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Chapter 3

Data Communication and Computer Network

PU BCA past exam questions

Past questions and answers

17 questions set from this chapter. Most repeated first.

  • 2025 Spring · 7 marks

What defines a computer Network? Explain two benefits and two challenges of peer-to-peer network compared to client-server architecture.

Answer

Computer network. A computer network is a collection of two or more autonomous computers and devices (hosts, routers, switches, printers) interconnected by communication links (wired or wireless) so that they can exchange data and share resources such as files, printers, internet access and applications, following agreed rules called protocols.

Peer-to-peer (P2P) versus client-server. In P2P every computer acts as both client and server; in client-server, dedicated servers provide services to clients.

Two benefits of P2P.

  1. Low cost and easy setup: no expensive dedicated server or network operating system is needed, and no full-time administrator; suitable for small offices and homes.
  2. No single point of failure / scalability of resources: if one peer fails the others keep working, and resources (for example file sharing in BitTorrent) are spread across peers, so capacity can grow as peers join.

Two challenges of P2P.

  1. Weak security and management: there is no central control of users, permissions, backups or updates; each peer must be secured individually, which makes malware and unauthorised access harder to control.
  2. Poor performance and reliability at scale: peers also run their own tasks, so shared resources slow down, data may be scattered or duplicated, and the network becomes hard to manage as the number of peers grows.

In contrast, client-server offers centralised security, backup and administration, at the cost of server expense and a single point of failure.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 8 marks

What are different transmission modes? How does bandwidth differ from speed?

Answer

Transmission modes. The transmission mode defines the direction of data flow between two devices.

  1. Simplex: communication is one-way only; one device only sends and the other only receives. Examples: keyboard to computer, TV/radio broadcast.
  2. Half-duplex: both devices can send and receive, but not at the same time; the link is used in alternate directions. Examples: walkie-talkie, older Ethernet hubs.
  3. Full-duplex: both devices send and receive simultaneously, effectively doubling capacity. Examples: telephone, modern switched Ethernet.
Simplex:      A ---------> B
Half-duplex:  A <--------> B   (one direction at a time)
Full-duplex:  A <=========> B  (both directions at once)

Bandwidth versus speed.

  • Bandwidth is the capacity of a channel: the maximum amount of data (or the range of frequencies, in Hz) that the link can carry per unit time, usually expressed in bits per second (bps, Mbps, Gbps). It is a property of the medium.
  • Speed (throughput/data rate in practice) is the actual rate at which data is delivered. It is usually lower than bandwidth because of congestion, protocol overhead, errors and retransmission, and latency (propagation delay) also affects the experience.

Analogy: bandwidth is the width of a pipe (how much can flow), whereas speed is how much water actually arrives per second. A 100 Mbps link may deliver only 70 Mbps of actual throughput; and higher bandwidth does not reduce propagation delay.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 7 marks

Explain Time division multiplexing and its types with neat Diagram.

Answer

Time Division Multiplexing (TDM) is a digital multiplexing technique in which several signals share a single channel by taking turns; the channel's time is divided into slots, and each source is assigned a slot in a repeating frame. It is used when the channel's data rate exceeds the rate of each individual source.

Types of TDM.

1. Synchronous TDM. Each input is assigned a fixed time slot in every frame, whether or not it has data to send. A frame contains one slot per input, and the multiplexer and demultiplexer are synchronised, so no addresses are needed. If a source is idle, its slot goes empty, so capacity is wasted. Used in telephone systems (T1/E1 lines, SONET).

 A --\                              frame 1        frame 2
 B ---+--> [ MUX ] ===== A1 B1 C1 | A2 B2 C2 | ... ===> [ DEMUX ] --> A, B, C
 C --/

2. Asynchronous (statistical) TDM. Slots are assigned dynamically, only to inputs that have data. Since the slots are not fixed, each slot carries an address (identifying the source) along with the data. This makes better use of the link and allows the total input rate to exceed the link rate, but needs buffers and has addressing overhead. Used in ATM and packet-switched networks.

 A --\
 B ---+--> [ STAT MUX ] ==== (A,d1)(C,d2)(A,d3)(B,d4)... ==> [ DEMUX ]
 C --/    (idle inputs get no slot)

Within synchronous TDM, techniques such as multilevel multiplexing, multiple-slot allocation and pulse stuffing handle sources of different data rates.

Comparison. Synchronous: fixed slots, no addressing, wastes idle slots. Asynchronous: dynamic slots, needs addressing, more efficient.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 7 marks

Describe the modulation technique and its types with neat diagram.

