TCP IP Illustrated

Chapter 11 111 since there are 8 additional bytes of

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 04.0, the other networks to which gateway is directly connected. 10.2 Sixty seconds will pass before the 25 routes advertised in the lost datagram are updated. This isn't a problem because RIP normally requires 3 minutes without an update before it declares a route dead. 10.3 RIP runs on top of UDP, and UDP provides an optional checksum for the data portion of the UDP datagram (Section 11.3). OSPF, however, runs on top of IP. The IP checksum covers only the IP header, so OSPF must add its own checksum field. 10.4 Load balancing increases the chances of packets being delivered out of order, and possibly distorts the round-trip times calculated by the transport layer. 10.5 This is called simple split horizon. 10.6 In Figure 12.1 we show that each of the 100 hosts processes the broadcast UDP datagram through the device driver, IP layer, and UDP layer, where it'll finally be discarded when it's discovered that UDP port 520 is not in use. Chapter 11 11.1 Since there are 8 additional bytes of header when IEEE 802 encapsulation is used, 1465 bytes of user data is the smallest size that causes fragmentation. 11.3 There are 8200 bytes of data for IP to send, the 8192 bytes of user data and the 8byte UDP header. Using the tcpdump notation, the first fragment is 1480@0+ (1480 bytes of data, offset of 0, with the "more fragments" bit set). The second is 1480@1480+, the third is 1480@2960+, the fourth is 1480@4440+, the fifth is 1480@5920+, and the file:///D|/Documents%20and%20Settings/bigini/Docu...homenet2run/tcpip/tcp-ip-illustrated/append_d.htm (7 of 20) [12/09/2001 14.48.04] Appendix D: Solutions to Selected Exercises sixth is 800@7400. 1480 x 5+ 800 = 8200, which is the number of bytes to send. 11.4 Each 1480-byte fragment is divided into three pieces: two 528-byte fragments and one 424-byte fragment. The largest multiple of 8 less than 532 (552 - 20) is 528. The 800byte fragment is divided into two pieces: a 528-byte fragment and a 272-byte fragment. Thus, the original 8192-byte datagr...
View Full Document

Ask a homework question - tutors are online