n devices is n(n − 1)/2. For six computers that is (6 × 5)/2 = 15. The division by two matters: multiplying six devices by five links each counts every link twice, once from each end.
• Each device in a mesh needs n − 1 input-output ports, which is five here. This is a per-device figure and must not be confused with the network-wide link total; examiners alternate between the two quantities in successive papers.
• The link count of a full mesh is always a triangular number — 1, 3, 6, 10, 15, 21 for two to seven devices — so any offered answer outside that series can be rejected without computation. Neither twelve nor thirteen is a member of it.
• Cabling in a mesh grows with the square of the number of devices: twenty devices would need one hundred and ninety links and nineteen ports each. That quadratic growth is why a full mesh is reserved for small, critical networks and why switching topologies replaced it at scale.
• The comparison to hold beside it: a star uses n links and a central hub or switch, which is a single point of failure; a ring uses n links, each device joined to two neighbours; a bus uses one backbone cable with drop lines, so a break in the backbone disables the segment.
• A mesh's advantages follow from the dedicated links: traffic between a pair is not shared with other devices, so throughput and privacy are protected, and a failed link isolates only itself, which makes fault identification and fault isolation easy.

• Forgetting to halve. Six devices with five links each gives thirty only because each link is counted from both of its ends; the number of distinct links is fifteen.
• Answering with the number of ports instead of the number of links. A device in this mesh needs five ports, which is a per-device figure, while the network as a whole has fifteen links.
• Picking the number printed in the stem. Six is the number of computers, and it is also the correct link count for a mesh of four devices, which is what makes it look familiar.
• Counting links by hand from a sketch. Fifteen lines across six points cross repeatedly near the centre and are easy to duplicate or drop; the formula exists precisely to avoid that.
• Applying the mesh formula to a topology that is not a mesh. A ring of six devices has six links and a star of six devices has six links plus a hub, so the same device count gives three different answers depending on the topology named.

Topology questions in this exam come in three shapes, all short. The first is this one: a device count and a demand for the number of links, sometimes for the number of ports instead, and occasionally run backwards — a link total is given and the device count is asked for, which means solving n(n − 1)/2 for n. The second is comparative: which topology needs the least cabling, which has a single point of failure, which allows a device to be added without disturbing the rest. The third is definitional, asking what distinguishes a point-to-point link from a multipoint connection or what a hub does that a bus does not. Preparation for all three is the same and it is small: memorise the four link counts and the port count, know that the mesh total is a triangular number, and be able to state one advantage and one disadvantage of each topology in a single clause. Nothing in this area rewards elaboration, and every item in it is designed to be finished in well under a minute.

No directly related past PYQ was found.

• practice — not a real PYQ

  A full mesh network is built with eight devices. How many links does the network need, and how many input-output ports must each device have?

  • (a)8 links and 8 ports per device
  • (b)16 links and 7 ports per device
  • (c)28 links and 7 ports per device
  • (d)56 links and 8 ports per device

  Answer(c) 28 links and 7 ports per device — the link count is n(n − 1)/2 = (8 × 7)/2 = 28, and each device needs a dedicated link to each of the other seven, so it needs seven ports. The 56 in the last option is the undivided product, which counts every link twice, and 28 is a triangular number as any mesh total must be.

• practice — not a real PYQ

  Six computers are to be connected so that every computer is joined only to its two immediate neighbours and the connection closes on itself. How many links does this arrangement need?

  • (a)5
  • (b)6
  • (c)12
  • (d)15

  Answer(b) 6 — the arrangement described is a ring, and a ring of n devices needs exactly n links, because each device contributes one link to the neighbour on one side and the last link closes the loop. Fifteen would be the answer for a full mesh of the same six devices, which is the contrast the question is built on: the same device count gives very different cabling depending on the topology.