2016 question paper

Computer Networks

26 questions

1. Q1a. In the following pairs of OSI protocol layer/sub-layer and its Functionality, the INCORRECT pair is (i) Network layer and Routing (ii) Data Link Layer and Bit synchronization (iii) Transport layer and End-to-end process communication (iv) Medium Access Control sub-layer and Channel sharing20162m

   Module 1: Physical Layer & Data Communication

   In the following pairs of OSI protocol layer/sub-layer and its Functionality, the INCORRECT pair is
   (i) Network layer and Routing
   (ii) Data Link Layer and Bit synchronization
   (iii) Transport layer and End-to-end process communication
   (iv) Medium Access Control sub-layer and Channel sharing

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   Worked Solution

   (ii) Data Link Layer and Bit synchronization

   Bit synchronization (ensuring sender and receiver clocks agree on bit timing/boundaries) is a Physical layer function, not a Data Link layer function. The Data Link layer's actual jobs are framing, error control, flow control, and media access control — not raw bit-level timing.

   The other pairs are correct: Network layer does routing; Transport layer does end-to-end communication; the MAC sub-layer handles channel sharing.

2. Q1b. Which of the following is NOT true with respect to a transparent bridge and a router? (i) Both bridge and router selectively forward data packets (ii) A bridge uses IP addresses while a router uses MAC addresses (iii) A bridge builds up its routing table by inspecting incoming packets (iv) A router can connect between a LAN and WAN20162m

   Module 2: Data Link Layer & MAC Sublayer

   Which of the following is NOT true with respect to a transparent bridge and a router?
   (i) Both bridge and router selectively forward data packets
   (ii) A bridge uses IP addresses while a router uses MAC addresses
   (iii) A bridge builds up its routing table by inspecting incoming packets
   (iv) A router can connect between a LAN and WAN

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   Worked Solution

   (ii) A bridge uses IP addresses while a router uses MAC addresses — this statement is NOT true (it has the roles reversed).

   The correct relationship is the opposite: a bridge operates at the Data Link layer and forwards frames using MAC addresses; a router operates at the Network layer and forwards packets using IP addresses.

   The other statements are all true: both devices selectively forward traffic (rather than broadcasting everything); a transparent bridge does build its forwarding table by inspecting (learning from) the source addresses of incoming frames; and a router, working at the Network layer, is capable of connecting a LAN to a WAN (crossing different network technologies), unlike a bridge which stays within the same LAN/broadcast domain type.

3. Q1c. Which one of the following statements is FALSE? (i) TCP guarantees a minimum communication rate (ii) TCP ensures in-order delivery (iii) TCP reacts to congestion by reducing sender window size (iv) TCP employs retransmission to compensate for packet loss20162m

   Module 4: Transport Layer & Congestion Control

   Which one of the following statements is FALSE?
   (i) TCP guarantees a minimum communication rate
   (ii) TCP ensures in-order delivery
   (iii) TCP reacts to congestion by reducing sender window size
   (iv) TCP employs retransmission to compensate for packet loss

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   Worked Solution

   Answer: (i) TCP guarantees a minimum communication rate

   Why: TCP provides reliable, in-order, congestion-controlled byte-stream delivery, but it never reserves bandwidth or guarantees any minimum data rate — the achievable rate depends on network conditions and the congestion window. Options (ii), (iii) and (iv) are all standard TCP services (in-order delivery via sequence numbers, window reduction on congestion, and retransmission on loss).

4. Q1d. Which one of the following statements is FALSE? (i) HTTP runs over TCP (ii) HTTP describes the structure of web pages (iii) HTTP allows information to be stored in a URL (iv) HTTP can be used to test the validity of a hypertext link20162m

   Module 5: Application Layer & Security

   Which one of the following statements is FALSE?
   (i) HTTP runs over TCP
   (ii) HTTP describes the structure of web pages
   (iii) HTTP allows information to be stored in a URL
   (iv) HTTP can be used to test the validity of a hypertext link

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   Worked Solution

   Answer: (ii) HTTP describes the structure of web pages

   Why: HTTP is only a request/response transfer protocol; the structure (markup) of web pages is described by HTML, not by HTTP. The other statements are true — HTTP runs over TCP (port 80/443), it can carry data inside the URL/query string (iii), and status codes such as 404 Not Found can be used to test whether a hypertext link is valid (iv).

5. Q1e. In a sliding window ARQ scheme, the transmitter's window size is N and the receiver's window size is M. The minimum number of distinct sequence numbers required to ensure correct operation of the ARQ scheme is (i) min(M, N) (ii) max(M, N) (iii) M + N (iv) M × N20162m

   Module 2: Data Link Layer & MAC Sublayer

   In a sliding window ARQ scheme, the transmitter's window size is N and the receiver's window size is M. The minimum number of distinct sequence numbers required to ensure correct operation of the ARQ scheme is
   (i) min(M, N)
   (ii) max(M, N)
   (iii) M + N
   (iv) M × N

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   Worked Solution

   (iii) M + N

   In a sliding window ARQ scheme with sender window size N and receiver window size M, the minimum number of distinct sequence numbers required is N + M.

   Why: Sequence numbers must be large enough that the sender's window and the receiver's window can never simultaneously refer to overlapping sequence numbers in a way that causes ambiguity between old and new frames. If the total range of usable sequence numbers were smaller than N + M, it would become possible for a frame from a previous window cycle to be mistaken for a frame in the current cycle, since the sender's and receiver's windows could "wrap around" and overlap incorrectly.

