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Token ring
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computer-networks
token-ring
asked
Dec 3, 2018
in
Computer Networks
by
Na462
Loyal
(
8.6k
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|
57
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answer
comment
0
I am getting B as an answer
0
Propagation Speed :: $200m/μsec$
Bw :: $20Mbps = 20 * 10^6$
Distance :: $20km = 200 * 10 ^ 3m$
propagation delay : $(200 * 10 ^ 3m)/(200*10^6m/sec)$
L=2*Propagationsdelay*Bw
$L = 20,000bits$
0
Are u taking The L = Bandwidth * RTT ?
It should be simply L = Bandwidth * Propagation delay becasue here in token ring only one request can happen at a time.
0
What i did is in 200km it will require 1000micro second
So total capacity is = 20Mbps * 10,000 = 20,000.
Assuming it to be half duplex
0
@Na462
I did efficiency = 1
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Token ring from computer network
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Token Ring - ISRO | ISRO CS 2016
In a token ring network the transmission speed is 107 bps and the propagation speed is 200 meters/ μs. The 1-bit delay in this network is equivalent to: (A) 500 meters of cable. (B) 200 meters of cable. (C) 20 meters of cable. (D) 50 meters of cable.
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Dec 4, 2017
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Token Ring
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Token Ring
Suppose each station in IEEE 802.5 with 1bit delay is 48m apart from its neighbouring stations. Token is of 3 Bytes and propagation speed is 2.4 * 10^8 m/sec. To avoid overlapping in the 4Mbps token ring the monitor must insert 15 bits of artificial delay into ring. How many stations are there in the ring? A.2 B.3 C.4 D.5
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