- 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
- Channel capture effect =
a phenomenon where
one user
of a shared (= broadcast) medium
"captures" (= hogs)
the medium for a
significant time.
- 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 !!!
- The likelihood that
node A will
transmit
before
the node C
is higher !!!
- Node A transmits the
frame for the
first time, so:
- Nodes A and
C
transmits
at the same time:
- 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
Answer:
- 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
- If node A
picks from [0..1] and
node C
picks from [0..3], then:
- 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
- The probability that
node A transmits
before node
C =
0.625
Conclusion:
- This is not fair !!!
- If node A
picks from [0..1] and
node C
picks from [0..3], then:
- 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"
- This problem
is very well-known, and
it is called the: