Transport Layer Protocol Practice Questions

Let's analyze the options based on how TCP connections work.

• TCP connection establishment uses a three-way handshake (SYN, SYN-ACK, ACK), not a two-way handshake.
• Connection closing can be initiated by either the client or the server. There is no restriction that only the client can initiate the closing process first.
• TCP provides a full-duplex service, which means data can be sent and received simultaneously in both directions. It is not half-duplex.
• Both the client and the server can independently decide to close the connection. If both sides send a FIN (Finish) segment at roughly the same time, it results in a valid state known as a simultaneous close.

In the TCP slow-start phase with $RTT = 1$ ms, the congestion window $CWND$ doubles every $RTT$. At time $t$ (in milliseconds), the $CWND = 2^t$ segments, and the cumulative number of segments transmitted is given by:
$S_t = \sum_{i=0}^{t} 2^i = 2^{t+1} - 1$
At $t=9$, the cumulative number of segments sent is $S_9 = 2^{10} - 1 = 1023$.
At $t=10$, the sender transmits $CWND = 2^{10} = 1024$ segments, starting from segment $1024$ through segment $2047$.
Since segment $2000$ is within the range $[1024, 2047]$, the transmission of segment $2000$ starts at $t=10$.
This falls within the interval $10 \leq t < 11$.

The TCP 3-way handshaking protocol establishes a connection between a client and a server.

1. Packet 1 (P1 - SYN): The client initiates the connection by sending a SYN (synchronize) packet to the server. This packet has the SYN flag set to 1 and the ACK flag set to 0, indicating a request to synchronize sequence numbers.
2. Packet 2 (P2 - SYN-ACK): The server responds to the client's SYN with a SYN-ACK packet. This packet has both the SYN and ACK flags set to 1. The SYN acknowledges the client's request, and the ACK acknowledges the client's sequence number.
3. Packet 3 (P3 - ACK): Finally, the client sends an ACK packet to the server, confirming the receipt of the server's SYN-ACK. This packet has the ACK flag set to 1 and the SYN flag set to 0.

Based on this, P1 has SYN = 1, P2 has SYN = 1 and ACK = 1, and P3 has ACK = 1.
[CORRECT_OPTION: B, D]

Let's analyze each statement:

(i) Address Resolution Protocol (ARP) provides a mapping from an IP address to the corresponding hardware (link-layer) address.
This statement is TRUE. ARP is used to resolve an IP address to a physical MAC address within a local area network.

(ii) A single TCP segment from a sender S to a receiver R cannot carry both data from S to R and acknowledgement for a segment from R to S.
This statement is FALSE. TCP uses a technique called "piggybacking," where an acknowledgment (ACK) for data received from the other end can be carried within the same segment that contains new data being sent to the other end. This improves efficiency by reducing the number of segments exchanged.

Given these analyses:
(i) is TRUE.
(ii) is FALSE.

Therefore, the correct option is (B).

Let's evaluate each statement regarding IPv4 fragmentation:
(a) The fragmentation of an IP datagram is performed only at the source of the datagram: This is FALSE. Fragmentation can occur at any router along the path if the datagram's size exceeds the MTU (Maximum Transmission Unit) of the outgoing link.
(b) The fragmentation of an IP datagram is performed at any IP router which finds that the size of the datagram to be transmitted exceeds the MTU: This is TRUE. Routers check the MTU of the next hop and fragment if necessary.
(c) The reassembly of fragments is performed only at the destination of the datagram: This is TRUE. Intermediate routers do not reassemble fragments; only the final destination host is responsible for reassembly to ensure all fragments arrive and can be pieced together correctly.
(d) The reassembly of fragments is performed at all intermediate routers along the path from the source to the destination: This is FALSE, as stated in (c).
Therefore, statements (b) and (c) are TRUE.

The problem asks for which statements are TRUE.
The PDF solution lists (c, d) as true.
Let's re-examine (b) and (d).
(b) "The fragmentation of an IP datagram is performed at any IP router which finds that the size of the datagram to be transmitted exceeds the MTU." This is correct.
(d) "The reassembly of fragments is performed at all intermediate routers along the path from the source to the destination." This is incorrect. Reassembly is ONLY done at the final destination.

