2022 question paper

Computer Networks

18 questions

Q1a. A proxy server is used as the computer (i) with external access (ii) acting as a backup (iii) performing file handling (iv) accessing user permissions20222m
Module 5: Application Layer & Security
A proxy server is used as the computer

(i) with external access
(ii) acting as a backup
(iii) performing file handling
(iv) accessing user permissions

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Answer: (i) with external access

Why: A proxy server sits between the internal clients and the outside internet, fetching resources on their behalf — so it is the computer with external access that works on behalf of internally-restricted hosts. It is not a backup machine, a file-handling server, or a permission-management host.
Q1b. The term 'WAN' stands for (i) Wide Area Net (ii) Wide Access Network (iii) Wide Area Network (iv) Wide Access Net20222m
Module 1: Physical Layer & Data Communication
The term 'WAN' stands for

(i) Wide Area Net
(ii) Wide Access Network
(iii) Wide Area Network
(iv) Wide Access Net

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Worked Solution
(iii) Wide Area Network

A WAN spans large geographical areas — cities, countries, or continents — typically connecting multiple LANs, often using leased telecom lines or satellite links. The internet itself is the largest example of a WAN.
Q1c. Which layer of the TCP/IP stack corresponds to the OSI model transport layer? (i) Host-to-host (ii) Application (iii) Internet (iv) Network access20222m
Module 1: Physical Layer & Data Communication
Which layer of the TCP/IP stack corresponds to the OSI model transport layer?

(i) Host-to-host
(ii) Application
(iii) Internet
(iv) Network access

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(i) Host-to-host

The Host-to-Host layer in TCP/IP provides the same end-to-end, process-to-process delivery service as the OSI Transport layer, using TCP (reliable) or UDP (unreliable).
Q1d. What is the maximum header size of an IP packet? (i) 32 bytes (ii) 64 bytes (iii) 30 bytes (iv) 60 bytes20222m
Module 3: Network Layer & Routing
What is the maximum header size of an IP packet?

(i) 32 bytes
(ii) 64 bytes
(iii) 30 bytes
(iv) 60 bytes

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Answer: (iv) 60 bytes

Why: The IPv4 header length field (IHL) is 4 bits, giving $2^4 - 1 = 15$ words of 4 bytes = $15 \times 4 = 60$ bytes total header, i.e. a maximum header size of 60 bytes (the minimum is 20 bytes with no options). 32/64/30 bytes do not correspond to any IHL-derived value; 64 would be the header limit only under protocols like IPv6's usual 40-byte base + extensions pattern.
Q1e. The address which is used to identify a process on a host is (i) physical address (ii) port address (iii) logical address (iv) specific address20222m
Module 4: Transport Layer & Congestion Control
The address which is used to identify a process on a host is

(i) physical address
(ii) port address
(iii) logical address
(iv) specific address

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Answer: (ii) port address

Why: A port number (port address) identifies a specific process/application on a host, enabling transport-layer process-to-process (end-to-end) delivery — e.g. port 80 → web server. The physical address identifies a NIC, the logical (IP) address identifies the host itself, and "specific address" is not a recognised addressing category.
Q1f. Which sublayer of the information link layer performs circuit functions that depend on the kind of medium? (i) Media access control sublayer (ii) Logical link control sublayer (iii) Network interface control sublayer (iv) Both (i) and (ii)20222m
Module 2: Data Link Layer & MAC Sublayer
Which sublayer of the information link layer performs circuit functions that depend on the kind of medium?

(i) Media access control sublayer
(ii) Logical link control sublayer
(iii) Network interface control sublayer
(iv) Both (i) and (ii)

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Worked Solution
(i) Media Access Control (MAC) sublayer

The Data Link layer is commonly split into two sublayers:

LLC (Logical Link Control): medium-independent — handles flow control, error control, and framing in a way that's the same regardless of the underlying physical medium (Ethernet, Wi-Fi, etc.).

MAC (Media Access Control): medium-dependent — handles the specific rules for accessing the shared physical medium, which vary by technology (e.g. CSMA/CD for Ethernet, CSMA/CA for Wi-Fi). It also handles physical/MAC addressing.

