1. Prepared by: Eng. Ali H. Elaywe1 Arab Open University - AOU T209 Information and Communication Technologies: People and Interactions Twelfth Session – Part1

2. Prepared by: Eng. Ali H. Elaywe2  This part is based on the following reference: Book N: Networks Part1: Local Area Network Reference Material

3. Prepared by: Eng. Ali H. Elaywe3  Ethernet Although the early mainframe and minicomputer systems provided experience of connecting terminals to computers, the first standard for a local area network of autonomous computers was Ethernet, developed by a consortium of companies (Xerox, DEC and Intel) in the mid-1970s  Frame We tend to use packet when talking generally (in the Internet) about Data Communication concepts, using the more precise terms when appropriate, for example, frame when talking about Ethernet Continue Topic 1: Ethernet

4. Prepared by: Eng. Ali H. Elaywe4  Collision If two computers send a data frame at the same time, and they are sharing a common channel, then the signals interfere and the data is corrupted – this is described as a collision A protocol is needed to specify how the computers are to share a common communication channel. This is described as Medium Access Control  Collision Domain Ethernet LANs use a shared communication channel, which can either be a bus, a star arrangement based around a hub, or combinations of the two In all cases the shared channel is referred to as a collision domain, an intuitive label reflecting the fact that frames sent within the collision domain can collide with other frames Continue Sub-Topic 1.1: Ethernet Fundamentals

5. Prepared by: Eng. Ali H. Elaywe5  Twisted pair Is one type of medium that can be used to connect computers together The name ‘twisted pair’ is derived from the way the wires are twisted together within the cable Each cable contains four separate wires, one pair (two wires) being used by each computer to send signals. Each send pair then becomes the receive pair for the other computer So each computer is connected to two pairs, sending frames on one, and receiving frames on the other The ‘twisting’ helps to increase the maximum length that can be used between two computers ? Continue

6. Prepared by: Eng. Ali H. Elaywe6  Crosstalk When electrical signals are sent over an electrical medium, such as twisted pair, the signal is attenuated (weakened) and distorted by the characteristics of that medium. In particular, signals from one wire spilling over into the other cause an undesirable effect known as crosstalk The twisting of the pairs helps to reduce crosstalk Currently, most local area networks used twisted pair cables and operate in half-duplex half-duplex is a mode in which each computer can either transmit or receive data, but cannot do both simultaneously Full duplex is a mode in which each computer can transmit and receive data, simultaneously Continue

7. Prepared by: Eng. Ali H. Elaywe7  Carrier Sense Multiple Access with Collision Detection (CSMA/CD) The computers in an Ethernet LAN use a number of procedures to increase the likelihood of successful transmission. The Ethernet protocol is called Carrier Sense Multiple Access with Collision Detection (CSMA/CD) and this name encompasses the main procedures. Now let’s see How CSMA/CD work? ‘Carrier sense’ means that computers do not just transmit at any time. Instead, they first monitor the communication channel to see if it is in use. Once a computer detects that the channel is free, then it can transmit Collision detection: Once a computer in an Ethernet LAN has started to transmit, it again monitors the channel, to see whether its frame has collided with a frame from another computer. This is the ‘collision detection’ part of the protocol. If a collision does occur, the computers stop transmitting. Each waits for a short, randomly chosen, time interval (so that they do not simply collide again), and then re-transmits Continue

8. Prepared by: Eng. Ali H. Elaywe8  Propagation delay One characteristic of networks that the CSMA/CD protocol has to overcome is the time it takes for a frame to travel across the network For example, it takes a finite time for each frame to travel along cables and through repeaters The time it takes the beginning of a packet to travel between two points (or two nodes) is called the propagation delay, often shortened to just delay  Repeater A repeater makes a copy of any packet received on one of its ports and sends it out on all other ports Continue

9. Prepared by: Eng. Ali H. Elaywe9  Round Trip Delay It is a time required for data to reach the extremity of the network and to come back to the sender For the ‘collision detect’ part of the protocol (CSMA/CD) to work properly, it is necessary that the transmitting computer ‘knows’ when it can stop monitoring for a collision. Once this point is reached, it can send the rest of its frame without worrying about collisions In practice, the worst case scenario occurs when a frame transmitted at one end of a collision domain collides with another frame at the farthest end. This is when a transmitted frame from one computer collides with a frame that has just been transmitted by the computer that is the furthest network distance away This is the worst case because this is the longest period of time for a collision to be detected by the transmitting computer, which has to wait for notice of the collision to propagate back. This time period is called the round trip delay, because it is the time it takes for a frame to take a ‘round trip’ to the furthest limit of the network Continue

