EE450-Discussion12-Fall10

EE450-Discussion12-Fall10 - Discussion #12 EE450,...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Discussion #12 EE450, 11/10/2010 -IP Fragmentation - Addressing and  Subnetting
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 Problem#1: Description Suppose that a TCP message that contains 2048 bytes of data and 20 bytes of TCP header is passed to IP for delivery across two networks of the Internet (i.e. from the source host to a router to the destination host ). The first network uses 14-byte headers and has an MTU of 1024 bytes, the second uses 8-byte headers with an MTU of 512 bytes. Each network’s MTU gives the total packet size that may be sent including the network header.
Background image of page 2
3 Problem#1: Question Give the sizes of the fragments delivered to the network layer at the destination host, assuming all IP headers are 20 bytes. S R D MTU=1024 (Header=14) MTU=512 (Header=8) TCP Data=2048 TCP Header=20 MAC IP TCP IP Header=20 MTU
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
4 Problem#1: Solution Across the first network: Packets have room for 1024-14-20=990 bytes of IP- level data. Since 990 is not multiple of 8, each fragment can contain at most 8 x 990/8 = 984 bytes. We need to transfer 2048 +20 bytes of such data. This would be fragmented into fragments of size 984,984 and 100 bytes. For Packet sizes across the first network , add 14+20 bytes to each fragment.
Background image of page 4
5 Problem#1: Solution Across the second network: Packets have room for 512-8-20=484 bytes of IP-level data. So the 100-byte packet will not be fragmented but the other two 984-byte packets will be fragmented. Since 484 is not multiple of 8, each fragment can contain at most 8 x 484/8 = 480 bytes. Each 984-byte fragment would be fragmented into fragments of size 480,480 and 24 bytes. For Packet sizes across the second network, add 8+20 bytes to each fragment.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
6 Problem#2: Description We have an IP datagram of size 1420 bytes (including the IP header) traversing the sequence of physical networks with different MTUs as follow: Ethernet : MTU = 1500 bytes FDDI : MTU = 4500 bytes PPP : MTU = 532 bytes R1 R2 Ethernet FDDI R3 A B Ethernet PPP
Background image of page 6
7 Problem#2: Question Show the unfragmented datagram as well as the fragmented datagrams and clearly identify the Header fields used in IP fragmentation. MAC IP IP Header=20, Payload=1400 bytes MTUs: 1500, 4500, 532
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
8 Problem#2: Solution The datagram will not be fragmented across the first two networks. Since the MTU of the PPP network is less than the size of the datagram (without the IP header), it will be fragmented to 3 datagrams. The fragments of the original datagram travel across the 4 th network with no further fragmentation and they will reassembled at the endpoint. R1 R2 Ethernet FDDI R3 A B Ethernet PPP     ETH IP   (1400)                                  FDDI IP   (1400)                                PPP IP   (512)                                PPP IP   (512)                                PPP IP   (376)                  
Background image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/25/2011 for the course EE 450 taught by Professor Zahid during the Fall '06 term at USC.

Page1 / 29

EE450-Discussion12-Fall10 - Discussion #12 EE450,...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online