   This is a well-known general result (a classic GATE-style question) that applies to any generic sliding window ARQ with independent sender/receiver window sizes — note that this differs from the more commonly quoted Go-Back-N-specific rule (sequence numbers ≥ N+1) because that assumes a receiver window of size 1 (M=1), giving N+M = N+1, consistent with the general formula.

6. Q1f. Which one of the following protocols is NOT used to resolve one form of address to another one? (i) DNS (ii) ARP (iii) DHCP (iv) RARP20162m

   Module 3: Network Layer & Routing

   Which one of the following protocols is NOT used to resolve one form of address to another one?
   (i) DNS
   (ii) ARP
   (iii) DHCP
   (iv) RARP

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   Worked Solution

   Answer: (iii) DHCP

   Why: DNS resolves a domain name to an IP address, ARP resolves an IP address to a MAC address, and RARP resolves a MAC address to an IP address — each maps one address form to another. DHCP instead dynamically assigns/leases an IP address and configuration parameters (subnet mask, default gateway, DNS server) to a host; it does not translate one address form into another.

7. Q1g. Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a webpage from a remote server, assuming that the host has just been restarted. (i) HTTP GET request, DNS query, TCP SYN (ii) DNS query, HTTP GET request, TCP SYN (iii) DNS query, TCP SYN, HTTP GET request (iv) TCP SYN, DNS query, HTTP GET request20162m

   Module 5: Application Layer & Security

   Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a webpage from a remote server, assuming that the host has just been restarted.
   (i) HTTP GET request, DNS query, TCP SYN
   (ii) DNS query, HTTP GET request, TCP SYN
   (iii) DNS query, TCP SYN, HTTP GET request
   (iv) TCP SYN, DNS query, HTTP GET request

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   Worked Solution

   Answer: (iii) DNS query, TCP SYN, HTTP GET request

   Why: The freshly restarted host has an empty DNS cache, so it must first resolve the server's name into an IP address (DNS query). It then opens a TCP connection to the server's port 80 (three-way handshake initiated by the TCP SYN) and only finally sends the HTTP GET request inside that established connection.

8. Q1h. How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame? (i) 10,000 bytes (ii) 12,000 bytes (iii) 15,000 bytes (iv) 27,000 bytes20162m

   Module 1: Physical Layer & Data Communication

   How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame?
   (i) 10,000 bytes
   (ii) 12,000 bytes
   (iii) 15,000 bytes
   (iv) 27,000 bytes

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   Worked Solution

   (ii) 12,000 bytes

   In asynchronous mode, each character frame = 1 start bit + 8 data bits + 1 parity bit + 2 stop bits = 12 bits per byte.

   Baud rate        = 9600 bits/sec
   Bits per byte     = 1 (start) + 8 (data) + 1 (parity) + 2 (stop) = 12
   
   Bytes/sec  = 9600 / 12 = 800 bytes/sec
   In 15 sec  = 800 × 15 = 12,000 bytes
   

9. Q1i. Which one of the following is TRUE interior Gateway routing protocols - Routing Information protocol (RIP) and Open Shortest Path First (OSPF)? (i) RIP uses distance vector routing and OSPF uses link state routing (ii) OSPF uses distance vector routing and RIP uses link state routing (iii) Both RIP and OSPF use link state routing (iv) Both RIP and OSPF use distance vector routing20162m

   Module 3: Network Layer & Routing

   Which one of the following is TRUE interior Gateway routing protocols - Routing Information protocol (RIP) and Open Shortest Path First (OSPF)?
   (i) RIP uses distance vector routing and OSPF uses link state routing
   (ii) OSPF uses distance vector routing and RIP uses link state routing
   (iii) Both RIP and OSPF use link state routing
   (iv) Both RIP and OSPF use distance vector routing

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   Worked Solution

   Answer: (i) RIP uses distance vector routing and OSPF uses link state routing

   Why: RIP periodically exchanges entire routing tables with its neighbours and selects paths by hop count (Bellman-Ford / distance-vector algorithm), whereas OSPF floods link-state advertisements so every router builds a common topology map and runs Dijkstra's SPF algorithm to compute least-cost paths.

10. Q1j. In the slow start phase of the TCP congestion control algorithm, the size of the congestion window: (i) does not increase (ii) increases linearly (iii) increases quadratically (iv) increases exponentially20162m

    Module 4: Transport Layer & Congestion Control

    In the slow start phase of the TCP congestion control algorithm, the size of the congestion window:
    (i) does not increase
    (ii) increases linearly
    (iii) increases quadratically
    (iv) increases exponentially

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    Worked Solution

    Answer: (iv) increases exponentially

    Why: In slow start the congestion window grows by one MSS for every ACK received, which doubles cwnd every RTT (1 -> 2 -> 4 -> 8 ...), i.e. exponential growth. Linear (additive) increase of +1 MSS per RTT begins only after cwnd reaches ssthresh, in the congestion-avoidance phase.

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12. Q2a. Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. What is the minimum frame size in bytes to achieve a link utilization of at least 50%.20167m

    Module 4: Transport Layer & Congestion Control

    Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. What is the minimum frame size in bytes to achieve a link utilization of at least 50%.

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    Worked Solution

    Minimum Frame Size for 50% Utilization on a Stop-and-Wait Link

    Given: bit rate $R = 64$ kbps, one-way propagation delay $T_p = 20$ ms, ACK transmission time and node processing delay are negligible, target utilization $U \ge 50\%$.

