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TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.
Correct answer is between '320,321'. Can you explain this answer?
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TCP opens a connection using an initial sequence number of 3500 and se...
Wrap around time for the first sequence number:
X = 320000 Bytes per sec
X = 320 KB
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TCP opens a connection using an initial sequence number of 3500 and se...
Given information:
- TCP opens a connection using an initial sequence number of 3500.
- TCP sends data at a rate of 5 MBps.
- The other party opens the connection with a sequence number of 1200.
- The wrap around time for both sequence numbers differs by 12562.77 sec.
Calculating the data rate for the second party:
To calculate the data rate for the second party, we need to determine the number of bytes that have been transmitted from the initial sequence number to the sequence number of the other party.
1. Calculate the difference in sequence numbers:
- Sequence number of the other party = 1200
- Initial sequence number = 3500
- Difference in sequence numbers = 1200 - 3500 = -2300
2. Calculate the number of bytes transmitted:
- The wrap around time for both sequence numbers is given as 12562.77 sec.
- The wrap around time is the time taken for the sequence number to wrap around from the maximum value to the initial value.
- The maximum value of the sequence number is 2^32 - 1, which represents the maximum number of bytes that can be transmitted.
- The wrap around time is equal to the difference between the maximum value and the initial value divided by the data rate.
- Wrap around time = (2^32 - 1 - 3500) / (5 * 10^6)
- 12562.77 = (2^32 - 1 - 3500) / (5 * 10^6)
- Solving this equation will give us the value of 2^32 - 1, which represents the number of bytes transmitted.
3. Calculate the number of bytes transmitted from the initial sequence number to the sequence number of the other party:
- Number of bytes transmitted = (2^32 - 1) - (-2300)
4. Convert the number of bytes to KB:
- Data rate (in KB) = Number of bytes transmitted / 1024
Result:
The data rate for the second party is between 320,321 KB.
- TCP opens a connection using an initial sequence number of 3500.
- TCP sends data at a rate of 5 MBps.
- The other party opens the connection with a sequence number of 1200.
- The wrap around time for both sequence numbers differs by 12562.77 sec.
Calculating the data rate for the second party:
To calculate the data rate for the second party, we need to determine the number of bytes that have been transmitted from the initial sequence number to the sequence number of the other party.
1. Calculate the difference in sequence numbers:
- Sequence number of the other party = 1200
- Initial sequence number = 3500
- Difference in sequence numbers = 1200 - 3500 = -2300
2. Calculate the number of bytes transmitted:
- The wrap around time for both sequence numbers is given as 12562.77 sec.
- The wrap around time is the time taken for the sequence number to wrap around from the maximum value to the initial value.
- The maximum value of the sequence number is 2^32 - 1, which represents the maximum number of bytes that can be transmitted.
- The wrap around time is equal to the difference between the maximum value and the initial value divided by the data rate.
- Wrap around time = (2^32 - 1 - 3500) / (5 * 10^6)
- 12562.77 = (2^32 - 1 - 3500) / (5 * 10^6)
- Solving this equation will give us the value of 2^32 - 1, which represents the number of bytes transmitted.
3. Calculate the number of bytes transmitted from the initial sequence number to the sequence number of the other party:
- Number of bytes transmitted = (2^32 - 1) - (-2300)
4. Convert the number of bytes to KB:
- Data rate (in KB) = Number of bytes transmitted / 1024
Result:
The data rate for the second party is between 320,321 KB.
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TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer?
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TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer?.
TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer?.
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TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer?, a detailed solution for TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer? has been provided alongside types of TCP opens a connection using an initial sequence number of 3500 and sends data at 5 MBps. The other party opens the connection with a sequence number of 1200. Wrap around time for both the sequence number differs by 12562.77 sec. Calculate the data rate(in KB) for the second party.Correct answer is between '320,321'. Can you explain this answer? theory, EduRev gives you an
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