1 00:00:00,000 --> 00:00:03,000 MAC addresses are once again, 48 bits in length 2 00:00:03,000 --> 00:00:07,000 but rather than showing MAC addresses as 48 bit values 3 00:00:07,000 --> 00:00:15,000 in this demonstrations I’m gonna represent MAC addresses 4 00:00:15,000 --> 00:00:17,000 by letter such as A, B, C and D and I’m doing that just for simplicity sake. 5 00:00:17,000 --> 00:00:20,000 So what is a hub do with received traffic. 6 00:00:20,000 --> 00:00:24,000 So in this example, let's assume that A is sending traffic to C. 7 00:00:24,000 --> 00:00:31,000 So the source address of the frame is A and the destination address of the frame is C. 8 00:00:31,000 --> 00:00:35,000 A sends that frame to the hub what will a hub do with the frame? 9 00:00:35,000 --> 00:00:42,000 now because a hub is a multi port repeater in other words it's simply a repeater 10 00:00:42,000 --> 00:00:47,000 with multiple ports and it has no understanding of the traffic it receives 11 00:00:47,000 --> 00:00:54,000 it will simply amplify the signal and send the traffic or frames out of all ports. 12 00:00:54,000 --> 00:00:58,000 So it literally receives a frame, amplifies it 13 00:00:58,000 --> 00:01:01,000 and sends it out of all other ports except the port on which it was received. 14 00:01:01,000 --> 00:01:07,000 so every device in this topology will receive the frame sent from A to C. 15 00:01:07,000 --> 00:01:11,000 so once again A is sending a frame to C 16 00:01:11,000 --> 00:01:16,000 but all devices except A have received the frame. 17 00:01:16,000 --> 00:01:20,000 The network interface cards or NICs of B and D will receive the frame 18 00:01:20,000 --> 00:01:20,000 and read the destination MAC address, they will see in this example 19 00:01:20,000 --> 00:01:26,000 that the destination MAC address is C and therefore the frame is not destined 20 00:01:26,000 --> 00:01:33,000 to themselves and the Network Interface Cards will therefore drop the frame. 21 00:01:33,000 --> 00:01:37,000 So the frames sent to D and B will be dropped 22 00:01:37,000 --> 00:01:41,000 by the Network Interface Cards or NICs of those PCs 23 00:01:41,000 --> 00:01:45,000 Host c however will accept the frame because the frame is destined to it. 24 00:01:45,000 --> 00:01:51,000 So the Network Interface Card or NIC on PC C will read the destination MAC address 25 00:01:51,000 --> 00:01:55,000 and we'll see that the destination MAC address of the frame is at self 26 00:01:55,000 --> 00:01:59,000 and it will therefore received the frame, strip the Layer 2 headers 27 00:01:59,000 --> 00:02:04,000 and pass the packet to the higher layer protocols on the machine 28 00:02:04,000 --> 00:02:07,000 in other words if this is an IPv4 packet it will send 29 00:02:07,000 --> 00:02:13,000 the packet to the IPv4 process running on the machine for further processing 30 00:02:13,000 --> 00:02:18,000 Now let's assume that A ping C, so it requires return traffic 31 00:02:18,000 --> 00:02:23,000 so C replies with the frame with source Mac address being C 32 00:02:23,000 --> 00:02:26,000 and the destination MAC address being A. 33 00:02:26,000 --> 00:02:30,000 C sends that frame to the hub and what does the hub do with the frame? 34 00:02:30,000 --> 00:02:34,000 Now once again a hub is simply a multi port repeater 35 00:02:34,000 --> 00:02:41,000 and it will therefore just amplify the signal 2:37 -0> 2:40 36 00:02:41,000 --> 00:02:44,000 without understanding of the data in the frames. 37 00:02:44,000 --> 00:02:47,000 So the frame is sent to both D and B 38 00:02:47,000 --> 00:02:51,000 which drop a frame because the destination MAC address is not themselves 39 00:02:51,000 --> 00:02:56,000 A will accept the frame because it destined to it, it will then strip the layer 2 40 00:02:56,000 --> 00:03:01,000 headers and send the data to higher layer protocols for further processing. 