10. Prepared by: Eng. Ali H. Elaywe10  Slot time Since computers connected to Ethernet only detect collisions while transmitting (or sending) frame So any frame must be able to cover the round trip distance within the time it takes a computer to send the minimum length frame. This period is called the slot time, and is specified at 512 bits (in CSMA/CD protocol), the number of bits that can be transmitted in the worst-case round-trip delay The slot time is defined in bits(512 bits), rather than time (in µs) because Ethernet can operate at different bit-rates Continue

11. Prepared by: Eng. Ali H. Elaywe11 The round trip delay can be calculated from the minimum frame length (512bits) and the bit rate Round Trip Delay = number of bits sent / bit rate When designing a network, it is the slot time that determines the maximum distance between any two computers Note: Slot time is less of a determining factor in a modern network; but still has to be met !! Continue

12. Prepared by: Eng. Ali H. Elaywe12  Activity 14 (reflection) What is the maximum round trip delay for an Ethernet system operating at 10 Mbit/s? The round trip delay can be calculated from the minimum frame length and the bit rate Delay = number of bits sent / bit rate = 512 / 10 × 10 6 seconds = 51.2 µs You should be able to see that the round trip delay for an Ethernet system operating at 100 Mbit/s is 5.12 µs Continue

13. Prepared by: Eng. Ali H. Elaywe13  Activity 16 (self-assessment) Why is it that, once a computer has monitored for the slot time, and detected no collisions, it can transmit the rest of its frame without risk of collision? If no collision has been detected in the slot time, then no other computer, even the one at the furthest limits of the collision domain, has tried to transmit. Once this point is reached, then all other computers will detected the transmitting computer’s frame through the carrier-sense (CS) part of the protocol, as by this time the frame will have reached all nodes within the same collision domain Continue

14. Prepared by: Eng. Ali H. Elaywe14  Activity 17 (exploratory) What would happen if this computer was sending a very large frame? Once a computer has passed the slot time without detecting a collision, it has gained control of the channel, and any other computer trying to access the channel will have to wait. In principle this would contradict the equal access (or equitable access) principle underlying the ‘multiple access’ part of the protocol, also there could be long delays while one computer with a long message to transmit ‘hogs’ the channel You will see later (in Activities 24 and 26), that for this reason Ethernet frames also have a maximum size (= 1518 bytes = 12 144 bits) Continue

15. Prepared by: Eng. Ali H. Elaywe15  Backoff algorithm In order to minimize repeat collisions, Ethernet uses an algorithm commonly known as the backoff algorithm, where each computer waits a random integral number of slot times before trying again The CSMA/CD protocol ensures that access to the channel is shared fairly amongst the computers within a collision domain Collisions are a normal part of operation, and with the correct operation of the backoff algorithm, frames are normally transmitted successfully Traffic is a term used to describe the quantity of data on a network over a given time However, if the total traffic being carried in a collision domain reaches the point where there are excessive collisions, and the backoff algorithm regularly reaches its limits (if it tries 16 times without success), then frames will be discarded !! Continue

16. Prepared by: Eng. Ali H. Elaywe16  Activity 18 (self-assessment) – (Important) Why does separating the attempts to transmit, of two computers, by at least one slot time ensure they do not collide? The slot time is chosen to allow a frame to propagate across the extremes of a collision domain and back again; called the round trip delay Therefore, if one computer waits for one slot time before beginning to transmit, it is certain to detect if another computer had started to transmit one slot time earlier Continue

17. Prepared by: Eng. Ali H. Elaywe17 Frames that are discarded are not re-transmitted by the Ethernet protocol. As traffic increases towards this critical situation, users may notice delays as their networking software tries to cope with the high collision rate and lost frames  Best effort service Ethernet will tend to share the misery evenly amongst all the computers. This mode of operation, where each computer competes on an even playing field, without rules to ‘guarantee’ delivery is referred to as a best effort service Continue

