GATE CSE Computer network fragmentation
| Q1➡ | GATE 2021 Set-1 Consider two hosts P and Q connected through a router R. The maximum transfer unit (MTU) value of the link between P and R is 1500 bytes, and between R and Q is 820 bytes. A TCP segment of size 1400 bytes was transferred from P to Q through R, with IP identification value as 0x1234. Assume that the IP header size is 20 bytes. Further, the packet is allowed to be fragmented, i.e., Don’t Fragment (DF) flag in the IP header is not set by P. Which of the following statements is/are correct? |
| i ➥ If the second fragment is lost, R will resend the fragment with the IP identification value 0x1234. |
| ii ➥ TCP destination port can be determined by analysing only the second fragment. |
| iii ➥ Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes. |
| iv ➥ If the second fragment is lost, P is required to resend the whole TCP segment. |
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Answer: III, IV
Explanation:
Given,
The maximum transfer unit (MTU) value of the link between P and R = 1500 bytes
The maximum transfer unit (MTU) value of the link between R and Q = 820 bytes.
Segment of size = 1400 bytes
IP identification value = 0x1234
IP header size = 20 bytes
Don’t Fragment (DF) flag value = 0
Logic
Since Segment size is 1400 B and MTU value of the link between P and R is 1500 bytes. So, Segment is successfully arrive at router R. MTU value of the link between R and Q is 820 bytes. Now here at router R, Fragmentation is required.
Segment size 1400 Byte is divided into two parts or two fragment
1400 = 800 + 600
1st fragment size = Data + Header = 800 + 20 = 820 Bytes
2nd fragment size = Data + Header = 600 + 20 = 620 Bytes
Option(I): If the second fragment is lost, R will resend the fragment with the IP identification value 0x1234.(False)
If second or any fragment is lost, then whole TCP segment or entire fragment has to be sent again. If any fragment is lost, the sender need to retransmit, But sender does not know about any fragment. Because fragmentation is done at intermediate router. So, sender has to retransmit entire fragment or whole TCP segment and for the next time in fragmentation, IP identification value will be changed.
Option(II): TCP destination port can be determined by analysing only the second fragment.(False)
Each fragment of TCP Segement is send seperately.Each fragment contains TCP destination port. So, TCP destination port can be determined by analysing any fragment.
Option(III): Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes.(True)
1st fragment size = Data + Header = 800 + 20 = 820 Bytes
2nd fragment size = Data + Header = 600 + 20 = 620 Bytes
Option(IV): If the second fragment is lost, P is required to resend the whole TCP segment. (True)
If second or any fragment is lost, then whole TCP segment or entire fragment has to be sent again.
So, Option(III) and Option(IV)are correct.
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| Q2➡ | GATE 2018 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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Answer: 144
Explanation:
Given,
IP Packet Length = 4500 Bytes
IP Header = 20 Bytes
MTU = 600 Bytes
Data,
IP Packet = IP Payload + IP Header
IP Payload = IP Packet – IP Header
IP Payload = 4500 – 20 = 4480 Bytes
MTU Payload = 600 – 20 = 580 Bytes
MTU Payload should be multiple of 8 But 580 is not multiple of 8. So, the nearest value of 580 which is multiple of 8 is 576.
Calculation,
The fragmentation offset value stored in the third fragment is 144.
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| Q3➡ | GATE 2016 Set-1 An IP datagram of size 1000 bytes arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is 100 bytes. Assume that the size of the IP header is 20 bytes. The number of fragments that the IP datagram will be divided into for transmission is __________. |
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Answer: 13
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| Q4➡ | GATE 2015 Set-2 Host A sends a UDP datagram containing 8880 bytes of user data to host B over an Ethernet LAN. Ethernet frames may carry data up to 1500 bytes (i.e. MTU = 1500 bytes). Size of UDP header is 8 bytes and size of IP heard is 20 bytes. There is no option field in IP header. How many total number of IP fragments will be transmitted and what will be the contents of offset field in the last fragment? |
| i ➥ 6 and 925 |
| ii ➥ 6 and 7400 |
| iii ➥ 7 and 1110 |
| iv ➥ 7 and 8880 |
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Answer: III
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| Q5➡ | GATE 2014 Set-3 An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are |
| i ➥ MF bit: 0, Datagram Length: 1444; Offset: 370 |
| ii ➥ MF bit: 1, Datagram Length: 1424; Offset: 185 |
| iii ➥ MF bit: 1, Datagram Length: 1500; Offset: 370 |
| iv ➥ MF bit: 0, Datagram Length: 1424; Offset: 2960 |
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Answer: I
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GATE CSE Computer network fragmentation
| Q6➡ | GATE 2013 In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are |
| i ➥ Last fragment, 2400 and 2789 |
| ii ➥ First fragment, 2400 and 2759 |
| iii ➥ Last fragment, 2400 and 2759 |
| iv ➥ Middle fragment, 300 and 689 |
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Answer: III
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