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Consider transferring an enormous file of \(L\) bytes from Host A to Host B. Assume an MSS of 536 bytes. a. What is the maximum value of \(L\) such that TCP sequence numbers are not exhausted? Recall that the TCP sequence number field has 4 bytes. b. For the \(L\) you obtain in (a), find how long it takes to transmit the file. Assume that a total of 66 bytes of transport, network, and data-link header are added to each segment before the resulting packet is sent out over a \(155 \mathrm{Mbps}\) link. Ignore flow control and congestion control so A can pump

Short Answer

Expert verified
a. Maximum \( L = 2^{32} \times 536 \) bytes; b. Time = \( \frac{L \times 602/536}{155 \times 10^6/8} \) seconds.

Step by step solution

01

Understanding TCP Sequence Number Limit

The TCP sequence number is 4 bytes, meaning there are a maximum of \( 2^{32} \) possible sequence numbers. This is the limit for the unique range of sequence numbers TCP can use before needing to wrap around.
02

Calculate Maximum File Size Without Sequence Number Exhaustion

Given that the Maximum Segment Size (MSS) is 536 bytes, we want to find the maximum file size \( L \) that can be transferred without exhausting sequence numbers. The maximum number of segments is \( 2^{32} \), so multiply by MSS to get: \[ L = 2^{32} \times 536 \text{ bytes} \]
03

Calculate Maximum File Size

Compute \[ L = 2^{32} \times 536 \] bytes. This gives the maximum size of the file that can be transferred without reusing a sequence number.
04

Calculate Total Data Size Including Headers Per Segment

For each 536-byte data segment (MSS), there are additional 66 bytes of headers, making the total per segment: \[ 536 + 66 = 602 \text{ bytes per packet} \]
05

Calculate Total Transfer Time for File

The transmission rate is 155 Mbps. Convert this to bytes per second: \[ 155 imes 10^6 ext{ bits/sec} = \frac{155 imes 10^6}{8} ext{ bytes/sec} \] Calculate the time to send the entire file size including headers: \[ \text{Time} = \frac{L \times \frac{602}{536}}{\frac{155 imes 10^6}{8}} \]
06

Solve for Transfer Time

Substitute in the values calculated from previous steps to find the actual transfer time required.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Maximum Segment Size (MSS)
The Maximum Segment Size (MSS) is a crucial concept in the Transmission Control Protocol (TCP) used in networking for defining the largest segment of data that can be sent in a TCP packet. Think of MSS as the chunk size for transmission over the network. For the problem we're discussing, the MSS is specified as 536 bytes. This size is chosen to optimize transmission efficiency while ensuring compatibility across various networks.
  • Why MSS is 536 bytes: This value is often used because it prevents IP fragmentation when accounting for typical TCP/IP header sizes.
  • MSS vs MTU: The MSS is slightly smaller than the Maximum Transmission Unit (MTU), which includes headers. MSS specifically refers to the payload maximum.
Choosing the right MSS can significantly affect network performance, helping to avoid issues like fragmentation and inefficient data transmission. By understanding and utilizing MSS appropriately, data streams can be more efficiently managed, minimizing delays and maximizing throughput.
TCP sequence number field
TCP sequence numbers play a critical role in data transmission, ensuring that packets arrive in the correct order. The TCP sequence number field is 4 bytes long, which provides each number with a range from 0 to \( 2^{32} - 1 \). This capacity means that there are roughly 4.29 billion unique sequence numbers available.
  • Purpose: Sequence numbers identify the byte order, helping the receiving host correctly reassemble data segments in the original sequence.
  • Wrapping around: When sequence numbers reach their maximum, they wrap around to 0, continuing sequentially. This wrap-around process is known as sequence number rollover.
For large data transfers similar to our exercise, understanding sequence number capacity is vital to prevent issues related to number exhaustion. Managing sequence numbers efficiently ensures the sender and receiver remain synchronized, maintaining communication integrity.
File transfer calculations
Calculating the time it takes to transfer a file is essential for understanding network efficiency. In our scenario, we consider the MSS of 536 bytes with additional headers, leading to a total packet size of 602 bytes. The goal here is to estimate the time required to transfer an enormous file over a network capable of 155 Mbps.
  • Conversion to bytes: First, convert the transfer rate from bits to bytes by dividing by 8, resulting in approximately 19.375 million bytes per second.
  • Information included in headers: The 66 additional bytes serve as headers, encapsulating data essential for the integrity and routing of the packets.
  • Calculating total time: Using the formula \(\text{Time} = \frac{L \times \frac{602}{536}}{\frac{155 \times 10^6}{8}}\), you can calculate how long it takes to complete the file transfer, factoring in the header overhead.
This calculation is critical for network planning and resource allocation, as it ensures that data can be sent in a timely manner while also being robust against possible data transmission issues like congestion or data loss.

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Most popular questions from this chapter

Compare GBN, SR, and TCP (no delayed ACK). Assume that the timeout values for all three protocols are sufficiently long such that 5 consecutive data segments and their corresponding ACKs can be received (if not lost in the channel) by the receiving host (Host B) and the sending host (Host A) respectively. Suppose Host A sends 5 data segments to Host B, and the 2 nd segment (sent from \(\mathrm{A}\) ) is lost. In the end, all 5 data segments have been correctly received by Host B. a. How many segments has Host A sent in total and how many ACKs has Host B sent in total? What are their sequence numbers? Answer this question for all three protocols. b. If the timeout values for all three protocol are much longer than 5 RTT, then which protocol successfully delivers all five data segments in shortest time interval?

In Section 3.5.4, we saw that TCP waits until it has received three duplicate ACKs before performing a fast retransmit. Why do you think the TCP designers chose not to perform a fast retransmit after the first duplicate ACK for a segment is received?

Consider a modification to TCP's congestion control algorithm. Instead of additive increase, we can use multiplicative increase. A TCP sender increases its window size by a small positive constant \(a(0

In our rdt protocols, why did we need to introduce sequence numbers?

Suppose two TCP connections are present over some bottleneck link of rate \(R\) bps. Both connections have a huge file to send (in the same direction over the bottleneck link). The transmissions of the files start at the same time. What transmission rate would TCP like to give to each of the connections?

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