18. Prepared by: Eng. Ali H. Elaywe18  Activity 20 (self-assessment) How long does it take an Ethernet network operating at 10 Mbit/s to send 64 bytes? 64 bytes = 64 × 8 bits = 512 bits time taken = number of bits to be transferred / bit rate = 512 / 10 × 10 6 seconds = 0.0000512 s = 51.2 µs Continue

19. Prepared by: Eng. Ali H. Elaywe19  Activity 21 (self-assessment) Two computers are joined by 300 meters of cable. If a packet travels along the cable at a speed of 1.77×10 8 m/s, how long will it take for a packet to travel between the two computers? The time taken is the distance traveled divided by the speed. (If you are not sure about this, think of a journey of 120 km at 60 km per hour: it would take 2 hours, which is calculated by dividing 120 by 60.) So time = distance / speed which can be written very briefly as t = d / v where I have written v for speed because mathematicians often use the term ‘velocity’ and hence use v. So here t = 300 / (1.77 ×10 8 )s = 0.00000169 s = 1.69 µs Continue

20. Prepared by: Eng. Ali H. Elaywe20  Buffer The original choice of slot time was based upon what was believed to be a sensible limit for a collision domain using the cables and repeaters of the time Slot time can be calculated for any network by adding together the delay due to cable runs and network devices within a single collision domain Devices such as repeaters use a buffer to store frames while new frames are prepared for transmission on all ports. Storage in a buffer introduces a delay Continue

21. Prepared by: Eng. Ali H. Elaywe21  Activity 22 (self-assessment) – (Important) A repeater introduces a delay of 3.5 µs. What length of cable is this equivalent to, if the propagation speed is 1.77×10 8 m/s? Here we need to treat the repeater’s delay as a time when otherwise the signal would have been traveling at 1.77 × 10 8 meters per second, and find out how far it would have traveled. This distance can be found from distance = speed × time or d = v × t (Think of a journey of 2 hours at 40 km per hour: you would travel 80 km, which is calculated by multiplying 40 by 2.) Before I can do the calculation I have to convert 3.5 µs into seconds (because the speed is in meters per second): 3.5 × 10): 3.5 × 10 -6 seconds. Then d = 1.77 × 10 8 × 3.5 × 10 -6 m = 619.5 m Continue

22. Prepared by: Eng. Ali H. Elaywe22  Activity 23 (self-assessment) – (Important) Given a propagation velocity of 1.77×10 8 m/s and a delay of 3.5 µs introduced by each repeater, what is the round trip delay for two computers at either end of a 2500 m link that includes three repeaters? We will do the calculation by finding the one-way time and then doubling it for the round trip 1- The signal travels 2500 m (cable). Using the formula  t = d / v  (as in Activity 21) gives  t = 2500 / (1.77×10 8 ) s = 0.0000141 s = 14.1 µs 2- The three repeaters add a further 3 × 3.5 µs delay, which is 10.5 µs. So the total one-way time is  14.1 + 10.5 µs = 24.6 µs 3- The round-trip delay is twice this value: 49.2 µs, or just under 50 µs

23. Prepared by: Eng. Ali H. Elaywe23  Framing and Frame The principal function of Ethernet is to transfer data between the computers that it serves Framing: is the process of adding extra bits, for the purposes of controlling the transmission of data between computers Frame: is the combination of the data and the control bits In an Ethernet frame the control information is contained in a header at the beginning of the frame, and includes the address of the computer the data is being sent to – the destination address The Ethernet frame is shown in Figure 1 The frame if it is divided into sections, which are called fields Continue Sub-Topic 1.2: The Ethernet frame

24. Prepared by: Eng. Ali H. Elaywe24 Figure 1 An Ethernet frame Header Trailer Data Whereas the minimum Ethernet frame size = 64 bytes (or 512 bits), The maximum Ethernet frame size = 1518 bytes (or 12144 bits) Continue

25. Prepared by: Eng. Ali H. Elaywe25  Fields of Ethernet frame The length of each field is given in bits or bytes, and all the fields before the data make up the header. The field after the data is called the trailer A- Preamble field: The first field is called the preamble, and it allows a little time for the electronics of the receiver to recognize that it is receiving a valid frame. For this reason these bits are not considered to be available for carrier sense, and are not usually counted when specifying the frame length Continue