    Step 1: Time for one stop-and-wait cycle

    One cycle = transmit the frame + propagate to receiver + propagate the ACK back:

    $$T_{cycle} = T_t + T_p + T_p = T_t + 2T_p$$

    Step 2: Utilization formula

    $$U = \frac{\text{time the link is busy sending data}}{\text{total cycle time}} = \frac{T_t}{T_t + 2T_p}$$

    Step 3: Apply the 50% condition

    $$0.5 \le \frac{T_t}{T_t + 0.040}$$

    $$0.5\,(T_t + 0.040) \le T_t \;\Longrightarrow\; 0.5T_t + 0.020 \le T_t$$

    $$0.020 \le 0.5\,T_t \;\Longrightarrow\; T_t \ge 0.040\ \text{s} = 40\ \text{ms}$$

    So the frame must occupy the link for at least 40 ms, i.e. $T_t \ge 2T_p$.

    Step 4: Convert transmission time into frame size

    $$L = R \times T_t = 64{,}000\ \text{bits/s} \times 0.040\ \text{s} = 2560\ \text{bits}$$

    $$L = \frac{2560}{8} = 320\ \text{bytes}$$

    Check

    $T_t = 320 \times 8 / 64000 = 0.04$ s, so $U = 0.04/(0.04 + 0.04) = 0.5$, i.e. exactly 50%.

    Final answer: the minimum frame size is 320 bytes. Any frame of 320 bytes or more yields a utilization of at least 50%.

13. Q2b. Consider an instance of TCP's Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.20167m

    Module 4: Transport Layer & Congestion Control

    Consider an instance of TCP's Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

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    Worked Solution

    TCP AIMD: Congestion Window at the End of the 10th Transmission

    Given: $cwnd = 2$ MSS at the start of slow start, $ssthresh = 8$ MSS at the first transmission, a timeout occurs during the 5th transmission.

    Rules used:

    • Slow start (SS): cwnd doubles each RTT (2 -> 4 -> 8 ...) but stops doubling once it reaches ssthresh.
    • Congestion avoidance (CA): cwnd grows additively by 1 MSS per RTT.
    • Timeout: $ssthresh = cwnd/2$, then $cwnd$ resets to 1 MSS and slow start restarts.

    Step 1: Behaviour up to the timeout

    Transmission	1	2	3	4	5
    cwnd (MSS)	2	4	8	9	10
    Phase	SS	SS	SS, reaches threshold	CA	CA, timeout occurs

    At transmission 3 the window hits $8 = ssthresh$, so from transmission 4 onward congestion avoidance adds 1 MSS per RTT: $8 \to 9 \to 10$.

    Step 2: Effect of the timeout in the 5th transmission

    $$ssthresh_{new} = \frac{10}{2} = 5\ \text{MSS}, \qquad cwnd = 1\ \text{MSS}$$

    Step 3: Behaviour after the timeout

    Transmission	6	7	8	9	10
    cwnd (MSS)	1	2	4	5	6
    Phase	SS	SS	SS, clamped at threshold 5	CA	CA
    cwnd at end of transmission	2	4	5	6	7

    Transmission 8 doubles $4 \to 8$ but is capped at the new threshold, $\min(8, 5) = 5$ MSS; with $cwnd = ssthresh = 5$ the connection switches to congestion avoidance, which adds exactly 1 MSS per RTT: $5 \to 6 \to 7$.

    Final answer: the congestion window size at the end of the 10th transmission is 7 MSS.

14. Q3a. The address of a class B host is to be split into subnets with a 6-bit subnet number: What is the maximum number of subnets and the maximum number of hosts in each subnet? What is the number of networks allowed under Class B addresses?20167m

    Module 3: Network Layer & Routing

    The address of a class B host is to be split into subnets with a 6-bit subnet number: What is the maximum number of subnets and the maximum number of hosts in each subnet? What is the number of networks allowed under Class B addresses?

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    Worked Solution

    Subnetting a Class B Address with a 6-bit Subnet Field

    Given: the address of a Class B host is to be divided into subnets using a 6-bit subnet number.

    A classful Class B address has 16 network bits and 16 host bits:

    | network id : 16 bits | host id : 16 bits |
    | network id : 16 bits | subnet : 6 bits | host : 16 - 6 = 10 bits |
    

    1. Maximum number of subnets

    Subnet bits are borrowed from the host id, so

    $$N_{subnets} = 2^{6} = \mathbf{64}\ \text{subnets}$$

    (If the all-zeros subnet and the all-ones subnet are not usable, as in the old classful rules, only $64 - 2 = 62$ subnets are usable, but the maximum count is $2^6 = 64$.)

    2. Maximum number of hosts in each subnet

    Host bits left after borrowing 6 bits:

    $$h = 16 - 6 = 10\ \text{bits}$$

    $$N_{hosts} = 2^{10} - 2 = 1024 - 2 = \mathbf{1022}\ \text{hosts per subnet}$$

    The subtraction of 2 removes the subnet's network address (all host bits 0) and its broadcast address (all host bits 1).

    3. Number of networks allowed under Class B addresses

    Class B reserves a 2-bit class prefix (10), leaving a 14-bit network id:

    $$N_{networks} = 2^{14} = \mathbf{16384}\ \text{Class B networks}$$

    Final answers: maximum subnets = 64, maximum hosts per subnet = 1022, number of Class B networks = 16384.

15. Q3b. Assume that source S and destination D are connected through two intermediate routers labelled R. Determine how many times each packet has to visit the network layer and the data link layer during a transmission from S to D. 20167m

    Module 2: Data Link Layer & MAC Sublayer

    Assume that source S and destination D are connected through two intermediate routers labelled R. Determine how many times each packet has to visit the network layer and the data link layer during a transmission from S to D.