Since the MAC sublayer's functions specifically depend on the type of medium in use, it is the correct answer.
Q1g. Transport layer protocol deals with (i) application-to-application communication (ii) node-to-node communication (iii) the process-to-process communication (iv) Both (i) and (ii)20222m
Module 4: Transport Layer & Congestion Control
Transport layer protocol deals with

(i) application-to-application communication
(ii) node-to-node communication
(iii) the process-to-process communication
(iv) Both (i) and (ii)

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Answer: (iii) the process-to-process communication

Why: The transport layer uses port numbers to deliver data from a sending process on one host to the correct process on the destination host (it also performs multiplexing/demultiplexing). Node-to-node communication is the data link layer's job, and "application-to-application" is not the standard term — the application layer connects user-facing software, while the transport layer beneath it handles processes.
Q1h. FTP server (i) maintains state information (ii) is stateless (iii) has single TCP connection for a file transfer (iv) has UDP connection for file transfer20222m
Module 5: Application Layer & Security
FTP server

(i) maintains state information
(ii) is stateless
(iii) has single TCP connection for a file transfer
(iv) has UDP connection for file transfer

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Worked Solution
Answer: (i) maintains state information

Why: FTP is stateful: the server tracks the control connection, the logged-in user, the current working directory and the transfer mode. It is not stateless (HTTP, by contrast, is), it uses two TCP connections (control port 21 + data port 20) rather than a single one, and file transfer runs over TCP, never UDP.
Q1i. Which of the following is the multiple access protocol for channel access control? (i) CSMA/CD (ii) CSMA/CA (iii) Both CSMA/CD and CSMA/CA (iv) HDLC20222m
Module 2: Data Link Layer & MAC Sublayer
Which of the following is the multiple access protocol for channel access control?

(i) CSMA/CD
(ii) CSMA/CA
(iii) Both CSMA/CD and CSMA/CA
(iv) HDLC

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Worked Solution
(iii) Both CSMA/CD and CSMA/CA

Both are multiple access protocols in the "random access with carrier sense" family, used to control which station gets to use a shared channel at a given moment:

CSMA/CD (Collision Detection) — used in traditional (shared-medium) Ethernet; detects collisions after they occur and stops transmission early.

CSMA/CA (Collision Avoidance) — used in Wi-Fi (802.11); since wireless collisions are hard to detect, it proactively tries to avoid them (via random backoff and optional RTS/CTS handshaking).

HDLC (High-Level Data Link Control) is not a multiple access protocol at all — it's a bit-oriented, point-to-point/point-to-multipoint data link protocol for framing and reliable transfer, unrelated to shared-channel access control.
Q1j. Physical layer provides (i) mechanical specifications of electrical connectors and cables (ii) electrical specification of transmission line signal level (iii) specification for IR over optical fiber (iv) All of the above20222m
Module 1: Physical Layer & Data Communication
Physical layer provides

(i) mechanical specifications of electrical connectors and cables
(ii) electrical specification of transmission line signal level
(iii) specification for IR over optical fiber
(iv) All of the above

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Worked Solution
(iv) All of the above

The Physical layer covers everything needed to physically move raw bits across a medium: mechanical specs (connector/cable shape and pin layout), electrical specs (voltage levels, signal timing), and specifications for the transmission medium itself, including optical/IR links. So all three listed items fall under its scope.
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Q2. Explain the duties of different layers in TCP/IP model.202214m
Module 1: Physical Layer & Data Communication
Explain the duties of different layers in TCP/IP model.

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Worked Solution
The TCP/IP protocol suite is organized into four layers, each with distinct duties:

+-------------------------------+ | Application Layer | HTTP, FTP, SMTP, DNS, Telnet +-------------------------------+ | Transport Layer (Host-Host) | TCP, UDP +-------------------------------+ | Internet Layer | IP, ICMP, ARP, IGMP +-------------------------------+ | Network Access Layer | Ethernet, Wi-Fi, PPP +-------------------------------+

1. Network Access (Link) Layer
Combines the OSI Physical + Data Link layers. Responsible for:

Physical addressing (MAC addresses)

