Idle sense

1
Idle sense
2
Idle sense

• Is doubling CW optimal?
• http//conferences.sigcomm.org/sigcomm/2005/paper-
  HeuRou.pdf

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 PtNPe (1-Pe)N-1, where N is the number of
  nodes in the network.
• The probability that there is a collision is
  Pc1-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 802.11b, a Tt / Ts 68.17 (about 1200 /
  20)
• So, Pe 0.1622/N.

9
Idle sense

• Very interestingly,
• Also note that the average number of idle periods
  is niPi/(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
802.11 channels

• In 802.11 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
  http//www.moonblinkwifi.com/80211a_frequency_chan
  nel_map.cfm

15
802.11a/g modulation

• From wiki http//en.wikipedia.org/wiki/IEEE_802.1
  1a-1999