Question

a) Discuss in detail the fields of the IPv4 header that deal with the fragmentation of IPv4 packets. Using these fields show how an IPv4 packet containing 4000 bytes of user data is fragmented to be carried using Ethernet frames.

b) The fragmented IPv4 packet in the Ethernet frames must go over a Token Ring Wide-Area Network where the Maximum Transmission Unit (MTU) is 820 bytes. Show how the IPv4 packet will be re-fragmented to go over the Token Ring Network.

c) In order to reach its final destination, the re-fragmented IPv4 packet, after going over the Token Ring, must now go over a new IEEE 802.11p vehicular network with a Maximum Transmission Unit (MTU) of 1024 bytes. Show how the IPv4 packet will go over this new network to reach its final destination.

Answer 1

a) The fields of the IPv4 header that deal with the fragmentation of IPv4 packets is highlighted shown below.

IPv4 packet containing 4000 bytes of user data is fragmented to be carried using Ethernet frames.

there are filled like DF , MF ,Fragment offset ,which deals with fragmentation

IPv4 header:

minimum size is 20 Byte and maximun size is 60 Byte. here we have assume min 20 byte of ip header

20 byte of IP header

4000 bytes of user data

since MTU (maximun transmittable unit) of Network 1 (Token Ring WAN) is 820 byte

MTU (maximun transmittable unit) of Network 2 (IEEE 802.11p vehicular )is 1024 byte

when router R1 receive ip v 4 packet containing 4000 byte of user data.it understant that MTU of Network 1 is 820 byte. so Router R1 divide the packet into 5 fragments .each of size 20 byte header + 800 byte data

800 byte of data must be multiple of 8 according to the Ip v 4header.

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(b) ip v 4 packet containing 4000 byte of user data is fragmented by Router R1 because MTU of Network 1 is 820 Byte. so Router R1 divide the packet into 5 fragments .each of size 20 byte header + 800 byte data

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IF DF=1 means do not fragment the packet   IF DF=0 means fragmentation is allow   

IF MF=1 means more fragment are coming IF MF=0 means No fragmentation is coming

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Fragment1: In the first fragment DF=0 and MF=1 fragment offset field  = 0

Fragment2: In the second fragment DF=0 and MF=1 fragment offset field  = 100 (means 100 * 8 = 800 bytes are ahead of this fragment)

Fragment3: In the third fragment DF=0 and MF=1 fragment offset field  = 200 (means 200 * 8 = 1600 bytes are ahead of this fragment)

Fragment4: In the fourth fragment DF=0 and MF=1 fragment offset field  = 300 (means 300 * 8 = 2400 bytes are ahead of this fragment)

Fragment5: In the five fragment DF=0 and MF=0 fragment offset field = 400(means 400 * 8 = 3200 bytes are ahead of this fragment)

----------------in this way 3200 bytes are ahead of this fragment + current fragement 800 byte become ==4000 bytes

20 byte of IP header

800 bytes data

like that 5 fragments

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(c) Since MTU of Network 2 is 1024 byte. so the fragments coming from network 1 need not be fragmented as they can easily pass on the second network.

so here in Network 2 when router R2 receive fragment 1 it will pass it directly as fragment size is just 820 byte and capacity of network 2 is more ie 1024 byte.

here all fragment will be pass as packets on second network.

Fragment1: directly pass as  packet with   DF=1 and MF=0 fragment offset field  = 0 // means do not fragment the packet

Fragment2: directly pass as  packet with   DF=1 and MF=0 fragment offset field  = 0

Fragment3: directly pass as  packet with   DF=1 and MF=0 fragment offset field  = 0

Fragment4: directly pass as  packet with DF=1 and MF=0 fragment offset field  = 0

Fragment5:directly pass as  packet with    DF=1 and MF=0   fragment offset field  = 0

this all 05 packet will be reach to deatination.

KQ0-607 1P14 Header Data / PO Byte 4000 Byle El min 20 Byte VER(U) Hlenlu) Servicetype(8) Total lengthakl 6 ) Identificahon No.(16) fragment offset (13 bija) TTL (8) bits protocol field T (8) bits Source IP Address (32) bit Destionation ip address (32) bits k ophons & Padding ( 40' Bules)

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