Communication Medium and Network Topologies | Advanced Computer Architecture | CS501_Lecture44

Asalamuaikum. Welcome to the series of lectures on advanced computer architecture. In our previous lecture, we started our discussion on computer networks. Computer networks provide an opportunity to connect different computers together or different computing facilities together and share the resources. For a computer architect, the knowledge of computer networks would be helpful in designing and understanding the modern computer networks. We classified the computer networks based on the distance that is covered and on the number of computers connected to the network. We discussed the features of the LAN and VAN. We also looked into the performance of a computer network. We said that the quality of a computer network in terms of data transmission would be judged by two factors. First one is the bit error rate. Lower the bit error rate, better would be the quality of data transmission. Secondly, we discussed the bandwidth. Now, raw bandwidth just corresponds to the number of bits transmitted per second or the maximum rate at which data could be transmitted through the network. However, the effective throughput is reduced because of the latency because of the additional delays. These delays could be either propagation delay or it could be in the form of overheads on the sender side or on the receiver side. Additionally, for wide area networks, the intermediate routers could also incorporate delays. Therefore, the effective latency would be higher and therefore the effective bandwidth would reduced. The simple formula which we discussed was effective bandwidth is equal to the total message size divided by the latency. Now after that we discussed some of the physical media which are available in practice to interconnect different computers with each other. The simplest medium is the copper medium. It could be either in the form of a twisted pair or coaxial cable. Based on the twisted pairs, the existing telephone networks could be utilized to interconnect different computers. And that could be done by using modm cards, which is simply a modulator and a demodulator combination. And this card could be physically used in one of the slots in PC. And therefore the data would be transmitted by using a modem on a copper pair through an exchange. Now some cases could be there where dedicated copper pair could be utilized. However, if we want to utilize the present switching network which is pretty vast and is meant primarily for telephony or for voice communication then this bassband of 4 kohz could provide us an opportunity by using modems in the computers and you know that it is used in the present day internet as a very common mode of data transmission. We will not go into the details of the functionality of the modem. However, one thing should be kept in mind and that is the speed of previous modms was lesser and there are some physical limitations on the availability of maximum bandwidth which could be achieved by using a modem. Now you remember that we have modms which could go at present up to 90 kilobits per second or it could be 56 kilobits or even less. You might have noticed that when you try to send an email on internet you are connected through an internet service provider then the connectivity varies from time to time. Sometimes when you are lucky then you get a good quality telephone line and you have a higher data rate. On the other hand when there's a lot of noise on the line the data rate could be as low as just 16 or even less kilobits per second. Now we discussed that if we want to increase the throughput and we want to have a local area network implementation within a building or even in the case of a van we want to have a backbone then we could utilize a medium which is optical fiber. For the case of optical fiber we just said that basically it's just glass. It could also be implemented by plastic but plastic would have a lesser bandwidth and the glass would have a higher bandwidth. Now there are different types of fibers available. Typically you could remember that we could have a multiode fiber or a monomode fiber. In the case of multiode fiber, usually the diameter of the fiber is relatively large and when light is injected on the transmitting end then a lot of dispersion or spreading takes place and because of this dispersion the effective data rate decreases. On the other hand, for a monode fiber, the diameter is small and you could imagine that there is usually just one mode and the light is transmitted axially along the axis of the fiber. Dispersion is much lesser and we could have a very large data rate through the fiber. Now at present even in Pakistan we have a fiber running from Karachi up to Islamabad and maybe onwards on spars also we could have the optical fiber to increase the capacity of this fiber additional techniques could be utilized like we have a wave division multiplexing and then using wave Wave division multiplexing different wavelengths are simultaneously transmitted on the fiber and therefore the effective throughput is increased. At present on the fiber available as a backbone we have data rates up to 10 gabits per second. Now apart from this optical fiber we have other media also available. Wireless is a very useful medium. Now in this case the transmission is used or electromagnetic waves are used. The range could be in gigahertz like in a microwave link or it could be on mobile phones in the range of 800 or 900 MGHertz or later on we could also have 1,800 MGHertz as the carrier frequencies. Now wireless appears to be a very attractive medium as far as the flexibility is concerned. The mobility is large. You could just move around anywhere and plug in your laptop with your mobile phone or you could just have even just a PDA in your hand and still transmit and send your email through your uh handheld computer. However, the bandwidth issue is important. When we compare the bandwidth available on optical fiber, this