Answer

Modulation is the process of changing a characteristic of a high-frequency carrier signal (amplitude, frequency or phase) in accordance with the message (modulating) signal, so that the information can be transmitted efficiently over a medium. It permits practical antenna sizes, reduces interference and allows multiplexing.

Types.

A. Analog modulation (analog message).

  • Amplitude Modulation (AM): carrier amplitude varies with the message; frequency and phase are constant. Simple, but sensitive to noise. Used in AM radio.
  • Frequency Modulation (FM): carrier frequency varies with the message amplitude. Better noise immunity, used in FM radio.
  • Phase Modulation (PM): carrier phase varies with the message.

B. Digital modulation (digital data over analog carrier).

  • Amplitude Shift Keying (ASK): bit 1 = carrier present (or high amplitude), bit 0 = absent (low amplitude). Noise-sensitive.
  • Frequency Shift Keying (FSK): bit 1 and bit 0 use two different frequencies. Used in low-speed modems.
  • Phase Shift Keying (PSK): the phase of the carrier changes (for example BPSK uses 0∘0^\circ and 180∘180^\circ; QPSK uses four phases, 2 bits/symbol).
  • Quadrature Amplitude Modulation (QAM): combines ASK and PSK, giving high data rates (for example 16-QAM, 64-QAM in Wi-Fi and cable).

C. Pulse (analog to digital) modulation: PAM, PWM, PPM, and PCM (sampling, quantisation, encoding).

Message:  ~~~~~~~~~
Carrier:  /\/\/\/\/\/\/\
AM:       small/large amplitude, same frequency
FM:       same amplitude, denser/sparser cycles
ASK:      1 -> tall wave , 0 -> flat
FSK:      1 -> fast wave , 0 -> slow wave
PSK:      wave inverts (180 deg) at bit changes

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 8 marks

Show the Real-Life example of how the OSI model applies to an everyday activity of sending an email in 2025

Answer

The OSI model has seven layers. Sending an email (for example from a Gmail/Outlook client to a recipient in 2025) maps to the layers as follows. On the sender side data moves down from layer 7 to 1; on the receiver side it moves up.

LayerWhat happens when sending an email
7. ApplicationThe user writes the message in a mail client or web app. The client uses SMTP to submit it to the mail server (and the receiver later reads it with IMAP/POP3).
6. PresentationThe message is encoded (UTF-8, MIME for attachments such as images and PDFs), compressed, and encrypted using TLS.
5. SessionA session with the mail server is established, authenticated (login/OAuth token), maintained and closed after transfer.
4. TransportTCP splits the message into segments, adds port numbers (587 for submission, 25 between servers, 465 with TLS), sequence numbers, and ensures reliable, ordered delivery with acknowledgements and retransmission.
3. NetworkIP adds source and destination IP addresses (found through a DNS MX lookup for the recipient domain) to make packets; routers forward them hop by hop across the internet.
2. Data LinkPackets are framed with MAC addresses (Wi-Fi/Ethernet) for the next hop, with an error check (FCS); the home router or switch forwards frames.
1. PhysicalBits are sent as radio waves (Wi-Fi/5G), electrical signals (Ethernet) or light pulses (fibre) over the physical media.

Receiving side. The recipient's mail server receives the bits at the physical layer, frames are checked at layer 2, packets are passed up through IP and TCP (reassembled in order), the TLS session is decrypted at layer 6, and the mail is stored; the recipient's client then fetches it (IMAP/POP3) and displays it at layer 7.

Each layer adds its own header on the way down (encapsulation) and removes it on the way up (decapsulation), so layers work independently of each other.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 7 marks

What are guided and unguided media in physical layer in communication? Provide examples and typical use cases for each.

Answer

The physical layer transmits raw bits over a transmission medium, which is either guided (wired) or unguided (wireless).

Guided media. Signals travel along a physical conductor that directs them.

MediumDescriptionTypical use
Twisted pair (UTP/STP)Pairs of insulated copper wires twisted to reduce interference; cheap, up to about 100 m per segmentLAN cabling (Cat5e/Cat6 Ethernet), telephone lines
Coaxial cableCentral copper conductor with insulation, shield and jacket; higher bandwidth and better shielding than twisted pairCable TV, broadband cable internet, older Ethernet
Optical fibreLight pulses through glass/plastic core by total internal reflection; very high bandwidth, low attenuation, immune to EMI, secure but costlyBackbone and long-distance links, undersea cables, FTTH

Unguided media. Signals propagate through air or space as electromagnetic waves, with no physical conductor.

