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Official PYQ · 2006Q44Transport Layer: Flow Control, Congestion Control, UDP, TCP, SocketsMCQ1 mark

GATE CS 2006 Q44 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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Station AA uses 32  byte32\; \text{byte} packets to transmit messages to Station BB using a sliding window protocol. The round trip delay between A and BB is 80  milliseconds80\; \text{milliseconds} and the bottleneck bandwidth on the path between AA and BB is 128  kbps128\; \text{kbps} . What is the optimal window size that AA should use?

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Answer: B

To achieve maximum (100%) utilization in a sliding window protocol, the sender must be able to transmit continuously during the round-trip time (RTT). The optimal window size WW (in number of packets) is given by:

W=Bandwidth×RTTPacket SizeW = \frac{\text{Bandwidth} \times \text{RTT}}{\text{Packet Size}}

Given data:

  • Packet size (LL) = 32 bytes=32×8 bits=256 bits32 \text{ bytes} = 32 \times 8 \text{ bits} = 256 \text{ bits}
  • Round Trip Time (RTT) = 80 ms=80×10−3 s80 \text{ ms} = 80 \times 10^{-3} \text{ s}
  • Bandwidth (BB) = 128 kbps=128×103 bps128 \text{ kbps} = 128 \times 10^3 \text{ bps}

Substituting these values into the formula:

W=128×103 bps×80×10−3 s256 bits=10240256=40W = \frac{128 \times 10^3 \text{ bps} \times 80 \times 10^{-3} \text{ s}}{256 \text{ bits}} = \frac{10240}{256} = 40

Thus, the optimal window size is 40.

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