Data Link Layer Practice Questions

Packet switching is more efficient as bandwidth is shared dynamically. It can handle bursty traffic better. Circuit switching has lower latency but is less efficient and complex to implement.

A: TRUE - VLANs partition a physical switch into multiple logical switches, each with its own broadcast domain. Hosts in different VLANs cannot communicate at Layer 2 (their broadcasts don't reach each other). This improves security, reduces broadcast traffic, and enables logical grouping independent of physical location. D: TRUE - IEEE 802.1Q adds a 4-byte VLAN tag to the Ethernet frame header, containing a 12-bit VLAN ID (0-4095), allowing up to 4094 distinct VLANs. Tagged frames (trunk links) carry VLAN IDs between switches; untagged frames (access links) connect to end hosts. B: FALSE - VLANs are purely logical. C: FALSE - inter-VLAN communication requires a Layer 3 router or Layer 3 switch.

The probability that a single bit is received correctly = 1 - BER = 1 - 10⁻³ = 0.999. Assuming bit errors are independent, the probability that all 1000 bits are received correctly = (1 - 10⁻³)^1000 = (0.999)^1000 ≈ e^(-1) ≈ 0.368. So there is approximately a 36.8% chance of error-free reception for a 1000-bit frame with BER = 10⁻³. This shows why error detection is critical - with 10⁻³ BER, about 63% of 1000-bit frames contain at least one error.

Probability of a single bit being correct = $1 - 10^{-3}$. For 1000 independent bits all being correct = $(1 - 10^{-3})^{1000} \approx e^{-1} \approx 0.368$.

Binary exponential backoff in Ethernet: after the k-th collision, the node randomly selects a backoff time from {0, 1, 2, ..., 2^min(k,10) - 1} slot times. After k=10 collisions: range = {0, 1, ..., 2^10 - 1} = {0, 1, ..., 1023}. The range is capped at k=10 (2^10 = 1024 slots) and does not increase further for collisions 11-15. After 16 consecutive collisions, Ethernet gives up and reports an error. The range after 10 collisions = 0 to 1023.

A code with minimum Hamming distance d_min: Error detection: can detect up to d_min - 1 bit errors. If up to d_min-1 bits flip, the corrupted codeword is not a valid codeword (since any two valid codewords differ in at least d_min positions). Error correction: can correct up to ⌊(d_min - 1)/2⌋ bit errors. The receiver corrects to the nearest valid codeword - this works as long as errors don't push a codeword closer to a different valid codeword, requiring errors ≤ ⌊(d_min-1)/2⌋. Example: d_min=3 → detect 2 errors, correct 1 error (Hamming codes).

TDMA (Time Division Multiple Access) divides time into slots and assigns specific slots to each station, effectively a reservation scheme. ALOHA and CSMA are contention-based protocols.

Simple parity check adds one parity bit to make the total number of 1-bits even (or odd). It detects all single-bit errors (one bit flip changes parity) but fails for double-bit errors (two bits flip, parity remains unchanged). In general, parity detects any odd number of bit errors but misses all even numbers of errors. CRC with good generators detects all single-bit, double-bit, odd-bit errors, and all burst errors ≤ CRC length. Two-dimensional (2D) parity can detect and correct single-bit errors. Checksum (Internet) detects most errors but can miss some multi-bit error patterns.

The 64-byte minimum ensures that a frame takes at least $2 \times$ propagation delay (round trip) to transmit, so the sender is still transmitting when a collision signal arrives, allowing collision detection.

Store-and-forward packet switching involves all three types of delays: transmission, propagation, and queuing.

First calculate the total transmission time: 100 MB = 800 Mb, Transmission time = 800 Mb / 10 Mbps = 80 seconds. Total time = 80 seconds + 100 ms = 80.1 seconds. Effective data rate = 800 Mb / 80.1 seconds = 9.9875 Mbps ≈ 9.9 Mbps

For Selective Repeat, both sender and receiver window sizes must be at most $2^{n-1}$ to avoid ambiguity between old and new frames. This is half of Go-Back-N's maximum window size ($2^n - 1$).

Multiple access protocols control how multiple nodes share a single broadcast channel: TDMA: divides time into slots, each node assigned a fixed slot - deterministic, no collisions, but wastes capacity if a node has no data. CSMA/CD: used in Ethernet - sense before transmit, detect collisions and retransmit after random backoff. ALOHA (Pure and Slotted): simple random access - transmit when data arrives; retransmit after random delay on collision. Slotted ALOHA achieves 37% efficiency; Pure ALOHA 18%. OSPF (Open Shortest Path First) is a network layer routing protocol, not a medium access control protocol.

Bit stuffing inserts a 0-bit after every five consecutive 1-bits in the data to prevent the flag pattern (01111110) from appearing inside the frame. The receiver removes the stuffed 0-bits.

When dividing a polynomial by $G(x)$ of degree $r=4$, the remainder has degree at most $r-1 = 3$, i.e., the CRC is $r = 4$ bits long (remainder has at most $r$ bits).

First, calculate the transmission time: 1000 bytes * 8 bits/byte / 10^6 bits/second = 0.008 seconds = 8 milliseconds. Total time = 8 ms + 10 ms + 5 ms = 23 ms. Throughput = (Data size in bytes)/(Total time) = 1000 bytes / 0.023 seconds = 43478.26 bytes/second ≈ 80 KB/s

The maximum throughput of slotted ALOHA is $1/e \approx 0.368$ (36.8%), achieved when the offered load $G = 1$. Pure ALOHA has half this efficiency at $1/(2e) \approx 18.4\%$.

A: TRUE - MAC addresses are 48 bits (6 bytes), typically written as 6 pairs of hexadecimal digits (e.g., 00:1A:2B:3C:4D:5E). B: TRUE - the first 3 bytes (OUI - Organizationally Unique Identifier) are assigned by IEEE to manufacturers; the last 3 bytes are assigned by the manufacturer to each NIC, making each MAC globally unique (in practice). D: TRUE - the broadcast MAC address FF:FF:FF:FF:FF:FF causes all stations on the LAN to receive the frame. C: FALSE - routers use IP addresses (network layer) to forward packets between networks. MAC addresses are only used within a single broadcast domain (LAN) by switches.

In CSMA/CD, the worst-case collision scenario: Station A starts transmitting, and just before A's signal reaches station B at the far end, B starts transmitting. A collision occurs near B, and the collision signal must propagate back to A. The round-trip propagation time = 2 × propagation delay. For A to still be transmitting when the collision signal arrives, the frame transmission time must be at least 2 × propagation delay. Minimum frame size = 2 × propagation delay × bandwidth (channel capacity). For Ethernet (10 Mbps, 500m segment, propagation speed ≈ 2×10⁸ m/s): min frame = 2 × (2500m/2×10⁸) × 10⁷ bps = 64 bytes.