    

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    Worked Solution

    Topology: Source S and Destination D are connected via two intermediate routers R1 and R2:

      S ---- R1 ---- R2 ---- D
    

    There are 3 links (hops) total: S-R1, R1-R2, R2-D.

    Layer visit count

    Data Link Layer — visited once per link/hop the packet crosses, because a new frame (with fresh MAC addressing) must be created for each physical link:

    S -> R1 : 1 visit (frame created at S, processed at R1)
    R1 -> R2: 1 visit (new frame created at R1, processed at R2)
    R2 -> D : 1 visit (new frame created at R2, processed at D)
    
    Total Data Link layer visits = 3 (once per link)
    

    Network Layer — visited once at each node the packet passes through (source, every intermediate router, and destination), since routing/forwarding decisions (or final delivery) happen at the Network layer at every node:

    At S : Network layer processes the packet once (originates it)
    At R1: Network layer processes it once (routing decision)
    At R2: Network layer processes it once (routing decision)
    At D : Network layer processes it once (final delivery, passed up to Transport layer)
    
    Total Network layer visits = 4 (once per node: S, R1, R2, D)
    

    Summary: For a path with k intermediate routers (here k=2), a packet visits the Network layer (k+2) times (source + k routers + destination) and the Data Link layer (k+1) times (once per link). Here: Network layer = 4 times, Data Link layer = 3 times.

16. Q4a. In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. Find the position of the datagram, the sequence numbers of the first and the last bytes of the payload respectively.20167m

    Module 3: Network Layer & Routing

    In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. Find the position of the datagram, the sequence numbers of the first and the last bytes of the payload respectively.

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    Worked Solution

    IPv4 Fragment: Position and Payload Byte Numbers

    Given: M bit = 0, HLEN = 10, Total Length = 400, Fragment Offset = 300.

    Step 1: Header size

    HLEN is counted in 32-bit (4-byte) words:

    $$\text{Header size} = \text{HLEN} \times 4 = 10 \times 4 = 40\ \text{bytes}$$

    Step 2: Payload carried by this fragment

    $$\text{Payload} = \text{Total Length} - \text{Header} = 400 - 40 = 360\ \text{bytes}$$

    Step 3: Position of the fragment in the original datagram

    The Fragment Offset field is expressed in units of 8 bytes:

    $$\text{start offset} = 300 \times 8 = 2400\ \text{bytes}$$

    So the payload of this datagram begins at byte 2400 of the original datagram's data field.

    Step 4: Interpretation of M = 0

    M = 0 (no more fragments) means this is the last fragment of the original datagram.

    Step 5: Sequence numbers of the first and last payload bytes

    Counting bytes from 0 at the start of the original data field:

    $$\text{first byte} = 2400$$

    $$\text{last byte} = 2400 + 360 - 1 = 2759$$

    Hence the original datagram's payload was $2759 + 1 = 2760$ bytes long (bytes 0 to 2759), and this fragment covers bytes 2400 to 2759. (If numbered from 1, the answers are 2401 and 2760.)

    Final answer: position = last fragment starting at byte offset $300 \times 8 = 2400$; payload sequence numbers: first byte = 2400, last byte = 2759 (0-based).

17. Q4b. An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are?20167m

    Module 3: Network Layer & Routing

    An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are?

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    Worked Solution

    Header Fields of the Third IP Fragment (MTU 1500)

    Given: MTU = 1500 bytes, received IP packet = 4404 bytes, IP header = 20 bytes.

    Step 1: Data that must be fragmented

    $$\text{Data} = 4404 - 20 = 4384\ \text{bytes}$$

    Step 2: Maximum data per fragment

    $$\text{Max data} = \text{MTU} - \text{header} = 1500 - 20 = 1480\ \text{bytes}$$

    The offset is counted in 8-byte blocks, so every non-final fragment must carry a multiple of 8 bytes: $1480 \bmod 8 = 0$, so 1480 is legal and is used as the data size of fragments 1 and 2.

    Step 3: Fragmentation plan

    Fragment	Data bytes (original data field)	Data size	Total length	Offset (value = bytes/8)	M flag
    1st	0 - 1479	1480	$1480 + 20 = 1500$	$0/8 = 0$	1
    2nd	1480 - 2959	1480	$1480 + 20 = 1500$	$1480/8 = 185$	1
    3rd	2960 - 4383	$4384 - 2960 = 1424$	$1424 + 20 = 1444$	$2960/8 = 370$	0

    Step 4: Fields in the header of the third fragment

    Field	Value
    Total Length	1444 bytes (1424 data + 20 header)
    Header Length (HLEN)	5 (i.e. 20 bytes)
    Identification	same as the original packet
    Flags	MF = 0, DF = 0 (this is the last fragment)
    Fragment Offset	370 (means $370 \times 8 = 2960$ bytes of data precede this fragment)
    Protocol	copied unchanged from the original header

    Final answer: Total Length = 1444 bytes, MF flag = 0, Fragment Offset = 370 (header length stays 20 bytes, identification unchanged).

18. Q5a. A serial transmission T1 uses 8 information bits, 2 start bits, 1 stop bit and 1 parity bit for each character. A synchronous transmission T2 uses 3 eight bit sync characters followed by 30 eight bit information characters. If the bit rate is 1200 bits/second in both cases, what are the transfer rates of T1 and T2?20167m

    Module 1: Physical Layer & Data Communication

    A serial transmission T1 uses 8 information bits, 2 start bits, 1 stop bit and 1 parity bit for each character. A synchronous transmission T2 uses 3 eight bit sync characters followed by 30 eight bit information characters. If the bit rate is 1200 bits/second in both cases, what are the transfer rates of T1 and T2?

