CH.1. Lecture # 2. Computer Networks and the Internet. Eng. Wafaa Audah. Islamic University of Gaza. Faculty of Engineering

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1 Islamic University of Gaza Faculty of Engineering Computer Engineering Department Networks Discussion ECOM 4021 Lecture # 2 CH1 Computer Networks and the Internet By Feb 2013

2 (Theoretical material: page 2-7, Review questions and problems :8-end) 11 What is the Internet Networks Discussion End systems (hosts): computing devices that are interconnected by the network like PC, server, cell phones, laptops End Systems are connected together by communication links and packet switches Packet switches come in many shapes, but the two most famous types in today's Internet are routers and link-layer switches Bandwidth mean transmission rate of data, measured by bit/sec End systems access the Internet through Internet Service Providers (ISPs) End systems and packet switches run protocols that control the sending and receiving of information within the Internet The Transmission Control Protocol (TCP) and the Internet Protocol (IP) are two of the most important protocols in the Internet Protocol: A protocol defines the format and the order for messages exchanged between two or more communicating entities, as well as the actions taken on the transmission and/or receipt of the message Computer network protocol 12 Network Edge Network edges mean the applications and hosts Access network is the physical links that connect an end system to the first router (also known as the "edge router") on a path from the end system to any other distant end system 2

3 Network core includes interconnected routers and network of networks Information about the lasting of this section which include: Internet access networks and physical media, are available at Dr slides :lecture 1 _slides and textbook 13 Network Core There are two fundamental approaches for moving data through a network of links and switches: circuit switching and packet switching Circuit Switching: - Dedicated link that represents end-end resources reserved for "call" and dedicated resources which mean "no share" - Network resources (eg, bandwidth) divided into "pieces" ***** - Every user has a bandwidth as the following equation show: Bandwidth one user = Total bandwidth / Number of users - Represent guaranteed transmission (ensure: data arrival, correct arrival and ordered data) - Call setup required - Resource piece idle if not used by owning call Disadvantage 3

4 ***** Dividing link bandwidth into "pieces": - Frequency division: Time is taken completely and frequency is divided between users ( Every user take a part of frequency all the time) - Time division: Frequency is taken completely and time is divided into slots that every user take a slot that is repeated periodically ( Every user take the total frequency periodically) Packet Switching: - Data stream is divided into packets - No dedicated link that represents No end-end resources reserved - No dedicated resources which means "sharing" (resources used as needed) - Network resources (bandwidth) are not divided into "pieces", each packet uses full link bandwidth - No guarantee of transmission (No ensuring for : data arrival, correct arrival and ordered data) 4

5 14 Delay Loss and Throughput in Packet-Switched Networks Dealing with packet switching means: probability of many types of delays which are: Networks Discussion - Nodal processing: Time taken by the node itself in order to (eg) check bit errors or determine output link - Queuing delay: When packets arrival rate to the link exceeds output link capacity, packets queue in router buffers period of time, wait for their transmission chance - Transmission delay: The time needed to load all the packet into the link, which is L/R (R=link bandwidth (bps), L=packet length (bits) - Propagation delay: The time needed for a packet to across the link, which is d/s (d = length of physical link, s = propagation speed in medium - Store and forward delay: The time needed for the node (router) to receive complete packet before forwarding it to another node Nodal Delay (End-to-End Delay) Note: When packets arrival rate to the link exceeds output link capacity: queuing happenes, but if the queuing buffer is full, then packet loss will happen 5

6 Throughput Rate (bits/time) at which bits transferred between sender/receiver and it measured by getting the Minimum **** of links rates between sender and receiver Note: Minimum link rate will get the control of all transmission links because it can allow transmission just at its rate even if there are higher rated links, that means getting the throughput as its rate - The following figure describe the meaning of how minimum link rate describe the throughput, suppose that every shape is a link and its area describe its rate as a figure shows, smaller shape control whole system because data are passed through it just according to its area which represent rate Data Data Data link 1 link 2 link 3 15 Protocol layers, service models - Internet protocol stack model provides reliable byte-stream service between client and service processes 6

7 - ISO/OSI reference model Note: According to Internet protocol stack model - Switch works at link and physical layers - Router works at network, link and physical layers For sections: 16, 17 see Dr slides (Lecture 3_11-25) and textbook 7

8 Review Questions and Problems Review Questions R11 What advantage does a circuit-switched network have over a packet-switched network? What advantages does TDM have over FDM in a circuit-switched network? - A circuit-switched network can guarantee a certain amount of end-toend bandwidth for the duration of a call, but packet-switched networks cannot make any end-to-end guarantees for bandwidth - In TDM each signal uses all of the bandwidth some of the time, TDM provides greater flexibility and efficiency, by dynamically allocating more time periods to the signals that need more of the bandwidth, while reducing the time periods to those signals that do not need it FDM lacks this type of flexibility, as it cannot dynamically change the width of the allocated frequency R12 Why is it said that packet switching employs statistical multiplexing? Contrast statistical multiplexing with the multiplexing that takes place in TDM In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern In TDM circuit switching, each host gets the same slot in a revolving TDM frame R15 Suppose users share a 2 Mbps link Also suppose each user transmits continuously at I Mbps when transmitting, but each user transmits only 20 percent of the time a When circuit switching is used, how many users can be supported? b For the remainder of this problem, suppose packet switching is used Why will there be essentially no queuing delay before the link if two or fewer users transmit at the same time? Why will there be a queuing delay if three users transmit at the same time? c Find the probability that a given user is transmitting 8