26. Prepared by: Eng. Ali H. Elaywe26 B- Destination and source address fields: The next field gives the destination address, immediately followed by the computer’s own address (the source address) Both fields are of 48 bits and allow any receiver reading the frame to determine whether the frame is intended for it (destination), and where it has come from (source) Addresses in Ethernet are divided into two parts:  1- The first 24 bits are organizationally unique identifiers (OUI) allocated to manufacturers of network interfaces by the IEEE standards association  2- The second 24 bits are used by a manufacturer to identify each of the interfaces they produce This gives each Network Interface Card (NIC) a unique address; this is also called the Media Access Control (MAC) address Continue

27. Prepared by: Eng. Ali H. Elaywe27  Activity 24 (self-assessment) An Ethernet frame has a maximum frame size of 1518 bytes (excluding preamble). How many bits is this? There are 8 bits in a byte, so the maximum sized frame is: 1518 × 8 bits = 12 144 bits  Activity 25 (self-assessment) How many different addresses are available to each manufacturer? Each manufacturer is allocated 24 bits. This provides for a theoretical maximum of 2 24 = 16 777 216 unique addresses Continue

28. 24 bits OUI Personal to the manufacturer MAC Address Ethernet Frame OUI: Organizationally Unique Identifier (24 bits) 24 bits  2 24 = 16 777 216 MAC (Media Access Control) address = Ethernet Address Prepared by: Eng. Ali H. Elaywe28 Continue

29. Prepared by: Eng. Ali H. Elaywe29 C- Type / length field: The next field is called type/length. The binary value of this field determines whether this is used for type or length:  1- In type mode, the value indicates which type of packet is being carried in the data field. For example, a unique value is used to indicate when an IP packet is carried in the Ethernet frame  2- When the type is not given, then the ‘value’ of the field indicates the length of the data in the data field D- Data field: The data field is the longest field in the frame, and can contain between 46 and 1500 bytes of data The maximum size has been chosen to ensure that each computer in a domain has equal access to the channel After a computer has transmitted one frame, it has to release the channel and compete with any other computer trying to send data. If a computer has more information to transmit than can be contained in the data field of one frame, then it will need to divide this information between a number of frames Continue

30. Prepared by: Eng. Ali H. Elaywe30  Activity 26 (revision) Explain why it is necessary to have a maximum frame size when using the CSMA/CD protocol, if each computer is to have equal access to the channel Once a computer has transmitted for a period in excess of the slot time, the CSMA/CD protocol does not allow another computer to access the channel until the first has finished sending its frame. There has to be a maximum frame size to allow all computers a chance to compete for access to the channel periodically, especially when the network is carrying a lot of traffic Continue

31. Prepared by: Eng. Ali H. Elaywe31 E- Frame check sequence field: The last field in each frame contains a frame check sequence called CRC (Cyclic Redundancy Check) This is a special arithmetic procedure applied to each frame at the transmitter. By applying a similar procedure at the receiver, and comparing it with the contents of this last field, it is possible to determine if any errors have occurred during transmission. If errors are detected the frame is discarded Continue

32. Prepared by: Eng. Ali H. Elaywe32  Activity 27 (revision) How does a computer connected to Ethernet by twisted pair detect a collision? Computers using twisted pair cables operate in half-duplex mode. In this configuration frames do not actually collide on the channel, but they do pass each other at some point between the two transmitting computers. Detection that more than one computer has been active is made at the computer, which can sense data on its transmit and receive paths simultaneously. Collisions can only be detected when a computer is sending a frame Continue

33. Prepared by: Eng. Ali H. Elaywe33  Activity 28 (exploratory) If the minimum frame length is 512 bits (64 bytes), why is the minimum data field size only 46 bytes? Looking back at Figure 1 we can see that the data field is the only one with a variable length. We'll add together the fixed parts starting at the left hand end We do not include the 64 bit preamble as this is used by the receiver to detect a frame. This leaves a 48 bit destination address, a 48 bit source address, a 16 bit length and a 32 bit frame check sequence. This gives a total of: 48 + 48 + 16 + 32 = 144 bits With a slot time of 512 bits, this means the data field must have a minimum length of 512 – 144 = 368 bits = 46 bytes

34. Prepared by: Eng. Ali H. Elaywe34  In this sub-topic we are going to consider when it becomes necessary to extend or grow LANs beyond a single collision domain, and how this can be achieved  However there are limits to how far a simple LAN can grow The two main limits on the growth of a LAN are the 1- physical distances involved and the 2- slot time defined in the Ethernet standard  Computers sharing a collision domain must meet the delay requirements of the slot time. As networks cover larger distances the propagation delay will increase, and new frames will need to be copied by repeaters, as the signals that represent the bits become attenuated or weak Continue Sub-Topic 1.3: Extended LANs