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    Worked Solution

    The transfer rate (useful information rate) is the fraction of the total bit rate that actually carries information bits, excluding framing/overhead bits.

    T1 (Asynchronous transmission)

    Each character frame = 2 start bits + 8 information bits + 1 parity bit + 1 stop bit = 12 bits total, of which 8 are useful information bits.

    Useful fraction = 8 / 12
    
    Transfer rate (T1) = Bit rate x (useful bits / total bits per frame)
                        = 1200 x (8/12)
                        = 1200 x 0.667
                        = 800 bits/sec
    

    T2 (Synchronous transmission)

    Each frame = 3 sync characters (8 bits each) + 30 information characters (8 bits each)

    Overhead bits = 3 x 8 = 24 bits
    Information bits = 30 x 8 = 240 bits
    Total frame size = 24 + 240 = 264 bits
    
    Useful fraction = 240 / 264 = 0.9091
    
    Transfer rate (T2) = 1200 x (240/264)
                        = 1200 x 0.9091
                        ≈ 1090.9 bits/sec
    

    Result: T1 ≈ 800 bps, T2 ≈ 1090.9 bps. Synchronous transmission (T2) achieves a higher effective transfer rate because its fixed per-frame overhead (3 sync bytes) is amortized over more information bytes (30), unlike asynchronous transmission where every single character carries its own start/stop/parity overhead.

19. Q5b. Compare and contrast the Go-Back-N ARQ Protocol with Selective-Repeat ARQ.20167m

    Module 2: Data Link Layer & MAC Sublayer

    Compare and contrast the Go-Back-N ARQ Protocol with Selective-Repeat ARQ.

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    Worked Solution

    Aspect	Go-Back-N (GBN)	Selective Repeat (SR)
    Receiver behavior	Accepts only frames arriving strictly in order; discards any out-of-order frame, even if correctly received	Accepts and buffers out-of-order frames, filling gaps later as missing frames arrive
    Acknowledgement type	Cumulative ACK — ACK(n) means "all frames up to n-1 received correctly"	Individual/selective ACK — each correctly received frame is acknowledged separately, regardless of order
    Timers	Single timer for the oldest unacknowledged frame	Multiple timers — one per outstanding frame
    Retransmission on loss	Resends the lost frame and every frame sent after it (even ones already successfully received)	Resends only the specific lost/corrupted frame
    Maximum window size (n-bit sequence numbers)	2^n - 1	2^(n-1) (must be half the sequence space to avoid ambiguity between old and new frames)
    Receiver buffering needed	None (only tracks "next expected" sequence number)	Requires buffer space to hold out-of-order frames until gaps are filled
    Bandwidth efficiency	Lower — wastes bandwidth resending already-correct frames after a loss	Higher — minimal wasted retransmission, only the actually lost frame is resent
    Implementation complexity	Simpler (minimal receiver logic/state)	More complex (per-frame timers, receiver-side buffering and reordering logic)
    Best suited for	Low-error-rate links, or where simplicity matters more than bandwidth efficiency	High-bandwidth or higher-error-rate links, where the retransmission overhead of GBN would be too costly

    Summary: Go-Back-N trades bandwidth efficiency for simplicity (a "dumb" receiver, minimal state), while Selective Repeat trades implementation complexity for much better bandwidth efficiency (a "smart" receiver that buffers and reorders) — making SR the better choice on links where retransmitting entire windows would be too costly, at the cost of more sender/receiver bookkeeping.

20. Q6a. What are the advantages of fiber optics over copper as a transmission medium? Is there any downside of using fiber optics over copper?20167m

    Module 1: Physical Layer & Data Communication

    What are the advantages of fiber optics over copper as a transmission medium? Is there any downside of using fiber optics over copper?

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    Worked Solution

    Advantages of fiber optics over copper

    1. Much higher bandwidth — supports data rates from Gbps into the Tbps range, far beyond copper's practical limits.
    2. Lower attenuation — signals travel tens of kilometers before needing regeneration, versus ~100 m for copper Ethernet.
    3. Immunity to electromagnetic interference and crosstalk — since it carries light, not current, it's unaffected by nearby electrical noise or other cables.
    4. Better security — tapping a fiber without detection (via measurable signal loss) is far harder than tapping copper.
    5. Lighter and thinner — easier to route through conduits, especially for high-density cabling.
    6. No electrical/spark hazard — safe for use in flammable or electrically noisy industrial environments.

    Downsides of fiber optics over copper

    1. Higher cost — both cable and the optical transceivers (transmitters/receivers) are more expensive than copper equivalents.
    2. Installation complexity — splicing and terminating fiber requires precision tools and trained technicians; copper (especially UTP with RJ-45) is far simpler to terminate.
    3. Fragility — glass fiber can crack or break if bent too sharply, unlike more flexible copper wire.
    4. No power delivery — fiber can't carry electrical power (unlike copper, which supports Power over Ethernet for devices like IP cameras or APs).
    5. Harder/costlier to repair — a break requires specialized fusion splicing equipment, whereas copper can often be re-terminated with basic tools.

    Overall, fiber wins decisively on performance and distance, but copper remains preferable for cost-sensitive, short-distance, or power-over-cable scenarios.

21. Q6b. A computer on a 6-Mbps network is regulated by a token bucket. The token bucket is filled at a rate of 1 Mbps. It is initially filled to capacity with 8 megabits. How long can the computer transmit at the full 6 Mbps?20167m

    Module 4: Transport Layer & Congestion Control

    A computer on a 6-Mbps network is regulated by a token bucket. The token bucket is filled at a rate of 1 Mbps. It is initially filled to capacity with 8 megabits. How long can the computer transmit at the full 6 Mbps?

