csma/cd

Question

A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1 km cable with no repeaters. The signal speed in the cable is 2 x 10⁸ m/sec. The  frame size for this network 10000 bits what is the total time requires to send this frame completly ?

Comments

is it 15 micro sec ??

Answer 1

i think this depends on number of collisions occur while transmitting the packet. As they did not mention anything regarding that and there is a possibility for collision to occur, the collision might occur once, twice, thrice,...and so on..before the receiver actually receives it.
if we assume there are no collisions then time taken is Tt+Tp only, but if there are collisions , on average Tt+6.44Tp  will be the total time.

Comments

please explain how do u conclude Tt+6.44Tp?

---

https://gateoverflow.in/27636/formula-for-calculating-efficiency-of-ethernet

---

Answer should be 25us in case no collision is happening.

Answer 2

total time = Transmission time + propagation time 
TT= 10000 bits/10*10⁹
     =10 microsec
PT= 1000/2*10⁸
     =5 microsec 

total time =10+5
                =15 microsec

Comments

TAG this question and ask from expert ??

---

@Arjun sir ??

---

ok.....

Answer 3

bandwidth = 1Gbps = 10^9 bps

distance ,d=1km

velocity , v= 2 *10^8 m/s

frame size ,l=10000 bits

total time required to send the frame = t(totlal) = ???????????????????

______________________________________________________________________________

tp= d/v= 1km / (2 * 10^5)km/s = .5 * 10^(-5) sec

tf=10000/10^9=10^(-5)

total time ,t=tp+tf=10+5 microsec=15 micro sec

Answer 4

Tf= length / bandwidth = 10,000/1000,000,000

tp=distance / velocity =1000/200,000,000

total time = tf +tp = 15 micro second