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Question

Consider a 8Mbps token LAN with a ring latency of 256µsec. A host need to transmit seizes the token, and then it sends a frame of 1024 bytes removes the frame after it has circulated all around the ring and finally releases the token. This process is repeated for every frame. Assuming that only a single host wishes to transmit, then the effective data rate (in Mbps) is?

4.53

5.36

6.7

9.4

Comments

@Samaa.qudah,

you've to add question source in title

Answer 1

removes the frame after it has circulated all around the ring and finally releases the token. 

Means Delay Token Reinsertion 

$RL = 256 μsec $

$L= 1024 Bytes = 1024 * 8 bits$

$B= 8*10^6 bps$

Transmission Time $(Tt)= L/b  = 1024*8 / 8*10^6 = 1024 μsec$

$a=RL/Tt 256/1024 = 1/4$

$η=1/1+a[(N+1)/N]  (N=1 For single host)

$η= 2/3$

and $ Effective Data Rate  =η * BW  = 8Mbps * 2/3  = 5Mbps$

 

Comments

thank you very much