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    Token Bucket: Time the Host Can Transmit at Full Rate

    Given: peak transmission rate $R_{out} = 6$ Mbps, token generation rate $r = 1$ Mbps, bucket capacity $B = 8$ megabits, bucket initially full.

    Reasoning

    The host may send at the full 6 Mbps only while the bucket still holds tokens. During transmission the bucket is simultaneously being refilled, so tokens are consumed at the net rate:

    $$r_{net} = R_{out} - r = 6 - 1 = 5\ \text{Mbit/s}$$

    The initial 8 Mbit of tokens therefore lasts:

    $$t = \frac{B}{r_{net}} = \frac{8\ \text{Mbit}}{5\ \text{Mbit/s}} = 1.6\ \text{s}$$

    Verification

    In $t = 1.6$ s:

    • tokens used = $6 \times 1.6 = 9.6$ Mbit
    • tokens available = initial $8$ Mbit + refilled $1 \times 1.6 = 1.6$ Mbit = $9.6$ Mbit

    Both match, so the bucket empties exactly at $t = 1.6$ s. After that the host is limited to the sustained rate of 1 Mbps.

    Final answer: the computer can transmit at the full 6 Mbps for 1.6 seconds (sending 9.6 Mbit), after which it is throttled to 1 Mbps.

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23. Q7a. DNS uses UDP instead of TCP. If a DNS packet is lost, there is no automatic recovery. Does this cause a problem, and if so, how is it solved?20167m

    Module 5: Application Layer & Security

    DNS uses UDP instead of TCP. If a DNS packet is lost, there is no automatic recovery. Does this cause a problem, and if so, how is it solved?

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    DNS over UDP: Does Packet Loss Cause a Problem?

    Short answer: Yes, a lost query or reply would stall name resolution if nothing else acted, but DNS deliberately places recovery in the application layer (the stub resolver) instead of in the transport layer, so the loss is handled without ever opening a TCP connection.

    Why UDP is preferred anyway

    • A lookup is normally one small request/response pair. Setting up TCP first (SYN, SYN-ACK, ACK) would add a full handshake plus teardown, roughly doubling the messages and adding 1-2 extra RTTs to every web access.
    • DNS needs no long-lived stream, no ordering across messages, and each message is self-contained.
    • UDP lets one name server answer thousands of clients concurrently with no per-client connection state.

    How the loss is solved

    1. Client retransmission with timeout (primary fix): every query is sent with a timer. If no reply arrives, the stub resolver retransmits the same query (typically after 1 s, 2 s, 4 s, ... up to 5 attempts) and finally returns SERVFAIL/time-out to the application. This is essentially stop-and-wait with timeout implemented by DNS itself.
    2. Transaction ID + source-port matching: each query carries a random 16-bit ID (and is sent from an ephemeral port), so the client can match the eventual reply to the outstanding query and discard stray, duplicate or forged responses. Late duplicates are harmless because replies are idempotent.
    3. TCP fallback for large responses: if an answer exceeds the message-size limit (512 bytes classically) the server sets the TC (truncation) bit and the resolver retries the same query over TCP; zone transfers (AXFR/IXFR) always use TCP. EDNS0 also allows larger UDP payloads.
    4. Redundant name servers: each zone is served by at least two authoritative servers (and there are multiple root/TLD servers), so the query can simply be sent to another server.

    Conclusion

    Loss only costs one retransmission timeout of delay, not correctness. Reliability is recovered by application-level retries and ID matching, and by TCP whenever responses are large - so using UDP costs nothing in accuracy while saving connection setup on the critical path of every request.

24. Q7b. Explain the responsibility of network and transport layers of OSI model with example.20167m

    Module 3: Network Layer & Routing

    Explain the responsibility of network and transport layers of OSI model with example.

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    Responsibilities of the Network and Transport Layers in OSI

    Network layer (Layer 3) - packet delivery between hosts

    Responsibilities:

    • Logical addressing: defines a universal, hierarchical address (IP address) that identifies a host regardless of the underlying LAN technology.
    • Forwarding: moves each packet hop-by-hop from the correct input interface to the output interface of every router.
    • Routing: computes the path using routing protocols - distance vector (RIP), link state (OSPF), path vector (BGP).
    • Packetization and fragmentation/reassembly: encapsulates transport segments into packets and fragments them when a link's MTU is smaller, reassembling only at the destination host.
    • Internetworking and diagnostics: routers interconnect dissimilar networks; ARP maps IP to MAC and ICMP reports errors ("destination unreachable", "TTL expired", echo request/reply).

    Service: connectionless best-effort datagram service (IP).

    Example: a packet from 192.168.1.10 to 8.8.8.8 is handed to the default gateway, routed hop-by-hop using OSPF-derived tables, and if a router's TTL expires it returns an ICMP "Time Exceeded" message.

    Transport layer (Layer 4) - end-to-end delivery between processes

    Responsibilities:

    • Segmentation and reassembly: splits the message into segments and reassembles them in order at the receiver.
    • Process multiplexing/demultiplexing: port numbers deliver data to the correct application (80 = web, 53 = DNS, 22 = SSH).
    • Connection management: TCP performs the three-way handshake (SYN / SYN-ACK / ACK) and orderly close (FIN/ACK); UDP is connectionless.
    • Flow control: the receiver's advertised window prevents the sender from overrunning the receiver's buffer.
    • Error control and reliability: sequence and acknowledgment numbers, checksum, ACKs and retransmission on timeout give exactly-once, in-order delivery (TCP).
    • Congestion control: slow start, congestion avoidance (AIMD), fast retransmit and fast recovery protect the network itself.
    • Low-overhead delivery: UDP provides minimal encapsulation with low delay for real-time applications.

