We have demand-paged memory with single-level paging and the page-table is stored in memory. It takes 6 milliseconds to service a page if an empty frame is available or if the replaced page is clean. If the replaced is dirty, servicing the page fault takes 10 milliseconds. Memory access time is 100 ns. TLB access time is 10ns. The probability of a TLB hit is 90%. The probability for a page-fault is 0.1%. It is given that the page to be replaced is dirty twenty percent of the time. Answer the following: What is the effective memor

We have demand-paged memory with single-level paging and the page-table is stored in memory. It takes 6 milliseconds to service a page if an empty frame is available or if the replaced page is clean. If the replaced is dirty, servicing the page fault takes 10 milliseconds. Memory access time is 100 ns. TLB access time is 10ns. The probability of a TLB hit is 90%. The probability for a page-fault is 0.1%. It is given that the page to be replaced is dirty twenty percent of the time. Answer the following: What is the effective memor

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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For each of the following page replacement policies, list the total number of page faults and fill in the contents ofthe page frames of memory after each memory reference.
Solve it ,if you are 100% sure , else i will downvote.
Consider a system that implements demand paging without a TLB. The single-level page table is held in memory, and we assume that there are no free frames. Each memory access requires 100 nanoseconds. It takes 6 additional milliseconds to handle a page fault if the victim page is not “dirty” or 12 additional milliseconds if the victim page is “dirty”. Experiments show that the probability of page faulting is 0.00003 (i.e., 0.003% chance) and that the average memory access time is 400ns. What is the percent chance that a page is dirty at the time it is evicted? Hint: Call the “being dirty” probability d and write a simple equation that gives the average data access time as a function of d. Solve for d (the result should be a number between 0 and 1, which you then convert to a percentage).
4. Consider a system which has logical address is 7 bits, physical address is 6 bits calculate the number of pages and no.of frames. 5. Given Memory partition of 110,510K,210K and310kand600kin that order show how each of the best-fit, worst-fit and first-fit algorithms place process of221,471,112and 426 in that order in memory.
A demand - paging uses a TLB (Translation Look - aside Buffer) and single level page table stored in main memory. The memory access time is 5 us and the page fault service time is 25 ms. If 70% of access is in TLB with no page fault and of the remaining, 20% is not present in main memory. The effective memory access time in ms is (Write answer up to two decimal places)
Give a comparison of FIFO, LRU and OPTIMAL algorithms in term of Page faults with three and four frames in each case for the reference string of pages duly demanded in Main Memory. Reference String: 2, 3, 3, 2, 1, 4, 0, 0, 3, 2, 1, 1, 2, 3, 4
A computer has four page frames. The time of loading, time of last access, and the R and M bits for each page are as show (the times are in clock ticks): Page Loaded Last ref. R M 0 126 280 1 0 1 230 265 0 1 2 140 270 0 0 3 110 285 1 1 At the time of page fault which page will be replaced by following algorithms: 1. FIFO 2. LRU 3. Second chance
Consider a system that implements demand paging without a TLB. The single-level page table is held in memory, and we assume that there are no free frames. Each memory access requires 100 nanoseconds. It takes 6 additional milliseconds to handle a page fault if the victim page is not "dirty" or 12 additional milliseconds if the victim page is "dirty". Experiments show that the probability of page faulting is 0.00003 (i.e., 0.03% chance) and that the average memory access time is 400ns. What is the percent chance that a page is dirty at the time it is evicted? Hint: Call the "being dirty" probability d and write a simple equation that gives the average data access time as a function of d. Solve for d (the result should be a number between 0 and 1, which you then convert to a percentage).
Assume we have a demand-paged memory. The page table is held in registers. It takes 9 milliseconds to service a page fault if an empty page is available or the replaced page is not modified, and 22 milliseconds if the replaced page is modified. Memory access time is 100 nanoseconds. Assume that the page to be replaced is modified 80 percent of the time. What is the maximum acceptable page-fault rate for an effective access time of no more than 250 nanoseconds?