would be much larger than what is available on wireless. We will consider a very simple example to illustrate the effective throughput over different physical media. In the example he has discussed first a twisted pair then a quaxial cable thirdly a fiber and just for the sake of illustration. He has also discussed a physical transportation. The data is available in the form of tapes. The MAC tapes are available and you could imagine that physically a number of tapes could be taken in a car and transported. It depends on the distance and sometimes that could be even a faster way of transporting large amounts of data. Additionally, you could imagine that you could use satellite or a wireless. However, in this example, uh this wireless medium is not considered. Suppose we have 25 magnetic tapes each containing 40 gabytes of data. Assume that we have enough tape readers to keep any network busy. We want to calculate how long will it take to transmit the data over a distance of 1 kilometer for the sake of illustration. Assume that the choices are either category 5 twisted pair which can transmit data at a rate of 100 megabits per second. multiode fiber at a rate of 1,000 mgabits per second or a single mode fiber at a rate of 2.5 GBs per second. Now compare the time of this delivery for the sake of illustration. If a car was used at a given speed of 30 kilo m hour, the amount of total data in this example is 25 into 40 which is 1,000 GB. Just multiply with 8 to get gabits. For the case of twisted pair, just divide the total message size by the rate which is bandwidth and it turns out to be 22.8 hours which is a tremendous time. For a multiode fiber the time is 2.3 hours and for single mode fiber it is.9 hours. For car we assume 300 seconds of time to load the car and 300 seconds to unload the car and the speed of the car is 30 kilometers per hour and this results in just.3 hours. It is very interesting. You see that physical transportation is in this case the quickest. However, if we increase the distance, this may not be the case. Now, for a computer network, we could have two options for the medium. Whether it is copper, coaxial cable or fiber, in all cases, there are two possibilities. We could have a shared medium or we could have a switched medium. Now in the case of a shared medium just imagine that we have a coaxial cable running in a laboratory or in the campus or it could be a fiber running in the campus and a number of computers are connected. Now this is an example of a shared medium. There would be a maximum throughput that is available. Now if we connect a number of computers to the same medium there should be some mechanism that the collisions do not occur. What is a collision? When more than one pair of computers simultaneously try to communicate with each other or transfer data that would not be possible. So some mechanism has to be evolved and there are protocols for that where you just say that like you send a carrier and all others are sensing and at one time only one of the computer is allowed to transmit or one is transmitting and others are listening. Now if at all the two computers try to transmit simultaneously and a collision does occur it can be resolved. Both the computers as an example would be asked to remain silent and for a random time of interval then one of these would retransmit. So the synchronism has to be broken that simultaneously two computers do not uh simultaneously transmit. Now the shared medium would have a maximum throughput which would depend on the number of computers which are connected on that medium. Naturally the effective throughput would decrease as the number of computers increase. Nevertheless, the maximum throughput cannot exceed what is specified as a raw bandwidth. As an example, if we have a local area network with a maximum uh rate of 100 megabits per second, then the maximum rate at which we could get the throughput would be 100 mgabits per second. That means if we increase the number of users the throughput would decrease but nevertheless it would never exceed the raw bandwidth. Now if we want to increase the throughput of a network there there is the second option we could use a switch. As an example if we have a switch with four nodes then this switch could connect any two computers with each other. And take an example. Out of these four computers, one is connected to three and two is connected to four. The effective bandwidth would double up. And if each link through the switch provides 100 megabits per second, effective or aggregate bandwidth would be double that is 200 mgabits per second. Let us illustrate this concept of increasing the aggregate bandwidth by using a switch uh in the next slide. Ethernet was originally a shared medium but Ethernet switches are now available. All nodes on the shared media must share 100 megabit per second interconnection. However, the switches can support multiple 100 megbit per second transfers simultaneously. Lowcost Ethernet switches are sometimes implemented with an internal bus. Let us consider the example with 16 nodes connected three ways. The first one using a single 100 megabit per second shared medium. Secondly, a switch connected via a cat five. Each segment running at 100 megabits per second. And thirdly, a an optical fiber which runs at 1,000 mgabit per second. For the case of the shared medium, the length is 500 m and average length of each segment to the switch is 50 m. Both switches can support the full bandwidth. Assume each switch adds five microscond to the latency. Calculate the aggregate bandwidth and transport latency. Assume the average message size of 125 bytes. Ignore the overheads of the sending or receiving a message and contention for the network. The solution is shown in this slide. The aggregate bandwidth of each example is calculated in a very simple way. For the shared medium, the maximum throughput would be 100 mgabits per second. For the case of the switch using cat 5 connectivity, the total throughput would be 16 divided