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Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The sender and receiver window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 μs.
Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 μs. Calculate the maximum achievable throughput.
- a)15.00 x 106 bps
- b)12.33 x 106 bps
- c)11.11 x 106 bps
- d)7.96 x 106 bps
Correct answer is option 'C'. Can you explain this answer?
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Host A is sending data to host B over a full duplex link. A and B are ...
Total size = 5 × 1000 = 5000 bytes
Total time = (5 × 50 + 200)10-6 = 450 × 10-6
Maximum achievable throughput,
Total time = (5 × 50 + 200)10-6 = 450 × 10-6
Maximum achievable throughput,
= 11.11 x 106 bps
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Host A is sending data to host B over a full duplex link. A and B are ...
Milliseconds. The link has a propagation delay of 10 milliseconds. Host A sends the first five packets to Host B and starts its timer. Host B receives the first five packets and sends an acknowledgement (ACK) back to Host A. The ACK reaches Host A after a propagation delay. Host A receives the ACK and stops its timer. How much time does it take for Host A to send the next five packets to Host B?
To answer this question, let's break down the time taken for each step:
1. Host A sends the first five packets to Host B:
- Each packet takes 50 milliseconds to transmit.
- Total transmission time for 5 packets = 5 * 50 milliseconds = 250 milliseconds.
2. Host B receives the first five packets and sends an ACK back to Host A:
- The ACK packet takes 50 milliseconds to transmit.
- The propagation delay is 10 milliseconds.
- Total transmission time for the ACK = 50 milliseconds + 10 milliseconds = 60 milliseconds.
3. The ACK reaches Host A after a propagation delay:
- The propagation delay is 10 milliseconds.
4. Host A receives the ACK and stops its timer:
- No significant time is taken for this step.
Now, let's calculate the total time taken for Host A to send the next five packets:
- After Host A receives the ACK, it knows that the first five packets have been successfully received by Host B. Since the window size is 5 packets, Host A can now send the next five packets without waiting for any additional ACKs.
- The transmission time for each packet is 50 milliseconds, and the total number of packets to be sent is 5.
- Therefore, the total transmission time for the next five packets = 5 * 50 milliseconds = 250 milliseconds.
Thus, it takes 250 milliseconds for Host A to send the next five packets to Host B.
To answer this question, let's break down the time taken for each step:
1. Host A sends the first five packets to Host B:
- Each packet takes 50 milliseconds to transmit.
- Total transmission time for 5 packets = 5 * 50 milliseconds = 250 milliseconds.
2. Host B receives the first five packets and sends an ACK back to Host A:
- The ACK packet takes 50 milliseconds to transmit.
- The propagation delay is 10 milliseconds.
- Total transmission time for the ACK = 50 milliseconds + 10 milliseconds = 60 milliseconds.
3. The ACK reaches Host A after a propagation delay:
- The propagation delay is 10 milliseconds.
4. Host A receives the ACK and stops its timer:
- No significant time is taken for this step.
Now, let's calculate the total time taken for Host A to send the next five packets:
- After Host A receives the ACK, it knows that the first five packets have been successfully received by Host B. Since the window size is 5 packets, Host A can now send the next five packets without waiting for any additional ACKs.
- The transmission time for each packet is 50 milliseconds, and the total number of packets to be sent is 5.
- Therefore, the total transmission time for the next five packets = 5 * 50 milliseconds = 250 milliseconds.
Thus, it takes 250 milliseconds for Host A to send the next five packets to Host B.
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Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The sender and receiver window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 μs.Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 μs. Calculate the maximum achievable throughput.a)15.00 x 106 bpsb)12.33 x 106 bpsc)11.11 x 106bpsd)7.96 x 106 bpsCorrect answer is option 'C'. Can you explain this answer?
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Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The sender and receiver window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 μs.Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 μs. Calculate the maximum achievable throughput.a)15.00 x 106 bpsb)12.33 x 106 bpsc)11.11 x 106bpsd)7.96 x 106 bpsCorrect answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The sender and receiver window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 μs.Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 μs. Calculate the maximum achievable throughput.a)15.00 x 106 bpsb)12.33 x 106 bpsc)11.11 x 106bpsd)7.96 x 106 bpsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The sender and receiver window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 μs.Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 μs. Calculate the maximum achievable throughput.a)15.00 x 106 bpsb)12.33 x 106 bpsc)11.11 x 106bpsd)7.96 x 106 bpsCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The sender and receiver window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 μs.Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 μs. Calculate the maximum achievable throughput.a)15.00 x 106 bpsb)12.33 x 106 bpsc)11.11 x 106bpsd)7.96 x 106 bpsCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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ample number of questions to practice Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The sender and receiver window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 μs.Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 μs. Calculate the maximum achievable throughput.a)15.00 x 106 bpsb)12.33 x 106 bpsc)11.11 x 106bpsd)7.96 x 106 bpsCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
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