Token Ring

Question

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

Answer 1

1 bit length = (1/bandwidth)*propagation speed

                = (1/4*10⁶)*2.4*10⁸

              = 60 metre

now , to hold complete token of 24bit ,minimum length of the ring =24*1 bit length

                                                                                           =24*60=1440 m

now , one every station is providing 1 bit delay , so length covered by bit delay of all stations = n*60m  ...............(1)

monitor is providing extra 15bit  artificial delay ,i.e. 15*60 =900m  .............................(2)

also , each station is 48m apart, hence physical length of ring = n*48...................(3)

now ,(1)+(2)+(3) shold be greater than or equal to 24 bit length i.e.1440

n*60+900m+ n*48 >=1440

so n>=5

Answer 2

Solution:

token size = 3 bytes

bandwidth = 4 Mbps = 4 * 10^6 bits per sec

so, ring latency = token size / bandwidth = 3 * 8 /  4 * 10^6

length = 48 km

no. of stations = N

so, propagation delay = 48N/(2.4 * 10^8) sec

now, in 1 sec -----------> 4 * 10^6 bits

1/4 * 10^6 sec <----------- 1 bit

(N+15)/4 * 10^6 sec <--------------- (N+15) bits

interface  delay = (N+15)/(4 * 10^6)

therefore, apply ring latency = propagation delay + interface delay

                  3 * 8 /  4 * 10^6 = 48N/(2.4 * 10^8) + (N+15)/(4 * 10^6)

on solving, N = 5.

so, no. of stations required are 5.

Comments

is this a gate question??

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With refrence to your answer how can the ring latency be equal to the transmission time of token ?

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No this is not gate question it is asked in Public Sector exams....