GATE CS 2016 Set 1 Q55 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To find the sender throughput in Stop-and-Wait ARQ, we need to calculate the total time ($T_{\text{total}}$) required to successfully transmit one frame and receive its acknowledgment.

1. Identify the given parameters:

• Frame size ($L_f$): $1000\text{ bytes} = 1000 \times 8 = 8000\text{ bits}$
• Sender transmission rate ($B_s$): $80\text{ Kbps} = 80,000\text{ bits/second}$
• Acknowledgment size ($L_a$): $100\text{ bytes} = 100 \times 8 = 800\text{ bits}$
• Receiver transmission rate ($B_r$): $8\text{ Kbps} = 8000\text{ bits/second}$
• One-way propagation delay ($T_p$): $100\text{ ms} = 0.1\text{ seconds}$

2. Calculate the components of the total cycle time ($T_{\text{total}}$):

• Frame transmission time ($T_{tf}$): $$T_{tf} = \frac{L_f}{B_s} = \frac{8000\text{ bits}}{80,000\text{ bps}} = 0.1\text{ seconds} = 100\text{ ms}$$
• Propagation delay for the frame: $T_p = 100\text{ ms}$
• Acknowledgment transmission time ($T_{ta}$): $$T_{ta} = \frac{L_a}{B_r} = \frac{800\text{ bits}}{8000\text{ bps}} = 0.1\text{ seconds} = 100\text{ ms}$$
• Propagation delay for the acknowledgment: $T_p = 100\text{ ms}$

3. Calculate the total time ($T_{\text{total}}$):

4. Calculate the sender throughput:

Throughput is defined as the amount of useful data transmitted per unit time: