efficiency

Question

Answer 1

--> Each node transmit 50 frames per second (given).

--> Frame length = 2500 bits , Number of nodes = 100 , Transmission rate = 10^8 bps (given).

--> So, Bits transmit by 100 node = 2500*100*50 = 125*10^5 bps

--> Efficiency(%) of Protocol  = (125*10^5 / 10^8 )*100 = 12.50% .

Answer 2

 in 1 sec a node can transmit 50*2500 bits = 125*10^3 bits

but its transmission rate in general is 10^8 bits in 1 sec

so efficiency is .................125*10^3 / 10 ^ 8

.this is for 1 node .

.For 100 node it will be 12.5%

Comments

Approach 2-

(I'm not sure where I'm making mistake in this one)

Tp = 4.762 usec
TT = 25 usec

1 cycle for successful transmission is TT + 2Tp = 34.524 usec

So in one such cycle we can transmit just 1 frame ie 2500 bits 
And in that cycle total bits generated by all nodes is 
In 1 sec 50*2500*100 bits = 12.5*10^6 bits
So in 34.524 usec bits generated by all is 431.55 bits 

But in 1 TT no of frames generated by a node is 
50 frames in 1 sec so 
50 *34.524 / 10^6 = 0.0017262 frames 
Bits generated in 1 cycle by a node is 
2500* ans = 4.3155 
So efficiency is 

4.3155/ 431.55 = 1%

---

@Vertika Srivastava

How did you calculate the propagation time?

& this propagation time is for one node to one node or for the entire transmission through 100 nodes?