Data transmission: IP addressing, media, packet switching and routing – IB Computer Science exam questions with mark scheme

1Standard4 marks

Describe how a large file is sent across the internet using packet switching. [4]

The file is broken into (small) packets(1)
Each packet has a header containing the destination IP address (allow: source address, sequence/packet number)(1)
Packets are sent independently and may take different routes, chosen by routers according to current traffic/availability(1)
At the destination the packets are reassembled in the correct order (using sequence numbers)(1)
Answer: Split into numbered packets, routed independently, reassembled at destination.
Examiner tip: Mention sequence numbers; they are what makes reassembly possible when packets arrive out of order.

2Standard4 marks

A 450 MB file is downloaded over a link with a bandwidth of 80 Mbit/s. (Use 1 MB = 8 Mbit.)

(a) Calculate the minimum time to download the file. [2]

(b) Protocol overhead means 12% more data than the file itself must be transmitted. Calculate the new download time. [2]

450 × 8 = 3600 Mbit(1)
3600 ÷ 80 = 45 s (unit required)(1)
3600 × 1.12 = 4032 Mbit (allow 45 × 1.12, or 504 MB)(1)
4032 ÷ 80 = 50.4 s (follow through from (a); unit required)(1)
Answer: (a) 45 s (b) 50.4 s
Examiner tip: Convert bytes to bits before dividing by a rate given in bits per second.

3Core4 marks

A network technician records three addresses from a configuration file:

A: 192.168.4.256 B: 2001:db8:0:12::4f C: 203.0.113.9

(a) For each address, state whether it is a valid IPv4 address, a valid IPv6 address or not valid. For any address that is not valid, give the reason. [3]

(b) An IPv4 address is 32 bits long and an IPv6 address is 128 bits long. State, as a power of 2, how many times more addresses the IPv6 address space contains than the IPv4 address space. [1]

A: not valid, because 256 is outside the range 0–255 for an 8-bit octet (both the verdict and the reason are needed for the mark)(1)
B: valid IPv6 (accept: hexadecimal groups separated by colons; :: shows omitted groups of zeros)(1)
C: valid IPv4(1)
$2^{96}$ (accept $2^{128}/2^{32}$ simplified to $2^{96}$; accept $\approx 7.9 \times 10^{28}$ if $2^{96}$ also seen; do not accept 4 or 96)(1)
Answer: (a) A not valid (256 > 255); B valid IPv6; C valid IPv4. (b) $2^{96}$
Examiner tip: Students often say IPv6 has ‘4 times’ as many addresses because 128 is 4 × 32: the number of addresses grows as a power of 2, so the ratio is $2^{128-32}$.

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4Standard5 marks

The diagram shows a home network connected to the internet through a router that performs network address translation (NAT). Part of the router's NAT table is shown below.

Inside (private) address : port	Outside (public) address : port
192.168.1.20 : 52100	203.0.113.45 : 40001
192.168.1.34 : 52100	203.0.113.45 : 40002
192.168.1.51 : 61822	203.0.113.45 : 40003

(a) Explain why the laptop's address, 192.168.1.20, cannot be used as the source address of a packet travelling across the internet. [2]

(b) The web server sends a reply addressed to 203.0.113.45, port 40002. State the IP address and port to which the router forwards this reply. [1]

(c) The laptop and the phone are both using port 52100. Explain why this does not cause replies to be delivered to the wrong device. [1]

(d) State one security benefit of NAT for the devices on the home LAN. [1]

(a) 192.168.x.x is a private address range (reserved for use inside local networks) …(1)
… so it is not unique / the same address is reused in many other LANs, so internet routers will not route to it / replies could not be returned to it (accept: private addresses are not routable on the public internet)(1)
(b) 192.168.1.34, port 52100 (both needed)(1)
(c) The router gives each connection a different public port (40001 and 40002) and uses this unique public port to look up the correct private address in the NAT table (accept: the public port, not the private port, identifies the connection)(1)
(d) Devices' private addresses are hidden from the internet / an outside host cannot start a connection to an inside device unless there is a matching NAT entry (accept: unsolicited inbound packets are dropped; reject ‘NAT encrypts data’)(1)
Answer: (a) Private range, not unique/routable on internet. (b) 192.168.1.34 : 52100. (c) Distinct public ports 40001/40002 identify each connection. (d) Internal addresses hidden; unsolicited inbound traffic has no NAT entry so is blocked.
Examiner tip: In (b) many students give the public port (40002) or forget the port entirely; the router rewrites the destination back to the inside address and port.

5Standard4 marks

A secondary school network has 600 student devices that connect to the wireless network each day, a network printer in the library, and a server that hosts the school's intranet.

