GO Classes DPP | CN | STOP AND WAIT

Answer 1

Frame size $L_{\text {frame }}=2000 ~\text{bytes}$

Sender data rate $R_{\text {data }}=125 \mathrm{Kbps}=125 \times 1000=125000 \mathrm{bps}$

ACK size $L_{\text {ack }}=100 ~\text{bytes}$

ACK transmission rate $R_{\text {ack }}=10 \mathrm{Kbps}=10 \times 1000=10000 \mathrm{bps}$

One-way propagation delay $T_p=50 \mathrm{~ms}=0.05 \mathrm{~s}$

$$
\begin{gathered}
T_{\text {tx-data }}=\frac{L_{\text {frame }} \times 8}{R_{\text {data }}}=\frac{2000 \times 8}{125000}=\frac{16000}{125000} \mathrm{~s} \\\\
T_{\text {tx-data }}=0.128 \mathrm{~s}
\end{gathered}
$$
$$
\begin{gathered}
T_{\mathrm{tx}-\mathrm{ack}}=\frac{L_{\mathrm{ack}} \times 8}{R_{\mathrm{ack}}}=\frac{100 \times 8}{10000}=\frac{800}{10000} \mathrm{~s} \\\\
T_{\mathrm{tx}-\mathrm{ack}}=0.08 \mathrm{~s}
\end{gathered}
$$

Data frame propagation delay to receiver: $T_p=0.05 \mathrm{~s}$

ACK propagation delay back to sender: $T_p=0.05 \mathrm{~s}$

Total propagation delay in cycle: $2 T_p=0.1 \mathrm{~s}$
 

$$
\begin{aligned}
& T_{\text {total }}=T_{\text {tx-data }}+T_p+T_{\text {tx-ack }}+T_p \\\\
& T_{\text {total }}=0.128+0.05+0.08+0.05 \\\\
& \qquad T_{\text {total }}=0.308 \mathrm{~s} \\\\
& \text { Throughput }=\frac{\text { Frame size in bytes }}{T_{\text {total }}}
\end{aligned}
$$
\begin{aligned}
&\text { Throughput }=\frac{2000}{0.308} \text { bytes } / \mathrm{s}\\\\
&\text { Throughput } \approx 6493.51 \text { bytes/s }
\end{aligned}

Answer 2

6493.5

Answer 3

HERE IS SIMPLE AND DETAILS EXPLANATION