24
ELEC6219: Wireless and Mobile Networks Session 6: Local Area Networks - Case study: Ethernet (Part II) Adriana Wilde and Jeff Reeve

ELEC6219: Wireless and Mobile Networks Session 6: Local Area Networks - Case study: Ethernet (Part II) Adriana Wilde and Jeff Reeve

Embed Size (px)

Citation preview

ELEC6219: Wireless and Mobile Networks

Session 6:Local Area Networks - Case study:

Ethernet (Part II)

Adriana Wilde and Jeff Reeve

At the end of this session you will be able to…

… evaluate CSMA/CD efficiency

… appraise the Ethernet Retransmission algorithm (exponential backoff )

… differentiate between frames and packets

… apply the backwards learning algorithm in Ethernet network switches

What

1 2 3 4 5

100%

0% 0%0%0%

The total time required to transmit the message is:

1. 3*RTT2. 5*RTT3. 7*RTT4. 11*RTT5. None of the above

These answers refer to the exercise presented in the OHP, from a past exam question

Yesterday we discussed the minimum frame size requirement in Ethernet

We want to avoid sending frames that are “too short”.

Where were we?

p. 302-303

Independent Traffic (station model)Single ChannelObservable Collisions (collision assuption)Continuous or Slotted TimeCarrier Sense or No Carrier Sense

Assumptions for Dynamic Channel Allocation

p. 249-250 (T.4e)p. 278-279 (T.5e)

For a 10Mbit/s Ethernet, with a maximum length of 2500 m and 4 repeaters, the RTT is set at about 50 µs (rounded up to 51.2), i.e. a minimum frame size of 512 bits.

The maximum round-trip delay of about 51.2 microseconds also imposes a restriction on the physical size of the LAN: D = RTT/2 * v (where ‘v’ is the signal propagation speed)

This results in a maximum value for D of 5120 m – if the network was larger than this then it would be impossible to rely on the RTT limit of about 51.2 µs (and in practice the cable losses restrict the overall network size to be much smaller).

The “small print”

p. 277-278 (T.4e)p. 302-303 (T.5e)

Talking point

CSMA/CD Efficiency (I)

CSMA/CD Efficiency (I)

CSMA/CD Efficiency (II)

CSMA/CD Efficiency (III)

Ethernet retransmission algorithm (I)

This is the exponential back-off algorithm – it has been highly successful:When a collision is detected, both nodes wait and try again (until successful)

The basic concept is that the wait time is doubled at every attempt (hence the term ‘exponential’)

This would not work if the algorithm was deterministic: both nodes would simply double their times in step and suffer repeated collisions

Ethernet retransmission algorithm (II)

Talking Point

Talking Point

Ethernet retransmission algorithm (III)

LAN Terminology

Unit Function

Repeater a simple amplifier (collision aware)

(rarely used in modern systems)

Hub essentially a multi-port repeater

Switch (or ‘Bridge’, or ‘Transparent Bridge’) S&F Frames (‘S&F’ = ‘store and forward’)

Router S&F Packets (typically IP packets)

(i.e. Network (L3) Level

LAN

NETWORK

Ethernet Switches (I)

Consider a four-port switch, connected as below:

A Port 0

B Port 1

C Port 2

D Port 3

When the switch receives a frame, it must decide what to do with it.

p. 281-283 (T.4e)p. 306-308 (T.5e)

• Is it a broadcast (is the destination address = FF…FF)? If so, the frame is flooded (i.e. forwarded on every port - this frame has to go everywhere, so there is no choice)

• Is the destination address location known? (i.e. the port that it is connected to)– No: flood the frame (i.e. transmit on every port)– Yes: filter the frame (i.e. transmit only on the

port connected to the destination address)

How do we know the location of the destination address?

Ethernet Switches (II)

p. 306-308

How do we know the location of the destination address?

From an internal table: we just look it up.

However, this table is initially empty (e.g. on switch-on), the switch needs to build it up by examining the traffic through the switch. This is the backwards learning algorithm: the switch inspects all the source addresses and updates the internal table.

Ethernet Switches (III)

p. 306-308

Consider an example: after reset, a frame is sent from A to B: this is flooded onto ports 1,2 and 3

Next a frame is sent from D to A: this frame is filtered (transmitted on port 0 only) because the switch is now aware (from the first frame) of the location of A.

The ‘time’ field is to allow a timeout on every entry. This allows the connections between the nodes and the switch to be changed without reconfiguration: after a short delay the existing entries will timeout and be removed, allowing the new locations to be discovered and entered

into the table.

Ethernet Switches (IV)

p. 306-308

At the end of this session you should be able to…

… note the limitations of CSMA/CD

… apply the Ethernet Retransmission algorithm (exponential backoff ), appraising Ethernet as a successful LAN technology

… differentiate between frames and packets

… note the advantages of switches for network management

Enter question text...

1 2 3 4

13%

37%

21%

29%

Study strategies

1. I have a partner and/or group study to revise the material

2. I devote a couple of hours every week to revise the lecture material

3. Both of the above4. None of the above (yet)

What next?

This is the end of the first two layers of our model.

We have just “loosely” covered the first 4 chapters in the Tanenbaum