(a) Explain why the network printer should be given a static IP address. [2]

(b) Outline two advantages of using dynamic IP addressing for the student devices. [2]

(a) Other devices / print queues are configured to send print jobs to one fixed address (accept: DNS record or users' computers point to its IP address) …(1)
… if the address changed (e.g. after a lease expired) devices would no longer find the printer / printing would fail until reconfigured(1)
(b) Any two different advantages from: (i) addresses are assigned automatically by DHCP, so the 600 devices need no manual configuration; (ii) no risk of human error giving duplicate / conflicting addresses; (iii) addresses are only leased while a device is connected and are returned to the pool, so fewer addresses are needed than the total number of devices that ever connect / addresses are reused efficiently; (iv) devices can move between networks and automatically receive a correct address – first advantage(1)
Second, different advantage from the list above (do not credit the same advantage twice in different words)(1)
Answer: (a) Clients are configured with a fixed address; a changing address would break printing. (b) Automatic configuration with no duplicates; addresses reused from a pool.
Examiner tip: ‘A static address is more secure’ is not credited; the reason for a static address is that other devices must always find the device at the same address.

6Hard6 marks

A farm business is upgrading its network. The table gives typical data for three transmission media.

Medium	Typical bandwidth	Maximum segment length	Cost to install	Effect of electrical interference
Twisted pair (Cat 6)	1–10 Gbit/s	100 m	Low	Some effect
Fibre optic	10–100 Gbit/s	Several km	High	None
Wireless (Wi-Fi)	Up to about 1 Gbit/s, shared	About 30–50 m indoors	Low	Affected by interference and obstacles

For each of the following links, identify the most suitable medium and justify your choice using the data and the context.

(a) A link between the farm office and a grain-processing barn 600 m away. The cable must run next to large electric motors. [2]

(b) Eight fixed desktop computers in the farm office, each within 20 m of the office switch. [2]

(a) Fibre optic(1)
Any one justification linked to the context: 600 m is far beyond the 100 m maximum segment length of twisted pair (and beyond Wi-Fi range) / fibre is unaffected by the electrical interference from the motors (the higher installation cost is acceptable for a single link; ignore ‘fibre is faster’ alone)(1)
(b) Twisted pair (Cat 6)(1)
20 m is within the 100 m limit, it is cheap to install and gives a reliable/fast (1–10 Gbit/s) dedicated connection for fixed devices; fibre's extra cost not needed (any one context-linked reason)(1)
(c) Wireless (Wi-Fi)(1)
Any one context-linked reason: customers are mobile and bring their own phones, which have no Ethernet port, so a cable is impractical / a shop-sized area is within the 30–50 m indoor range / the lower, shared bandwidth is enough for browsing or payments(1)
Answer: (a) Fibre: 600 m > 100 m and immune to motor interference. (b) Twisted pair: within 100 m, cheap, fast and reliable. (c) Wi-Fi: mobile customer devices, within range.
Examiner tip: A justification must use the scenario: ‘fibre is faster’ alone does not earn the mark; link it to the 600 m distance or the motors.

7Standard6 marks

A video file of 5 850 000 bytes is sent across a network using packet switching. Each packet can carry at most 1460 bytes of the file as its payload, and every packet also has a 40-byte header.

(a) Calculate the number of packets needed to send the file. [2]

(b) Calculate the total number of bytes transmitted, including headers. [2]

(c) The link has a bandwidth of 50 Mbit/s. Calculate the minimum time to transmit all the packets. Give your answer in seconds to 2 significant figures. (Take 1 Mbit = $10^6$ bits.) [2]

(a) $5\,850\,000 \div 1460 = 4006.8\ldots$(1)
rounded up to 4007 packets (the last packet carries the remaining 1240 bytes; 4006 scores this mark 0)(1)
(b) header bytes $= 4007 \times 40 = 160\,280$ (follow through from their (a))(1)
total $= 5\,850\,000 + 160\,280 = 6\,010\,280$ bytes (accept 6.01 MB)(1)
(c) $6\,010\,280 \times 8 \div (50 \times 10^6)$ (conversion of bytes to bits and division by bandwidth; follow through from (b))(1)
0.96 s (accept 0.9616 s; follow through from (b) correctly rounded to 2 s.f.)(1)
Answer: (a) 4007 (b) 6 010 280 bytes (c) 0.96 s
Examiner tip: The packet count must be rounded up: a partly filled last packet is still a whole packet with its own header.

8Core6 marks

A message is split into six packets, numbered 1 to 6 in their headers, and sent across the internet. The table shows the packets received so far, in the order they arrived.

