1. Final Examination of Internet Communication Protocol

2. Problem one Consider a transmission from your computer (140.127.114.103; 00-50-18-0E-7C-84) to a destination host in purdue University (128.10.2.1). A TCP end-to-end connection will be established between these hosts. Assume the ARP cache in your computer is empty, hence an ARP request should be sent. Please describe the frames of the protocol operations for establishing a TCP connection. 35%

3. Answer for Problem one (1) your computer (140.127.114.103; 00-50-18-0E-7C-84) destination host in purdue University (128.10.2.1). –DNS (Domain Name Server) request and reply –ARP request and reply (ARP cache in your computer is empty) –TCP three way handshake to establish a TCP connection 140.127.114.103 -> 128.10.2.1

4. Answer for Problem one (2) PreambleDASATypeDataCRC 8 octets6 octets 2 octets64-1500 octets4 octets ARP Request Hardware TypeProtocol Type HlenPlenOperation Sender HA (Octets 0-3) Sender HA (Octets 4-5)Sender IP (Octets 0-1) Sender IP (Octets 2-3)Target HA (Octets 0-1) Target HA (Octets 2-5) Target IP (Octets 0-3)

5. Answer for Problem one (3) PreambleFFFFFF 005018 0E7C84 0806DataCRC 8 octets6 octets 2 octets64-1500 octets4 octets ARP Request 0001 (Ethernet)0800 (IP) 06040001 (ARP Request) 00-50-18 -0E-7C-84140.127.114.103 0000 00000000 128.10.2.1 ARP Broadcast

7. Answer for Problem one (5) Internet purdue 00-50-18-0E-7C-84 140.127.114.103 HA 1 128.10.2.1 HA 1 HA 2 HA 3 HA 4 HA 5 HA 6 HA 7 HA 1 140.127.114.103 HA 2 128.10.2.1 HA 2 140.127.114.103 HA 3 128.10.2.1 HA 3 140.127.114.103 HA 4 128.10.2.1 HA 4 140.127.114.103 HA 5 128.10.2.1 HA 5 140.127.114.103 HA 6 128.10.2.1 HA 6 140.127.114.103 HA 7 128.10.2.1 Source and Destination address

8. Answer for Problem one (5) Send SYN seq = x Receive SYN segment Send SYN seq =y, ACK x+1 Receive ACK segment Receive SYN+ACK segment Send ACK y+1 Three-way handshake to establish a end-to-end TCP connection

9. Problem two (1) (a) Consider a fixed subnet partition of a class B network number that will accommodate at least 76 networks. How many hosts can be on each network? How to write its subnet mask? –The number 76 is greater than 64, and smaller than 128. Therefore we should select the number 128 (using 7 bits for subnet mask). Hence, each network can assign 512 – 2 = 510 hosts, and its subnet mask is 255.255.254.0 11111111 1111111000000000

10. Problem two (2) (b) Consider a site that chooses to subnet a class B network address, but decides that some physical nets will use 6 bits of the local portion to identify the physical net while others will use 8. Find an assignment of host addresses that makes destination addresses ambiguous. 35% –Choice more than one subnet length in a network may cause the ambiguous address. It might happen without hierarchical address assignment. –For example, the physical net A using 6 subnet bits can hold 64 subnets, and the physical net B using 8 subnet bits can hold 256 small subnets. They share the same address space, so a address assigned at intersection address space will be ambiguous.

11. Problem two (3) Address space 12345……….64 1234567…….. 127 128 ambiguous Can not use

12. Problem three Define the terms as following: 30% –Fragmentation The process of dividing an IP datagram into smaller pieces when they must travel across a network that cannot handle the original datagram size. –ICMP Internet control message protocol; an integral part of the Internet protocol that handles error and control message. Routers and hosts use it to send reports of problems about datagrams back to the original source. –Circuit switching A method of communicating in which data are sent as a continuous stream of bits. Bandwidth is guaranteed, and delay is limited to propagation time. The telephone system uses circuit switching.