Q2 (4 marks) Consider 100 Mbps Ethernet CSMA/CD operating over a coaxial cable t
ID: 3737046 • Letter: Q
Question
Q2 (4 marks) Consider 100 Mbps Ethernet CSMA/CD operating over a coaxial cable that is 800 metres long. Suppose the signal propagation velocity along the cable is 3c/4, where c is the speed of light, C-3 108 m/s. Find the following i. What is the time needed for a signal to propagate from one end of the cable to the other? i What is the maximum number of bits on the cable at any one time? iii. What is the length of a bit on the cable in metres? iv. How many bits does a round-trip time correspond to?Explanation / Answer
I. Propogation time = total distance/propogation velocity
distance=800mts, propagation velocity =3c/4 m/s
propogation time=(800*4)/(3*3*10^8)=3.55*10^-6 s
II. Maximum number of bits on the cable at any one time = distance / bit length
= 800/2.25=(aprox) 355 bits
the maximum number of bits on cable at any one time is 355 bits.
III. Bit length = propagation speed / Bandwidth
propagation speed =3c/4 m/s Band width = 100Mbps=10^8 bps
Bit lenght = (3*3*10^8)/(4*10^8) = 2.25 mts/bit
therefore the length of a bit on the cable is 2.25 mts
IV. Round Trip Time = 2*Propogation time
= 2*3.55*10^-6=(aprox)7.11*10^-6 s
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