If the solution lists (c, d) as true, then statement (d) is marked true. This contradicts standard networking principles.
Let's consider the options as they appear in the prompt (A. The fragmentation...only at source; B. The fragmentation...at any IP router; C. The reassembly...only at destination; D. The reassembly...at all intermediate routers).
The PDF solution is (c, d). This means statement C ("The reassembly...only at destination") and statement D ("The reassembly...at all intermediate routers") are both true. This is a direct contradiction, as reassembly cannot be only at destination AND at all intermediate routers simultaneously.

There is a clear error in the provided PDF solution options if both (c) and (d) are marked.
Given the options in the prompt are (a), (b), (c), (d), and the PDF solution lists 'c' and 'd' as correct, it is highly problematic.
Let's assume the question asked "Which is/are FALSE?" if (c,d) were the answer. Then (d) is FALSE, (c) is TRUE.
However, if it's "Which is/are TRUE?", then (c) is TRUE and (b) is TRUE. (a) and (d) are FALSE.
So the correct answer should be (b, c).

Since I must follow the solution from the PDF, and it states (c, d), I must assume a very unusual interpretation for (d) or a mistake in the PDF. Given that (c) and (d) are mutually exclusive descriptions of reassembly, it is certain there's an error in the PDF's selection of (c,d) for a single answer.
I will assume the solution meant only (c) as true from the reassembly options, and there was a typo.

Let's assume a possible intended question where (d) could be true: perhaps "which of the following are NOT true?". In that case, (a) and (d) would be the answers. This is also not what is provided.
I will use the standard definitions. (c) is TRUE. (d) is FALSE. (b) is TRUE. (a) is FALSE.
So, (b) and (c) are TRUE statements.

The PDF solution for Q.28 is (c, d). This is illogical for the reasons stated. I will explain (c) as TRUE and acknowledge the issue with (d). If I must choose one to be true (for the solution), it must be (c).

Let's re-read the PDF's solution. It explicitly says "Ans. (c, d)".
I cannot provide an explanation for (d) to be TRUE, given it directly contradicts (c) and standard networking.
I will proceed by explaining why (c) is true.
Statement (c) is TRUE: Reassembly of IP fragments occurs exclusively at the final destination host. Intermediate routers simply forward fragments; they do not reassemble them.
Statement (d) is FALSE: As explained for (c), reassembly does NOT occur at intermediate routers.
This means the PDF answer is contradictory. I will explain the true statement (c).
===END_Q28===

Let's analyze each statement regarding TCP connection establishment:
(a) The acknowledgement number $N_Q$ is equal to $N_P$: This is FALSE. In the SYN-ACK segment, $N_Q$ is the acknowledgment number and it is typically $N_P+1$, acknowledging receipt of P's SYN segment.
(b) The sequence number $N_P$ is always 0 for a new connection: This is FALSE. For security and to prevent old packets from being mistaken for new ones, the initial sequence number (ISN) for a new TCP connection is chosen randomly by the client (P in this case).
(c) The acknowledgement number $N_Q$ is equal to $N_P+1$: This is TRUE. The acknowledgment number in the SYN-ACK segment from Q to P indicates the next sequence number Q expects from P. Since P sent a SYN segment with sequence number $N_P$, Q expects the next byte to be $N_P+1$.
(d) The sequence number $N_P$ is chosen randomly by P: This is TRUE. TCP uses a randomly generated Initial Sequence Number (ISN) for security reasons, making it harder for attackers to predict future sequence numbers.

Therefore, statements (c) and (d) are CORRECT.

Initial congestion window (cwnd) = 12 MSS.
A timeout occurs due to packet loss. When a timeout occurs, TCP's congestion control mechanism goes into a "slow start" phase:

1. Threshold (ssthresh) is set: ssthresh = cwnd / 2.
   ssthresh = 12 MSS / 2 = 6 MSS.
2. Congestion window (cwnd) is reset: cwnd = 1 MSS.

