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Consider a link $(A,B)$ with bandwidth $\text{1 Kbps}$ and propagation delay $\text{300 msec}$ and stop and wait protocol is used from $A$ to $B$ and Acknowledgements are piggybacked. Frame size is same for $A$ and $B$ and it is $\text{100 bits}$. Assume that $B$ always have the data to send to $A$.

The Throughput for $A$ in this model is ___________ $\text{Kbits/sec}$?

1. $200$
2. $250$
3. $333$
4. $125$
+6

@Shaik Masthan

throughput  = eff * BW

eff = [Tt]  /  [Tt +Tp+Tack+Tp]

Tt = 100 / 1000 = 0.1 sec

Tp = 300 msec = 0.3 sec

eff = (0.1) / (0.1+0.3+0.1+0.3) ==> 0.1 / 0.8 = 1/8

Throughput = (1/8) * 1Kbps = 125 bps

what is wrong in this calculation ?

+1
It is correct
+1
k bits /sec ??

it should be bits/sec

then 125 bits/sec will be the answer