Name : Syed Shahzad Ali Course # : CS 742 Date : 9/11/2001 Instructor : Ge Assignment#: 1 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ Q1. Download the following RFCs from IETF's RFC page at http://www.ietf.org/rfc.html or from Internet FAQ Consortium's RFC page at http://www.faqs.org/rfcs/: RFC 791, RFC 826, RFC 768, RFC 793, RFC 792 For each RFC, your only need to write down the title of the RFC, the author names, and the date of publication. Ans(1) RFC 791 - Internet Protocol (IP) by Information Sciences Institute University of Southern California 4676 Admiralty Way Marina del Rey, California 90291 Date Published : September 1981 RFC 826 - An Ethernet Address Resolution Protocol (ARP) -- or -- Converting Network Protocol Addresses to 48.bit Ethernet Address for Transmission on Ethernet Hardware by David C. Plummer Date Published : November 1982 RFC 768 - User Datagram Protocol (UDP) by J. Postel Date Published : 28 August 1980 RFC 793 - TRANSMISSION CONTROL PROTOCOL by Information Sciences Institute University of Southern California 4676 Admiralty Way Marina del Rey, California 90291 Date Published : September 1981 RFC 792 - INTERNET CONTROL MESSAGE PROTOCOL (ICMP) by J. Postel Date Published : September 1981 Q(2) Use one of the two search engines, Google (http://www.google.com) and Altavista (http://www.altavista.com/), to locate usful, general, and noncommercial information about the following topics: MBone, ATM, MPEG, IPv6, and Ethernet. You only need to give a concise definition for each term above in no more than 2 sentences A(2) MBone - Multicast Backbone The MBONE is an outgrowth of the first two IETF "audiocast" experiments in which live audio and video were multicast from the IETF meeting site to destinations around the world. The MBONE is a virtual network. Mbone is also called the Multicast Internet In which the data or message is send to a group of multiple users rather sending as broadcast of unicasr. Q(3) ATM - Asynchronus Transfer Mode ( only this Acronym is defined here) ATM - is also an acronym for Automated Teller Machine Asynchronous Transfer Mode (ATM) has been defined as a networking technology to support data, voice, and video transmission. In particular, ATM is being developed to support numerous high-end applications, such as high-end video, which includes the emerging MPEG and MPEG2 video standards, one key reason for the adoption of ATM is its ability to scale to higher bandwidth and to provide control for latency and jitter in the network and enable numerous emerging video applications. MPEG : Motion Picture Experts Group This technology brief discusses (MPEG) video codecs and their usage when streaming video across enterprise networks, which is a group of compression standards for audio, video, and data established by the International Telecommunications Union and International Standards Organization. The basic job of MPEG is to convert and compress analog or digital video signals into packets of digital information that are more efficient to transport on modern networks. IPv6 : Internet Protocl version 6 IPv6 is the new version of IP protocol that will replace the old verion of IP Protocol ver 4 due to the reason that the IPv4 doesn't have enough IP Addresses left to offer to the Enterprises and Devices that are coming into the market rapidly. IPv6 is 128 bits as compare to the 32 bits of ver 4 which besides adding exponentially more IP addresses, offers other benefits, such as automatic configuring. Ethernet The term Ethernet refers to the family of local area network (LAN) implementations that includes three principal categories as Ethernet and IEEE 802.3, 100-Mbps Ethernet and 1000 Mbps Ethernet. Ethernet is a baseband LAN specification that operates at 10 Mbps using carrier sense multiple access collision detect (CSMA/CD) to run over coaxial cable. It was created by Xerox in the 1970s, but the term is now often used to refer to all CSMA/CD LANs. Ex. (5) Calculate the total time required to transfer a 1000kb file in the following cases assuming RTT = 100msec Packet Size = 1 KB and initial 2*RTT of handshaking beofore data is sent. 2*RTT = 2*100msec = 200msec (a) Delay(msec) = Time to propogate + time to transmit + time to Handshake = 1/(2RTT) + 8000*10^3/(1.5*10^6Sec) + 2RTT = (1/2)100msec + 5.333msec + 200msec = 50 msec + 5333.33msec + 200msec = 5583.33 msec (b) Total # of Packets = 1000/1 = 1000 Packets RTT delay for a file = (1000 -1) * RTT = 999 * 100 msec = 99900 msec Trip Time = 99900 + 5583.33 + 200 = 105683.33 msec = 105.68 sec (c) Bandwidth = unlimited Wait for one RTT before sending bunch of 20 Packets => (1000/20) -1 = 50 - 1 = 49 Complete (t) = 49*100msec + 0.5*100 msec + 200 msec = 5150 msec = 5.15 sec Ex.(13) Bandwidth = 100 Mbps Distance in between = 385,000 Km Speed of Data Transfer =3 * 10^8 m/sec (a) 3*10^8 meters travelled -------- 1 sec 1 meters travelled -------- 1 / 3*10^8 sec 385,000 * 10^3 meters travelled -------- 1.28333 sec considering uplink and downlink RTT = 2*1.28333 =2.5667 sec (b) Delay * Bandwidth = 2.56 *100*10^6 = 256.666 Mbits (c) This result tells us that it is equivalent to the number of pipes. We require 3 pipes to transmit 256.666 Mbits each having 100 Mbps Capacity. (d) 25*10^6 / 100*10^6 + 2*RTT = 0.25 + 5.1334 = 5.3834 sec Ex(14): (a) File Opening : Delay Sensitive. (b) Read the content of a file : Bandwidth Sensitive (c) List the content of a Directory : Delay Sensitive (d) Display attributes of a file Delay Sensitive. Ex(15): (a) Bandwidth = 10Mbps Packet Size = 5000 bits Propagation Delay from Source to Switch = 10 micro sec Propagation Delay from Switch to Destination = 10 micro sec => Total Prop. Delay = 20 micro sec Latency(t) = 20*10^-6 + 2(5000/10*10^6) Latency(t) = 1.02 msec (b) Latency(t) = 40*10^-6 + 4(5000/10*10^6) Latency(t) = 2.04msec (c) For Cut through switch Latency(t) = 20*10-6 + 2 ( 200/10*10^6) + 1* (4800/10*10^6) Latency(t) = 0.54msec Hence the latency is reduced in case of using the Cut Through Switch Ex(17): Bandwidth = 10 Mbps Delay = 10 micro sec (a) Bandwidth * Delay = 10*10^6 * 10*10^-6 = 100 bits (b) Bandwidth * Delay = {10*10^6} * {2*( 5000/10 *10^6) + 20* 10^ -6} => = 10*10^6 * 1020*10^-6 = 10200 bits (c) Bandwidth * Delay = 1.5*10^6 * (50*10-3) = 75000 bits (d) Bandwidth * Delay = 1.5*10^6 * 359000*10^3 / 3*10^8 = 1795 Kbits Ex(21): (a) Bandwidth = 100 Mbps Propagation Speed = 2*10^8 m/sec Circumference = ? S = Vt t= 250*8/100*10^6 = 20 * 10-6 S = 2*10^8 * 20* 10-6 S = 4000m total nodes = z/100 If 100 Mbits -------- 1 sec 1 bit -------- 10 nano sec 10 bits -------- 100 nano sec Therefore 250* 8 = 100* 10^6 * [ z/2*10^8 + (z/100) * 100 nano sec ] 20 * 10^-6 = [ 5nsec z + 1nsec Z ] z = 3333.333 m