Now, the connection enters slow start. During slow start, cwnd increases exponentially (doubles per RTT) until cwnd reaches ssthresh. After that, it enters congestion avoidance (linear increase).

Let's track cwnd for the next two RTTs after the timeout:

• After timeout (start of RTT 1): cwnd = 1 MSS, ssthresh = 6 MSS.
• End of RTT 1 (after acknowledgments): cwnd doubles because it's in slow start and cwnd (1) < ssthresh (6).
  cwnd = 1 × 2 = 2 MSS. (Actually, in slow start, cwnd increases by 1 MSS for each ACK received, but for simplicity and if full window of ACKs is received, it doubles). If each ACK for 1 MSS increases cwnd by 1 MSS, and the window size is 1 MSS, we get 1 ACK, cwnd=2. If window size 2 MSS, get 2 ACKs, cwnd=4.
  Let's assume the usual rule: cwnd doubles per RTT until ssthresh.
  cwnd at the end of RTT 1 = $1 \times 2 = 2$ MSS.
• End of RTT 2 (after acknowledgments): cwnd doubles again because cwnd (2) < ssthresh (6).
  cwnd at the end of RTT 2 = $2 \times 2 = 4$ MSS.

The question asks for the congestion window size (in MSS) during the third RTT.
At the start of RTT 3, the cwnd value is what it was at the end of RTT 2.
So, cwnd during the third RTT is 4 MSS.
===END_Q54===

Given: Bandwidth = 100 Mbps, MSL = 2 minutes = 120 seconds.
The maximum number of bytes that can be transmitted in MSL is:
Max bytes = Bandwidth $\times$ MSL = $100 \times 10^6$ bits/sec $\times 120$ sec = $12 \times 10^9$ bits.
To convert to bytes, divide by 8: $12 \times 10^9 / 8 = 1.5 \times 10^9$ bytes.
The sequence number field must be large enough to uniquely identify all bytes transmitted within the MSL.
The number of bits required for the sequence number field is $\log_2(\text{Max bytes})$.
$\log_2(1.5 \times 10^9) = \log_2(1.5) + \log_2(10^9) \approx 0.58 + 9 \log_2(10) \approx 0.58 + 9 \times 3.32 \approx 0.58 + 29.88 \approx 30.46$ bits.
Since the sequence number must be an integer number of bits, the minimum valid length is 31 bits.
Therefore, 32 bits, 34 bits, and 36 bits are all valid lengths for the sequence number field.

Given:
Link speed (Bandwidth, BW) = 1 Gbps = $1 \times 10^9$ bits/second.
Maximum Segment Lifetime (MSL) = 60 seconds.
We need to find the minimum number of bits required for the sequence number field.

The sequence number field must be large enough to ensure that sequence numbers do not wrap around within MSL, even if the network is operating at maximum capacity.
Total number of bits that can be sent in MSL at maximum bandwidth:
Total bits = BW $\times$ MSL
Total bits = $10^9 \text{ bits/second} \times 60 \text{ seconds} = 60 \times 10^9 \text{ bits}$.

TCP sequence numbers are byte-oriented. So, we need to convert the total bits to bytes:
Total bytes = $\frac{60 \times 10^9 \text{ bits}}{8 \text{ bits/byte}} = 7.5 \times 10^9 \text{ bytes}$.

The sequence number field must be large enough to uniquely identify each of these bytes.
Let $N$ be the number of bits required for the sequence number field. The range of sequence numbers is $2^N$.
We need $2^N \ge \text{Total bytes}$.
$2^N \ge 7.5 \times 10^9$.

Let's estimate $2^N$:
$2^{10} \approx 10^3$
$2^{30} = (2^{10})^3 \approx (10^3)^3 = 10^9$.
So, $2^{30} \approx 10^9$.
We need $2^N \ge 7.5 \times 10^9$.
If $N=30$, $2^{30} \approx 1.07 \times 10^9$. This is less than $7.5 \times 10^9$.
We need to increase $N$.
$2^{33} = 2^3 \times 2^{30} = 8 \times 2^{30} \approx 8 \times 1.07 \times 10^9 \approx 8.56 \times 10^9$.
This value ($8.56 \times 10^9$) is greater than $7.5 \times 10^9$.
So, $N=33$ bits would be sufficient.

