Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Frames of 20 kb are sent over a 10 Mbps duple...
Start Learning for Free
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?
| FREE This question is part of | Download PDF Attempt this Test |
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.
Attention Computer Science Engineering (CSE) Students!
To make sure you are not studying endlessly, EduRev has designed Computer Science Engineering (CSE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Computer Science Engineering (CSE).
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Similar Computer Science Engineering (CSE) Doubts
Top Courses for Computer Science Engineering (CSE)View all
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?
Question Description
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?.
Solutions 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? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
Download more important topics, notes, lectures and mock test series for Computer Science Engineering (CSE) Exam by signing up for free.
Here you can find the meaning of 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? defined & explained in the simplest way possible. Besides giving the explanation of
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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
ample number of questions to practice 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? tests, examples and also practice Computer Science Engineering (CSE) tests.
|
|
Explore Courses for Computer Science Engineering (CSE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup