1. Idle sense

2. Idle sense Is doubling CW optimal?

3. Idle Sense
The problem is that DCF is responding to a packet loss by doubling the contention window.
But in wireless, a packet is lost may also (actually may mainly) because of random interference and noise.
DCF is responding to packet loss by assuming the same cause as Ethernet – loss is caused by a packet collision.
(doubling CW does not solve the hidden terminal problem!)

4. Idle Sense
So, the idea is to adjust the size of the contention window based on contention, i.e., the number of stations with data to send.
CW:
if too small, may result in a lot of contention.
If too large, may result in unnecessary longer delay before a packet is transmit.

5. Idle Sense
If given Pe, the probability that a station attempts to transmit,
the probability that a slot is idle is Pi=(1-Pe)N, where N is the number of nodes in the network.
the probability that only one station transmits is Pt=NPe (1-Pe)N-1, where N is the number of nodes in the network.
The probability that there is a collision is Pc=1-Pi-Pt.

6. Idle sense
If Pe is too large, the probability of contention will be large. Else the number of idle slots will be large.
What is the optimal Pe?
Assuming the data packets are of the same size and the data rates are the same, the normalized throughput is
Pt Tt / (Pt Tt + Pc Tc + Pi Ts)

7. Idle sense To maximize throughput is to minimize (Pc a + Pi ) / Pt ,
where a = Tt / Ts.
Therefore, we should take a derivative over Pe and see when it becomes 0.
It leads to The value that satisfies this equation is the optimal Pe.

8. Idle Sense To get the optimal Pe, Let We have
This can then be solved numerically.
If in b, a = Tt / Ts = (about 1200 / 20)
So, Pe = /N.

9. Idle sense Very interestingly,
Also note that the average number of idle periods is ni=Pi/(1-Pi).
So, the optimal average number of idle period is roughly a constant, around 5, regardless of the number of nodes in the network!

10. Idle Sense
But, how can the nodes automatically set their Pe to be the optimal value?

11. Idle sense
You can observe the average number of idle time slots between two transmission attempts.
If it is below 5, it means that Pe is too large, otherwise it is too small.

12. Idle Sense If you believe Pe is not optimal, you should adjust it.
How?

13. Idle Sense
Using AIMD.

14. channels
In b/g, there are 11 channels, starting at 2.412GHz at a spacing of 5MHz.
Each channel owns a bandwidth of 22MHz.
So, only 3 non-overlapping channels, 1,6,11.
802.11a has more channels and you may check at

15. 802.11a/g modulation
From wiki: