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Suppose the stop and wait protocol is employed over an asymmetric link $A$ to $B$. The $A$ to $B$ link bandwidth is $8\text{ Mbps}$ with a propagation delay of $20\;ms$, however the $B$ to $A$ link bandwidth is $800\;\text{Kbps}$ with a propagation delay of $10\;ms$ (note the link is asymmetric i.e., the reverse link characteristics are different from forward link). Assume a data packet size of $1000$ byte and an $\text{ACK}$ size of $100$ byte. What is the throughput achieved by the protocol? Ignore other delays. Express answer in kbps. ($1\text{ Mbps} = 10^6 \text {bps and } 1\text{ Kbps} = 10^3 \text{bps})$

Channel utilization should be calculated beforehand to calculate throughput

Answer is 210 kbps(mistakenly wrote ‘kps’ in pic).