Q 3 Project 2: Carrier Sense Multiple Access: CSMA (20pt) In WiFi systems, sodes
ID: 3718181 • Letter: Q
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Q 3 Project 2: Carrier Sense Multiple Access: CSMA (20pt) In WiFi systems, sodes adopt CSMA MAC questions. (All answers limit to 1-2 sentences, please.) 1) For three active tusers, n 3, use an example to illustrate how two consecutive collisions oecurred protocol to access to the shared media through contention ning that there are 32 slots available for uses Answer the following 2) For two active users, n 2, say node A and node B. If uode A selected a backoff counter (BC) of 25 but it failed to content for the channel after 15 slots since it detected node B begatn to send. What was the BC of node B? What is the probability for node A to win the contention for the channel for next round 3) Calculate the minimal and maximum values of throughput when there is one active node, n-i The packet size is 1024 bytes and date rate 12M bits s. Explain how throughout varies based on the project and write a formula to calculate the average of throughput. (hint: the average throughput equals to the summation of cach throughput multiplying corresponding probability.) 4) If node A has a message with the size of 10K B, hoe may packets needed to seud it out completely. 5) If there are 1200 packets that were sent through the channel, how many packets was sent by node A. Assume there were n8 active users 6) When does the protocol double its content window? Why ?Explanation / Answer
Given that in wifi systems nodes adopt the CSMA( carrier sense multiple access) MAC(media access control) protocol to access shared media through contention.
At the begining CWMin = 31 i.e there are 31 slots available for users.
In this protocol node itself verifies the traffic before transmitting.
(a)
At the begining, if node A pick up the back off counter BO=10 , the probability to win the channel contention means the channel should not be activated with a formal disagreement. Given that after 3slots node A detected that some other node began to sent package. Hence the probability to win the channel actication is 10 /(31-3) = 0.357
Node A Has better chance to win the channel contention in the next round, as the probablity is going to increase if we increase the number of rounds.
(B)
Node A won the channel content and sent package but did not get the ACK after SIFS (Short interframe spacing) Then the Node A should make a request for the ACK soon as the packet has already been sent. Without ACK of the sent package, the Node A has no access to send the other packet in the next round.
(c)
Assuming that there is one active user n=1, The packet size = 1024 bytes and the data rate is 24Mbits/sec
generally speaking 8bits = byte, hence the data rate is 24/8 = 3Mbytes/sec
Hence the range of throughput is packet size/ data rate
Pcket size = 1024bytes = 1Kbyte
hence range = 1/3*1024= 3.25*10-4 s = 0.325 ms
(d)Given that there are 12 active nodes .
Hence there are 12 packets delivered by the users.
totally there are 31 nodes .
hence from the given definition, fairness coefficient = 12/31 = 0.387
In IEEE 802.11 is a WLAN network in which medium access control layer employs the distribution coordination function(DCF). hence the IEEE 802.11 has been effected by the fairness problem. The enhanced CS algorithm is used in this network to improve the fairness of the network.
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