35. Prepared by: Eng. Ali H. Elaywe35  As you will see in the next section, loss of signal can be overcome with repeaters, but delay is less easily dealt with !!  The impact of delay has become more significant as the speed at which Ethernet operates has increased, from the dominant 10 Mbit/s used on bus technology, to the 100 Mbit/s and 1 Gbit/s systems that are common with hub operation  So Ethernet LANs can operate at different data rates. Each data rate can be carried on different types of cables, including twisted pair, co-axial cable and fibre optic  The use of Ethernet on each medium is identified by a code, 10baseT, for example. This code can be interpreted as 10 Mbit/s (10), operating over twisted pair cable (baseT) Continue

36. Prepared by: Eng. Ali H. Elaywe36  For each media type and bit-rate the IEEE standards organization has laid down guidelines for the maximum delay for cable segments and network devices For example, the standard specifies that a maximum of 100 meters of cable should be used between a computer and a hub using 10baseT Continue

37. Prepared by: Eng. Ali H. Elaywe37  Activity 29 (self-assessment) Each frame contains two addresses. Explain how these are used, within a single collision domain, for computer A to reach B, and for B to know who to send a reply to. Just like the post office mail, each computer has a unique address, and when A sends a frame to B it writes B’s address in the header; this is called the destination address. Every computer receives a copy of the frame, but only B’s address matches the destination address in the header. As well as writing the destination address in the header, A also writes its own address; called the source address. If B wishes to send frames to A then it uses the source address in the header as its destination address Continue

38. Prepared by: Eng. Ali H. Elaywe38  Activity 30 (self-assessment) If the minimum frame length is kept at 512 bits, what is the slot time for 10, 100 and 1000 Mbit/s? The slot time is calculated by working out how long it takes to send 512 bits At 10 Mbit/s:  slot time = 512 / 10 × 10 6 s = 51.2 µs At 100 Mbit/s:  slot time = 512 / 100 × 10 6 s = 5.12 µs At 1000 Mbit/s:  slot time = 512 / 1000 × 10 6 s = 0.512 µs Continue

39. Prepared by: Eng. Ali H. Elaywe39  A- Extended LAN by using repeaters: (Important) A single continuous section of LAN cable, called a segment The LAN can be extended by using repeaters to connect segments as shown in Figure 2 A repeater is a device which regenerates digital signals. Provided the input signal is strong enough for the ‘1’s and ‘0’s to be recognizable, a repeater will re-create a digital signal at its original level In terms of the OSI Reference Model, a repeater is at the physical layer. However there are limits to the number of repeaters which can be used in a LAN, mainly because they introduce delay Continue

40. Prepared by: Eng. Ali H. Elaywe40 Figure 2(a) shows a simple repeater with just two connections (often called ports). It lies in a length of cable between two computers and regenerates a signal before passing it on Figure 2(b) shows a slightly different sort of repeater. It lies at the ‘hub’ of a star network and not only regenerates a signal coming in from one computer but broadcasts it to all the other computers A hub has a finite number of ports, and this limits the number of computers that can be connected to it. Once the limit is reached, two (or more) hubs can simply be connected together to form a larger network Continue

41. Prepared by: Eng. Ali H. Elaywe41 Continue Figure 2 Extending a LAN using repeaters Segment

42. Prepared by: Eng. Ali H. Elaywe42  B- Using a bridge to divide a LAN: (Important) The reasons for dividing a LAN by using the bridge: Increasing the size of a LAN by adding more computers usually increases the amount of data being sent over the LAN. This will degrade the performance of the LAN, as it becomes more and more congested This problem can be alleviated by using a bridge, rather than a repeater, to divide the Local Area Network into separate collision domains A bridge filters frames by reading the destination address in the frame. The bridge will only forward a frame onto a connected collision domain if the frame is addressed to a computer on the opposite side of the bridge Continue