Framing data into frames for the transmission medium

Media access control and error detection at the link level

Interfacing directly with the physical hardware (NIC, cabling, wireless medium)

2. Internet Layer
Equivalent to the OSI Network layer. Responsible for:

Logical (IP) addressing of hosts

Routing packets across interconnected networks, choosing best paths

Fragmentation and reassembly of packets to fit different network MTUs

Handled mainly by IP, with ICMP for error reporting and ARP for address resolution

3. Transport (Host-to-Host) Layer
Equivalent to the OSI Transport layer. Responsible for:

End-to-end, process-to-process delivery between applications

TCP: connection-oriented, reliable, ordered delivery with flow and congestion control

UDP: connectionless, best-effort, low-overhead delivery

Segmentation of application data and reassembly at the receiver

4. Application Layer
Combines OSI's Application, Presentation, and Session layers. Responsible for:

Providing network services directly to user applications

Data formatting, encryption/compression (presentation concerns)

Session management between communicating processes

Implemented by protocols like HTTP (web), FTP (file transfer), SMTP (email), DNS (name resolution)

Each layer only interacts with the layers directly above and below it, passing data downward by encapsulation (adding headers) on the sending side, and upward by decapsulation (removing headers) on the receiving side.

Q3. What do you mean by the term 'ethernet'? What are the types of cable commonly used for ethernet?202214m

Module 1: Physical Layer & Data Communication

What do you mean by the term 'ethernet'? What are the types of cable commonly used for ethernet?

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Ethernet is a family of wired LAN technologies, standardized as IEEE 802.3, that defines how devices on a local network frame, address, and access a shared or switched medium. Originally based on CSMA/CD (Carrier Sense Multiple Access with Collision Detection) over a shared bus, modern Ethernet mostly runs over switched, full-duplex links where collisions don't occur.

An Ethernet frame includes fields for destination/source MAC address, a type field, the data payload, and a trailing CRC for error detection.

Common cable types used for Ethernet

Cable type	Standard	Typical speed	Notes
Coaxial (Thicknet/Thinnet)	10BASE5 / 10BASE2	10 Mbps	Legacy, bus-topology Ethernet, largely obsolete
Twisted pair (UTP), Cat 5e/6/6a	100BASE-TX, 1000BASE-T, 10GBASE-T	100 Mbps – 10 Gbps	Most common in modern LANs; uses RJ-45 connectors
Fiber optic (single-mode/multi-mode)	1000BASE-SX/LX, 10GBASE-SR/LR	1 Gbps – 100+ Gbps	Used for backbones, long-distance, or high-speed links; immune to EMI

Twisted-pair copper (Cat 5e/6) is by far the most common choice for desktop/office connections today, while fiber is preferred for data-center backbones and long-haul links.

Q4. Differentiate between pure aloha and slotted aloha with examples.202214m

Module 2: Data Link Layer & MAC Sublayer

Differentiate between pure aloha and slotted aloha with examples.

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Pure ALOHA

In pure ALOHA, a station transmits a frame the moment it has one ready, without checking whether the channel is idle or coordinating with other stations in any way. If two stations' transmissions overlap in time even partially, both frames are destroyed (collision), and each station must wait a random backoff time before retrying.

Vulnerable time = 2 x Tfr (frame transmission time), since a collision can happen if any other station starts within one frame-time before or after this station's own transmission begins.
Maximum throughput ≈ 18.4% (at G = 0.5), where G is the offered load.

Example: Station A sends a frame at t=5ms. If Station B starts transmitting anytime between t=4ms and t=6ms (assuming Tfr=1ms), both frames collide and are lost.

Slotted ALOHA

Slotted ALOHA improves on pure ALOHA by dividing time into discrete slots, each exactly equal to one frame transmission time. A station is only permitted to begin transmission at the start of a slot — it must wait if its data becomes ready mid-slot.

Vulnerable time = 1 x Tfr, since collisions can now only occur if two stations pick the exact same slot (partial overlaps across slot boundaries are eliminated by the synchronization).
Maximum throughput ≈ 36.8% (at G = 1) — exactly double pure ALOHA's maximum.

Example: With 1ms slots, if Station A transmits in the slot starting at t=5ms, a collision only occurs if Station B also chooses to transmit in that same [5ms, 6ms) slot; B cannot start mid-slot at, say, t=5.5ms.

Key contrast

Aspect	Pure ALOHA	Slotted ALOHA
Transmission timing	Immediate, unsynchronized	Must align to slot boundaries
Vulnerable time	2 x Tfr	1 x Tfr
Max throughput	~18.4%	~36.8%
Needs global clock sync	No	Yes

Halving the vulnerable window by requiring slot-synchronized transmission is what allows slotted ALOHA to double pure ALOHA's maximum achievable throughput, at the cost of needing all stations to share a common time reference.

Q5. How data link layers provide flow and error control? Explain with examples.202214m
Module 2: Data Link Layer & MAC Sublayer
How data link layers provide flow and error control? Explain with examples.

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The Data Link layer provides both flow control and error control to ensure reliable, well-paced delivery of frames across a single link.

Flow Control

Flow control prevents a fast sender from overwhelming a slower receiver's buffer capacity. The Data Link layer achieves this primarily through the sliding window mechanism:

The receiver's available buffer space determines how many frames the sender is allowed to have outstanding (unacknowledged) at once — the window size.

As the receiver processes and acknowledges frames, the window "slides" forward, permitting the sender to transmit more.

Example: In Go-Back-N with window size N=4, the sender can have at most 4 frames in flight; it must pause once 4 are unacknowledged, resuming only as ACKs arrive.

Error Control

Error control detects (and where possible corrects) frames that were corrupted or lost during transmission, and arranges for recovery:

Error detection — redundant bits (parity, checksum, or CRC) let the receiver verify whether a received frame matches what was sent.

Example: CRC using polynomial division catches virtually all common burst errors.

Acknowledgement (ACK) — the receiver confirms successful, error-free receipt of a frame.

Timeout and retransmission (ARQ — Automatic Repeat reQuest) — if the sender doesn't receive an ACK within a set time, it assumes the frame was lost or corrupted and retransmits.

Go-Back-N: retransmits the lost frame and every frame sent after it.

Selective Repeat: retransmits only the specific lost/corrupted frame, using individual timers and receiver-side buffering for out-of-order frames.

Together: flow control governs how much data can be sent before pausing for acknowledgement, while error control governs what happens when a frame doesn't arrive correctly — both working through the same underlying sliding-window/ACK/timeout machinery to guarantee reliable link-level delivery.
Q6. Describe RIP routing protocol with an example. Explain count to infinity problem and its countermeasures.202214m
Module 3: Network Layer & Routing
Describe RIP routing protocol with an example. Explain count to infinity problem and its countermeasures.

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RIP Routing Protocol, Example, Count-to-Infinity Problem and Countermeasures

RIP (Routing Information Protocol) is a distance-vector (Bellman-Ford) interior gateway protocol in which every router periodically shares its entire routing table with its neighbours and computes routes as the minimum of the costs reported by its neighbours plus the cost of the link to each neighbour.

1. Key characteristics

Metric: hop count; maximum usable metric = 15 hops, 16 = infinity (unreachable).

Update: full table every 30 s over UDP port 520 (RIPv1 broadcast 255.255.255.255; RIPv2 multicast 224.0.0.9).

RIPv1 is classful (no subnet mask, no VLSM); RIPv2 supports classless masks, authentication and next-hop field.

Simple to configure; used in small networks; limited by the 15-hop limit and slow convergence.

2. Example

A(1)---(1)B---(1)C | / |----(1)D----/

Link costs = 1 each.

Round 1 (D advertises): D tells C "to A via me, cost 2" (D→A cost 1 + link 1). C computes A cost = 2+1 = 3.
Round 2 (B advertises): B tells C "to A cost 1", so C recomputes A cost = 1+1 = 2 and installs route A → next hop B, metric 2 (better than via D = 3).
Round 3: C shares with B and D; every router converges to its least-hop route:

Router Destination Next hop Metric

B A direct 1

C A B 2

D A direct 1

C D direct 1

Routers apply split horizon (never advertise a route back out the interface it was learned from) and delete stale entries after 180 s without an update (garbage-collection timer).

