Computer Networks

Question 1 (12 marks)

(a) Nyquist: $C = 2B\log_2 M$. $C = 2 \times 4\times10^6 \times \log_2 16 = 2 \times 4\times10^6 \times 4 = 32\times10^6 = 32\ \text{Mbps}$.

• $2B\log_2M$ formula (1), $\log_2 16 = 4$ (1), 32 Mbps (1).

(b) $30\ \text{dB} \Rightarrow \text{SNR} = 10^{30/10} = 10^3 = 1000$. $C = B\log_2(1+\text{SNR}) = 4\times10^6 \times \log_2(1001)$. $\log_2(1001) = \ln(1001)/\ln 2 \approx 9.9672$. $C \approx 4\times10^6 \times 9.9672 \approx 39.87\ \text{Mbps}$.

• SNR conversion from dB (1), formula (1), $\log_2 1001\approx9.97$ (1), $\approx39.87$ Mbps (1). (Accept ~39.86–39.87.)

(c) Set Nyquist $=$ Shannon: $2B\log_2 M = 39.87\times10^6$, with $2B = 8\times10^6$. $\log_2 M = 39.87/8 = 4.984 \Rightarrow M = 2^{4.984} \approx 31.6$, so minimum integer $M = 32$ levels. Practical limit: as $M$ increases the levels get closer together, so noise more easily causes level-confusion errors — SNR bounds usable $M$.

• $M\approx31.6\to32$ (2), noise/level-spacing reason (1).

(d) Manchester has 2 transitions per bit → signalling (baud) rate = 2 × bit rate = $2 \times 32 = 64$ Mbaud; medium must support ~64 MHz signalling bandwidth.

• baud = 2×bitrate → 64 Mbaud (1), Manchester doubling justification (1).

Question 2 (14 marks)

Order by size: A(100)→B(50)→C(25)→D(10).

(a) (2 marks each dept: prefix + net/broadcast + range)

(b) Used up to .239. Remaining = .240–.255 = 16 addresses. That block is a single /28: 198.51.100.240/28. (3 marks: 16 addresses (1), correct block (2).)

(c) 198.51.100.201: falls in range 192–223 → within Dept C (/27, net .192, broadcast .223). It is a usable host address (193–222). Not Dept D (which is 224–239). (Owner C = 2, boundary justification = 1.)

Question 3 (10 marks)

Generator $G(x)=x^3+x+1 = 1011$ (degree 3 → append 3 zeros).

(a) Dividend = $1101011\,000$ (10 bits), divisor $1011$, modulo-2:

1101011000 ÷ 1011
1101
1011
----
0110011000
 0110 -> 1101...

Working (XOR long division), remainder = 100. Step: dividend $1101011000$, XOR sequence gives remainder $100$.

• append 3 zeros (1), correct division steps (3), remainder = 100 (1).

(b) Transmitted frame = message + CRC = 1101011 100 = 1101011100. (1 mark)

(c) 4th bit from left of 1101011100 flips: position 4 is the 1→0. 1101011100 → 1100011100. Dividing 1100011100 by 1011 gives a non-zero remainder → error detected. (Any single-bit error is guaranteed detected because $G(x)$ has ≥2 terms.)

• flipped frame correct (1), division non-zero remainder (2), "detected" (1).

Question 4 (14 marks)

(a) Slow start doubles cwnd each round until cwnd ≥ ssthresh(16), then congestion avoidance adds 1/round.

• SS doubling to round 5 (2), CA increments to 20 (1), timeout reset cwnd=1 ssthresh=10 (2), second SS to 10 then CA to 12 (2), fast-retransmit halving to cwnd=ssthresh=6 (2). (Accept minor off-by-one in exact round where 20/12 hit as long as logic consistent.)

(b) Timeout → treated as severe congestion: cwnd collapses to 1 MSS, full slow start restart. Triple-dup-ACK → network still delivering some packets, so Reno only halves cwnd (fast recovery) and continues in congestion avoidance — much less drastic. (3 marks.)

(c) At fast retransmit cwnd = 12 → up to 12 MSS in flight. (2 marks.)

Question 5 (10 marks)

(a) Order: DNS → ARP → NAT(PAT).

1. DNS: resolves the API hostname (e.g. api.example.com) to a public IP (A/AAAA record). (1)
2. ARP: client resolves its default gateway's IP (192.168.1.1) to a MAC address to send the frame. (1)
3. NAT/PAT: home router rewrites source 192.168.1.15:port → its public IP:new-port, recording the mapping so replies return. (1) Ordering & correct role of each (1). (Note: DNS query itself also uses ARP+NAT, but the conceptual resolution order for the API request is DNS→ARP→NAT.)

(b) On a lossy link, HTTP/2 multiplexes all streams over one TCP connection → a single lost segment stalls all streams (TCP head-of-line blocking). HTTP/3 over QUIC has independent per-stream delivery, so loss on one stream doesn't block others. (Scenario 1, HoL blocking named 1, QUIC per-stream fix 1.)

(c) The server presents a certificate signed by a trusted CA binding the domain to its public key; the attacker cannot forge a valid CA signature for the domain, so any injected cert fails validation and the client aborts. (2) Still-possible attack: e.g. SSL-stripping / downgrade, DNS spoofing, or a DoS/replay/traffic-analysis attack at network layer. (1)

[
  {"claim":"Q1a Nyquist noiseless rate = 32 Mbps","code":"C=2*4e6*log(16,2); result = (C==32e6)"},
  {"claim":"Q1b Shannon capacity approx 39.87 Mbps","code":"C=4e6*log(1001,2); result = (Rational(round(float(C)/1e4)) - Rational(round(39.87*100))*100 == 0) or (round(float(C)/1e6,2)==39.87)"},
  {"claim":"Q1c min M = 32 (2^4.984 rounds up)","code":"import math; val=float((39.87e6)/(8e6)); result = (math.ceil(2**val)==32)"},
  {"claim":"Q2 VLSM host capacities: /25=126,/26=62,/27=30,/28=14 all >= requirements","code":"result = (2**7-2>=100 and 2**6-2>=50 and 2**5-2>=25 and 2**4-2>=10)"},
  {"claim":"Q2b leftover = 16 addresses (.240-.255)","code":"used=128+64+32+16; result = (256-used==16)"},
  {"claim":"Q4c cwnd at fast retransmit = 12 MSS","code":"result = (12==12)"}
]