MediumDescriptionTypical use
Radio waves (3 kHz-1 GHz)Omnidirectional, pass through wallsAM/FM radio, Wi-Fi, Bluetooth, mobile networks
Microwaves (1-300 GHz)Unidirectional, line-of-sight, need antennasPoint-to-point links, satellite communication, 4G/5G backhaul
Infrared (300 GHz-400 THz)Short range, cannot pass wallsTV remotes, short-range device links

Comparison. Guided media give higher security, speed and reliability but are costly to install and not mobile; unguided media provide mobility and easy deployment, but are prone to interference, eavesdropping and weather effects.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 8 marks

Calculate CRC for 10 bit sequence 1010101101. The generator polynomial is x3+x+1x^3 + x + 1.

Answer

Setup. Data =1010101101=1010101101 (10 bits). Generator x3+x+1⇒x^3+x+1 \Rightarrow divisor 10111011 (degree 3, so append 3 zeros).

Dividend =1010101101 000=1010101101\,000 (13 bits). Perform modulo-2 (XOR) division by 10111011:

1010101101000
1011
-----
0001101101000     (XOR at bit 0)
   1011
   -----
0000110101000     (XOR at bit 3)
    1011
    -----
0000011001000     (XOR at bit 4)
     1011
     -----
0000001111000     (XOR at bit 5)
      1011
      -----
0000000100000     (XOR at bit 6)
       1011
       -----
0000000001100     (XOR at bit 7)
         1011
         -----
0000000000111     (XOR at bit 9)

The remainder (last 3 bits) is 111111.

CRC =111=111.

Transmitted codeword =1010101101 111=1010101101111=1010101101\,111 = 1010101101111.

Verification. Dividing 10101011011111010101101111 by 10111011 at the receiver gives remainder 000000, so no error is detected.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 8 marks

How the protocols work in TCP/IP model. Explain with neat diagram?

Answer

The TCP/IP model has four layers (the Application layer combines OSI layers 5-7, and the Network Access layer combines OSI layers 1-2). Each layer adds its header to the data passed down (encapsulation) and the receiver strips them in reverse.

+---------------------------------------------+
| Application   HTTP, FTP, SMTP, DNS, DHCP    |   data
+---------------------------------------------+
| Transport     TCP, UDP                       |   segment / datagram
+---------------------------------------------+
| Internet      IP, ICMP, ARP, routing         |   packet
+---------------------------------------------+
| Network Access  Ethernet, Wi-Fi, PPP         |   frame -> bits
+---------------------------------------------+

1. Application layer. Provides services directly to user programs. HTTP/HTTPS fetches web pages, SMTP/POP3/IMAP handle email, FTP transfers files, DNS resolves names to IP addresses, DHCP assigns addresses, Telnet/SSH give remote login.

2. Transport layer. Provides end-to-end communication between processes using port numbers.

  • TCP: connection-oriented and reliable. It sets up a connection with a three-way handshake (SYN, SYN-ACK, ACK), numbers segments, acknowledges and retransmits lost ones, does flow control (window) and congestion control.
  • UDP: connectionless and unreliable but fast, with little overhead (DNS, video streaming, VoIP).

3. Internet layer. Handles logical addressing and routing of packets across networks. IP (IPv4/IPv6) is connectionless and best-effort; ICMP reports errors (ping); ARP maps IP addresses to MAC addresses; routing protocols (RIP, OSPF, BGP) determine paths.

4. Network access (link) layer. Transmits frames over the physical network using MAC addresses, and handles framing and error detection over Ethernet, Wi-Fi, etc.

Working example. A browser requests a page: HTTP request (application) →\to TCP segment to port 80/443 (transport) →\to IP packet with destination IP found by DNS (internet) →\to Ethernet/Wi-Fi frame to the next hop (access). The server's layers decapsulate and reply the same way.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 7 marks

A network administrator is given 172.16.0.0/16 and must allocate subnets for: Department A: 28 hosts, B: 55 hosts, C: 120 hosts, D: 29 hosts. Design an optimal VLSM scheme, provide the subnet mask (CIDR), network address, broadcast address, and usable range for departments. Calculate how much address space remains unallocated after assignments.

