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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.
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The bandwidth is 1 Gbps or 109 bits , 

1 bit = 1/ 109 sec = 1/ 103 microsecond 

RTT = 30  microsecond 

1/ 103 microsec for 1 bit , 1 microsec for  103 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/(109)  = 1/ 103 microsecond .

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

----------------

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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