Computer Networks

Question 1

(a) Shannon capacity (5 marks)

SNR linear: $30\text{ dB} = 10^{30/10} = 10^{3} = 1000$. (1) Shannon–Hartley: $C = B\log_2(1+\text{SNR})$. (1) $C = 4\times10^{6}\cdot\log_2(1001).$ $\log_2(1001) = \ln(1001)/\ln 2 \approx 9.9672$. (1) $C \approx 4\times10^{6}\times 9.9672 = 39.87\times10^{6}\text{ bit/s} \approx 39.87\text{ Mbit/s}.$ (2) (Accept $\approx 39.9$ Mbit/s; using $\log_2 1000 = 9.966$ gives essentially the same.)

(b) Nyquist M-ary (6 marks)

Nyquist noiseless capacity: $C = 2B\log_2 M$. (1) Max signalling (Nyquist) rate $= 2B = 8\times10^{6}$ symbols/s. (1) Set $2B\log_2 M = C_{\text{Shannon}}$: $\log_2 M = \frac{C}{2B} = \frac{39.87\times10^{6}}{8\times10^{6}} \approx 4.984.$ (2) So $M \approx 2^{4.984} \approx 31.6$. Round up to smallest usable power of two $\ge$ this that keeps $\log_2 M$ integer for clean encoding: $M = 32$ ($=2^5$). (1) Bounding explanation: Nyquist assumes a noiseless channel and lets $M$ grow arbitrarily; in a real channel, noise limits reliable distinguishable levels, and Shannon gives the upper bound on error-free rate. Hence any Nyquist scheme's reliable throughput cannot exceed the Shannon capacity. (1)

(c) Manchester baud/bandwidth (4 marks)

Manchester encodes each bit with a mid-bit transition ⇒ up to 2 transitions per bit ⇒ signalling (baud) rate is twice the bit rate. For $R = 10$ Mbit/s the transition rate is $20$ Mbaud. (2) Minimum bandwidth $\approx$ baud rate: Manchester needs $\approx 10$ MHz whereas NRZ needs $\approx 5$ MHz for the same 10 Mbit/s — Manchester requires double the bandwidth of NRZ. (2)

(d) Manchester pros/cons (5 marks)

• Advantage (any one, 3): Guaranteed transition in every bit period ⇒ self-clocking / built-in clock recovery, and it is DC-balanced (no baseline wander), so long runs of identical bits don't cause loss of sync. (3)
• Disadvantage (2): Requires twice the bandwidth of NRZ (higher signalling rate), limiting maximum bit rate on a given medium. (2)

Question 2

(a) VLSM allocation (9 marks) — 2.25 per subnet, allocate sequentially

Requirements → host bits: need $2^h - 2 \ge$ hosts.

Allocation reasoning: /20 spans 16 third-octet values (0–15) → next starts at .16.0; /21 spans 8 (16–23) → next .24.0; /22 spans 4 (24–27) → next .28.0; /30 spans 4 addresses. (marks: correct prefix + network each subnet)

(b) Proof of $2^h-2$ (5 marks)

With $h$ host bits, the host portion ranges over all $2^{h}$ bit patterns → $2^{h}$ total addresses in the subnet. (2) Two are reserved:

• All-zero host bits = the network (subnet) address, used to identify the network itself; not assignable to an interface. (1)
• All-one host bits = the directed broadcast address for that subnet; not assignable to a host. (1) Hence usable $= 2^{h} - 2$. (0.5) Exception: For a /31 ($h=1$, point-to-point link, RFC 3021) the "$-2$" is waived — both addresses are usable as the two endpoints, since broadcast is meaningless on a P2P link. (A /32 is a single host route.) (0.5)

(c) CRC (6 marks)

Degree of $G$ is 4 → append 4 zeros: dividend $= 1101011011\,0000$. Divide by $10011$ (XOR long division):

1101011011 0000 ÷ 10011
110101...
 10011
 ------
 10001 1011 0000   (running remainder as we bring bits)

Performing the full XOR division, the remainder (FCS) is:

$\boxed{R = 1110}$

Transmitted frame $= D \,\|\, R = 1101011011\,1110$. (4)

Verification: the transmitted frame $1101011011\,1110$ divided by $10011$ gives remainder $0000$ → divisible, confirming FCS. (2)

(Standard textbook result: $D=1101011011$, $G=10011$ ⇒ remainder $1110$.)

