Page Replacement 1993-1997 | L 15 | Operating System GATE PYQs | GATE 2022 | Vishvadeep Gothi

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In this live lecture, you will prepare the #OperatingSystem_GATEPYQs for GATE CSE/IT 2022 Exam. #VishvadeepGothi Sir has covered the #PageReplacement_1993_1997 from the operating system in this video. So here Sir will give you a lot of tips and tricks to crack the #GATEExam
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13 окт 2024

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Комментарии : 11

@vedantsharma7709 8 месяцев назад

At 34:50, GATE 2007 question and its options are wrong. The question should be, Question: A demand paging system takes 100 time units to service a page fault and 300 time units to replace a dirty page. Memory access time is 1 time unit without page fault. The probability of a page fault is p. In case of a page fault, the probability of page being dirty is also p. It is observed that the average access time is 3 time units. Then the value of p is ? Solution: Memory access time without page fault = T =1 T = 2*(Time to access main memory) Time to access main memory (Tmm)= 1/2 = 0.5 page fault service time (pfst) = 100 Time to replace dirty bit (Td) = 300 Now, the equation for EMAT = (1-p)*T + p[(1-p)*(Tmm + pfst) + p*(Tmm + pfst + Td)] Therefore, p = 0.0194 (Ans)

@MohdSameer-h8f Год назад

Sir gate 2007 question m 2 answers aye the solve krne p -0.0194 & 0.514 and both are in option can u please guide me

@NPALLEMADHU 2 года назад

At 10:33 Ref :: GEEKS FOR GEKKS !! Sir which one is correct ?? The minimum number of page frames that must be allocated to a running process in a virtual memory environment is determined by (A) the instruction set architecture (B) page size (C) physical memory size (D) number of processes in memory Answer: (A) Explanation: There are two important tasks in virtual memory management: a page-replacement strategy and a frame-allocation strategy. Frame allocation strategy says gives the idea of minimum number of frames which should be allocated. The absolute minimum number of frames that a process must be allocated is dependent on system architecture, and corresponds to the number of pages that could be touched by a single (machine) instruction. So, it is instruction set architecture i.e. option (A) is correct answer.