A sender uses the Stop-and-Wait ARQ protocol for reliable transmission of frames. Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps = 1000 bits/second). Size of an acknowledgement is 100 bytes and the transmission rate at the receiver is 8 Kbps. The one-way propagation delay is 100 milliseconds. Assuming no frame is lost, the sender throughput is ____________ bytes/second.

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Official answer: 2500

Data Link Layer

Frame transmission time ( $T_{t,\text{data}}$ ):
$\text{Frame size} = 1000\text{ bytes} = 8000\text{ bits}$
$\text{Sender transmission rate} = 80\text{ Kbps} = 80{,}000\text{ bps}$
$T_{t,\text{data}} = \frac{8000}{80{,}000} = 0.1\text{ s} = 100\text{ ms}$
Propagation delay ( $T_p$ ):
$T_p = 100\text{ ms}$
Acknowledgement transmission time ( $T_{t,\text{ack}}$ ):
$\text{ACK size} = 100\text{ bytes} = 800\text{ bits}$
$\text{Receiver transmission rate} = 8\text{ Kbps} = 8{,}000\text{ bps}$
$T_{t,\text{ack}} = \frac{800}{8{,}000} = 0.1\text{ s} = 100\text{ ms}$
Total cycle time for one frame ( $T_{\text{total}}$ ):
$T_{\text{total}} = T_{t,\text{data}} + T_p + T_{t,\text{ack}} + T_p = 100 + 100 + 100 + 100 = 400\text{ ms} = 0.4\text{ s}$
Sender throughput:
$\text{Throughput} = \frac{\text{Data transferred per cycle}}{T_{\text{total}}} = \frac{1000\text{ bytes}}{0.4\text{ s}} = 2500\text{ bytes/second}$

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This Data Link Layer previous year question is Q55 of the GATE CSE 2016 Set 1 paper. Find more GATE CSE PYQs on Data Link Layer in Computer Networks, or browse all GATE CSE PYQs.

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Data Link Layer in GATE CSE

Data Link Layer has 27 GATE CSE PYQs across 17 papers from 2013 to 2026. 12 of them are 2-mark NAT questions like this one, and it is the only one in GATE CSE 2016 Set 1.

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