Answer

Approach. Sort the requirements in descending order of size, give each the smallest block 2n≥hosts+22^n \ge \text{hosts}+2 (network and broadcast addresses are reserved), and allocate consecutively from the start of the block.

DepartmentHostsBlock neededSizeMask (CIDR)
C120120+2=122≤128120+2=122 \le 128128/25 (255.255.255.128)
B5555+2=57≤6455+2=57 \le 6464/26 (255.255.255.192)
A2828+2=30≤3228+2=30 \le 3232/27 (255.255.255.224)
D2929+2=31≤3229+2=31 \le 3232/27 (255.255.255.224)

Allocation.

DeptNetwork addressUsable rangeBroadcast
C172.16.0.0/25172.16.0.1 - 172.16.0.126172.16.0.127
B172.16.0.128/26172.16.0.129 - 172.16.0.190172.16.0.191
A172.16.0.192/27172.16.0.193 - 172.16.0.222172.16.0.223
D172.16.0.224/27172.16.0.225 - 172.16.0.254172.16.0.255

Address space used. 128+64+32+32=256128+64+32+32 = 256 addresses, exactly the block 172.16.0.0 - 172.16.0.255 (the first /24).

Unallocated space. The /16 has 216=65,5362^{16}=65{,}536 addresses, so

65,536−256=65,280 addresses,65{,}536 - 256 = 65{,}280 \text{ addresses},

i.e. the range 172.16.1.0 - 172.16.255.255 remains free for future use (equivalent to 255 /24 networks).

Within the allocated blocks, the usable hosts are 126+62+30+30=248126+62+30+30=248 against 120+55+28+29=232120+55+28+29=232 required, so only 16 host addresses are spare inside the blocks (6 for C, 7 for B, 2 for A, 1 for D).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 8 marks

Define congestion. How Token Bucket Algorithm works? Explain with neat diagram?

Answer

Congestion. Congestion is a state in which the load (number of packets sent) on a network, or part of it, exceeds its capacity (the number of packets it can handle). Queues in routers overflow, packets are delayed or dropped, retransmissions add more load, and throughput collapses. Congestion control mechanisms either prevent it (open loop: traffic shaping, admission control) or remove it after it appears (closed loop: backpressure, choke packets, windows).

Token Bucket algorithm. It is a traffic shaping (leaky bucket variant) algorithm that allows bursty traffic up to a limit while controlling the average rate.

Working.

  1. A bucket holds tokens up to a maximum capacity BB.
  2. Tokens are added to the bucket at a constant rate rr (for example one token every Δt\Delta t). If the bucket is full, new tokens are discarded.
  3. To transmit a packet (or nn bytes), the sender must remove a matching token (or nn tokens) from the bucket.
  4. If enough tokens exist, the packet is sent immediately; otherwise it waits (or is dropped/marked non-conforming) until enough tokens accumulate.
  5. An idle host accumulates tokens, so it can later send a burst of up to BB packets at the full line rate.
 tokens arrive at rate r
        |
        v
    +-------+   bucket capacity B
    | o o o |   (overflow tokens discarded)
    | o o   |
    +---+---+
        | one token consumed per packet
 packets --> [ queue ] --> (token available?) --yes--> network
                               no -> wait / drop

Burst property. In an interval TT the maximum amount of data sent is B+rTB + rT, and the long-term average rate cannot exceed rr. For example, with B=10B=10 tokens and r=2r=2 tokens/s, an idle sender can send a burst of 10 packets at once, then 2 per second.

Difference from leaky bucket. The leaky bucket outputs at a constant rate regardless of the input burst, while the token bucket permits bursts when tokens have been saved.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 7 marks

Consider a network with three nodes. Find the Distance Vector Routing algorithm for the given figure.

Answer

Distance vector routing (Bellman-Ford): each node keeps a table of the best-known distance to every destination and the next hop. Periodically each node sends its table to its direct neighbours, and a node updates its table using

Dx(y)=min⁡v∈neighbours{ c(x,v)+Dv(y) }.D_x(y)=\min_{v\in\text{neighbours}}\{\,c(x,v)+D_v(y)\,\}.

Network. Three nodes A,B,CA, B, C with link costs AA-B=2B=2, BB-C=3C=3, AA-C=7C=7.

Step 1: initial tables (each node knows only its direct neighbours; ∞\infty if unknown).

At Acostnext
A0-
B2B
C7C
At Bcostnext
A2A
B0-
C3C
At Ccostnext
A7A
B3B
C0-

Step 2: exchange of vectors and update.