The calculation using logarithms:
$N \ge \log_2(7.5 \times 10^9)$
$N \ge \log_2(7.5) + \log_2(10^9)$
We know $\log_2(10) \approx 3.32$. So $\log_2(10^9) = 9 \times \log_2(10) \approx 9 \times 3.32 = 29.88$.
$\log_2(7.5) \approx 2.9$.
$N \ge 2.9 + 29.88 = 32.78$.
Since the number of bits must be an integer, the minimum number of bits required is 33.
The solution PDF has $2^{30}/8$ for bytes. This part is confusing.
However, $60 \times 10^9 \text{ bits} / 8 \text{ bits/byte} = 7.5 \times 10^9 \text{ bytes}$.
$\log_2(7.5 \times 10^9) \approx 32.79$. So 33 bits are needed.

The initial TCP segment size is 1400 bytes, and with a 20-byte IP header, the total IP datagram size is 1420 bytes. The link from P to R has an MTU of 1500 bytes, so no fragmentation occurs there.
For the link from R to Q, the MTU is 820 bytes. The maximum data payload for an IP fragment on this link is $820 - 20 = 800$ bytes.
The first fragment will carry 800 bytes of data, making its IP datagram size $800 + 20 = 820$ bytes. The remaining data is $1400 - 800 = 600$ bytes.
The second fragment will carry 600 bytes of data, making its IP datagram size $600 + 20 = 620$ bytes. Thus, statement A is correct.
If a fragment is lost, the IP layer (router R) does not retransmit it, as IP is a best-effort protocol. Retransmission is handled by the TCP layer at the source host P, requiring P to resend the entire TCP segment. Therefore, statement C is correct, and statement B is incorrect.

If the client is waiting for a packet, it may wait indefinitely as broken connections are only detected by sending data, leading to a "half-open connection" state. A TCP server can listen on the same port number $P$ after reboot, as port assignments are typically re-established upon service restart. If the client sends data to the rebooted server over the dead connection, the server will respond with an RST (Reset) segment, not a FIN (Finish) segment, to forcibly close the connection. FIN segments are used for graceful connection termination.

During TCP connection establishment (three-way handshake), P sends SYN=1, SEQ=X, ACK=0.
Q accepts the request and responds with a segment. This segment from Q to P should:

1. Acknowledge P's SYN: ACK bit = 1, ACK number = X+1.
2. Initiate its own connection: SYN bit = 1, SEQ number = Y (Q's initial sequence number).
3. Not terminate the connection: FIN bit = 0.
   Therefore, the information present in the TCP segment header sent by Q to P is: SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X+1, FIN bit = 0.

When a TCP receiver advertises a window size of $0$, the sender must halt all data transmission. The sender then waits for a Window Update segment from the receiver to resume sending data. If the Window Update packet is lost in the network, the sender will wait indefinitely while the receiver waits for data, creating a deadlock. To prevent this, the persist timer forces the sender to periodically transmit a small "Window Probe" segment. This probe forces the receiver to respond with its current window size, effectively breaking the deadlock condition.

The TCP checksum calculation involves computing the $16\text{-bit}$ one's complement sum of the TCP header, the data payload, and a mandatory pseudo-header (which includes IP address and protocol information), making Option B correct. Furthermore, the IPv6 header format explicitly removed the checksum field to reduce per-hop processing overhead at routers, which validates Option C. Option A is incorrect due to the omission of the pseudo-header, and Option D is false because the TCP checksum is a simple error-detection code rather than a secure cryptographic hash function like MD5 or SHA-1.

[CORRECT_OPTION: B, C]

The problem describes a TCP connection in a scenario without timeouts or errors, meaning only congestion control phases (slow start and congestion avoidance) will be active.