9 a) according to equation: Bandwidth one user = Total bandwidth / Number of users I M= 2 M/N N = 2 users can be supported b) Each user requires 1Mbps when transmitting, if two or fewer users transmit simultaneously, a maximum of 2Mbps will be required Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link Whereas, if three users transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link In this case, there will be queuing delay before the link c) Probability that a given user is transmitting = 02 (each user transmits only 20 percent of the time) R16 Consider sending a packet from a source host to a destination host over a fixed route List the delay components in the end-to-end delay Which of these delays are constant and which are variable? processing delay, transmission delay, propagation delay and queuing delay All of these delays are fixed, except for the queuing delays, which are variable according to the pressure of the packets on link R19 Suppose Host A wants to send a large file to Host B The path from Host A to Host B has three links, of rates RI =500 kbps, R2 =2 Mbps, and R) = I Mbps a Assuming no other traffic in the network, what is the throughput for the file transfer b Suppose the file is 4 million bytes Dividing the file size by the through put, roughly how long will it take to transfer the file to Host B? c Repeat (a) and (b), but now with R2 reduced to 100 kbps 9

10 a) throughput according to the previous information above_ page 6: min (500k,2M,1M) bit/sec =500k bit/sec b) 500*10^3 sec 8*4*10^6? : 64 second c) a-100k bit/sec (the same "min" approach), b- 100*10^3 sec 8*4*10^6? : 320 second Problems P3 Consider the circuit-switched network in Figure 112 Recall that there are n circuits on each link a What is the maximum number of simultaneous connections that can be in progress at any one time in this network? b Suppose that all connections are between the switch in the upperleft-hand corner and the switch in the lower-right-hand corner What is the maximum number of simultaneous connections that can be in progress? 10

11 a) We can have n connections between each of the four pairs of adjacent switches This gives a maximum of 4n connections Networks Discussion n n n n b) We can n connections passing through the switch in the upper-right-hand corner and another n connections passing through the switch in the lowerleft-hand corner, giving a total of 2n connections n 2 n n n 2 n n P4 Review the car-caravan analogy in Section l4 Assume a propagation speed of 100 km/hour a Suppose the caravan travels 150 km, beginning in front of one tollbooth, passing through a second tollbooth, and finishing just after a third tollbooth What is the end-to-end delay? 11

12 Example at section 14 Tollbooths are 75 km apart, and the cars propagate at 100km/hr A tollbooth services a car at a rate of one car every 12 seconds a) There are ten cars It takes 120 seconds, or 2 minutes, for the first tollbooth to service the 10 cars Each of these cars has a propagation delay of 45 minutes (travel 75 km) before arriving at the second tollbooth Thus, all the cars are lined up before the second tollbooth after 47 minutes The whole process repeats itself for traveling between the second and third tollbooths It also takes 2 minutes for the third tollbooth to service the 10 cars Thus the total delay is 96 minutes Explanation: - Time _ start -stage1: tr_10 cars loaded to link 1 (10*12sec=120sec=2min)+ tp (75k/100k/hr=45min)= 47min - Time _ stage1-stage 2 (10 cars loaded to link 2): the same of the previous work =47min - Time_ stage2-end tr_10 cars passed tollbooth and loaded out of it: 10*12sec=120sec=2min - Total end to end delay = (min) = 96 min The following figure explain the work

13 start stage 1 stage 2 end caravan Link 1 Link 2 tp tp 3 tr _10cars tr _10cars tr _10cars End-to-end delay P6 In this problem we consider sending real-time voice from Host A to Host B over a packet-switched network (VoIP) Host A converts analog voice to a digital 64 kbps bit stream on the fl y Host A then groups the bits into 56-byte packets There is one link between Host A and B; its transmission rate is 2 Mbps and its propagation delay is 10 msec As soon as Host A gathers a packet, it sends it to Host B As soon as Host B receives an entire packet, it converts the packet's bits to an analog signal How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B)?