Question 3

(a) Bandwidth–delay product (5 marks)

$\text{BDP} = \text{capacity}\times\text{RTT} = 10^{9}\text{ bit/s}\times 0.02\text{ s} = 2\times10^{7}\text{ bit} = 2.5\times10^{6}\text{ bytes} = 2.5\text{ MB}.$ (3) The receive window must be $\ge 2.5$ MB to keep the pipe full. Since the base TCP window field is 16 bits (max 65 535 B), the TCP Window Scale option (RFC 7323) is required to advertise windows this large. (2)

(b) cwnd table (8 marks)

Slow start doubles each round until reaching ssthresh = 8; then additive increase (+1/round).

Loss event at end of round 10 (cwnd = 14): $\text{ssthresh} = 14/2 = 7$, $\text{cwnd} = 7$ (Reno fast recovery → set to new ssthresh). Continue additive increase:

Mark points: correct doubling 1→2→4→8 (2), correct transition at 8 (1), correct CA increments to 14 (2), correct halving to ssthresh=7, cwnd=7 (2), correct rounds 11–13 (1).

(c) TCP echo server (7 marks)

import socket
 
srv = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
srv.setsockopt(socket.SOL_SOCKET, socket.SO_REUSEADDR, 1)
srv.bind(('', 5000))
srv.listen(1)
 
conn, addr = srv.accept()          # completes 3-way handshake
with conn:
    while True:
        data = conn.recv(1024)
        if not data:               # client closed
            break
        conn.sendall(data.upper())
conn.close()
srv.close()

Marks: socket()+bind(5000)+listen (2), accept (1), recv/upper/sendall loop (2).

Handshake explanation (2): Before recv() returns application data, TCP completes the 3-way handshake: the client sends SYN (its ISN), the server replies SYN-ACK (its ISN + ack of client's SYN), and the client sends ACK. Only after this does accept() return an established connection ready for data transfer.

[
  {"claim":"Shannon capacity ~39.87 Mbit/s for B=4MHz, SNR=1000", "code":"import sympy as sp; C=4*10**6*sp.log(1001,2); result = abs(float(C)-39.869e6) < 0.05e6"},
  {"claim":"Nyquist log2 M ~4.984 -> M rounds up to 32", "code":"C=39.869e6; B=4*10**6; L=C/(2*B); result = (abs(L-4.9836)<0.01) and (2**5==32)"},
  {"claim":"BDP = 2.5 MB for 1Gbps x 20ms", "code":"bdp_bits=10**9*0.02; bdp_bytes=bdp_bits/8; result = bdp_bytes==2.5*10**6"},
  {"claim":"CRC remainder of 1101011011 with G=10011 is 1110 and frame divisible", "code":"D=0b1101011011; G=0b10011; msg=D<<4\nrem=msg\nfor i in range(len(bin(msg))-2-1, 4-1, -1):\n    if rem>>i & 1:\n        rem ^= G<<(i-4)\nfcs=rem & 0b1111\nframe=(D<<4)|fcs\nchk=frame\nfor i in range(len(bin(frame))-2-1,4-1,-1):\n    if chk>>i &1: chk^=G<<(i-4)\nresult = (fcs==0b1110) and ((chk & 0b1111)==0)"},
  {"claim":"cwnd sequence reaches 14 at round 10, halves to 7", "code":"cwnd=1; ssthresh=8; seq=[]\nfor r in range(1,11):\n    seq.append(cwnd)\n    if cwnd<ssthresh: cwnd=cwnd*2\n    else: cwnd=cwnd+1\nloss_cwnd=seq[-1]; new_ss=loss_cwnd//2\nresult = (seq[9]==14) and (new_ss==7)"}
]