41 00:03:01,000 --> 00:03:06,000 So A and C are communicating with one another but it’s important to realize 42 00:03:06,000 --> 00:03:12,000 that the hub is a physical layer device that is simply a multi port repeater 43 00:03:12,000 --> 00:03:15,000 and will therefore amplify frames out of all interfaces. 44 00:03:15,000 --> 00:03:21,000 So B and D will see all the frames sent between A and C. 45 00:03:21,000 --> 00:03:27,000 Physically this topology is a star topology but logically it doesn’t work that way. 46 00:03:27,000 --> 00:03:33,000 The physical topology of a hub is a star but logically it's a bus. 47 00:03:33,000 --> 00:03:36,000 It’s very important to realize that there’s a difference 48 00:03:36,000 --> 00:03:41,000 between a physical and logical topology in networks. 49 00:03:41,000 --> 00:03:44,000 The way the network is physically cabled 50 00:03:44,000 --> 00:03:47,000 isn’t necessarily the way the network is going to operate. 51 00:03:47,000 --> 00:03:52,000 It is important to remember that when a device sends traffic in a hub environment 52 00:03:52,000 --> 00:03:59,000 all devices receive a frame, that's exactly the way it works in 10base2 or 10base5. 53 00:03:59,000 --> 00:04:06,000 A hub operates in the same way is 10base2 because when A sends a frame 54 00:04:06,000 --> 00:04:13,000 unto the network all devices receive the frame in the same way as 10base2. 55 00:04:13,000 --> 00:04:17,000 Just like in 10base2 environment when there's a collision on the network 56 00:04:17,000 --> 00:04:20,000 it will affect all devices in the network. 57 00:04:20,000 --> 00:04:22,000 This is a single collision domain. 58 00:04:22,000 --> 00:04:28,000 A collision anywhere will cause devices to back off, send a jamming signal 59 00:04:28,000 --> 00:04:30,000 and then attempt to transmit again. 60 00:04:30,000 --> 00:04:34,000 As you increase the number of devices in a hub environment 61 00:04:34,000 --> 00:04:39,000 the number of collisions increases and your network throughput goes down. 62 00:04:39,000 --> 00:04:45,000 In addition broadcast are received by everyone as this is a single broadcast domain. 63 00:04:45,000 --> 00:04:48,000 A broadcast sent by B is received by everyone. 64 00:04:48,000 --> 00:04:53,000 It’s a single broadcast domain because all devices need to process broadcast 65 00:04:53,000 --> 00:04:56,000 sent by every other device in the network. 66 00:04:56,000 --> 00:04:59,000 Broadcast traffic will flood through the entire network 67 00:04:59,000 --> 00:05:04,000 and interrupt the CPU of every device which is obviously not ideal. 68 00:05:04,000 --> 00:05:08,000 From a bandwidth point of view this maybe 10baseT 69 00:05:08,000 --> 00:05:16,000 where 10 means 10 Mbps but its 10 Mbps shared between all devices. 70 00:05:16,000 --> 00:05:20,000 So assuming that we have 10 Mbps like we do in this example. 71 00:05:20,000 --> 00:05:26,000 And they are four devices in the network with a maximum utilization of 30% 72 00:05:26,000 --> 00:05:31,000 that means that each device only gets 0.75 Mbps throughput 73 00:05:31,000 --> 00:05:40,000 its not 10 Mbps dedicated its 10 Mbps shared between all the devices. 74 00:05:40,000 --> 00:05:44,000 Once again because it's shared you need to divide the bandwidth 75 00:05:44,000 --> 00:05:48,000 by the number of devices in a shared Ethernet environment. 76 00:05:48,000 --> 00:05:52,000 And because you’re not generally getting more than 30 to 40% utilization 77 00:05:52,000 --> 00:05:56,000 because of collisions on the network you need to multiply that 78 00:05:56,000 --> 00:05:59,000 by 30%, 30% being a conservative value. 79 00:05:59,000 --> 00:06:08,000 So your bandwidth is 10 divided by 4*30% which equates to 0.75 Mbps 80 00:06:08,000 --> 00:06:13,000 which is obviously not very good.