Arrival order	Sequence number	Payload
1st	3	`"THE "`
2nd	1	`"MEET"`
3rd	5	`"H GA"`
4th	2	`" AT "`
5th	6	`"TE"`

(a) Outline why packets from the same message can arrive in a different order from the order they were sent. [1]

(b) Write out the message as reassembled so far, marking where data is missing. [1]

(c) Explain how the receiving computer detects that data is missing and how the complete message is eventually obtained. [2]

(d) Distinguish between the role of a switch and the role of a router in delivering these packets. [2]

(a) Each packet is routed independently, so packets may take different routes with different lengths / delays / congestion(1)
(b) MEET AT THE ____H GATE (gap shown between “THE ” and “H GA”; accept any clear indication of the gap)(1)
(c) The header shows sequence numbers 1–6 (or the total number of packets) and sequence number 4 has not arrived (accept: gap in sequence numbers detected)(1)
The missing packet is requested again / is not acknowledged, so after a timeout the sender retransmits packet 4 and it is put in position (accept: TCP retransmission)(1)
(d) A switch forwards data between devices within the same local network, using the destination MAC/hardware address(1)
A router forwards packets between different networks, reading the destination IP address and using its routing table to choose the next hop(1)
Answer: (a) Packets routed independently via different paths. (b) MEET AT THE ____H GATE. (c) Sequence number 4 absent; retransmitted after request/timeout. (d) Switch: within LAN by MAC address; router: between networks by IP address and routing table.
Examiner tip: For (d), state what each device uses to decide: MAC address for a switch, destination IP address and routing table for a router.

9HardHL only7 marks

The diagram shows six routers, R1 to R6, and the cost of each link. A lower total cost is a better route.

(a) State the lowest-cost route from R1 to R6 and its total cost. [2]

(b) The link between R2 and R4 fails. State the new lowest-cost route from R1 to R6 and its total cost. [2]

(c) Before the failure, every router used static routing, with routes set up to follow your answer to (a). Explain what happens to packets sent from R1 to R6 after the R2–R4 link fails, and how this would differ if the routers used dynamic routing. [3]

(a) R1 → R3 → R2 → R4 → R5 → R6(1)
total cost 13 (follow through: cost correctly added for a valid stated route)(1)
(b) R1 → R3 → R5 → R6(1)
total cost 14(1)
(c) Static: the routing tables are not updated automatically, so R2 still forwards the packets to R4 over the failed link and they are dropped / lost(1)
This continues until an administrator manually changes the routing tables (accept: the network has no automatic recovery even though another route exists)(1)
Dynamic: routers detect the failure and share updated route information with each other, recalculating routes so packets are automatically sent via R3 → R5 → R6 (accept: after a short delay while routers update / converge)(1)
Answer: (a) R1–R3–R2–R4–R5–R6, cost 13. (b) R1–R3–R5–R6, cost 14. (c) Static: packets dropped at R2 until manual reconfiguration; dynamic: routers exchange updates and automatically reroute via R3–R5–R6.
Examiner tip: The direct-looking route R1–R2–R4 is not cheapest: add the costs of every candidate route rather than counting hops.

10StretchHL only6 marks

A logistics company has a head office and 40 depots, connected by a wide area network. At present, a technician configures every router with static routes. The company plans to add extra links between some depots so that traffic can still flow if a link fails, and it expects to open 15 new depots over the next two years.

Discuss whether the company should change from static routing to dynamic routing. [6]

Static routing: no route-update traffic between routers, so no bandwidth or router processing used for routing information(1)
Static routing: routes are fixed and predictable, and are not advertised to other routers, so they cannot be altered by false route updates (more control / security)(1)
Static routing: every route on every router must be set up by hand, which is time-consuming and error-prone for 40 (rising to 55) depots; each new depot needs many routers updating(1)
Static routing: does not adapt to failures, so the new backup links would not be used automatically when a link fails(1)
Dynamic routing: routers discover routes and update tables automatically, so a failed link is routed around using the extra links, and new depots are added with little manual configuration (scales well)(1)
Dynamic routing: uses bandwidth, CPU and memory for exchanging routing information, takes time to converge after a change, is more complex to set up, and route updates could be exploited if not secured(1)
Conclusion: dynamic routing is the better choice here because the redundant links and expansion need automatic failover and scalability; the overhead is small compared with the manual effort and downtime of static routes (accept a reasoned hybrid, e.g. static default routes at small depots with a single link and dynamic routing in the core)(1)

Level	Marks	What it looks like
Level 3	5–6	Balanced discussion of both static and dynamic routing, consistently applied to the backup links and expansion, with a reasoned conclusion that follows from the points made.
Level 2	3–4	Advantages and/or disadvantages of both approaches described with some link to the scenario; a conclusion may be stated but is weakly justified, or the discussion is one-sided.
Level 1	1–2	Isolated points about static or dynamic routing, mostly generic, with little or no application to the company and no judgement.
Level 0	0	Nothing worthy of credit.

Answer: Dynamic routing is justified: it makes use of the new redundant links automatically and scales to 55 depots, at the cost of routing overhead, complexity and some security risk; static routing is simpler and more predictable but would need manual reconfiguration after every failure or new depot.
Examiner tip: Answers that only list general advantages and disadvantages stay in the middle band: tie each point to the backup links or the 15 new depots, and finish with a judgement.

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