path: root/assignments/3/README.md

---
title: "COMP-347: Computer Networks"
author: "Munir Khan (ID: 3431709)"
date: "September 2025"
subtitle: "Assignment 3"
institute: "Athabasca University"
geometry: margin=1in
fontsize: 11pt
linestretch: 1.0
---

# Part 1: Short Answer Questions (30%)

## 1.1 Anchor MSC in GSM Networks (5%)

> (5%) What is the role of the anchor MSC in GSM networks?

The anchor MSC (Mobile Switching Center) keeps a call "anchored" to a single MSC for its entire duration while the mobile moves. It:
- Coordinates with HLR/VLR for authentication, location, and subscriber data.
- Manages intra- and inter-MSC handovers by setting up and tearing down trunks to visited MSCs while keeping the original call leg.
- Maintains continuity so numbering, charging, and signaling remain stable as the user roams.

## 1.2 LTE Network Characteristics (5%)

> (5%) What are the main characteristics of LTE radio access networks? How does LTE network differ from previous generations of cellular networks?

- E-UTRAN with a flat architecture: eNodeBs handle radio and many control functions; no RNC.
- Air interface: OFDMA (DL), SC-FDMA (UL); flexible channel bandwidths (1.4 to 20 MHz).
- MIMO and advanced scheduling; adaptive modulation and coding.
- All-IP core (EPC); low latency (short TTI), high throughput, QoS bearers.
- Compared to 3G (UMTS/WCDMA): LTE moves from wideband CDMA to OFDM-based access, eliminates circuit-switched core (CS services via IMS/VoLTE), reduces latency, increases spectral efficiency, and simplifies the RAN.

## 1.3 CSMA/CD Protocol (5%)

> (5%) What does CSMA/CD stand for? How does the protocol work? Explain why RTT on an Ethernet LAN is an important parameter for the CSMA/CD protocol to work properly.

- CSMA/CD: Carrier Sense Multiple Access with Collision Detection.
- Operation: Sense channel idle, transmit; detect collision (voltage anomalies), send jam signal; back off using binary exponential backoff (K slots; slot = 512 bit times); retry.
- RTT importance: To guarantee any collision is detected while transmitting, frames must be at least 512 bits so transmission lasts at least one round-trip propagation time (including repeater delays). If frames were shorter than the slot time, a station could finish sending without detecting a collision.

## 1.4 CSMA/CA Protocol (5%)

> (5%) What does CSMA/CA stand for? How does the protocol work? How can collisions be avoided in the protocol?

- CSMA/CA: Carrier Sense Multiple Access with Collision Avoidance (802.11 DCF).
- Operation: After sensing idle for DIFS, choose random backoff; decrement timer while idle; transmit when backoff reaches zero; receiver sends ACK after SIFS. Collisions inferred by missing ACK.
- Avoidance: Random backoff spaces transmissions; RTS/CTS reserves the channel and mitigates hidden terminals; interframe spacing (SIFS/DIFS) prioritizes control/ACK.

## 1.5 Data Link Layer Error Detection/Correction (5%)

> (5%) What techniques can be used for error-detection and error-correction, respectively, on the data link layer?

- Detection: Parity (single, two-dimensional), checksum (1's complement), CRC.
- Correction: Hamming codes, Reed-Solomon, convolutional/turbo/LDPC codes. ARQ/HARQ combine FEC with retransmissions.

## 1.6 Wi-Fi Network Standards (5%)

> (5%) What wireless (Wi-Fi) network standards are used in today’s industries? What are the characteristics of the link specified in each standard?

- 802.11n (Wi-Fi 4): 2.4/5 GHz, 20/40 MHz channels, MIMO, up to 600 Mbps PHY.
- 802.11ac (Wi-Fi 5): 5 GHz, 20/40/80/160 MHz, MU-MIMO (DL), higher-order QAM, multi-Gbps PHY.
- 802.11ax (Wi-Fi 6/6E): 2.4/5/6 GHz, OFDMA, MU-MIMO (UL/DL), BSS coloring, TWT; higher efficiency and capacity.
- 802.11be (Wi-Fi 7, emerging): 2.4/5/6 GHz, 320 MHz channels, multi-link operation, higher-order MIMO and QAM; very high throughput.

