GATE CS 2016 Set 1 — Question 65
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Computer Networks → Data Link Layer → Efficiency Numericals
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Question
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.
Assuming no frame is lost, the sender throughput is __________ bytes/second.
Correct answer
2500 to 2500
Solution
Given:
Frame size, bytes bits bits.
Sender transmission rate (Bandwidth), Kbps bps bps.
Acknowledgement size, bytes bits bits.
Receiver transmission rate, Kbps bps.
Propagation delay, ms s.In Stop-and-Wait ARQ, the total time to send one frame and receive an acknowledgement is:Calculate transmission times:Total cycle time:Throughput is the amount of data sent per unit time:
Frame size, bytes bits bits.
Sender transmission rate (Bandwidth), Kbps bps bps.
Acknowledgement size, bytes bits bits.
Receiver transmission rate, Kbps bps.
Propagation delay, ms s.In Stop-and-Wait ARQ, the total time to send one frame and receive an acknowledgement is:Calculate transmission times:Total cycle time:Throughput is the amount of data sent per unit time:
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