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Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?
- a)Number of fragments transmitted = 11
- b)Offset field in the last fragment = 2002
- c)Number of fragments transmitted = 12
- d)Offset field in the last fragment = 1820
Correct answer is option 'B,C'. Can you explain this answer?
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Most Upvoted Answer
Consider host A sends a UDP datagram containing 16060 bytes of user da...
Given Information:
- Host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN.
- Ethernet MTU is 1500 bytes.
- Size of the IP header is 40 bytes.
- Size of the UDP header is 8 bytes.
Calculating the Total Datagram Size:
To calculate the total size of the datagram, we need to consider the UDP user data, UDP header, and IP header.
UDP user data size = 16060 bytes
UDP header size = 8 bytes
IP header size = 40 bytes
Total Datagram Size = UDP user data size + UDP header size + IP header size
Total Datagram Size = 16060 + 8 + 40 = 16108 bytes
Calculating the Number of Fragments:
To determine the number of fragments, we need to divide the total datagram size by the MTU.
Number of fragments = Total Datagram Size / MTU
Number of fragments = 16108 / 1500 = 10.73867
Since the number of fragments cannot be in decimal, we round it up to the nearest whole number.
Number of fragments = 11
Calculating the Offset Field in the Last Fragment:
To calculate the offset field in the last fragment, we need to consider the size of the IP header and the UDP header.
Offset field in the last fragment = (Number of fragments - 1) * (MTU - IP header size) + UDP header size
Offset field in the last fragment = (11 - 1) * (1500 - 40) + 8
Offset field in the last fragment = 10 * 1460 + 8
Offset field in the last fragment = 14608 + 8
Offset field in the last fragment = 14616 bytes
Conclusion:
Based on the calculations, the correct statements are:
- Number of fragments transmitted = 11
- Offset field in the last fragment = 14616 bytes
- Host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN.
- Ethernet MTU is 1500 bytes.
- Size of the IP header is 40 bytes.
- Size of the UDP header is 8 bytes.
Calculating the Total Datagram Size:
To calculate the total size of the datagram, we need to consider the UDP user data, UDP header, and IP header.
UDP user data size = 16060 bytes
UDP header size = 8 bytes
IP header size = 40 bytes
Total Datagram Size = UDP user data size + UDP header size + IP header size
Total Datagram Size = 16060 + 8 + 40 = 16108 bytes
Calculating the Number of Fragments:
To determine the number of fragments, we need to divide the total datagram size by the MTU.
Number of fragments = Total Datagram Size / MTU
Number of fragments = 16108 / 1500 = 10.73867
Since the number of fragments cannot be in decimal, we round it up to the nearest whole number.
Number of fragments = 11
Calculating the Offset Field in the Last Fragment:
To calculate the offset field in the last fragment, we need to consider the size of the IP header and the UDP header.
Offset field in the last fragment = (Number of fragments - 1) * (MTU - IP header size) + UDP header size
Offset field in the last fragment = (11 - 1) * (1500 - 40) + 8
Offset field in the last fragment = 10 * 1460 + 8
Offset field in the last fragment = 14608 + 8
Offset field in the last fragment = 14616 bytes
Conclusion:
Based on the calculations, the correct statements are:
- Number of fragments transmitted = 11
- Offset field in the last fragment = 14616 bytes
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Community Answer
Consider host A sends a UDP datagram containing 16060 bytes of user da...
Data =16068 bytes
MTU =1500 bytes
Data + header =1500
⇒ Data =1500 − 40
∴ Data = 1460 bytes
Number of fragments = 16068/1460 = 12
Since initial fragment offset is not given assume it to be 0.
1460 in not divisible by remove 4 byte and add to last fragment
Fragment offset of 2nd fragment,
=1456/8
= 182
Fragment offset of 3rd fragment,
= 364
Fragment offset of 4th fragment,
= 546
Fragment offset of 12th fragment,
= 2002
As we know,
Since there are 12 fragments,
Fragment offset of 12th fragment,
= 2002
Hence, the correct options are (B) and (C).
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Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?a)Number of fragments transmitted = 11b)Offset field in the last fragment = 2002c)Number of fragments transmitted = 12d)Offset field in the last fragment = 1820Correct answer is option 'B,C'. Can you explain this answer?
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Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?a)Number of fragments transmitted = 11b)Offset field in the last fragment = 2002c)Number of fragments transmitted = 12d)Offset field in the last fragment = 1820Correct answer is option 'B,C'. 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 Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?a)Number of fragments transmitted = 11b)Offset field in the last fragment = 2002c)Number of fragments transmitted = 12d)Offset field in the last fragment = 1820Correct answer is option 'B,C'. 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 Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?a)Number of fragments transmitted = 11b)Offset field in the last fragment = 2002c)Number of fragments transmitted = 12d)Offset field in the last fragment = 1820Correct answer is option 'B,C'. Can you explain this answer?.
Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?a)Number of fragments transmitted = 11b)Offset field in the last fragment = 2002c)Number of fragments transmitted = 12d)Offset field in the last fragment = 1820Correct answer is option 'B,C'. 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 Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?a)Number of fragments transmitted = 11b)Offset field in the last fragment = 2002c)Number of fragments transmitted = 12d)Offset field in the last fragment = 1820Correct answer is option 'B,C'. 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 Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?a)Number of fragments transmitted = 11b)Offset field in the last fragment = 2002c)Number of fragments transmitted = 12d)Offset field in the last fragment = 1820Correct answer is option 'B,C'. Can you explain this answer?.
Solutions for Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?a)Number of fragments transmitted = 11b)Offset field in the last fragment = 2002c)Number of fragments transmitted = 12d)Offset field in the last fragment = 1820Correct answer is option 'B,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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ample number of questions to practice Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?a)Number of fragments transmitted = 11b)Offset field in the last fragment = 2002c)Number of fragments transmitted = 12d)Offset field in the last fragment = 1820Correct answer is option 'B,C'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
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