    Example: a web page arrives as thousands of TCP segments which may be lost or reordered by IP; TCP numbers, ACKs, retransmits and reorders them so the browser receives one contiguous byte stream.

    Comparison

    Aspect	Network layer	Transport layer
    Addressing unit	IP address (host)	Port number (process)
    PDU	Packet / datagram	Segment (TCP) / datagram (UDP)
    Scope	Host-to-host, hop-by-hop	End-to-end, process-to-process
    Reliability	Best effort	TCP reliable, UDP unreliable
    Protocols	IP, ICMP, ARP, OSPF, RIP	TCP, UDP, SCTP
    Devices / location	Routers, layer-3 switches	Implemented in end hosts (OS/kernel)

    Key idea: the network layer finds and forwards packets between machines, while the transport layer guarantees correct, ordered, congestion-free delivery between applications on those machines.

25. Q8a. What are the functions of a RIP message? List the RIP shortcomings and their corresponding fixes.20167m

    Module 3: Network Layer & Routing

    What are the functions of a RIP message? List the RIP shortcomings and their corresponding fixes.

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    Functions of a RIP Message, Shortcomings and Their Fixes

    Functions of a RIP message

    RIP (Routing Information Protocol) is a distance-vector protocol that exchanges routing information in UDP datagrams (port 522) with directly connected neighbours. Its message functions are:

    1. Request message: asks neighbours to send part or all of their routing tables. Sent at router startup (to discover neighbours) or when a router suspects an inconsistency.
    2. Response message (update): carries routing entries - address family, IP address, next hop and metric (hop count). Sent
       • every 30 s as a periodic full-table update,
       • as a triggered (partial) update immediately when a route changes or a link fails,
       • as a direct reply to a request message.
    3. Route lifetime management: an entry becomes invalid after 180 s without an update (enters hold-down) and is flushed after 240 s - this ages out stale routes.
    4. RIPv2 additions: carries subnet masks (VLSM/CIDR), next-hop address, route tags, multicast updates to 224.0.0.9 and authentication data.

    Shortcomings of RIP and their corresponding fixes

    Shortcoming	Corresponding fix
    Count-to-infinity - loops and extremely slow convergence when a link fails	Split horizon, poison reverse, hold-down timers, triggered updates, and the hard metric cap of 15 hops
    Metric is only hop count - a 64-kbps link can be preferred over a 1-Gbps link	Use a bandwidth/delay-aware metric: OSPF cost (based on bandwidth) or EIGRP's composite metric
    Periodic full-table flooding every 30 s wastes bandwidth	Triggered (event-driven) and partial updates
    Slow convergence due to periodic updates	Triggered updates + hold-down timers; otherwise migrate to link-state routing (OSPF)
    No authentication - a rogue router can poison routes	RIPv2 simple password / MD5 authentication
    Classful, no subnet mask / VLSM support	RIPv2 advertises subnet masks and supports CIDR; uses multicast instead of broadcast
    Poor scalability (15-hop limit, no route redistribution policy)	Move to OSPF/IS-IS (link state) or BGP (path vector)
    Single path only (no load balancing on unequal paths)	Equal-cost multi-path (ECMP) in OSPF, or adaptive routing

    Summary: RIP's simplicity makes it easy to configure for small networks, but its hop-count metric, periodic updates and count-to-infinity behaviour are fundamental limitations. Split horizon, poison reverse, hold-down and triggered updates mitigate them locally, while OSPF removes the causes by using link-state flooding with a cost metric.

26. Q8b. Describe the functions of the two FTP connections. What kinds of file types can FTP transfer?20167m

    Module 5: Application Layer & Security

    Describe the functions of the two FTP connections. What kinds of file types can FTP transfer?

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    FTP: The Two Connections and the File Types It Can Transfer

    FTP is unusual in that it uses two separate TCP connections between client and server: one long-lived control connection and one short-lived data connection.

    1. Control connection (well-known port 21)

    • Established when the session starts and kept open for the whole session, closed only with QUIT.
    • Carries ASCII commands and replies: USER, PASS, PWD, CWD, LIST, RETR (retrieve), STOR (store), PORT/PASV, TYPE, QUIT and the 3-digit replies (220, 200, 226, 550, ...).
    • Small, interactive, one line at a time, always in ASCII mode; it controls the session, authentication and the state of the data connection.

    2. Data connection (port 20 in active mode, negotiated ephemeral port in passive mode)

    • Opened only for one transfer (a file, a directory listing) and closed as soon as it finishes - one data connection per RETR/STOR/LIST.
    • Carries the actual file bytes or directory listing, in ASCII, EBCDIC or image (binary) mode.
    • Two variants:
      • Active mode (PORT): the server initiates the connection from its port 20 to the port the client advertised.
      • Passive mode (PASV): the server opens an ephemeral port and waits; the client connects to it - preferred today because it works through client-side NAT and firewalls.
    • Bulk data flows at full window size, independent of the interactive channel.

    Separating the two lets the user stay logged in and issue further commands while (or between) transfers, and lets each transfer be optimised independently.

