Question

Question 4: If the BW of the line is 1.5Mbps, round trip time is 45ms and packet size is 1kB, what is the link utilisation in stop and wait? Question 5: Consider a transmission line with CSMA/CD. Assume that the signals travel in the fibre optic line at 0.8 times the speed of light, with an round trip time of 3.2ms, and the temperature of the glass medium is 32 degrees Celsius. Considering a packet (frame) size of 1600 bits, what is the condition on the line bandwidth for collisions to occur? Question 6: What is the efficiency (n) in a CSMA/CD with 4 stations, if a = 0.1, BW - 4 Mbps and the distance between the stations is 1000 meters, when only 1 station ansmits at a given time?

Answer 1

Solution:

(4)

Given,

=>Bandwidth(B) = 1.5 Mbps

=>Round trip time(RTT) = 45 msec

=>Packet size(L) = 1 KB

Explanation:

=>% Link utilization in stop and wait = (TT/(1+RTT))*100

Calculating transmission time(TT):

=>TT = L/B

=>TT = 1 KB/1.5 Mbps

=>TT = 1*8 Kb/1.5 Mbps             (as 1 byte = 8 bits)

=>TT = 8 Kb/1.5*10^3 Kbps         (As 1 M = 1000 K)

=>TT = 5.33 msec

=>% link utilization = (5.33/(1+45))*100

=>% link utilization = 11.58 %

I have explained each and every step of the first question with the help of statements attached to it.