14 Consider the first bit in a packet Before this bit can be transmitted, all of the bits in the packet must be generated This requires 1 The time required to make analog- to digital conversion 56*8/(64*10^3)sec=7msec 2 The time required to transmit the packet is 56*8/(2*10^6)sec=224μsec 3 Propagation delay = 10 msec The delay until decoding is 7msec +μ224sec + 10msec = 17224msec A similar analysis shows that all bits experience a delay of msec P11 A packet switch receives a packet and determines the outbound link to which the packet should be forwarded When the packet arrives, one other packet is halfway done being transmitted on this outbound link and four other packets are waiting to be transmitted Packets are transmitted in order of arrival Suppose all packets are 1,500 bytes and the link rate is 2 Mbps What is the queuing delay for the packet? More generally, what is the queuing delay when all packets have length L, the transmission rate is R, x bits of the currently-being-transmitted packet have been transmitted, and n packets are already in the queue? Networks Discussion The arriving packet must first wait for the link to transmit 6,750 bytes or 54,000 bits Since these bits are transmitted at 2 Mbps, the queuing delay is 27 msec Generally, the queuing delay is - Waiting link to be free : (L-x)/R - Waiting n-queue packets: nl/r - Total delay = (nl + (L - x))/r n packets in queue packet L-x L X A Link B Queu New packet

15 P12 Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queued Each packet is of length L and the link has transmission rate R What is the average queuing delay for the N packets? The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)l/r for the nth transmitted packet Thus, the average delay for the N packets is (L/R + 2L/R + + (N-1)L/R)/N = L/(RN) * ( (N-1)) = L/(RN) * N(N-1)/2 = LN(N-1)/(2RN) = (N-1)L/(2R) Note that here we used the well-known fact that N = N(N+1)/2 P24 Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps Suppose the propagation speed over the link is 25 10H meters/sec a Calculate the bandwidth-delay product, R d prop ' b Consider sending a file of 800,000 bits from Host A to Host B Suppose the file is sent continuously as one large message What is the maximum number of bits that will be in the link at any given time? c Provide an interpretation of the bandwidth-delay product d What is the width (in meters) of a bit in the link? Is it longer than a football field? e Derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate R, and the length of the link m

16 a) R* (D/V) = 2M *20,000 k / 25*10^8 = 160,000 bits b) 160,000 bits because the link can only has a number of bits equals bandwidth delay product c) The bandwidth-delay product of a link is the maximum number of bits that can be in the link d) the width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters long e) s/r P26 Consider problem P24 but now with a link of R =1 Gbps a Calculate the bandwidth-delay product, R d prop b Consider sending a file of 800,000 bits from Host A to Host B Suppose the file is sent continuously as one big message What is the maximum number of bits that will be in the link at any given time? c What is the width (in meters) of a bit in the link? a) 80,000,000 bits b) 800,000 bits, this is because that the maximum number of bits that will be in the link at any given time = min(bandwidth delay product, packet size) = 800,000 bits (the same concept of throughput measurement) c) 25 meters

17 P30 In modern packet-switched networks, the source host segments long, application-layer messages (for example, an image or a music file) into smaller packets and sends the packets into the network The receiver then reassembles the packets back into the original message We refer to this process as message segmentation Figure 128 illustrates the end-to-end transport of a message with and without message segmentation Consider a message that is 8 * 106 bits long that is to be sent from source to destination in Figure 128 Suppose each link in the figure is 2 Mbps Ignore propagation, queuing, and processing delays a Consider sending the message from source to destination without message segmentation How long does it take to move the message from the source host to the first packet switch? Keeping in mind that each switch uses store-and-forward packet switching, what is the total time to move the message from source host to destination host? b Now suppose that the message is segmented into 4,000 packets, with each packet being 2,000 bits long How long does it take to move the first packet from source host to the first switch? When the first packet is being sent from the first switch to the second switch, the second packet is being sent from the source host to the first switch At what time will the second packet be fully received at the first switch? c How long does it take to move the file from source host to destination host when message segmentation is used? Compare this result with your answer in part (a) and comment d Discuss the drawbacks of message segmentation

18 a) Time to send message from source host to first packet switch = (8*10^6)/(2*10^6)=4sec With store-and-forward switching, the total time to move message from source host to destination host = 3*4 sec=12 sec (three links has the same rate so the same time will be needed) b) Time to send 1st packet from source host to first packet switch = (2*10^3)/(2*10^6) =1msec - Time at which 2nd packet is received at the first switch = time at which 1st packet is received at the second switch: 2*1msec=2msec c) Time at which 1st packet is received at the destination host = 1msec*3=3 msec After this, every 1msec one packet will be received; thus time at which last (4000th) packet is received = 3msec+3999*1msec=4002sec It can be seen that delay in using message segmentation is significantly less (almost 1/3rd) Detailed explanation **** d) Drawbacks: i Packets have to be put in sequence at the destination ii Message segmentation results in many smaller packets Since header size is usually the same for all packets regardless of their size, with message segmentation the total amount of header bytes is more **** Assume that we have 3-packets (small number instead of 4000) As it clear, 2 nd packet need just 1msec after the arrival of 1 st packet (4msec-3msec), third packet has the same thing Meaning for 4000 packets: 1 st packet needs 3msec to reach destination and all other packets need just 1msec because every packet needs 1msec after its previous packet reach S sw1 sw2 d 1msec 2msec 3msec 4msec 1 st packet 2 nd packet 3 rd packet 5msec See You at lecture 3 Best Wishes

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