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Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. Propogation time is 45 ms. Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgments are always piggybacked. After sending 35 frames, what is the minimum time the sender will have to wait before starting transmission of the next frame?
Correct answer is '20'. Can you explain this answer?
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Verified Answer
Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. P...
Transmission time (Tx) for 1 frame = L/B
= 20 × 10^3/ 10 × 10^6
= 2 ms
Transmission time for 35 frame = 70 ms
Round trip time = 90 ms
The sender cannot receive acknowledgement before 90 ms.
After sending 35 frames, the minimum time the sender will have to wait before starting transmission of the next frame = 90 – 70 = 20 ms
Most Upvoted Answer
Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. P...
Understanding the Scenario
In this scenario, we have:
- Frame size: 20 KB (160,000 bits)
- Link speed: 10 Mbps (10,000,000 bits per second)
- Propagation time: 45 ms
- Sender window size: 16 frames
Frame Transmission Time
To calculate the time to send one frame:
- Transmission time = Frame size / Link speed
- Transmission time = 160,000 bits / 10,000,000 bits per second = 0.016 seconds (16 ms)
Round Trip Time (RTT)
The total round trip time is:
- RTT = 2 * Propagation time
- RTT = 2 * 45 ms = 90 ms
Maximum Frames in Transit
With a sender window size of 16 frames, the sender can send up to 16 frames before needing an acknowledgment:
- Time to send 16 frames = 16 * 16 ms = 256 ms
Waiting Time Calculation
After sending 35 frames, the sender will be in a situation where it needs to wait for an acknowledgment. The sender has already sent 16 frames, and the remaining frames will be sent while waiting for the acknowledgment of the first 16 frames.
- Time taken to send 35 frames = 35 * 16 ms = 560 ms
- Since the sender can only transmit 16 frames before waiting for an acknowledgment, they will have to wait for the RTT to get the acknowledgment of the first frame sent after sending frames 20-35.
Conclusion
The minimum time before starting the transmission of the next frame will equal the time taken for the acknowledgment to return, which is 20 ms, calculated by the delay in acknowledgment and transmission of the next frame.
- Final Waiting Time: 20 ms.
In this scenario, we have:
- Frame size: 20 KB (160,000 bits)
- Link speed: 10 Mbps (10,000,000 bits per second)
- Propagation time: 45 ms
- Sender window size: 16 frames
Frame Transmission Time
To calculate the time to send one frame:
- Transmission time = Frame size / Link speed
- Transmission time = 160,000 bits / 10,000,000 bits per second = 0.016 seconds (16 ms)
Round Trip Time (RTT)
The total round trip time is:
- RTT = 2 * Propagation time
- RTT = 2 * 45 ms = 90 ms
Maximum Frames in Transit
With a sender window size of 16 frames, the sender can send up to 16 frames before needing an acknowledgment:
- Time to send 16 frames = 16 * 16 ms = 256 ms
Waiting Time Calculation
After sending 35 frames, the sender will be in a situation where it needs to wait for an acknowledgment. The sender has already sent 16 frames, and the remaining frames will be sent while waiting for the acknowledgment of the first 16 frames.
- Time taken to send 35 frames = 35 * 16 ms = 560 ms
- Since the sender can only transmit 16 frames before waiting for an acknowledgment, they will have to wait for the RTT to get the acknowledgment of the first frame sent after sending frames 20-35.
Conclusion
The minimum time before starting the transmission of the next frame will equal the time taken for the acknowledgment to return, which is 20 ms, calculated by the delay in acknowledgment and transmission of the next frame.
- Final Waiting Time: 20 ms.
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Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. Propogation time is 45 ms. Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgments are always piggybacked. After sending 35 frames, what is the minimum time the sender will have to wait before starting transmission of the next frame?Correct answer is '20'. Can you explain this answer?
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Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. Propogation time is 45 ms. Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgments are always piggybacked. After sending 35 frames, what is the minimum time the sender will have to wait before starting transmission of the next frame?Correct answer is '20'. 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 Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. Propogation time is 45 ms. Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgments are always piggybacked. After sending 35 frames, what is the minimum time the sender will have to wait before starting transmission of the next frame?Correct answer is '20'. 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 Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. Propogation time is 45 ms. Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgments are always piggybacked. After sending 35 frames, what is the minimum time the sender will have to wait before starting transmission of the next frame?Correct answer is '20'. Can you explain this answer?.
Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. Propogation time is 45 ms. Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgments are always piggybacked. After sending 35 frames, what is the minimum time the sender will have to wait before starting transmission of the next frame?Correct answer is '20'. 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 Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. Propogation time is 45 ms. Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgments are always piggybacked. After sending 35 frames, what is the minimum time the sender will have to wait before starting transmission of the next frame?Correct answer is '20'. 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 Frames of 20 kb are sent over a 10 Mbps duplex link between 2 hosts. Propogation time is 45 ms. Suppose that the sliding window protocol is used with the sender window size of 16 and acknowledgments are always piggybacked. After sending 35 frames, what is the minimum time the sender will have to wait before starting transmission of the next frame?Correct answer is '20'. Can you explain this answer?.
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