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In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page is
  • a)
    3.0 ns
  • b)
    68.0 ns
  • c)
    68.5 ns
  • d)
    78.5 ns
Correct answer is option 'C'. Can you explain this answer?

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Answers

AKSHIT MALHOTRA
May 15, 2020
Related In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page isa)3.0 nsb)68.0 nsc)68.5 nsd)78.5 nsCorrect answer is option 'C'. Can you explain this answer?
Hit ratio = 0.35

Time (secondary memory) = 100 ns

T(main memory) = 10 ns

Average access time = h(Tm) + (1 - h) (Ts)

= 0.35 x 10 +(0.65) x 100

= 3.5 + 65 

= 68.5 ns

Hence option (C) is correct

For basics of Memory management lecture click on the following link:

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In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page isa)3.0 nsb)68.0 nsc)68.5 nsd)78.5 nsCorrect answer is option 'C'. Can you explain this answer? for Computer Science Engineering (CSE) 2022 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 In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page isa)3.0 nsb)68.0 nsc)68.5 nsd)78.5 nsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2022 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page isa)3.0 nsb)68.0 nsc)68.5 nsd)78.5 nsCorrect answer is option 'C'. Can you explain this answer?.
Solutions for In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page isa)3.0 nsb)68.0 nsc)68.5 nsd)78.5 nsCorrect answer is option 'C'. 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.
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Hit ratio = 0.35Time (secondary memory) = 100 nsT(main memory) = 10 nsAverage access time = h(Tm) + (1 - h) (Ts)= 0.35 x 10 +(0.65) x 100= 3.5 + 65= 68.5 nsHence option (C) is correctFor basics of Memory management lecture click on the following link:Basic of Memory Managementhttps://edurev.in/studytube/basic-of-memory-management/292f6f74-5b2f-459f-8185-cfd91c03aadf_c