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Official PYQ · 2025Set 2 · Q26Transport Layer: Flow Control, Congestion Control, UDP, TCP, SocketsMCQ2 marks

GATE CS 2025 Set 2 Q26 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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​​​​Suppose we are transmitting frames between two nodes using Stop-and-Wait protocol. The frame size is 30003000 bits. The transmission rate of the channel is 20002000 bps (bits/second) and the propagation delay between the two nodes is 100100 milliseconds. Assume that the processing times at the source and destination are negligible. Also, assume that the size of the acknowledgement packet is negligible. Which ONE of the following most accurately gives the channel utilization for the above scenario in percentage?

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Answer: A

To find the channel utilization for the Stop-and-Wait protocol, we calculate the transmission delay of the frame and the total cycle time.

1. Given Parameters:

  • Frame size (LL) = 30003000 bits
  • Transmission rate (BB) = 20002000 bps
  • Propagation delay (TpT_p) = 100100 ms = 0.10.1 seconds
  • Acknowledgement size and processing times are negligible (Tack=0T_{ack} = 0, Tproc=0T_{proc} = 0)

2. Step-by-Step Calculation:

Transmission Delay (TtT_t):

Tt=Frame SizeTransmission Rate=3000 bits2000 bps=1.5 secondsT_t = \frac{\text{Frame Size}}{\text{Transmission Rate}} = \frac{3000 \text{ bits}}{2000 \text{ bps}} = 1.5 \text{ seconds}

Total Cycle Time (TtotalT_{total}):
In Stop-and-Wait protocol, the total time taken to send a frame and receive its acknowledgement is given by:

Ttotal=Tt+2⋅Tp+Tack+TprocT_{total} = T_t + 2 \cdot T_p + T_{ack} + T_{proc}

Substituting the given values:

Ttotal=1.5+2⋅(0.1)+0+0=1.7 secondsT_{total} = 1.5 + 2 \cdot (0.1) + 0 + 0 = 1.7 \text{ seconds}

Channel Utilization (Efficiency, η\eta):

η=TtTtotal=1.51.7≈0.88235\eta = \frac{T_t}{T_{total}} = \frac{1.5}{1.7} \approx 0.88235

In percentage:

η%≈88.23%\eta \% \approx 88.23\%

Thus, the channel utilization is approximately 88.23%88.23\%, which matches option A.

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