GATE 2022
December 5, 2023UGC NET CS 2014 June-paper-2
December 5, 2023IPv4-and-Fragmentation
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Question 4
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Consider an IP packet with a length of 4,500 bytes that includes a 20-byte IPv4 header and a 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600 bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20 bytes. Assume that the fragmentation offset value stored in the first fragment is 0.
The fragmentation offset value stored in the third fragment is __________.
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144
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145
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146
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147
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Question 4 Explanation:
MTU = 600 bytes, IP header = 20 bytes
Therefore Payload = 600 – 20 = 580 bytes.
As we know fragment size should be multiple of 8 but 580 bytes is not a multiple of 8, therefore fragment size is 576 bytes.
Offset value of kth fragment = Fragment size *( kth fragment – 1) / scaling factor
Offset value of third fragment = 576 * (3-1) / 8 = 144
Therefore Payload = 600 – 20 = 580 bytes.
As we know fragment size should be multiple of 8 but 580 bytes is not a multiple of 8, therefore fragment size is 576 bytes.
Offset value of kth fragment = Fragment size *( kth fragment – 1) / scaling factor
Offset value of third fragment = 576 * (3-1) / 8 = 144
Correct Answer: A
Question 4 Explanation:
MTU = 600 bytes, IP header = 20 bytes
Therefore Payload = 600 – 20 = 580 bytes.
As we know fragment size should be multiple of 8 but 580 bytes is not a multiple of 8, therefore fragment size is 576 bytes.
Offset value of kth fragment = Fragment size *( kth fragment – 1) / scaling factor
Offset value of third fragment = 576 * (3-1) / 8 = 144
Therefore Payload = 600 – 20 = 580 bytes.
As we know fragment size should be multiple of 8 but 580 bytes is not a multiple of 8, therefore fragment size is 576 bytes.
Offset value of kth fragment = Fragment size *( kth fragment – 1) / scaling factor
Offset value of third fragment = 576 * (3-1) / 8 = 144