Recent questions tagged paging

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"In a 32-bit machine we subdivide the virtual address into 4 segments as follows:[10-bit 8-bit 6-bit 8 bit]We use a 3-level page table, such that the ... clear my concept. should we take the entire page table size and not just the entry?
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I am not able to understand the concept of multilevel paging.plz can someone tell what is stored in 1st level and 2nd level page table and how do we access it??
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For the implementation of a paging scheme, suppose the average process size be $x$ bytes, the page size be $y$ bytes, and each page entry requires $z$ bytes. The optimum page size ... \frac{x}{2}$\frac{xz}{2}$\sqrt{2xz}$\frac{\sqrt{xz}}{2}$
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A job has four pages A, B, C, D and the main memory has two page frames only. The job needs to process its pages in following order:ABACABDBACDAssuming that ... in FIFO and LRU page replacement algorithms are9 and 77 and 69 and 88 and 6
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Let the page fault service time be $10$ ms in a computer with average memory access time being $20$ ns. If one page fault is generated for every $10^6$ memory ... is the effective access time for the memory?21.4 ns29.9 ns23.5 ns35.1 ns
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Dirty bit is used to indicate which of the following?A page fault has occurredA page has corrupted dataA page has been modified after being loaded into cacheAn illegal access of page
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A computer has 16 pages of virtual address space but the size of main memory is only four frames. Initially the memory is empty. A program references the ... 10. How many page faults occur if LRU page replacement algorithm is used?3578
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What is the page size taken if page size is less than frame size?Take an example page size =4kb and frame size = 8kb ,Then for calculating virtual address space , which page size value do we take?
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What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid and invalid bit) and a page size of 512 bytes?$2^{11}$2^{15}$2^{19}$2^{20}$
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A computer has 32-bit virtual addresses and 4-KB pages. The program and data together fit in the lowest page (0-4095). The stack fits in the highest page. ... page table entries are needed for two-level paging, with 10 bits in each part?
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Consider, 2 level paging , with Process P, Inner page table PTi, and outer page table PTo.Now we know that P and PTi are divided in various pages but PTo is in ... have used formula given in case 3 and not in case 2 in case of page faults?
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What is the name of the technique in which the operating system of a computer executes several programs concurrently by switching back and forth between them?PartitioningMulti-taskingWindowingPaging
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If the page size in a 32-bit machine is 4K bytes then the size of page table is1 M bytes2 M bytes4 M bytes4 K bytes
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how to calculate offset of page
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Using the page table shown below, translate the physical address 25 to virtual address. The address length is 16 bits and page size is 2048 words while the size of the physical memory ... – 25616920734121
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In a $64$- bit machine, with $2$ GB RAM, and $8$ KB page size, how many entries will be there in the page table if its is inverted?$2^{18}$2^{20}$2^{33}$2^{51}$
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Till now i have not got any exciting and detailed explanation of page tables and multi level page tables concept. if you have any resource on this please post the link ..
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suppose a paging system has 2^g+h virtual addresses & uses 2^h+k locations in primary memory for integers g,h & k. what is the page size of the ... the virtual & physical address sizes? how many bits are required to store a virtual address
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Represent the sentence "We are IT students" in physical memory if page size is 5 characters and the entries in page table are-061527310Assume physical memory size to be 55 characters
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Given answer: A. Please explain
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Given answer is A.I believe that B should be the answer because as the number of page tables are k. k memory access are required for each table and finally, one for accessing the data from main memory. Please check.
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In a two level paging environment a page table is divided in to 2^12 pages each of size 4KW. The memory is word addressable. The physical address space ... 128MW#frames = 2^14using the #pages and pagesize we can get VAS? how to approach.
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Problem 3In a 32-bit machine we subdivide the virtual address into 4 pieces as follows:8-bit 4-bit 8-bit 12-bitWe use a 3- ... a stack segment of 64K starting at address 0xf0000000 and growing upward (towards higher addresses)?
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The question I answered was 4096 bytes and what the test answered is in KB and both the answers are correct . But in GATE how do we come to know whether the answer should be given in bytes or Kb or MB or GB?
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Consider a machine with 64 MB physical memory and 34 bit virtual address space.If the page size is 4 KB ,the approximate size of conventional and inverted page table will be??
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I have a doubt which is accessed first TLB or cache?I think answer should be TLB as we need it for address translation to get frame no. and then access cacheIs my approach right?