3. Count-to-infinity problem

If the A–B link fails, B loses its direct route to A. But C still advertises "A via D, metric 2" (stale, reverse of the old path), so B learns "A via C, cost 3" and advertises it to C. C then computes via B = 4, advertises 4 to B, B computes 5, and so on. Because each neighbour believes the other's stale distance is good news, the metric increments by 1 on every exchange until it reaches 16 (infinity) — a classic count-to-infinity loop where packets circulate while the metric climbs. Convergence is very slow (could be minutes), during which time packets are routed in circles.

Root cause: distance-vector has no notion of "poisoned" routes — old stale information from a neighbour looks like new information.

4. Countermeasures

Split horizon: never advertise a route back the interface it came from — eliminates the basic A↔B↔C loop above.

Poison reverse: actively advertise the failed route with metric 16 (infinity) to the neighbour from which it was learned, poisoning it faster than silence.

Hold-down timer: on learning a route is unreachable, ignore updates about that route for a fixed period (typically 180 s), stopping stale bad news from being accepted.

Triggered (flash) updates: advertise a change immediately on topology change instead of waiting for the 30 s periodic update, speeding convergence.

Maximum metric 16: bounds the counting process — the count always terminates at infinity.

Conclusion: RIP's periodic distance-vector updates make it simple, but the count-to-infinity problem makes convergence slow; split horizon, poison reverse, hold-down timers and triggered updates together mitigate it (modern link-state protocols such as OSPF avoid the problem entirely).

Router	Destination	Next hop	Metric
B	A	direct	1
C	A	B	2
D	A	direct	1
C	D	direct	1

Q7. An ISP is granted the block 80.70.56.0/21. The ISP needs to allocate addresses for two organizations each with 500 addresses, two organizations each with 250 addresses, and three organizations each with 50 addresses. (a) Find the number and range of addresses in the ISP block. (b) Find the range of addresses for each organization and the range of unallocated addresses.202214m

Module 3: Network Layer & Routing

An ISP is granted the block 80.70.56.0/21. The ISP needs to allocate addresses for two organizations each with 500 addresses, two organizations each with 250 addresses, and three organizations each with 50 addresses.

(a) Find the number and range of addresses in the ISP block.
(b) Find the range of addresses for each organization and the range of unallocated addresses.

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(a) ISP Block 80.70.56.0/21 — Number and Range

Given: network address $= 80.70.56.0$ , prefix $= /21$ .

Step 1 — Number of addresses

$2^{32-21} = 2^{11} = 2048 \text{ addresses}$

Step 2 — Subnet mask: 21 bits = two full octets + 5 bits:

$255.255.248.0$

Step 3 — Range: the third octet advances in steps of $256 - 248 = 8$ , starting at 56, so it ends at $56 + 8 - 1 = 63$ .

First address: 80.70.56.0
Last (broadcast) address: 80.70.63.255
Check: $8 \times 256 = 2048$ ✓

Answer (a): 2048 addresses, from 80.70.56.0 to 80.70.63.255.

(b) Allocation to Organisations and Unallocated Range

Sizing rule: give each organisation the next power of two block that can hold the requested number of usable addresses (usable $= 2^h - 2$ for a normal subnet).

Demand	Needed	Smallest block	Prefix	Total addresses
2 × 500	500	/23	512	510 ✓ ≥ 500
2 × 250	250	/24	256	254 ✓ ≥ 250
3 × 50	50	/26	64	62 ✓ ≥ 50

Sequential allocation from 80.70.56.0 upward:

$2 \times 512 = 1024$ addresses
$2 \times 256 = 512$ addresses
$3 \times 64 = 192$ addresses

Total used $= 1024 + 512 + 192 = 1728$ addresses.