---

# Part 2: Long Answer Questions (70%)

## 2.1 Code Division Multiple Access (CDMA) (15%)

> (15%) Begin with reading about the simple CDMA protocol... choose one CDMA scheme and explain how it works. Describe the advantages that CDMA has over other coding schemes, such as TDM and FDM. Include in your answer the titles and sources of the articles/documents you consulted.

Chosen scheme: UMTS W-CDMA (FDD).
- Spreading: User data is spread by orthogonal variable spreading factor (OVSF) channelization codes to a common chip rate (3.84 Mcps). Scrambling codes separate cells/users at the cell and UE level.
- Power control: Fast closed-loop power control combats near-far and fading (e.g., 1500 Hz DL commands).
- Receiver: RAKE combining aligns and coherently combines energy from multipath components (fingers), exploiting multipath diversity.
- Handover: Soft/softer handover maintains simultaneous links to multiple cells/sectors; diversity improves robustness.
- Advantages over TDM/FDM: Interference averaging (soft capacity), graceful degradation as load increases, strong multipath resilience via RAKE, flexible rate adaptation by varying spreading factor and coding, inherent security through spreading.

Sources (examples):
- 3GPP TS 25.213: Spreading and Modulation (FDD).
- Holma & Toskala, WCDMA for UMTS.
- Proakis & Salehi, Digital Communications (CDMA chapters).

## 2.2 Two-Dimensional Checksum (15%)

> (15%) Suppose host A has payload 1011 0110 1010 1011 to send to host B, and A wants to use a two-dimensional checksum... even parity ... minimize length ... Show all your work and prove why shortest. Prove that any 1-bit error can be detected and corrected.

Arrange the 16 data bits in a 4x4 matrix (row-major):

Row1: 1 0 1 1 (sum 3) -> row parity pR1 = 1 (to make even)
Row2: 0 1 1 0 (sum 2) -> pR2 = 0
Row3: 1 0 1 0 (sum 2) -> pR3 = 0
Row4: 1 0 1 1 (sum 3) -> pR4 = 1

Column parities:
- Col1: 1,0,1,1 (sum 3) -> pC1 = 1
- Col2: 0,1,0,0 (sum 1) -> pC2 = 1
- Col3: 1,1,1,1 (sum 4) -> pC3 = 0
- Col4: 1,0,0,1 (sum 2) -> pC4 = 0

Overall parity p0: even parity over all 16 data bits plus 8 parity bits. Data ones = 10 (even). Row parity ones = 2. Column parity ones = 2. Total so far = 14 (even) -> p0 = 0.

Checksum bits (one possible ordering): [pR1 pR2 pR3 pR4 pC1 pC2 pC3 pC4 p0] = 1 0 0 1 1 1 0 0 0.

Why shortest: For an m x n arrangement, checksum length is m + n + 1. For 16 bits, possibilities include 1x16 (18), 2x8 (11), 4x4 (9), 8x2 (11), 16x1 (18). Thus 4x4 is minimal.

1-bit detect/correct: A single flipped data bit causes exactly one row parity and one column parity to fail, and toggles overall parity (odd). The intersection of the failing row and column identifies the bit to correct. If a parity bit flips, overall parity detects it while rows/columns localize which parity bit erred.

## 2.3 CSMA/CD Ethernet Analysis (20%)

> (20%) Assume a 1 Gbps Ethernet has two nodes, A and B, connected by a 180 m cable with three repeaters in between... propagation speed 2*10^8 m/s; backoff multiples of 512 bits; each repeater inserts 20-bit delay. At t=0 both transmit. After first collision, A draws K=0 and B draws K=1 after sending 48-bit jam.
> a) What is the one-way propagation delay (including all repeater delays) between A and B in seconds? At what time is A’s packet completely delivered at B?
> b) Now suppose only A has a packet and repeaters are replaced with switches. Each switch has an 8-bit processing delay in addition to a store-and-forward delay. At what time is A’s packet delivered at B? Include all delays and show work.