by two. Since there would be eight couples together interconnecting 16 computers with two each multiplied by 100, that makes 800 megabits per second. So for this second example, the throughput has increased or aggregate or effective bandwidth has increased 8fold. For the case of optical fiber, it would correspond to 8 now multiplied with 1,000 and therefore it is 8 GBs per second. Let us now consider the latency. The transport time is equal to time of flight plus the transmission time corresponding to the message which is given by message size divided by the bandwidth. For shared media, we just plug in the distance, bandwidth and the message size and corresponding value for this case is 12.5 microsconds out of which 10 microcond is the transmission time and 2.5 microcond is the propagation time. We have made the same assumption that the speed through the medium is 2/3 of the speed of light. For the switches, the distance is twice the average segment. Since there is one segment from the sender end to the switch and the other from the switch to the receiver, so we must also add the latency for the switch. Now going through this calculation which is very simple for the case of switch using the CAT 5 connectivity we have a total latency of 15.5 microsconds whereas for the case of optical fiber this is 6.5 microsconds. Although the bandwidth of the switch using optical fiber is many times that of the shared medium, the latency for the unloaded network in this case is comparable. For the case of circuit switched networks, we have a given path available at a time. Now this path may not be necessarily a physical path. This may be just a logical path. For example, if one medium is shared, we could use multiplexing which could be in the form of frequency division multiplexing where different frequency bands are allocated to different users. In this case, it is a logical division of diff of the same medium to different users. However, in the case of circuit switch networks, there is always a defined path at a given time and we say it this connection is the connection oriented established path. In the case of packet switching, we have two possibilities. Now the packets may go from the source up to the destination via any path which may not be defined and which may keep on changing from one packet to another packet. In one case it may go via one node in another case it may go via another route. However, in such a case where first a path is established from point A up to another point, let us say B, we say that the connection oriented strategy is followed. In some cases where it is not important or the delays are not critical, the second strategy where it is connectionless orientation that may be utilized. Now this concept of connection orientation or connectionless path is indicated in the next slide. This slide shows a connectivity from host A to the destination B. In between the cloud has got four routers. Now if we define a virtual path which may be a case using an ATM we use and define a path via router R1 R2 up to B. That means first through appropriate protocol a path is established and then the packets are transmitted from A to router R1. Router R1 would forward these packets to R2 and then R2 would deliver to the destination B. This would be a connectionoriented scheme. However, if A does not define such a scheme, for example, in the header of the packet, then the packets could be routed from R1 to R3 to R4 and R2. There are two alternate paths available and sometimes some packets may be coming out of order at B and through software these packets could be reordered. Let us now consider different network topologies available. There are three possible topologies usually used. The first one is the bus topology. This is an example of a shared medium. You could imagine that one has a coaxial cable or an optical fiber for that network and all computers are connected to the same bus. This is similar to a shared bus as we have discussed in computer architecture. Now as we said that at one time only one of the computers could be a source and could transmit data. So therefore in such a case we need to have a collisionless situation. So when there is a collision then collisions are to be resolved. Ethernet as an example is very typical to this particular topology. The second possible topology is the star. In the case of a star, different computers are interconnected through a central hub. Now this is central as far as connectivity is concerned. Now there could be different modes in which this hub could work. This hub could be a plain switch in which case the switch would just interconnect different computers. We have already discussed an example of such a switch. This could be an optical switch or it could be an electrical switch depending upon whether we are using twisted pair or we are using an optical fiber. However, in this case, the basic throughput would depend on the performance of the switch. There is no intelligence otherwise in this switch if it is just acting to interconnect different pairs of computers. Now, we could incorporate additional intelligence and have more intelligent switches. This would be illustrated later on in one of the examples. Now the third topology is the ring topology. In this topology all the computers are connected together in the form of a ring. At one time only one of the computers passes on the information in one direction to the neighboring computer and this permission is given in the form of what is called a token. Let us look at these three topologies given in the next slide and see how this ring topology works. The ring topology which is shown in this slide avoids the collisions that are possible in the bus topology. Each pair of computers has a pointto-point connection. The most common communication method used in the ring topology is the IBM token ring protocol. In this protocol, a data packet which is called a token is passed in one direction around the ring. When a particular computer receives a token and it does not