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, Propagation time = 400 us 

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

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

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

Thus, option (C) is correct. 

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