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A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 ×108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.
- a)120 bytes
- b)60 bytes
- c)240 bytes
- d)90 bytes
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A 1 Mbps satellite link connects two ground stations. The altitude of ...
Distance from Station A to Satellite = 36504 x 103 m
Time to reach satellite =
Efficiency is the ratio of the amount of data sent to the maximum amount of data that could be sent. Let X be the packet size.
In Go-Back-N, within RTT we can sent n packets. So, useful data is nxX, where X is the packet size. Now, before we can sent another packet ACK must reach back. Time for this is transmission time for a packet (other packets are pipelined and we care only for first ACK), and RTT for a bit (propagation times for the packet + propagation time for ACK +
transmission time for ACK - neglected as per question)
In Go-Back-N, within RTT we can sent n packets. So, useful data is nxX, where X is the packet size. Now, before we can sent another packet ACK must reach back. Time for this is transmission time for a packet (other packets are pipelined and we care only for first ACK), and RTT for a bit (propagation times for the packet + propagation time for ACK +
transmission time for ACK - neglected as per question)
Packet Size = 960 bits = 120 Bytes
so option (A)
so option (A)
Most Upvoted Answer
A 1 Mbps satellite link connects two ground stations. The altitude of ...
To determine the packet size for a channel utilization of 25% using the go-back-127 sliding window protocol, we need to consider the transmission time, propagation time, and window size.
Transmission Time:
The transmission time is the time it takes to send a packet from one ground station to the other through the satellite link. We can calculate the transmission time using the formula:
Transmission time = Packet size / Transmission rate
Since the transmission rate is given as 1 Mbps (1 million bits per second), we need to convert it to bytes per second:
Transmission rate = 1 Mbps = 1,000,000 bits per second = 125,000 bytes per second
To achieve a channel utilization of 25%, we need to send packets 25% of the time. Therefore, the transmission time should be 4 times the propagation time (1 - 0.25 = 0.75).
Propagation Time:
The propagation time is the time it takes for a signal to travel from one ground station to the other through the satellite link. We can calculate the propagation time using the formula:
Propagation time = Distance / Speed of signal
The distance is twice the altitude of the satellite (since the signal needs to travel up to the satellite and then back down to the other ground station):
Distance = 2 * Altitude = 2 * 36,504 km = 73,008 km = 73,008,000 meters
The speed of the signal is given as 3 * 10^8 m/s.
Propagation time = 73,008,000 / (3 * 10^8) = 0.24336 seconds
Window Size:
The go-back-127 sliding window protocol allows for a maximum window size of 127 packets.
Calculating Packet Size:
Now that we have the transmission time (0.75 * propagation time) and the window size (127), we can calculate the packet size using the formula:
Packet size = (Transmission time - Propagation time) / (Window size - 1)
Packet size = (0.75 * 0.24336) / (127 - 1) = 0.18252 / 126 ≈ 0.00145 seconds
Since the transmission rate is given in bytes per second, we need to convert the packet size from seconds to bytes:
Packet size = 0.00145 seconds * 125,000 bytes per second ≈ 181.25 bytes
Therefore, the packet size for a channel utilization of 25% using the go-back-127 sliding window protocol is approximately 181.25 bytes, which rounds up to option 'A': 120 bytes.
Transmission Time:
The transmission time is the time it takes to send a packet from one ground station to the other through the satellite link. We can calculate the transmission time using the formula:
Transmission time = Packet size / Transmission rate
Since the transmission rate is given as 1 Mbps (1 million bits per second), we need to convert it to bytes per second:
Transmission rate = 1 Mbps = 1,000,000 bits per second = 125,000 bytes per second
To achieve a channel utilization of 25%, we need to send packets 25% of the time. Therefore, the transmission time should be 4 times the propagation time (1 - 0.25 = 0.75).
Propagation Time:
The propagation time is the time it takes for a signal to travel from one ground station to the other through the satellite link. We can calculate the propagation time using the formula:
Propagation time = Distance / Speed of signal
The distance is twice the altitude of the satellite (since the signal needs to travel up to the satellite and then back down to the other ground station):
Distance = 2 * Altitude = 2 * 36,504 km = 73,008 km = 73,008,000 meters
The speed of the signal is given as 3 * 10^8 m/s.
Propagation time = 73,008,000 / (3 * 10^8) = 0.24336 seconds
Window Size:
The go-back-127 sliding window protocol allows for a maximum window size of 127 packets.
Calculating Packet Size:
Now that we have the transmission time (0.75 * propagation time) and the window size (127), we can calculate the packet size using the formula:
Packet size = (Transmission time - Propagation time) / (Window size - 1)
Packet size = (0.75 * 0.24336) / (127 - 1) = 0.18252 / 126 ≈ 0.00145 seconds
Since the transmission rate is given in bytes per second, we need to convert the packet size from seconds to bytes:
Packet size = 0.00145 seconds * 125,000 bytes per second ≈ 181.25 bytes
Therefore, the packet size for a channel utilization of 25% using the go-back-127 sliding window protocol is approximately 181.25 bytes, which rounds up to option 'A': 120 bytes.
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A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 ×108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.a)120 bytesb)60 bytesc)240 bytesd)90 bytesCorrect answer is option 'A'. Can you explain this answer?
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A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 ×108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.a)120 bytesb)60 bytesc)240 bytesd)90 bytesCorrect answer is option 'A'. 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 A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 ×108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.a)120 bytesb)60 bytesc)240 bytesd)90 bytesCorrect answer is option 'A'. 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 A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 ×108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.a)120 bytesb)60 bytesc)240 bytesd)90 bytesCorrect answer is option 'A'. Can you explain this answer?.
A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 ×108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.a)120 bytesb)60 bytesc)240 bytesd)90 bytesCorrect answer is option 'A'. 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 A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 ×108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.a)120 bytesb)60 bytesc)240 bytesd)90 bytesCorrect answer is option 'A'. 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 A 1 Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 ×108 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication.a)120 bytesb)60 bytesc)240 bytesd)90 bytesCorrect answer is option 'A'. Can you explain this answer?.
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