contain a data packet, it may attach a packet of data to the token and send it on to the next station. The token continues around the ring until it reaches its destination which is indicated as a destination address in the header. The token marked as empty and the empty token is sent on in the same way without any operation. Now whichever medium is available the overall objective of the communication system would be to provide a couple of important functions. The first thing is to define a an interface with the software. So a user should have an application interface from where the data could be passed on to the physical layer. Now it is important to realize that if everything has to be done in the form of bits and byes it would be difficult. The user would be just working in the application program. It would just be initiated. A session would be established. Now the software within the computer after getting a nice and a decent communication system would initiate the corresponding signals. It would provide an elegant flow control. It would also make a provision of error control in order to have a quality of the signal in terms of a desired bit error rate. And then it would also provide a possibility of having an interface from the application up to down to the physical layer and this would be done through a combination of the hardware and software. Nevertheless, some sort of abstraction is required. Different abstractions are provided and one of the abstractions is to have a seven layer OSI model. We will just have a look on this model in the next slide and see what each layer is doing in this model. This slide indicates the OSI model which is actually just an architectural model indicating the open system interconnection and this was developed by the international standards organization in 1974. Now this architectural model has seven layers. At the bottom we have the physical layer. Physical layer is concerned with the physical means used to transmit and receive data on the network. It consists of actual wiring, transmitting and receiving bits and byes, generation of appropriate signals involved in packet transmission and reception. The next higher layer is data link layer. This layer is concerned with the final preparation for transmission over the physical link. This layer performs the final framing, low-level synchronization, flow control, and error control. The next layer on top of data link layer is the network layer. The network layer manages the lower level network details. These include formatting the packet for transmission over a local area network. The corresponding addressing issues are resolved in the network layer. The next higher layer is the transport layer. The transport layer packetizes the data and ensures that all packets are received in the same order in which these packets were sent. The error control is also incorporated in the transport layer. This is accomplished by requesting retransmission of all dropped or erroneous packets and possibly by reordering correctly received packets. The next higher layer is the session layer. The session layer provides for the establishment, maintenance and termination of the connection. The next higher layer is the presentation layer. This layer is concerned with translation of syntax and intercommunication of two entities or modules. It would include such matters as encryption, decryption and translation between terminal protocols. The presentation layer is often missing for many network suits. The next layer at the top is the application layer. The application layer is the originator of transmitted data. It specifies the data in unpacketized form and often specifies destination address in symbolic form. Typical applications with a network access components include email file transfer, remote login and various distributed applications such as distributed databases. Although the connection is always through the physical layer through this abstraction if we have two terminals one let us say A the other one is B then at the top from user's point of view we would have a virtual connection between A and B that means actually the data would be originated in the application layer It would be passed on to the lower layers till it reaches the physical layer. From physical layer on the other side on the second terminal B, it would move upwards from physical layer up to the top application layer. Now we could say from functional point of view and this is shown in the slide as a dotted line between two ends from application program up to the application program and we say there is a virtual connection between the two sides through the cloud or through the network we have a connection from one layer to the corresponding layer that means physical layer to the physical layer network layer up to the network layer. There's no physical connection. Physical connection is only through the physical layer. But virtual connections are there on two ends uh on the corresponding layers from application layer to the application layer on the other end. So this abstraction gives us a very useful tool for understanding and analyzing the system protocols. Now physically when we look at the implementation we could say that the data generated by each layer is passed on to the lower layer with some additional bits in the header or possibly in the tail and this is called encapsulation. We could say up to the encapsulation, we go down to the physical layer and as we move upwards the header or corresponding tail would be stripped off and finally in the application layer we will get the original data which the terminal was supposed to send from one end to the other. Now basically the uh computer networks as we said utilize packet switching. So the packets are originated at one terminal and are to be passed on through the cloud on to another destination. Now the information about the sender and the receiver has to be placed in the header. Now this header would contain the addresses. There are specific ways and means to specify