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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 send and receive 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. What is the maximum achievable throughput in this communication?
  • a)
    7.69 x 106 Bps
  • b)
    11.11 x 106 Bps
  • c)
    12.33 x 106 Bps
  • d)
    15.00 x 106 Bps
Correct answer is 'B'. 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 ...
Network throughput ≈ Window size / roundtrip time

Roundtrip time = 2 x Packet delivery time + processing delay
= ransmission delay+2*propagation delay 
=50microsec+2*200microsec=450microsec

Now Throughput = ((5*1000*bytes)/450microsec) = 11.1111 * 106 bytes per second
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Host A is sending data to host B over a full duplex link. A and B are ...
To calculate the maximum achievable throughput in this communication, we need to consider the various factors that affect the data transfer rate. Let's break down the calculation step by step:

1. Window Size:
- Both the sender (host A) and the receiver (host B) have a window size of 5 packets each. This means that host A can send up to 5 packets without waiting for acknowledgments from host B.

2. Packet Size and Transmission Time:
- Each data packet is 1000 bytes long, and the transmission time for such a packet is 50 s.

3. Propagation Delay:
- The propagation delay over the link is 200 s. This is the time required for a packet to travel from host A to host B.

4. Effective Window Size:
- To calculate the effective window size, we need to consider the propagation delay. The effective window size is the maximum number of packets that can be in transit at any given time.
- The effective window size can be calculated using the formula:
Effective Window Size = Window Size + (Propagation Delay / Transmission Time)
= 5 + (200 / 50)
= 9 packets

5. Maximum Throughput:
- The maximum throughput can be calculated by multiplying the effective window size by the packet size and dividing it by the round trip time (RTT).
- The RTT can be calculated by adding the transmission time and the propagation delay.
RTT = Transmission Time + Propagation Delay
= 50 + 200
= 250 s
- Therefore, the maximum achievable throughput is:
Throughput = (Effective Window Size * Packet Size) / RTT
= (9 * 1000) / 250
= 36,000 / 250
= 144 Bps (Bytes per second)

6. Conversion to Bits:
- The throughput is given in bytes per second, but the answer choices are in bits per second. To convert the throughput to bits per second, we multiply it by 8.
Throughput (in bits per second) = Throughput (in bytes per second) * 8
= 144 * 8
= 1152 Bps

Hence, the maximum achievable throughput in this communication is 1152 Bps, which is equivalent to 11.11 x 106 Bps (rounded to two decimal places). Therefore, option B is the correct 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 send and receive 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. What is the maximum achievable throughput in this communication?a)7.69 x 106 Bpsb)11.11 x 106Bpsc)12.33 x 106Bpsd)15.00 x 106BpsCorrect answer is 'B'. 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 send and receive 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. What is the maximum achievable throughput in this communication?a)7.69 x 106 Bpsb)11.11 x 106Bpsc)12.33 x 106Bpsd)15.00 x 106BpsCorrect answer is 'B'. 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 send and receive 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. What is the maximum achievable throughput in this communication?a)7.69 x 106 Bpsb)11.11 x 106Bpsc)12.33 x 106Bpsd)15.00 x 106BpsCorrect answer is 'B'. 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 send and receive 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. What is the maximum achievable throughput in this communication?a)7.69 x 106 Bpsb)11.11 x 106Bpsc)12.33 x 106Bpsd)15.00 x 106BpsCorrect answer is 'B'. Can you explain this answer?.
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