Organisation	Block	Address range	Count
Org 1 (500)	80.70.56.0/23	80.70.56.0 – 80.70.57.255	512
Org 2 (500)	80.70.58.0/23	80.70.58.0 – 80.70.59.255	512
Org 3 (250)	80.70.60.0/24	80.70.60.0 – 80.70.60.255	256
Org 4 (250)	80.70.61.0/24	80.70.61.0 – 80.70.61.255	256
Org 5 (50)	80.70.62.0/26	80.70.62.0 – 80.70.62.63	64
Org 6 (50)	80.70.62.64/26	80.70.62.64 – 80.70.62.127	64
Org 7 (50)	80.70.62.128/26	80.70.62.128 – 80.70.62.191	64

Unallocated addresses:

$2048 - 1728 = 320 \text{ addresses} \Rightarrow \textbf{80.70.62.192 – 80.70.63.255}$

(as 80.70.62.192/26 → 64 addresses, plus 80.70.63.0/24 → 256 addresses; $64 + 256 = 320$ ✓).

Answer (b): Org1 → 80.70.56.0–80.70.57.255, Org2 → 80.70.58.0–80.70.59.255, Org3 → 80.70.60.0–80.70.60.255, Org4 → 80.70.61.0–80.70.61.255, Org5 → 80.70.62.0–80.70.62.63, Org6 → 80.70.62.64–80.70.62.127, Org7 → 80.70.62.128–80.70.62.191; unallocated range 80.70.62.192 – 80.70.63.255 (320 addresses).

Q8. How transport layer provides congestion control? Differentiate between leaky bucket and token bucket algorithms.202214m

Module 4: Transport Layer & Congestion Control

How transport layer provides congestion control? Differentiate between leaky bucket and token bucket algorithms.

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Transport-Layer Congestion Control, and Leaky Bucket vs Token Bucket

1. How the transport layer provides congestion control

Congestion means the network's routers/links are carrying more traffic than their buffers and bandwidth can absorb, causing queueing delay, packet loss and throughput collapse. IP itself has no congestion control, so the end hosts — specifically TCP at the transport layer — control the rate at which they inject data, treating losses and delay as feedback signals.

Window-based rate control. The sender may have at most $\min(\text{cwnd}, \text{rwnd})$ unacknowledged bytes in flight; increasing cwnd increases the injection rate.
Slow start. cwnd starts at 1 MSS and doubles once per RTT (exponential growth) until it reaches the slow-start threshold $ssthresh$ or a loss occurs.
Congestion avoidance. Beyond $ssthresh$ , cwnd grows additively (+1 MSS per RTT) — probing cautiously for spare capacity (AIMD).
Loss as congestion signal (multiplicative decrease).
- 3 duplicate ACKs → fast retransmit + fast recovery: $ssthresh = \text{cwnd}/2$ , $\text{cwnd} = ssthresh + 3$ , then additive increase.
- Timeout (severe congestion): $ssthresh = \text{cwnd}/2$ , $\text{cwnd} = 1$ MSS, back to slow start.
Explicit signals: ECN marks IP headers (CE bit) and the receiver echoes ECE so the sender reduces cwnd without loss; a receiver can also reduce its advertised rwnd, throttling the sender (receiver flow control).
Retransmission timers with RTT estimation keep the in-flight data bounded so buffers upstream are not overrun.

 cwnd  |        additive increase (+1/RTT)
 (MSS) |      ______________/
       |     /      × loss → half (multiplicative decrease)
 ssth. |/____/ 
       | slow start (exponential)
       +---------------------------→ time/RTT

2. Leaky bucket vs Token bucket

Both are traffic-shaping (policing) algorithms that convert bursty input into a smoother, rate-limited output.

Basis	Leaky bucket	Token bucket
Working idea	Packets enter a bucket (queue) and drain at a constant fixed rate; if full, arriving packets are dropped/rejected	Tokens are generated at a fixed rate; a packet may be sent only if it grabs enough tokens, otherwise it waits
Empty bucket	Bucket empty ⇒ nothing to output	Bucket empty ⇒ new packets may still be sent if burst capacity is available — the bucket accumulates surplus
Burst tolerance	No bursts — output is strictly uniform (any burst is delayed/dropped)	Allows bursts up to the bucket's token capacity
Output rate	Always exactly the leak rate	Average = token rate, but instantaneous bursts are permitted
Implementation	Simple FIFO queue with a timer	Counter of tokens + queue
Typical use	Constant bit-rate shaping, hard policing	Limiting average rate while permitting bursts (e.g. network access rate policing)
Analogy	Water leaks out of a punctured bucket at a fixed rate	Coins are minted at a fixed rate; spending them authorises sending
Burst after idle	Cannot send a burst immediately	Idle time accumulates tokens → burst up to bucket size can be sent instantly

Essence: the leaky bucket enforces a rigid, smooth output rate and treats any excess as error, while the token bucket enforces a long-term average rate but deliberately allows bursts up to the accumulated token count — which is why it is preferred for real, bursty application traffic.

Conclusion: congestion control is achieved end-to-end by TCP's AIMD window algorithms (slow start, congestion avoidance, fast recovery, ECN), and routers additionally police/shape traffic at entry points using leaky or token bucket algorithms.

Q9. Explain the difference between TCP and UDP. A client residing on a host with IP address 122.45.12.7 sends a message to the corresponding server residing on a host with IP address 200.112.45.90. If the well-known port is 161 and the ephemeral port is 51000, what are the pair of socket addresses used in this communication?202214m

Module 4: Transport Layer & Congestion Control

Explain the difference between TCP and UDP. A client residing on a host with IP address 122.45.12.7 sends a message to the corresponding server residing on a host with IP address 200.112.45.90. If the well-known port is 161 and the ephemeral port is 51000, what are the pair of socket addresses used in this communication?

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Worked Solution

Part 1 — Difference between TCP and UDP

Parameter	TCP	UDP
Nature	Connection-oriented (3-way handshake)	Connectionless
Reliability	Reliable — ACK, retransmission, sequencing	Unreliable best-effort
Ordering	Yes, via sequence numbers	No
Flow control	Sliding window (rwnd)	None
Congestion control	Slow start, congestion avoidance, fast recovery	None
Header	20–60 bytes	Fixed 8 bytes
Data unit	Byte-stream segment (no message boundaries)	Datagram (preserves message boundaries)
Multiplexing	Port numbers	Port numbers
Overhead / speed	Higher overhead, slower	Lightweight, faster
Use cases	HTTP, FTP, e-mail, file transfer	DNS, DHCP, VoIP, streaming, gaming

Part 2 — Socket addresses for the given communication

Given:

Client IP $= 122.45.12.7$ , ephemeral port $= 51000$
Server IP $= 200.112.45.90$ , well-known port $= 161$

Concept: a socket address = (IP address, port number) pair. The port selects the process on a host, while the IP selects the host itself. A TCP/UDP connection (or datagram exchange) is identified by the pair of socket addresses — source and destination.

Step 1 — Server (destination) socket:

$\text{Server socket} = (200.112.45.90,\ 161)$

Step 2 — Client (source) socket: the client uses its own IP with an ephemeral (temporarily assigned) port chosen above 49151.

$\text{Client socket} = (122.45.12.7,\ 51000)$

Step 3 — The pair of socket addresses:

$\boxed{(\text{Client: } 122.45.12.7:51000) \quad \longleftrightarrow \quad (\text{Server: } 200.112.45.90:161)}$

Destination socket = (200.112.45.90, 161) — the well-known port 161 (SNMP's standard port) identifies the server process.
Source socket = (122.45.12.7, 51000) — the ephemeral port 51000 identifies the client process and lets the server's reply be demultiplexed back to it.
Traffic from client to server is addressed to 200.112.45.90:161 with source 122.45.12.7:51000; the response simply reverses these two addresses.

Final answer: socket addresses = (122.45.12.7, 51000) and (200.112.45.90, 161), i.e. 122.45.12.7:51000 ↔ 200.112.45.90:161.

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RIP Routing Protocol, Example, Count-to-Infinity Problem and Countermeasures

1. Key characteristics

2. Example

3. Count-to-infinity problem

4. Countermeasures

(a) ISP Block 80.70.56.0/21 — Number and Range

(b) Allocation to Organisations and Unallocated Range

Transport-Layer Congestion Control, and Leaky Bucket vs Token Bucket

1. How the transport layer provides congestion control

2. Leaky bucket vs Token bucket

Part 1 — Difference between TCP and UDP

Part 2 — Socket addresses for the given communication

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