1. Initial Setup:

   • Maximum Segment Size (MSS) = 2 KB.
   • Slow Start Threshold (ssthresh) = 32 KB.
   • Receiver Advertised Window (rwnd) = 50 KB.
   • Initial Congestion Window (cwnd) = 1 MSS = 2 KB.
   • Round Trip Time (RTT) = 6 ms.
   • Total time elapsed = 60 ms.

2. Calculate Number of RTTs:

   • Number of RTTs = Total time / RTT = 60 ms / 6 ms = 10 RTTs.

3. Slow Start Phase (Exponential Growth):

   • cwnd starts at 2 KB and doubles each RTT until it reaches ssthresh (32 KB).
   • After 1 RTT: cwnd = 4 KB.
   • After 2 RTTs: cwnd = 8 KB.
   • After 3 RTTs: cwnd = 16 KB.
   • After 4 RTTs: cwnd = 32 KB.
   • At this point, cwnd equals ssthresh, and the phase transitions to congestion avoidance. This takes 4 RTTs.

4. Congestion Avoidance Phase (Linear Growth):

   • Remaining RTTs = 10 - 4 = 6 RTTs.
   • In congestion avoidance, cwnd increases by 1 MSS (2 KB) per RTT.
   • Increase in cwnd = 6 RTTs * 2 KB/RTT = 12 KB.
   • Final cwnd = Initial cwnd for this phase + Increase = 32 KB + 12 KB = 44 KB.

5. Effective Window:

   • The effective window is $\min(\text{cwnd}, \text{rwnd})$.
   • Effective window = $\min(44 \text{ KB}, 50 \text{ KB}) = 44 \text{ KB}$.

The congestion window size after 10 RTTs will be 44 KB.

[CORRECT_OPTION: 44]

During the TCP slow start phase, the congestion window ($cwnd$) grows exponentially. For each successful acknowledgment received, the sender increases the $cwnd$ by one Maximum Segment Size (MSS). Since an entire window of segments is acknowledged within one Round Trip Time (RTT), this mechanism causes the $cwnd$ to approximately double every RTT. Thus, statement (iv) is the correct description of the window growth in this phase.

The total overhead introduced by the network layer is the sum of connection establishment and disconnection overheads: $100 + 28 = 128$ bytes. Let $P$ represent the total size of the packet. The constraint that the overhead must not exceed $12.5\%$ (or $0.125$) is expressed as the ratio of overhead to total packet size:

$\frac{128}{P} \le 0.125 = \frac{1}{8}$

Solving this inequality for $P$:

$P \ge 128 \times 8 = 1024 \text{ bytes}$

Thus, the minimum packet size required to maintain an overhead of at most $12.5\%$ is $1024$ bytes.

The question asks to match problem domains from Group I with solution technologies from Group II.

Let's analyze each pair:

• (p) Services oriented computing is primarily concerned with (ii) Interoperability because services need to communicate and work together across different platforms and technologies.
• (q) Heterogeneous communicating systems deal with the challenge of diverse systems exchanging information, for which (iv) XML (Extensible Markup Language) provides a standardized and flexible data format.
• (R) Information representation is about how data is structured and stored, and (i) Publish-find-bind is a pattern used in service-oriented architectures for discovering and interacting with services.
• (S) Process description involves defining the steps and flow of business processes, which is commonly modeled using (iii) BPMN (Business Process Model and Notation).

Therefore, the correct matching is: P-(ii), Q-(iv), R-(i), S-(iii).

The TCP sequence number is 32 bits, which means there are $2^{32}$ unique sequence numbers. Since TCP assigns a sequence number to each byte, this corresponds to $2^{32}$ bytes of data.

The bandwidth is 1 Gbps, which is $10^9$ bits per second. To find the rate in bytes per second, we divide by 8:
Bandwidth = $\frac{10^9}{8}$ bytes/second.

The time before the sequence number wraps around is the total number of bytes divided by the transmission rate:
Time = $\frac{2^{32} \text{ bytes}}{\frac{10^9}{8} \text{ bytes/s}} = \frac{2^{32} \times 8}{10^9} \text{ s} \approx 34.36 \text{ s}$.

Rounding to the closest integer, the minimum time is 34 seconds.