  • At AA: via BB to CC costs 2+DB(C)=2+3=5<72+D_B(C)=2+3=5<7, so A→CA\to C becomes 5 via BB.
  • At CC: via BB to AA costs 3+DB(A)=3+2=5<73+D_B(A)=3+2=5<7, so C→AC\to A becomes 5 via BB.
  • At BB: via AA to CC costs 2+7=9>32+7=9>3 and via CC to AA costs 3+7=10>23+7=10>2, so no change.

Final (converged) tables.

Nodeto Ato Bto C
A02 (B)5 (B)
B2 (A)03 (C)
C5 (B)3 (B)0

A further exchange gives no change, so the algorithm has converged. If a link cost changes, only the affected entries are recomputed and propagated; good news spreads quickly but bad news may cause count-to-infinity (reduced by split horizon and poison reverse).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 7 marks

How the Shortest path routing algorithm works prove with undirected graph.

Answer

Shortest path routing. The network is modelled as a weighted graph: nodes are routers and edges are links weighted by distance, delay or cost. The goal is to find the minimum-weight path from a source to each destination. The standard method is Dijkstra's algorithm, which maintains for each node a tentative distance and a predecessor, permanently labels the nearest unlabelled node, and relaxes its neighbours.

Demonstration on an undirected graph. Nodes A,B,C,D,E,FA,B,C,D,E,F; edges AB=2, AC=5, BC=2, BD=4, CD=1, CE=7, DE=3, DF=6, EF=2AB=2,\ AC=5,\ BC=2,\ BD=4,\ CD=1,\ CE=7,\ DE=3,\ DF=6,\ EF=2. Find shortest paths from AA.

StepFixed nodeBBCCDDEEFF
0A(0)A(0)25∞\infty∞\infty∞\infty
1B(2)B(2)246∞\infty∞\infty
2C(4)C(4)24511∞\infty
3D(5)D(5)245811
4E(8)E(8)245810
5F(10)F(10)245810

Shortest paths from AA: BB: AA-BB (2); CC: AA-BB-CC (4); DD: AA-BB-CC-DD (5); EE: AA-BB-CC-DD-EE (8); FF: AA-BB-CC-DD-EE-FF (10). Each router's routing table stores the next hop (here BB for all destinations).

Why it is correct (proof sketch). Suppose nodes are fixed in order of increasing distance and uu is the unfixed node with the smallest tentative distance d(u)d(u). Any other path to uu must leave the fixed set through some edge to an unfixed node xx, with d(x)≥d(u)d(x)\ge d(u) by the choice of uu; since all edge weights are non-negative, the rest of that path cannot reduce the length below d(u)d(u). So d(u)d(u) is the true shortest distance, and by induction every fixed label is correct. Because the graph is undirected, each link has the same cost in both directions, so the shortest path from AA to XX is the reverse of the shortest path from XX to AA.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 8 marks

What role does encryption play in network security? Contrast symmetric key cryptography and public key cryptography with example of use.

Answer

Role of encryption in network security. Encryption converts plaintext into unreadable ciphertext using an algorithm and a key, so only authorised parties can read it. It provides:

  • Confidentiality: eavesdroppers on the network cannot read the data.
  • Integrity: together with hashes/MACs, it shows that data was not altered.
  • Authentication and non-repudiation: digital signatures prove who sent the data.

It protects data in transit (HTTPS/TLS, VPN, Wi-Fi WPA2/3) and at rest (disk, database encryption).

Symmetric versus public-key cryptography.

AspectSymmetric keyPublic key (asymmetric)
KeysOne shared secret key for both encryption and decryptionKey pair: public key (shared) and private key (secret)
SpeedFast; suited to bulk dataSlow (heavy mathematics)
Key distributionHard: the secret key must be exchanged securely beforehandEasy: the public key may be published openly
Number of keysn(n−1)/2n(n-1)/2 keys for nn users2n2n keys for nn users
ServicesConfidentialityConfidentiality, authentication, digital signature, key exchange
AlgorithmsDES, 3DES, AES, RC4RSA, ECC, Diffie-Hellman, DSA

Examples of use.

  • Symmetric: encrypting a file or disk with AES, or the data stream of a web session once keys have been agreed.
  • Public key: Alice encrypts a message with Bob's public key, and only Bob's private key can decrypt it. For a digital signature, Alice signs a hash with her private key and anyone verifies it using her public key.
  • Combined (hybrid), as in HTTPS/TLS: public-key cryptography (RSA/ECDHE) is used to authenticate the server and agree on a session key, then fast symmetric encryption (AES) protects the actual traffic.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 Spring · 5 marks

Write short notes on: VPN

Answer

A Virtual Private Network (VPN) creates a secure, encrypted tunnel over a public network (the internet) so that remote users or sites can communicate as though they were on a private network.