43. Prepared by: Eng. Ali H. Elaywe43 In Figure 3, a bridge separates the LAN into three collision domains, each containing three computers. Using a bridge to link collision domains means that a frame sent by a computer to another computer in the same collision domain is not transmitted onto any other collision domains A frame sent from a computer in one collision domain to a computer in another collision domain is only passed onto that collision domain by the bridge. So the amount of data on each collision domain is lower than it would have been without the bridge Bridges are often called switching hubs in commercial literature, which reflects the way a switching matrix is used to interconnect ports. These can have many ports, and can be referred to as multi- port bridges Continue

44. Prepared by: Eng. Ali H. Elaywe44 Continue Figure 3 Using a bridge to divide a LAN 1 Collision domain 1 3 Collision domain 3 2 Collision domain 2 Multi-port bridge Switching hub

45. Prepared by: Eng. Ali H. Elaywe45 A bridge needs a way of deciding, from the destination address of a frame, whether the frame should be forwarded to a connected collision domain and, in the case of a multi-port bridge, which collision domain to forward it to. This information is stored by the bridge in a forwarding table, which lists the addresses of all the computers on the LAN, and the port number which gives access to that destination address Bridges learn where to forward frames addressed to particular computers by building a forwarding table through the analysis of source and destination addresses The bridge in LAN uses the forwarding table to decide where the packet should be forwarded to Continue

46. Prepared by: Eng. Ali H. Elaywe46  Activity 31 (self-assessment) – (Important) Figure 4 shows a bridge connecting three collision domains, using ports A, B and C. Construct a forwarding table, after the bridge has received the following frames: [to 02 : from 01], [to 07 : from 02], [to 05 : from 08], [to 03 : from 02]  The bridge constructs the following forwarding table. It does this by recording the source address against the port number for any packet it receives  Destination, port  01, A  02, A  08, B Continue

47. Prepared by: Eng. Ali H. Elaywe47 Continue Figure 4 Bridge ports

48. Example: Construct the forwarding table after the bridge had received the following frames: To 02 : from 01 To 02 : from 07 To 07 : from 04 To 07 : from 06 To 03 : from 01 DestinationPort 1A 7B 4C 6B Prepared by: Eng. Ali H. Elaywe Continue 48

49. Prepared by: Eng. Ali H. Elaywe49  Broadcast and multicast addresses: 1- A broadcast address is recognized by all interfaces. This means that it is received and read by all computers, and forwarded by all bridges 2- A multicast address is a designated address that can be received by many computers  C- Router concepts: Routing is the process of finding a path through a network from source to destination So far we have discussed breaking up LANs into smaller collision domains, but often what is required is to join existing, separate LANs into a single larger one The complex routing of data, often needed in a larger LAN, can be carried out by devices called routers Continue

50. Prepared by: Eng. Ali H. Elaywe50 A router uses information about the structure of the network as a whole, and hence can choose an optimum route for sending frames from one computer to any other on the LAN Router operation is very similar to that of a bridge, and routing decisions are also based upon address information contained in packets that are read by the router  D- Gateways: Routers are also used to interface between a LAN and a WAN. When carrying out this function the routers are sometimes called gateways, reflecting their role of extending the reach of a user beyond the boundary of a LAN In practice the terminology for connecting devices (repeaters, bridges, routers, gateways), is not always consistent, as the combination of functions which can be carried out by a given device may depend on the manufacturer Continue

51. Prepared by: Eng. Ali H. Elaywe51  Relate devices to the layers of the OSI Reference Model: A helpful way to think about these devices is to relate them to the layers of the OSI Reference Model Repeaters and hubs operate at the physical layer Bridges (or switching hubs) are at the data link layer Routers are network layer devices, and make routing decisions Gateways can operate at the network layer or above

52. Prepared by: Eng. Ali H. Elaywe52  A landmark ‘Ethernet’ paper published by two Xerox PARC scientists in 1976 can be found in the Module 3 Companion, and its bibliographic reference is: Metcalfe, R. and Boggs, D. (1976): Ethernet: Distributed Packet Switching for Local Computer Networks, Communications of the ACM, volume 19, no 7 Continue Sub-Topic 1.4: A seminal paper on Ethernet

53. Prepared by: Eng. Ali H. Elaywe53  Activity 34 (exploratory) Would you describe the collision domains described in this book as more like a tree or a star?  This Book describes the topology of the networks as more like a star. This is due largely to the use of repeater hubs. This change in topology stems from the high reliability and low cost of hubs, which has negated some of the reliability concerns. It is testimony to the robust design of Ethernet that it has survived this shift in network design