these addresses in different protocols. Now how do we transfer the data or the packets from one end to the other? This mechanism would be called the routing or delivery of packets. How it is done? We will discuss that in a little bit more detail. But before that we should also see that if the packet is too large then the delays would be involved. In order to reduce the delay mechanism or the amount of delay we should limit the size of the packet and that can that can be done by what is called fragmentation. As an example, if we say that the maximum number of bytes in a packet could be 1,500, then at some stage when the total packet exceeds 1,500, let us say it is 2,000, then we divide it into 1500 and the next 500 bytes would be placed in another packet. Now the fragmented packets each would contain the same header as in the original packet. Now finally since the destination address would be the same at the destination these two fragmented packets would be reassembled. So the fragmentation is done at some point which may be done at the host itself or it could be done on way at some point maybe some router could fragment these packets right it would depend on the specification of the link or the bandwidth available on that link. So therefore the latency could be reduced by what is called the fragmentation. Uh routing is a very important aspect of this packet switching or the computer networking. By routing we mean that how the packets are passed on and delivered through the network to the final destination. The concept is very simple. It is based on store and forward concept. The packet is passed on from one node to another node and then from that node onto the other one. We assume that some memory or some buffer space is available at each node. Now these uh routing algorithms could be different and the implementation could be done done differently in different protocols but all of these would be primarily based on the concept of store and forward uh scenario. Let us look on the scheme of routing in the next slide. For a given path between the nodes, the navigation would depend on the topology of the system. The situation might be tackled in the form of a broadcast or pointto-point communication. Switch media could use three solutions for routing. It could be either sourcebased routing. The source base routing specifies the message path to the destination. Since the network merely follows the direction, it can be simpler. A second possibility is to establish a virtual circuit. In a virtual circuit established between source and destination, the message simply names the circuit to follow. Asynchronous transfer mode or ATM technology uses virtual circuits. The third approach is the destination based routing which is one of the most popular routing mechanisms. In this case, message merely contains a destination address and the switch must pick up a path to deliver the message. The IP packet uses destination routing scheme. In the case of ATM switches, the switches would be simpler once a virtual circuit is established. However, packet switching is much faster. In the case of IP router, the router should decide how to route each packet it receives by looking at the routing table. Destinationbased routing may be implemented either in a deterministic fashion or it could be adaptive which would allow the network to pick up different routes to avoid failures or congestion. Closely related to adaptive routing is randomized routing. In the case of randomized routing, the network will randomly pick between several equally good paths so that the traffic is uniformly spread over the network and this would avoid hotspots or traffic bottlenecks. Switches in the wide area networks route messages using a store and forward policy. Each switch waits for the full message to arrive in the switch before it is sent on to the next switch. Generally, store and forward can retry a message within the network in case of a network failure. Another popular protocol stack is the TCP IP protocol which is used on internet. TCP just stands for transmission control protocol and IP is the internet protocol. Now in TCP IP two layers the presentation layer and the session layer is not used. So therefore it is just consisting of five layers. Now the strategy used for routing in this case is the same as store forward. It is based on destination address and the encapsulation scheme is similar as we have already discussed. Just see in the next slide a comparison of the TCP IP protocol with the conventional OSI 7 layer model. This slide shows the correspondence between the OSI layer model and TCP IP combined with applications at the bottom and the various LAN protocols are also indicated. TCP depends on applications developed external to the protocol and the IP layer depends on the services of other lower level protocols to manage the actual physical packet delivery. Once again, vendors have developed a number of data link and physical layer implementations that adhere to the IP layer specification. Application level services such as email, FTP, TNET are not part of the TCP IP. These are developed to run on top of this protocol. Today we have uh looked at different network topologies and we have seen features of each topology. Now through the implementation of appropriate switches at different levels we have seen how to increase the effective throughput or the effective bandwidth in an overall network. Now we have also looked at the abstraction in the form of layers and we have seen the seven layer OSI model and the corresponding TCP IP model. We have also looked into how the packets are delivered from source to the destination through a network. I hope you understand these basic concepts regarding networking. More details of the networks would be available in a separate course and I hope you will see these details on a conventional course on computer networks. After discussing some of the basic concepts of computer networks, we stop at this point of time for today. Till next time, Allah hop.