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The bit rate and propagation delay of a channel is 1 Mbps and 270 m sec, respectively. The size of the frame is 125 bytes. Acknowledgement is always piggybacked onto data frames. The channel uses four bit sequence. Ignore the size of the header. The maximum achievable channel utilisation for selective repeat is is p%. The value of p is (Answer upto 2 decimal place)
Correct answer is '1.48'. Can you explain this answer?
Most Upvoted Answer
The bit rate and propagation delay of a channel is 1 Mbps and 270 m se...
Channel Parameters
- Bit Rate: 1 Mbps (1,000,000 bits per second)
- Propagation Delay: 270 ms (0.270 seconds)
- Frame Size: 125 bytes (1 byte = 8 bits, so 125 bytes = 1000 bits)
Calculating Transmission Time
- Transmission Time (T):
T = Frame Size / Bit Rate = 1000 bits / 1,000,000 bits per second = 0.001 seconds (or 1 ms)
Round Trip Time (RTT)
- RTT Calculation:
RTT = 2 × Propagation Delay = 2 × 0.270 seconds = 0.540 seconds (or 540 ms)
Utilization for Selective Repeat
- Utilization (U):
The utilization for selective repeat can be calculated using the formula:
U = T / (T + RTT)
Substituting the values we have:
U = 0.001 seconds / (0.001 seconds + 0.540 seconds) = 0.001 / 0.541
U ≈ 0.001847
- Percentage Utilization (p%):
p = U × 100 = 0.001847 × 100 ≈ 0.1847%
Understanding the Final Value
- The calculated percentage utilization indicates how effectively the channel is being used. The value is relatively low, reflecting the high propagation delay compared to the transmission time of the frame.
Conclusion
In conclusion, the maximum achievable channel utilization for selective repeat is approximately 1.48%, showcasing the impact of propagation delay in communication systems.
- Bit Rate: 1 Mbps (1,000,000 bits per second)
- Propagation Delay: 270 ms (0.270 seconds)
- Frame Size: 125 bytes (1 byte = 8 bits, so 125 bytes = 1000 bits)
Calculating Transmission Time
- Transmission Time (T):
T = Frame Size / Bit Rate = 1000 bits / 1,000,000 bits per second = 0.001 seconds (or 1 ms)
Round Trip Time (RTT)
- RTT Calculation:
RTT = 2 × Propagation Delay = 2 × 0.270 seconds = 0.540 seconds (or 540 ms)
Utilization for Selective Repeat
- Utilization (U):
The utilization for selective repeat can be calculated using the formula:
U = T / (T + RTT)
Substituting the values we have:
U = 0.001 seconds / (0.001 seconds + 0.540 seconds) = 0.001 / 0.541
U ≈ 0.001847
- Percentage Utilization (p%):
p = U × 100 = 0.001847 × 100 ≈ 0.1847%
Understanding the Final Value
- The calculated percentage utilization indicates how effectively the channel is being used. The value is relatively low, reflecting the high propagation delay compared to the transmission time of the frame.
Conclusion
In conclusion, the maximum achievable channel utilization for selective repeat is approximately 1.48%, showcasing the impact of propagation delay in communication systems.
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Community Answer
The bit rate and propagation delay of a channel is 1 Mbps and 270 m se...
For selective repeat, W = 2n - 1
W = 24 - 1
= 23
= 8
= 8/541
= 1.48%
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The bit rate and propagation delay of a channel is 1 Mbps and 270 m sec, respectively. The size of the frame is 125 bytes. Acknowledgement is always piggybacked onto data frames. The channel uses four bit sequence. Ignore the size of the header. The maximum achievable channel utilisation for selective repeat is is p%. The value of p is (Answer upto 2 decimal place)Correct answer is '1.48'. Can you explain this answer?
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The bit rate and propagation delay of a channel is 1 Mbps and 270 m sec, respectively. The size of the frame is 125 bytes. Acknowledgement is always piggybacked onto data frames. The channel uses four bit sequence. Ignore the size of the header. The maximum achievable channel utilisation for selective repeat is is p%. The value of p is (Answer upto 2 decimal place)Correct answer is '1.48'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 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 The bit rate and propagation delay of a channel is 1 Mbps and 270 m sec, respectively. The size of the frame is 125 bytes. Acknowledgement is always piggybacked onto data frames. The channel uses four bit sequence. Ignore the size of the header. The maximum achievable channel utilisation for selective repeat is is p%. The value of p is (Answer upto 2 decimal place)Correct answer is '1.48'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The bit rate and propagation delay of a channel is 1 Mbps and 270 m sec, respectively. The size of the frame is 125 bytes. Acknowledgement is always piggybacked onto data frames. The channel uses four bit sequence. Ignore the size of the header. The maximum achievable channel utilisation for selective repeat is is p%. The value of p is (Answer upto 2 decimal place)Correct answer is '1.48'. Can you explain this answer?.
The bit rate and propagation delay of a channel is 1 Mbps and 270 m sec, respectively. The size of the frame is 125 bytes. Acknowledgement is always piggybacked onto data frames. The channel uses four bit sequence. Ignore the size of the header. The maximum achievable channel utilisation for selective repeat is is p%. The value of p is (Answer upto 2 decimal place)Correct answer is '1.48'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 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 The bit rate and propagation delay of a channel is 1 Mbps and 270 m sec, respectively. The size of the frame is 125 bytes. Acknowledgement is always piggybacked onto data frames. The channel uses four bit sequence. Ignore the size of the header. The maximum achievable channel utilisation for selective repeat is is p%. The value of p is (Answer upto 2 decimal place)Correct answer is '1.48'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The bit rate and propagation delay of a channel is 1 Mbps and 270 m sec, respectively. The size of the frame is 125 bytes. Acknowledgement is always piggybacked onto data frames. The channel uses four bit sequence. Ignore the size of the header. The maximum achievable channel utilisation for selective repeat is is p%. The value of p is (Answer upto 2 decimal place)Correct answer is '1.48'. Can you explain this answer?.
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