Please help with this question, I will give a good rating, Thanks! Suppose an 80
ID: 2084075 • Letter: P
Question
Please help with this question, I will give a good rating, Thanks!
Suppose an 802.11 b station will be transmitting the same packet for the third time (due to unsuccessful delivery- of the previous two transmission attempts). (i) What are the minimum and maximum number backoff slots that could be used? (ii) How much time does the maximum backoff correspond to if 802.1 lb is being used? (iii) The formula given in the lecture notes for the backoff window size is given as CW(n) = min (CW_max, (CW_min + 1)*2^n - 1) Explain in detail how this formula is derived. (iv) What fields in the packet header does a staion use to determine the NAV time?Explanation / Answer
After each unsuccessful transmission the backoff windows size is doubled, up to a maximum value. Once the backoff window size reaches its maximum value it will stay at that value until it is reset.
A backoff occurs the backoff time is uniformly chosen in the range [0, W 1].Since W is used to control the backoff counter its value will affect the performance of the DCF protocol.
There is a range exists for backoff window and within that range only backoff window can be increased as per the requirement. So when the backoff window size reaches to its maximum value, it conflicts with condition where it needs to select the minimum of CWmax or the max range of backoff window as given the problem.
The stations in the same basic service set (BSS) uses this information to update its network allocation vector (NAV) that represents the amount of time it has to defer in accessing the medium.