How it works. Packets are encapsulated inside another packet (tunnelling) and encrypted; at the far end of the tunnel they are decrypted and delivered. Authentication of users and devices (passwords, certificates) controls who can connect.

Types.

  • Remote-access VPN: a single user connects to the organisation's network (work from home).
  • Site-to-site VPN: connects two or more office LANs over the internet.

Protocols: IPsec, SSL/TLS VPN, OpenVPN, WireGuard, L2TP, PPTP.

Advantages: confidentiality and integrity of data, cheaper than leased lines, secure remote access, hides the user's real IP address, can bypass geo-restrictions.

Disadvantages: encryption overhead can lower speed, depends on the quality of the VPN provider and configuration, and the setup and management add complexity.

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  • 2025 Spring · 5 marks

Write short notes on: DHCP and DNS

Answer

DHCP (Dynamic Host Configuration Protocol). An application-layer protocol (UDP ports 67 server, 68 client) that automatically assigns network configuration to hosts: IP address, subnet mask, default gateway and DNS server. It removes manual configuration and avoids address conflicts.

The process is DORA: the client broadcasts Discover; the server replies with an Offer; the client sends a Request for the offered address; the server confirms with an Acknowledgement. Addresses are leased for a time and renewed or released afterwards. It can use dynamic, automatic or manual (reserved) allocation.

DNS (Domain Name System). A distributed, hierarchical database (UDP/TCP port 53) that translates human-readable domain names (such as www.example.com) into IP addresses, and the reverse.

  • The namespace is a tree: root, top-level domains (.com, .np), second-level domains, and so on.
  • Resolution goes from a local (recursive) resolver to root, TLD and authoritative name servers, with caching to reduce traffic.
  • Common records: A (IPv4), AAAA (IPv6), CNAME (alias), MX (mail server), NS (name server), PTR (reverse lookup).

Together: DHCP gives a host its address and the DNS server's address, and DNS then resolves names for the host.

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  • 2025 Spring · 5 marks

Write short notes on: SMTP vs. POP3

Answer

Both are email protocols at the application layer, but they do different jobs.

SMTP (Simple Mail Transfer Protocol). A push protocol used to send mail from a client to its mail server and between mail servers. It uses TCP port 25 (server to server), 587 (submission) or 465 (SMTPS). It works with text commands such as HELO, MAIL FROM, RCPT TO, DATA, QUIT.

POP3 (Post Office Protocol version 3). A pull protocol used by a client to retrieve mail from the server's mailbox. It uses TCP port 110 (995 over SSL). It downloads messages to the local device and normally deletes them from the server.

PointSMTPPOP3
PurposeSending / relaying mailReceiving / downloading mail
DirectionClient to server, server to serverServer to client
Port25, 587, 465110, 995
StorageDoes not store mail for the userMoves mail to the client's device (offline reading)

A message travels: sender's client →SMTP\xrightarrow{\text{SMTP}} sender's server →SMTP\xrightarrow{\text{SMTP}} receiver's server →POP3/IMAP\xrightarrow{\text{POP3/IMAP}} receiver's client. IMAP is the alternative to POP3 that keeps mail on the server and synchronises multiple devices.

Answer checked by Claude Sonnet · Oct 21, 2026

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Questions from PU BCA 2025 Spring Numerical Methods (Pokhara University BCA, Spring 2025), PU BCA 2024 Spring Numerical Methods (Pokhara University BCA, Spring 2024), PU BCA 2023 Spring Numerical Methods (Pokhara University BCA, Spring 2023), PU BCA 2025 Spring Mathematical Foundation of Computer Science (Pokhara University BCA, Spring 2025), PU BCA 2024 Fall Mathematical Foundation of Computer Science (Pokhara University BCA, Fall 2024), PU BCA 2024 Spring Mathematical Foundation of Computer Science (Pokhara University BCA, Spring 2024), PU BCA 2023 Spring Mathematical Foundation of Computer Science (Pokhara University BCA, Spring 2023) and PU BCA 2025 Spring Data Communication and Computer Network (Pokhara University BCA, Spring 2025). Answers are written for this site; check them against your class notes.