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A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overload. Assume that IP overhead per packet is 26 bytes. What is the total IP overhead (in bytes) in the network for this transmission?
- a)156 bytes
- b)52 bytes
- c)208 bytes
- d)26 bytes
Correct answer is option 'C'. Can you explain this answer?
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A TCP message consisting of 2100 bytes is passed to IP for delivery ac...
To determine the total IP overhead in the network for this transmission, we need to consider the number of frames required to transmit the TCP message across both networks and calculate the IP overhead for each frame.
1. Calculate the number of frames required for the first network:
- Maximum payload per frame in the first network = 1200 bytes
- TCP message size = 2100 bytes
- Number of frames required in the first network = ceil(2100 / 1200) = 2 frames
2. Calculate the number of frames required for the second network:
- Maximum payload per frame in the second network = 400 bytes
- Number of frames required in the second network = ceil((2100 + IP overhead per frame) / 400)
3. Calculate the IP overhead for each frame in the first network:
- IP overhead per packet = 26 bytes
- IP overhead per frame in the first network = IP overhead per packet = 26 bytes
4. Calculate the IP overhead for each frame in the second network:
- IP overhead per packet = 26 bytes
- IP overhead per frame in the second network = IP overhead per packet = 26 bytes
5. Calculate the total IP overhead in the network:
- Total IP overhead = (IP overhead per frame in the first network * number of frames in the first network) + (IP overhead per frame in the second network * number of frames in the second network)
- Total IP overhead = (26 bytes * 2 frames) + (26 bytes * (ceil((2100 + 26) / 400)))
Simplifying the equation:
- Total IP overhead = 52 bytes + 26 bytes * ceil(2126 / 400)
- Total IP overhead = 52 bytes + 26 bytes * 6
- Total IP overhead = 52 bytes + 156 bytes
- Total IP overhead = 208 bytes
Therefore, the total IP overhead in the network for this transmission is 208 bytes, which corresponds to option 'C'.
1. Calculate the number of frames required for the first network:
- Maximum payload per frame in the first network = 1200 bytes
- TCP message size = 2100 bytes
- Number of frames required in the first network = ceil(2100 / 1200) = 2 frames
2. Calculate the number of frames required for the second network:
- Maximum payload per frame in the second network = 400 bytes
- Number of frames required in the second network = ceil((2100 + IP overhead per frame) / 400)
3. Calculate the IP overhead for each frame in the first network:
- IP overhead per packet = 26 bytes
- IP overhead per frame in the first network = IP overhead per packet = 26 bytes
4. Calculate the IP overhead for each frame in the second network:
- IP overhead per packet = 26 bytes
- IP overhead per frame in the second network = IP overhead per packet = 26 bytes
5. Calculate the total IP overhead in the network:
- Total IP overhead = (IP overhead per frame in the first network * number of frames in the first network) + (IP overhead per frame in the second network * number of frames in the second network)
- Total IP overhead = (26 bytes * 2 frames) + (26 bytes * (ceil((2100 + 26) / 400)))
Simplifying the equation:
- Total IP overhead = 52 bytes + 26 bytes * ceil(2126 / 400)
- Total IP overhead = 52 bytes + 26 bytes * 6
- Total IP overhead = 52 bytes + 156 bytes
- Total IP overhead = 208 bytes
Therefore, the total IP overhead in the network for this transmission is 208 bytes, which corresponds to option 'C'.
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A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overload. Assume that IP overhead per packet is 26 bytes. What is the total IP overhead (in bytes) in the network for this transmission?a)156 bytesb)52 bytesc)208 bytesd)26 bytesCorrect answer is option 'C'. Can you explain this answer?
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A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overload. Assume that IP overhead per packet is 26 bytes. What is the total IP overhead (in bytes) in the network for this transmission?a)156 bytesb)52 bytesc)208 bytesd)26 bytesCorrect answer is option '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 A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overload. Assume that IP overhead per packet is 26 bytes. What is the total IP overhead (in bytes) in the network for this transmission?a)156 bytesb)52 bytesc)208 bytesd)26 bytesCorrect answer is option '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 A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overload. Assume that IP overhead per packet is 26 bytes. What is the total IP overhead (in bytes) in the network for this transmission?a)156 bytesb)52 bytesc)208 bytesd)26 bytesCorrect answer is option 'C'. Can you explain this answer?.
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