Given:
- Rate = 1e9 bps. Frame length = 1024 bits. Jam = 48 bits. Slot = 512 bits.
- Cable propagation = 180 m / (2e8 m/s) = 0.9 microseconds.
- 3 repeaters x 20-bit times = 60 ns = 0.06 microseconds (treated as store-and-forward delays in a collision domain per problem statement).
- One-way propagation including repeaters: 0.9 + 0.06 = 0.96 microseconds.

a) Timeline with collision at t=0:
- Both start transmitting at t=0.
- Each detects collision when the other’s signal arrives: t = 0.96 microseconds.
- Each sends 48-bit jam: duration 48 ns = 0.048 microseconds. Local jam ends at 1.008 microseconds.
- The last bit of B’s jam arrives at A at 1.008 + 0.96 = 1.968 microseconds (channel becomes idle then).
- Backoff: A draws K=0, B draws K=1. A begins retransmission immediately when idle (ignoring IFG, as typical in homework timing unless specified). B would attempt after 1 slot (0.512 microseconds) but will sense busy and defer.
- A’s frame transmission time: 1024 bits / 1e9 = 1.024 microseconds. Last bit leaves A at 1.968 + 1.024 = 2.992 microseconds.
- Last bit arrives at B after one-way propagation: 2.992 + 0.96 = 3.952 microseconds.

Answer (a): one-way delay = 0.96 microseconds; A’s packet completely delivered at B at about 3.952 microseconds.

b) Replace 3 repeaters with 3 store-and-forward switches (8-bit processing per switch):
- Per switch delay = frame time + processing = 1.024 microseconds + 8 ns = 1.032 microseconds.
- End-to-end: A transmits 1.024 microseconds, then 3 switches each add 1.032 microseconds, plus cable propagation (0.9 microseconds). Delivery time (from t=0):
  1.024 + 3*(1.032) + 0.9 = 1.024 + 3.096 + 0.9 = 5.020 microseconds.

Answer (b): about 5.020 microseconds.

## 2.4 802.11 RTS/CTS Transmission (10%)

> (10%) Suppose an 802.11 station is configured to always reserve the channel with RTS/CTS. At t=0 it wants to transmit 1024 bytes. All other stations are idle. At what time will the station complete the transmission? At what time can the station receive the acknowledgement?

Assumptions (matching common textbook values and ignoring PHY preamble/PLCP overheads):
- Data rate = 11 Mbps. DIFS = 50 microseconds. SIFS = 10 microseconds.
- Frame sizes: RTS = 160 bits; CTS = 112 bits; ACK = 112 bits; DATA payload = 1024 bytes = 8192 bits; assume MAC header + FCS bring the data bits to 8416 bits (as commonly used in examples).

Timing (all at 11 Mbps except interframe spaces):
- DIFS: 50.000 microseconds.
- RTS: 160/11e6 = 14.545 microseconds.
- SIFS: 10.000 microseconds.
- CTS: 112/11e6 = 10.182 microseconds.
- SIFS: 10.000 microseconds.
- DATA: 8416/11e6 = 765.091 microseconds.
- SIFS: 10.000 microseconds.
- ACK: 112/11e6 = 10.182 microseconds.

Totals:
- Data transmission complete at: 50 + 14.545 + 10 + 10.182 + 10 + 765.091 = 859.818 microseconds.
- ACK received at: 859.818 + 10 + 10.182 = 880.000 microseconds.

## 2.5 Bluetooth Frame Format Analysis (10%)

> (10%) Conduct research about Bluetooth technology and describe and comment on the format of the Bluetooth frame. Focus on its features and limitations. Is there anything in the frame format that inherently limits the number of active nodes in a network to eight active nodes? Explain.

Classic Bluetooth baseband frame:
- Access Code (72 bits): preamble, sync word (based on lower address), trailer; used for timing and identification.
- Header (54 bits transmitted with 1/3 FEC over an 18-bit header): AM_ADDR (3 bits), Type (4), Flow (1), ARQN (1), SEQN (1), HEC (8); robust via FEC.
- Payload (0 to 2745 bits) with 16-bit CRC; optional FEC (2/3, 3/5) and whitening; various packet types define payload length and structure.

Features and limitations:
- Robust short-range links with FEC, ARQ, fast frequency hopping; low power design.
- Payload and slot structure constrain throughput; tight timing and hopping sequence complexity.
- Piconet concurrency: AM_ADDR is 3 bits, yielding up to 7 active slaves addressed by the master (plus the master itself). Additional devices can be parked (no AM_ADDR) and polled in/out, and multiple piconets can form scatternets, but a single piconet supports at most 7 active slaves concurrently due to the 3-bit address field in the frame header.