    File types FTP can transfer

    TYPE mode	Purpose
    ASCII (A)	Text files; default - end-of-line characters are converted to/from the local convention (CRLF <-> LF)
    EBCDIC (E)	Text transfer between EBCDIC mainframes
    Image / Binary (I)	Byte-for-byte copy of any file: executables, images, PDFs, archives, compressed files
    Local (L n)	Machine-dependent byte size between two hosts

    Conclusion: because image/binary mode transfers every bit unchanged, FTP can move any kind of file; the ASCII and EBCDIC modes exist only for character-set and newline conversion between dissimilar systems.

27. Q9a. Compare the TCP header and the UDP header. List the fields in the TCP header that are missing from UDP header. Give the reason for their absence.20167m

    Module 4: Transport Layer & Congestion Control

    Compare the TCP header and the UDP header. List the fields in the TCP header that are missing from UDP header. Give the reason for their absence.

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    TCP Header vs UDP Header

    TCP and UDP are both transport-layer protocols running over IP, but TCP's header is built for a reliable, connection-oriented byte stream while UDP's is a minimal 8-byte shim.

    Aspect	TCP header	UDP header
    Minimum size	20 bytes (up to 60 bytes with options)	8 bytes, always fixed
    Fields	Source port, Dest port, Sequence no, Acknowledgment no, Data offset, Reserved, Flags (URG, ACK, PSH, RST, SYN, FIN), Window, Checksum, Urgent pointer, Options/padding	Source port, Dest port, Length, Checksum
    Reliability	Seq/ack numbers, checksum, retransmission, duplicate detection	Checksum only (optional in IPv4), no retransmission
    Flow control	Receiver's advertised window	None
    Connection	SYN/FIN handshake and teardown	Connectionless
    Multiplexing	Ports + connection state	Ports only
    Overhead / use	Higher latency; file transfer, web, email	Lower latency; DNS, DHCP, RTP streaming, VoIP

    TCP fields missing from the UDP header, and the reason for their absence

    1. Sequence number - TCP numbers every byte of the stream so the receiver can reorder segments and detect gaps; UDP datagrams are independent and unordered, so there is nothing to sequence.
    2. Acknowledgment number - needed for cumulative ACKs and reliable delivery; UDP has no ACK mechanism (an application such as DNS must build its own retry logic).
    3. Data offset (header length) - TCP can grow to 60 bytes with options, so it must state how long it is; UDP's header is always exactly 8 bytes, so it carries an explicit Length field instead.
    4. Window size - implements receiver-side flow control; UDP does no flow control and leaves pacing entirely to the application.
    5. Control flags (SYN, ACK, FIN, RST, PSH, URG) - used for connection setup, orderly release, reset, push and urgent data; a connectionless protocol needs none of them.
    6. Urgent pointer - points to urgent out-of-band data, which is meaningful only within an ordered byte stream.
    7. Options (MSS, window scaling, SACK, timestamps) - parameters negotiated during connection setup; there is no connection to negotiate.

    Key idea: every field that UDP lacks exists to provide connection, ordering, reliability or flow control - exactly the services UDP deliberately omits in exchange for low overhead and low delay.

28. Q9b. What is the difference between open-loop congestion control and closed-loop congestion control?20167m

    Module 4: Transport Layer & Congestion Control

    What is the difference between open-loop congestion control and closed-loop congestion control?

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    Open-Loop vs Closed-Loop Congestion Control

    Congestion-control policies are classified by when they act. Open-loop policies are decided before traffic is sent and never use feedback from the network; closed-loop policies rely on feedback to detect congestion that has already occurred and then correct it.

    Basis	Open-loop congestion control	Closed-loop congestion control
    Principle	Prevent congestion by design; no feedback is used	Detect congestion, feed information back to the sender/routers, then react
    When applied	Before/during policy setup, ahead of transmission	Continuously during transmission, once per RTT
    Feedback	None - the policy is blind to current load	Implicit (ACK rate, RTT, packet loss) or explicit (router marking, ICMP)
    Goal	Avoid deadlock, starvation and queue overflow a priori	Keep the network in the comfortable operating region after congestion appears
    Typical mechanisms	Retransmission policy, acknowledgment policy, error-control policy, admission/flow control at connection setup, packet-discard policy (drop-old vs drop-new), FQ/WFQ scheduling, differentiated service classes, traffic shaping (leaky bucket/token bucket)	Slow start, congestion avoidance (AIMD), fast retransmit and fast recovery (TCP), RED/WRED active queue management, ECN marking, adaptive routing
    Response time	Slow - cannot adapt while traffic runs	Fast - adapts every RTT
    Complexity	Simple, predictable	More complex; needs signalling between routers, receiver and sender
    Where implemented	Mainly in routers and in the sender before sending	In the sender, driven by receiver/router feedback
    Example	Shaping traffic at the edge so the core never overloads	TCP hearing three duplicate ACKs and cutting cwnd by half

    Key differences in brief

    1. Feedback: open-loop is fire-and-forget; closed-loop is a classic control loop (measure -> decide -> act).
    2. Timing: open-loop acts at policy/setup time; closed-loop acts during the session and re-evaluates each RTT.
    3. Objective: open-loop prevents bad states (guarantees are made so queues cannot overflow); closed-loop repairs them once loss or delay rises.
    4. Adaptability: closed-loop can exploit changing conditions (grow the window while ACKs arrive quickly), whereas open-loop must stay conservative to remain safe.
    5. Cost of failure: open-loop failures are hard to detect (no feedback), while closed-loop can oscillate around the optimum as it reacts to delayed feedback.

    Rule of thumb: if a mechanism must hear from the network before it changes behaviour, it is closed-loop; if its behaviour is fixed in advance, it is open-loop.