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The bandwidth in a Stop n Wait ARQ is $1$ Gbps and $1$ bit delay to make round trip time $30$ microseconds, if data frames are $2000$ bits in length then the link is utilized in __________ percentage.

Now i understand what actually happened here ..

many thanks @papesh :)

yes @ dhruvkc123 you provided value is 6.67 is correct but you put that like this 6. 67 with a gap between decimal point and 6 so exam interface is not taking it :)

Hope you understand now..

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Best answer

The bandwidth is 1 Gbps or 10^{9} bits ,

1 bit = 1/ 10^{9} sec = 1/ 10^{3} microsecond

RTT = 30 microsecond

1/ 10^{3} microsec for 1 bit , 1 microsec for 10^{3} bits

30 microsec 30, 000 bit

data frme size 2000 bit

link utilization = (2,000 / 32,000) * 100 = 2/32 = 0.0625 * 100 = 6.25

in this question

1 bit delay to make round trip time 30 microseconds

means

Transmission delay for 1 bit t = 1/(10^{9}) = 1/ 10^{3} microsecond .

we calculate RTT by using RTT= 2 * Tp + Tt ..

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Another way of solving :

the efficiency formula for Stop&Wait ,

1/1+2a .... (i)

a = Tp / Tt

Tp = 15 Microseconds and Tt = 2 microseconds

now put them in (i) , a = 15/2 , 2a = 15 .

1/1+15 = 1/16 = 0.0625

0.0625 * 100 = 6.25 %

we get link utilization or efficiency is 0.0625 or 6.25%

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tusharhigh Yes right but that is Bandwidth utilization formula in exact value, to find percentage you would again have to divide answer by bandwidth and multiply by 100 which is nothing but

utilization in percentage =100 * Transmission Delay (TD) / Transmission Delay (TD) + 2 * Propagation Delay (TP)

utilization in value = L/TD+2TP

utilization in percentage =100 * utilization in value / Band width

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