All language subtitles for 13. How does a Hub forward traffic

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Original subtitles

1

MAC addresses are once again, 48 bits in length

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but rather than showing MAC addresses as 48 bit values

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in this demonstrations I’m gonna represent MAC addresses

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by letter such as A, B, C and D and I’m doing that just for simplicity sake.

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So what is a hub do with received traffic.

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So in this example, let's assume that A is sending traffic to C.

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So the source address of the frame is A and the destination address of the frame is C.

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A sends that frame to the hub what will a hub do with the frame?

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now because a hub is a multi port repeater in other words it's simply a repeater

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with multiple ports and it has no understanding of the traffic it receives

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it will simply amplify the signal and send the traffic or frames out of all ports.

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So it literally receives a frame, amplifies it

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and sends it out of all other ports except the port on which it was received.

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so every device in this topology will receive the frame sent from A to C.

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so once again A is sending a frame to C

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but all devices except A have received the frame.

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The network interface cards or NICs of B and D will receive the frame

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and read the destination MAC address, they will see in this example

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that the destination MAC address is C and therefore the frame is not destined

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to themselves and the Network Interface Cards will therefore drop the frame.

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So the frames sent to D and B will be dropped

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by the Network Interface Cards or NICs of those PCs

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Host c however will accept the frame because the frame is destined to it.

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So the Network Interface Card or NIC on PC C will read the destination MAC address

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and we'll see that the destination MAC address of the frame is at self

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and it will therefore received the frame, strip the Layer 2 headers

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and pass the packet to the higher layer protocols on the machine

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in other words if this is an IPv4 packet it will send

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the packet to the IPv4 process running on the machine for further processing

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Now let's assume that A ping C, so it requires return traffic

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so C replies with the frame with source Mac address being C

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and the destination MAC address being A.

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C sends that frame to the hub and what does the hub do with the frame?

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Now once again a hub is simply a multi port repeater

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and it will therefore just amplify the signal

36 2:37 -0> 2:40

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without understanding of the data in the frames.

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So the frame is sent to both D and B

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which drop a frame because the destination MAC address is not themselves

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A will accept the frame because it destined to it, it will then strip the layer 2

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headers and send the data to higher layer protocols for further processing.

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So A and C are communicating with one another but it’s important to realize

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that the hub is a physical layer device that is simply a multi port repeater

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and will therefore amplify frames out of all interfaces.

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So B and D will see all the frames sent between A and C.

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Physically this topology is a star topology but logically it doesn’t work that way.

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The physical topology of a hub is a star but logically it's a bus.

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It’s very important to realize that there’s a difference

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between a physical and logical topology in networks.

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The way the network is physically cabled

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isn’t necessarily the way the network is going to operate.

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It is important to remember that when a device sends traffic in a hub environment

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all devices receive a frame, that's exactly the way it works in 10base2 or 10base5.

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A hub operates in the same way is 10base2 because when A sends a frame

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unto the network all devices receive the frame in the same way as 10base2.

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Just like in 10base2 environment when there's a collision on the network

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it will affect all devices in the network.

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This is a single collision domain.

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A collision anywhere will cause devices to back off, send a jamming signal

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and then attempt to transmit again.

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As you increase the number of devices in a hub environment

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the number of collisions increases and your network throughput goes down.

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In addition broadcast are received by everyone as this is a single broadcast domain.

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A broadcast sent by B is received by everyone.

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It’s a single broadcast domain because all devices need to process broadcast

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sent by every other device in the network.

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Broadcast traffic will flood through the entire network

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and interrupt the CPU of every device which is obviously not ideal.

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From a bandwidth point of view this maybe 10baseT

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where 10 means 10 Mbps but its 10 Mbps shared between all devices.

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So assuming that we have 10 Mbps like we do in this example.

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And they are four devices in the network with a maximum utilization of 30%

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that means that each device only gets 0.75 Mbps throughput

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its not 10 Mbps dedicated its 10 Mbps shared between all the devices.

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Once again because it's shared you need to divide the bandwidth

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by the number of devices in a shared Ethernet environment.

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And because you’re not generally getting more than 30 to 40% utilization

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because of collisions on the network you need to multiply that

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by 30%, 30% being a conservative value.

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So your bandwidth is 10 divided by 4*30% which equates to 0.75 Mbps

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which is obviously not very good.

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