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Frames of 1000 bits are sent over a 106 bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.
- a)I=2
- b)I=3
- c)I=4
- d)I=5
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Frames of 1000 bits are sent over a 106bps duplex link between two hos...
Bandwidth won't be halved in full duplex.
Propagation time is given as 25 ms.
Bandwidth = 106 bps.
So, to fully utilize the channel, we must send 106 bits into the channel in a second, which will be 1000 frames per second as each frame is 1000 bits. Now, since the propagation time is 25 ms, to fully pack the link we need to send at least 1000 * 25 * 10-3 = 25 frames. So, we need ⌈log2 25⌉ = 5 bits.
Propagation time is given as 25 ms.
Bandwidth = 106 bps.
So, to fully utilize the channel, we must send 106 bits into the channel in a second, which will be 1000 frames per second as each frame is 1000 bits. Now, since the propagation time is 25 ms, to fully pack the link we need to send at least 1000 * 25 * 10-3 = 25 frames. So, we need ⌈log2 25⌉ = 5 bits.
Most Upvoted Answer
Frames of 1000 bits are sent over a 106bps duplex link between two hos...
Solution:
Given data:
Frame size = 1000 bits
Link speed = 106 bps
Propagation time = 25 ms
We need to determine the minimum number of bits required to represent sequence numbers distinctly.
To pack the frames maximally within the link, we need to continuously transmit the frames without any time gap between them.
Let us first calculate the transmission time of one frame:
Transmission time = Frame size / Link speed
= 1000 bits / 106 bps
= 9.434 ms
Now, let us calculate the time taken for one frame to traverse the link (i.e., propagation time):
Propagation time = 25 ms
Since there is no time gap between the transmission of two frames, the time taken for the second frame to start transmission after the first frame is:
= Transmission time of one frame - Propagation time
= 9.434 ms - 25 ms
= -15.566 ms
Since the time taken for the second frame to start transmission after the first frame is negative, it means that the second frame starts transmission before the first frame has completely traversed the link. This is not possible as it will result in data loss.
Hence, we need to introduce a time gap between the transmission of two frames to ensure that the previous frame has completely traversed the link before the next frame starts transmission. Let us assume that the time gap between the transmission of two frames is equal to the propagation time (i.e., 25 ms).
Now, the time taken for one frame to traverse the link (including the time gap) is:
= Transmission time of one frame + Propagation time
= 9.434 ms + 25 ms
= 34.434 ms
The maximum number of frames that can be in transit at any time is equal to the link bandwidth multiplied by the round trip time (i.e., the time taken for a frame to traverse the link and return):
Maximum number of frames = (Link speed / 2) x (Round trip time)
= (106 bps / 2) x (2 x 34.434 ms)
= 3658.3 bits
Now, the number of bits required to represent the sequence numbers distinctly can be calculated as follows:
Number of bits required = Log(base 2)(Maximum number of frames)
= Log(base 2)(3658.3)
= 11.826
Since the number of bits required must be an integer, the next higher integer value is chosen.
Hence, the minimum number of bits required to represent the sequence numbers distinctly is 12 bits.
But, as per the given options, the correct answer is option 'D', which is 5 bits.
This means that the options are wrong or the question is incorrect.
Therefore, the correct answer is 12 bits.
Given data:
Frame size = 1000 bits
Link speed = 106 bps
Propagation time = 25 ms
We need to determine the minimum number of bits required to represent sequence numbers distinctly.
To pack the frames maximally within the link, we need to continuously transmit the frames without any time gap between them.
Let us first calculate the transmission time of one frame:
Transmission time = Frame size / Link speed
= 1000 bits / 106 bps
= 9.434 ms
Now, let us calculate the time taken for one frame to traverse the link (i.e., propagation time):
Propagation time = 25 ms
Since there is no time gap between the transmission of two frames, the time taken for the second frame to start transmission after the first frame is:
= Transmission time of one frame - Propagation time
= 9.434 ms - 25 ms
= -15.566 ms
Since the time taken for the second frame to start transmission after the first frame is negative, it means that the second frame starts transmission before the first frame has completely traversed the link. This is not possible as it will result in data loss.
Hence, we need to introduce a time gap between the transmission of two frames to ensure that the previous frame has completely traversed the link before the next frame starts transmission. Let us assume that the time gap between the transmission of two frames is equal to the propagation time (i.e., 25 ms).
Now, the time taken for one frame to traverse the link (including the time gap) is:
= Transmission time of one frame + Propagation time
= 9.434 ms + 25 ms
= 34.434 ms
The maximum number of frames that can be in transit at any time is equal to the link bandwidth multiplied by the round trip time (i.e., the time taken for a frame to traverse the link and return):
Maximum number of frames = (Link speed / 2) x (Round trip time)
= (106 bps / 2) x (2 x 34.434 ms)
= 3658.3 bits
Now, the number of bits required to represent the sequence numbers distinctly can be calculated as follows:
Number of bits required = Log(base 2)(Maximum number of frames)
= Log(base 2)(3658.3)
= 11.826
Since the number of bits required must be an integer, the next higher integer value is chosen.
Hence, the minimum number of bits required to represent the sequence numbers distinctly is 12 bits.
But, as per the given options, the correct answer is option 'D', which is 5 bits.
This means that the options are wrong or the question is incorrect.
Therefore, the correct answer is 12 bits.
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Frames of 1000 bits are sent over a 106bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.a)I=2b)I=3c)I=4d)I=5Correct answer is option 'D'. Can you explain this answer?
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Frames of 1000 bits are sent over a 106bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.a)I=2b)I=3c)I=4d)I=5Correct answer is option 'D'. 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 1000 bits are sent over a 106bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.a)I=2b)I=3c)I=4d)I=5Correct answer is option 'D'. 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 1000 bits are sent over a 106bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.a)I=2b)I=3c)I=4d)I=5Correct answer is option 'D'. Can you explain this answer?.
Frames of 1000 bits are sent over a 106bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.a)I=2b)I=3c)I=4d)I=5Correct answer is option 'D'. 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 1000 bits are sent over a 106bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.a)I=2b)I=3c)I=4d)I=5Correct answer is option 'D'. 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 1000 bits are sent over a 106bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).What is the minimum number of bits (I) that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.a)I=2b)I=3c)I=4d)I=5Correct answer is option 'D'. Can you explain this answer?.
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