Q4. 2 marks) i. What is the maximum backoff value for Ethernet CSMA/CD on the 4\
ID: 3737803 • Letter: Q
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Q4. 2 marks) i. What is the maximum backoff value for Ethernet CSMA/CD on the 4'th retransmission attempt? ii Suppose two stations have a packet to send and it is the 2"d retransmission attempt for both of them. What is the probability that both stations attempt to transmit in the same slot? Q5. (2 marks) Packets of fixed size 2000 bytes are transmitted using slotted ALOHA over a link of data rate 4 Mbps i. What is the maximum possible throughput in bps (or kbps or Mbps) of the system? ii. What is the maximum possible throughput in packets per second of the system? Use the formula S -G exp(-G), G- offered traffic, S- throughputExplanation / Answer
Answer 4) Part i)
In Ethernet CSMA/CD,
For each retransmission, the transmitter constructs a set of numbers:
{0, 1, 2, 3, 4, 5, ... L} where L is ([2 to the power (K)]-1) and where K=N; K<= 10;
A random value R is picked from this set, and the transmitter waits (defers) for a period
R x (slot time) i.e. R x 51.2 Micro Seconds (Here 51.2 microseconds is the minimum slot time)
Hence on 4th retransmission, the set of values to choose from will be {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} and hence the maximum backoff value will be 15 * 51.2 = 768 micro seconds
Answer 4) Part ii)
On a 2nd retransmission each station can choose 1 out of 4 slots (i.e. {0,1,2,4}) so the probability of a collision will be (1/4) * (1/4) = 1/16
Answer 5)
In slotted ALOHA, the maximum possible throughput is when G=1 (Using the formula S= G exp(-G)).
Hence
Answer i) 4*e^(-4) = 0.07326 Mbps (approximately)
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