UNIVERSITY OF TORONTO ELECTRICAL AND COMPUTER ENGINEERING ECE 361 Test February 2, 2012

Transcription

1 Student Number: UNIVERSITY OF TORONTO ELECTRICAL AND COMPUTER ENGINEERING ECE 361 Test February 2, 2012 ECE 361 Computer Networks This test is Closed Book 75 Minutes Examiner: A. Leon-Garcia Question Mark 1 /10 2 /10 3 /10 4 /10 TOTAL /40 Page 1 of 6 pages

2 1. Provide a brief answer to each of the following questions. The following questions are worth 2 marks each. a. Explain the N 2 problem in telephone networks and explain how it is solved. The number of dedicated lines between pairs of users in the original telephone system was approximately N 2 or more specifically [N(N-1)/2] which grows very quickly. At first telephone switches with human operators, which connected users on demand, were introduced to solve the problem. Later automated electromechanical switches were introduced. With switches only N connections are required to the central office. b. Explain the difference between message switching and packet switching. Explain which one is better in terms of: i. End-to-end delay; ii. Header overhead. With message switching a message from an application is transmitted in its entirety. Packet switching networks long messages are segmented and transmitted using multiple packets. i. Packet switching networks are better in terms of E2E delay since long messages in a message switching network can impose wait times for other messages. ii. Message switching networks are better in terms of header overhead since each packet (multiple packets in a message) requires header information to be correctly delivered. Page 2 of 6 pages

3 c. Give two aspects in which the Data Link Layer and the Network Layer differ. i. Data link layer provides for transfer across a transmission link that directly connects two nodes while network layer provides transfer of data across a communication network. ii. Data link layer uses flat addressing structure while the network layer uses hierarchical, routable addressing structure d. What is the purpose of medium access control? Medium access control is required to coordinate the transmissions from machines into a broadcast medium to prevent collisions. e. What is the purpose of error control? Error control ensures reliable transfer of information either node-to-node or end-to-end. As an example check bits can be calculated from the information bits according to some algorithm that enables the receiver to detect whether the received information contains errors. Page 3 of 6 pages

4 2. Provide a brief answer to each of the following questions. The following questions are worth 2 marks each. a. Explain the notion of encapsulation. The layer n+1 PDU (protocol data unit) is encapsulated in the layer n PDU. Encapsulation narrows the scope of dependencies between adjacent layers to the service definition only. b. Explain why Ethernet and IP addresses are both needed. The Ethernet address is a globally unique addresses assigned by the hardware manufacturer which can be used to identify the host. The IP address is a hierarchical, routable address which can be used to reach a particular host. Ethernet addresses are needed for communication in the data link layer while IP addresses are used to communicate across the network layer. c. Explain the relationship between the IP address structure and the routing of IP packets in the Internet. The IP address structure was originally defined to have a two-level hierarchy: network ID and host ID. The network ID identifies the network the host is connected to. Routers can forward packets based on the network ID only thereby shortening the routing table significantly. More recently classless interdomain routing (CIDR) was adopted allowing arbitrary prefix length to indicate the network number. Page 4 of 6 pages

5 d. Applications are in the application layer, True or False? Explain. False. Applications are software programs that require the service of all layers including the application layer. Eg. Web browser application uses HTTP protocol in the application layer, TCP in transport layer, IP in network layer, etc. e. Explain why the entire four-tuple (Source IP Address, Source Port Number, Destination IP Address, Destination Port Number) is required to uniquely identify a TCP connection. A TCP connection provides a reliable stream service between two application processes across a network. Thus the source and destination IP addresses are required to uniquely identify the two hosts at the end-points whereas the source and destination port numbers are required to uniquely identify the application processes within the hosts. Mark breakdown: 2 marks for a reasonable explanation that the four-tuple is needed to identify the process-to-process connection. Page 5 of 6 pages

6 3. a. (5 marks) A web server breaks a 4 MB file into blocks of 1000 bytes each, which are sent to the client. The TCP header is 20 bytes, the IP header is 20 bytes, and the sum of the Ethernet header and trailer is 18 bytes. The distance between the hosting server and the client is 30 km. The client and the server are linked by a 1 Mbps link. Find the minimum possible delay from when the user at the client clicks on the URL to when the first packet of the file is received at the client. Repeat for the delay until the last byte in the file is received, assuming that the server sends packets back-to-back. Assume that the speed of light in this network is 2x10 8 meters/second. Solution syn syn ack ack get P 1 P 2 P 3 P 4000 Propagation Delay = (30,000m) / (2E8 m/s) = s Conn. Setup (3 way handshake + GET) = 3 * prop. Delay = s note: the size of the setup packets can be considered negligible. However if some nominal value for pkt size is used it s fine. Time to transmit 1 st packet = prop. delay + trans. delay Packet trans. delay = ( )*8 / 1E6 = s a) delay to first packet of file: = s b) 4E6 bytes / 1000 bytes = 4000 packets total delay = s * (trans. delay) = s Page 6 of 6 pages

7 3. b. (5 marks) A Voice-over-IP system takes the stream of digital voice signal of 8000 bytes/second and forms blocks that contain 80 voice bytes each. The system encapsulates each block inside: an RTP Application Layer message (which adds a 16 bytes header); then inside a UDP datagram (which adds a 12 byte header); an IP packet with header of 20 bytes; and finally an Ethernet frame with header and trailer is 18 bytes. Suppose the VoIP system has a 10 Mbps transmission link to its first Internet router, how many simultaneous voice calls can be supported by the system? Solution: 8000 bytes/second ð 80 voice byte blocks ð 100 blocks / second ð 1 block: 80 voice bytes + 16 RTP bytes + 12 UDP bytes + 20 IP bytes + 18 Ethernet bytes = 146 bytes ð 100 blocks / second * 146 bytes / block = 14,600 bytes / second = 116,800 bits/second ð Floor(10 Mbps / 116,800 bps) = floor(85.6) Therefore this system can support 85 simultaneous calls Page 7 of 6 pages

8 4. The diagram below shows two Ethernets connected by a router. The host and router interfaces have the indicated Ethernet address (ek) and the pair (I,J) denotes its IP Net ID (denoted by I) and IP Host ID (denoted by J). Suppose that each host has a table with entries that for each destination host identifies the pair (Ethernet and IP address) that the host uses to determine the addresses in the header of the packets and frames to the given host. Assume that all host and router tables have complete Ethernet and IP address information for all hosts in both networks. a. Suppose that Host 1 sends a packet to Host 2. Show the addresses in the headers of the Ethernet frame and IP packet that traverses Network 1. b. Suppose that Host 1 sends a packet to Host 4. Show the addresses of the headers of the Ethernet frames and IP packets that traverse Network 1 and Network 2. Host 1 Host 2 (1,1) e1 (1,2) e2 (1,3) (2,3) e5 Router e6 e3 (2,1) Host 3 (2,2) e4 Host 4 Solution a) Eth Src: e1 Eth Dest: e2 IP Src: 1,1 IP Dest: 1,2 b) Network 1: Eth Src: e1 Eth Dest: e5 IP Src: 1,1 IP Dest: 2,2 Network 2: Eth Src: e6 Eth Dest: e4 IP Src: 1,1 IP Dest: 2,2 Page 8 of 6 pages