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Official PYQ · 2003Q84Transport Layer: Flow Control, Congestion Control, UDP, TCP, SocketsMCQ1 mark

GATE CS 2003 Q84 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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Host AA is sending data to host BB over a full duplex link. AA and BB are using the sliding window protocol for flow control. The send and receive window sizes are 55 packets each. Data packets (sent only from AA to BB) are all 10001000 bytes long and the transmission time for such a packet is 5050 μsμs. Acknowledgment packets (sent only from BB to AA) are very small and require negligible transmission time. The propagation delay over the link is 200200 μsμs. What is the maximum achievable throughput in this communication?

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Answer: B

To find the maximum achievable throughput, we need to determine how much data the sender can transmit in one complete cycle (round-trip time).

1. Identify the Given Parameters

  • Send Window Size (WsW_s): 55 packets
  • Packet Size: 10001000 bytes
  • Transmission Time (TtT_t): 50 μs50\ \mu\text{s}
  • Propagation Delay (TpT_p): 200 μs200\ \mu\text{s}
  • Acknowledgment Transmission Time (TackT_{ack}): 00 (negligible)

2. Calculate the Total Cycle Time (TcycleT_{cycle})

The total time required for a packet to be transmitted and its acknowledgment to be received is:

Tcycle=Tt+2×Tp+TackT_{cycle} = T_t + 2 \times T_p + T_{ack} Tcycle=50 μs+2×200 μs+0=450 μsT_{cycle} = 50\ \mu\text{s} + 2 \times 200\ \mu\text{s} + 0 = 450\ \mu\text{s}

3. Determine the Maximum Data Sent in One Cycle

With a send window size of Ws=5W_s = 5, the sender can transmit at most 55 packets before waiting for an acknowledgment.

The total transmission time for 55 packets is:

5×Tt=5×50 μs=250 μs5 \times T_t = 5 \times 50\ \mu\text{s} = 250\ \mu\text{s}

Since the total transmission time (250 μs250\ \mu\text{s}) is less than the cycle time (450 μs450\ \mu\text{s}), the sender will finish transmitting all 55 packets and then wait for the acknowledgment of the first packet to arrive. Thus, the communication is window-limited, and the sender can transmit exactly 55 packets every 450 μs450\ \mu\text{s}.

4. Calculate the Throughput

Throughput=Total Data Sent in One CycleTcycle\text{Throughput} = \frac{\text{Total Data Sent in One Cycle}}{T_{cycle}} Throughput=5×1000 bytes450×10−6 seconds\text{Throughput} = \frac{5 \times 1000\ \text{bytes}}{450 \times 10^{-6}\ \text{seconds}} Throughput=5000450×10−6 Bps=1009×106 Bps≈11.11×106 Bps\text{Throughput} = \frac{5000}{450 \times 10^{-6}}\ \text{Bps} = \frac{100}{9} \times 10^6\ \text{Bps} \approx 11.11 \times 10^6\ \text{Bps}

Therefore, the maximum achievable throughput is 11.11×10611.11 \times 10^6 Bps.

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