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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) 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 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) 2024 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?.

Solutions 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? 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 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? defined & explained in the simplest way possible. Besides giving the explanation of 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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an ample number of questions to practice 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? tests, examples and also practice Computer Science Engineering (CSE) tests.