GATE | GATE-IT-2004 | Question 82

Consider a 10 Mbps token ring LAN with a ring latency of 400 µs. A host that needs to transmit seizes the token. Then it sends a frame of 1000 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, the effective data rate is
(A) 1Mbps
(B) 2Mbps
(C) 5Mbps
(D) 6Mbps


Answer: (C)

Explanation:
Data rate = 10 Mbps
Ring latency = 400 us
So, Propogation time = 400 us

Transmission time = (1000 * 8)/(10 * 10⁶) = 800us

Efficiency :
= Transmission time / (Transmission time + 2 * (Propogation time))
= 800 / (800 + 2 * 400)
= 0.5

Effective data rate = efficiency * data rate = 0.5 * 10 = 5 Mbps


Thus, option (C) is correct.


Please comment below if you find anything wrong in the above post.


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