MAC addresses are once again, 48 bits in length
but rather than showing MAC addresses as 48 bit values
in this demonstrations I’m gonna represent MAC addresses
by letter such as A, B, C and D and I’m doing that just for simplicity sake.
So what is a hub do with received traffic.
So in this example, let's assume that A is sending traffic to C.
So the source address of the frame is A and the destination address of the frame is C.
A sends that frame to the hub what will a hub do with the frame?
now because a hub is a multi port repeater in other words it's simply a repeater
with multiple ports and it has no understanding of the traffic it receives
it will simply amplify the signal and send the traffic or frames out of all ports.
So it literally receives a frame, amplifies it
and sends it out of all other ports except the port on which it was received.
so every device in this topology will receive the frame sent from A to C.
so once again A is sending a frame to C
but all devices except A have received the frame.
The network interface cards or NICs of B and D will receive the frame
and read the destination MAC address, they will see in this example
that the destination MAC address is C and therefore the frame is not destined
to themselves and the Network Interface Cards will therefore drop the frame.
So the frames sent to D and B will be dropped
by the Network Interface Cards or NICs of those PCs
Host c however will accept the frame because the frame is destined to it.
So the Network Interface Card or NIC on PC C will read the destination MAC address
and we'll see that the destination MAC address of the frame is at self
and it will therefore received the frame, strip the Layer 2 headers
and pass the packet to the higher layer protocols on the machine
in other words if this is an IPv4 packet it will send
the packet to the IPv4 process running on the machine for further processing
Now let's assume that A ping C, so it requires return traffic
so C replies with the frame with source Mac address being C
and the destination MAC address being A.
C sends that frame to the hub and what does the hub do with the frame?
Now once again a hub is simply a multi port repeater
and it will therefore just amplify the signal 2:37 -0> 2:40
without understanding of the data in the frames.
So the frame is sent to both D and B
which drop a frame because the destination MAC address is not themselves
A will accept the frame because it destined to it, it will then strip the layer 2
headers and send the data to higher layer protocols for further processing.
So A and C are communicating with one another but it’s important to realize
that the hub is a physical layer device that is simply a multi port repeater
and will therefore amplify frames out of all interfaces.
So B and D will see all the frames sent between A and C.
Physically this topology is a star topology but logically it doesn’t work that way.
The physical topology of a hub is a star but logically it's a bus.
It’s very important to realize that there’s a difference
between a physical and logical topology in networks.
The way the network is physically cabled
isn’t necessarily the way the network is going to operate.
It is important to remember that when a device sends traffic in a hub environment
all devices receive a frame, that's exactly the way it works in 10base2 or 10base5.
A hub operates in the same way is 10base2 because when A sends a frame
unto the network all devices receive the frame in the same way as 10base2.
Just like in 10base2 environment when there's a collision on the network
it will affect all devices in the network.
This is a single collision domain.
A collision anywhere will cause devices to back off, send a jamming signal
and then attempt to transmit again.
As you increase the number of devices in a hub environment
the number of collisions increases and your network throughput goes down.
In addition broadcast are received by everyone as this is a single broadcast domain.
A broadcast sent by B is received by everyone.
It’s a single broadcast domain because all devices need to process broadcast
sent by every other device in the network.
Broadcast traffic will flood through the entire network
and interrupt the CPU of every device which is obviously not ideal.
From a bandwidth point of view this maybe 10baseT
where 10 means 10 Mbps but its 10 Mbps shared between all devices.
So assuming that we have 10 Mbps like we do in this example.
And they are four devices in the network with a maximum utilization of 30%
that means that each device only gets 0.75 Mbps throughput
its not 10 Mbps dedicated its 10 Mbps shared between all the devices.
Once again because it's shared you need to divide the bandwidth
by the number of devices in a shared Ethernet environment.
And because you’re not generally getting more than 30 to 40% utilization
because of collisions on the network you need to multiply that
by 30%, 30% being a conservative value.
So your bandwidth is 10 divided by 4*30% which equates to 0.75 Mbps
which is obviously not very good.