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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 amaximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame,excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the secondnetwork for this transmission?
- a)40 bytes
- b)80 bytes
- c)120 bytes
- d)160 bytes
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A TCP message consisting of 2100 bytes is passed to IP for delivery ac...
Answer is C).
Question says TCP message is of size = 2100 B, hence header is encapsulated with the frame and send to IP layer, and IP
layer when receives this message thinks it as complete data to be send .
To be send across the first network, It is fragmented into 2 payloads of sizes 1200 and 900 .
No need to add IP Header size with 2100B, as fragmentation acts on data/Payload and not the packet.
Now, since, IP thinks 2100B as the data so, it will directly fragment it and hence, no need to add any header.
Payload / Fragment 1: 1200 B {1200 / 8 = 150, Except last fragment all the fragments should be divisible by 8}
Payload 2: 900 B
At the second network, 1200 B is fragmented into 400 B,400 B and 400 B .
Similarly, 900 B is fragmented into 400, 400 and 100 B.
Since, 2100 B is fragmented into 6 data payloads and Header is attached to all the payloads when they will be forwared to
DLL layer. So, Total Overhead = 6 * 20 = 120 B
Hence, DLL layer will receive total Data = 2100 + 120 = 2220 B and DLL will treat it as a complete data to be sent and
DLL header will be encapsulated in DLL frame.
Question says TCP message is of size = 2100 B, hence header is encapsulated with the frame and send to IP layer, and IP
layer when receives this message thinks it as complete data to be send .
To be send across the first network, It is fragmented into 2 payloads of sizes 1200 and 900 .
No need to add IP Header size with 2100B, as fragmentation acts on data/Payload and not the packet.
Now, since, IP thinks 2100B as the data so, it will directly fragment it and hence, no need to add any header.
Payload / Fragment 1: 1200 B {1200 / 8 = 150, Except last fragment all the fragments should be divisible by 8}
Payload 2: 900 B
At the second network, 1200 B is fragmented into 400 B,400 B and 400 B .
Similarly, 900 B is fragmented into 400, 400 and 100 B.
Since, 2100 B is fragmented into 6 data payloads and Header is attached to all the payloads when they will be forwared to
DLL layer. So, Total Overhead = 6 * 20 = 120 B
Hence, DLL layer will receive total Data = 2100 + 120 = 2220 B and DLL will treat it as a complete data to be sent and
DLL header will be encapsulated in DLL frame.
Most Upvoted Answer
A TCP message consisting of 2100 bytes is passed to IP for delivery ac...
Calculation:
To determine the total IP overhead in the second network for this transmission, we need to consider the maximum payload per frame for both networks and the IP overhead per packet.
Given:
TCP message size = 2100 bytes
Maximum payload per frame in the first network = 1200 bytes
Maximum payload per frame in the second network = 400 bytes
IP overhead per packet = 20 bytes
Calculation for the first network:
In the first network, the maximum payload per frame is 1200 bytes. Since the TCP message size is 2100 bytes, it will require multiple frames to transmit the entire message.
Number of frames required in the first network = ceil(2100 / 1200) = 2
Calculation for the second network:
In the second network, the maximum payload per frame is 400 bytes. Again, multiple frames will be required to transmit the remaining part of the TCP message after the first network.
Remaining payload after the first network = 2100 - (1200 * 1) = 900 bytes
Number of frames required in the second network = ceil(900 / 400) = 3
Total IP overhead:
Now, we need to calculate the total IP overhead, considering the IP overhead per packet.
Total IP overhead in the first network = 2 * (20 bytes) = 40 bytes
Total IP overhead in the second network = 3 * (20 bytes) = 60 bytes
Final Answer:
The total IP overhead in the second network for this transmission is 60 bytes. However, based on the given options, the correct answer is option 'C', which states 120 bytes. This appears to be an error because the actual calculation yields 60 bytes.
To determine the total IP overhead in the second network for this transmission, we need to consider the maximum payload per frame for both networks and the IP overhead per packet.
Given:
TCP message size = 2100 bytes
Maximum payload per frame in the first network = 1200 bytes
Maximum payload per frame in the second network = 400 bytes
IP overhead per packet = 20 bytes
Calculation for the first network:
In the first network, the maximum payload per frame is 1200 bytes. Since the TCP message size is 2100 bytes, it will require multiple frames to transmit the entire message.
Number of frames required in the first network = ceil(2100 / 1200) = 2
Calculation for the second network:
In the second network, the maximum payload per frame is 400 bytes. Again, multiple frames will be required to transmit the remaining part of the TCP message after the first network.
Remaining payload after the first network = 2100 - (1200 * 1) = 900 bytes
Number of frames required in the second network = ceil(900 / 400) = 3
Total IP overhead:
Now, we need to calculate the total IP overhead, considering the IP overhead per packet.
Total IP overhead in the first network = 2 * (20 bytes) = 40 bytes
Total IP overhead in the second network = 3 * (20 bytes) = 60 bytes
Final Answer:
The total IP overhead in the second network for this transmission is 60 bytes. However, based on the given options, the correct answer is option 'C', which states 120 bytes. This appears to be an error because the actual calculation yields 60 bytes.
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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 amaximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame,excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the secondnetwork for this transmission?a)40 bytesb)80 bytesc)120 bytesd)160 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 amaximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame,excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the secondnetwork for this transmission?a)40 bytesb)80 bytesc)120 bytesd)160 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 amaximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame,excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the secondnetwork for this transmission?a)40 bytesb)80 bytesc)120 bytesd)160 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 amaximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame,excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the secondnetwork for this transmission?a)40 bytesb)80 bytesc)120 bytesd)160 bytesCorrect answer is option 'C'. Can you explain this answer?.
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