CS485 Sylabus

### The unfairness of the Ethernet Backoff Algorithm

• Channel capture effect

• Channel capture effect:

 Channel capture effect = a phenomenon where one user of a shared (= broadcast) medium "captures" (= hogs) the medium for a significant time. Wikipedia: click here

• Unfairness of the Ethernet's backoff algorithm

• Suppose:

 The nodes A and C in an Ethernet has a large number of frames to transmit

• Consider the following scenario:

• Nodes A and C transmits at the same time:

• The transmissions of nodes A and C will collide and they will back off

Suppose:

 Node A picked 0 Node C picked 1

Then node A will transmits first:

Node A will complete its transmission

• When node A is done, A will transmit again

The node C will also transmit.

Their transmissions collide again and back off:

Notice that:

• Node A transmits the frame for the first time, so:

 Node A will pick a random number from the range range [0..1]

• Node C transmits the frame for the second time, so:

 Node C will pick a random number from the range range [0..3]

• Problem:

• The likelihood that node A will transmit before the node C is higher !!!

 The probability is calculated in the section below....

• This is unfair for node C !!!

• Unfairness calculation....

• Question:

• If node A picks from [0..1] and node C picks from [0..3], then:

 What is the probability that node A will transmit before node C ?

Pre-question:   when will node A transmit before node C:

• Node A transmits before node C if:

 Node A picks a smaller (random) value than node C

• A will transmit before C when:

 A picks 0 and C picks 1 A picks 0 and C picks 2 A picks 0 and C picks 3 A picks 1 and C picks 2 A picks 1 and C picks 3

• The probability that C will pick any one number is 0.25

The probability that B will pick any one number 0.5

• The chance that C will transmit before B is:

 (0.25 × 0.5) + (0.25 × 0.5) + (0.25 × 0.5) + (0.25 × 0.5) + (0.25 × 0.5)    =    0.625

• Summary:

• If node A picks from [0..1] and node C picks from [0..3], then:

 The probability that node A transmits before node C = 0.625 The probability that node C will transmit before node A = 0.125

Conclusion:

 This is not fair !!!

• And the bad news is:

 It will get even worse if C loses again ...

Suppose node B transmits first and during B's transmission the node A, scheduled a transmission.

Then:

The likelyhood that node C will transmit first is now very very smaller !!!

• NOTE:

• This problem is very well-known, and it is called the:

 "Ethernet Capture Effect"