43 votes 43 votes Indicate all the false statements from the statements given below: The amount of virtual memory available is limited by the availability of the secondary memory Any implementation of a critical section requires the use of an indivisible machine- instruction ,such as test-and-set. The use of monitors ensure that no dead-locks will be caused . The LRU page-replacement policy may cause thrashing for some type of programs. The best fit techniques for memory allocation ensures that memory will never be fragmented. Operating System gate1991 operating-system virtual-memory normal multiple-selects + – Kathleen asked Sep 12, 2014 • edited Apr 17, 2021 by Lakshman Bhaiya Kathleen 10.7k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Dileep kumar M 6 commented Dec 6, 2017 reply Follow Share if u consider the sequence like 12341234......... assume no of frames is 3 then here every page results page fault in LRU It leads to lot of page movements so it causes thrashing 15 votes 15 votes Kishan Kumar commented Jan 14, 2018 reply Follow Share Option b is true. Peterson solution cannot guarantee mutual exclusion. This line is mentioned in Galvin itself. In fact without the use of atomic instruction synchronization cannot be guaranteed. 0 votes 0 votes Rajsukh Mohanty commented Feb 1 reply Follow Share A is True because, the size of virtual storage is limited by the addressing scheme of the computer system and by the amount of secondary memory available. ref: see slide-1 in https://home.adelphi.edu/~siegfried/cs553/553l8.pdf 0 votes 0 votes Please log in or register to add a comment.
Best answer 36 votes 36 votes True. This is false. Example:- Peterson's solution is a purely software-based solution without the use of hardware.https://en.wikipedia.org/wiki/Peterson's_algorithm False. Reference: https://en.wikipedia.org/wiki/Monitor_(synchronization) True. This will happen if the page getting replaced is immediately referred to in the next cycle. False. Memory can get fragmented with the best fit. Akash Kanase answered Nov 20, 2015 • edited Apr 17, 2021 by Lakshman Bhaiya Akash Kanase comment Share Follow See all 21 Comments See all 21 21 Comments reply Show 18 previous comments swapnil1 commented Jan 4, 2019 reply Follow Share C should be true Since Monitors have an important property that only one process can be active in the monitor at any time. This will guarantee Mutual Exclusion and hence prevent inconsistency and deadlock. 1 votes 1 votes Abir Mazumder commented Aug 19, 2020 reply Follow Share Option A is TRUE , Kindly see this blog https://docs.microsoft.com/en-us/archive/blogs/ericlippert/out-of-memory-does-not-refer-to-physical-memory 0 votes 0 votes Abhineet Singh commented Dec 7, 2020 reply Follow Share is there a typo in E “Memory can get fragmented with First fit.”, should it be best fit as best fit is asked the question 0 votes 0 votes Please log in or register to add a comment.
14 votes 14 votes before answering some points to know:- Monitor is one of the ways to achieve Process synchronization. Monitor is supported by programming languages to achieve mutual exclusion between processes. Peterson's solution is the purely software-based solution without the use of hardware. if u consider the sequence like 12341234......... assume no of frames is 3 then here every page results page fault in LRU It leads to a lot of page movements so it causes thrashing now go and see selected answer learner_geek answered Jan 5, 2018 learner_geek comment Share Follow See all 2 Comments See all 2 2 Comments reply Karthik Kumar Mudr 1 commented Mar 25, 2018 reply Follow Share is test and set is it hardware implementation 0 votes 0 votes talha hashim commented May 29, 2018 reply Follow Share Yes test and set is a hardware approach 1 votes 1 votes Please log in or register to add a comment.
5 votes 5 votes A is true as amount of virtual memory is limited by length of MAR and available secondary storage. D is false as best fit technique causes internal fragmentation. Not sure about B and C. Rajarshi Sarkar answered Apr 21, 2015 Rajarshi Sarkar comment Share Follow See 1 comment See all 1 1 comment reply Subhajit Panday commented May 30, 2020 i edited by Subhajit Panday May 30, 2020 reply Follow Share "Best fit" is an algo to find the "holes" ( empty space) in Dynamic Partitioning . Dynamic Partitioning does not cause Internal Fragmentation. External Fragmentation is possible 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes a) True.Link b)False c) d)In any replacement policy thrashing could happen, but it mainly causes in time sharing system thrashing e) False. It minimize fragmentation onlyLink srestha answered Nov 30, 2017 srestha comment Share Follow See all 3 Comments See all 3 3 Comments reply Dileep kumar M 6 commented Dec 6, 2017 reply Follow Share @srestha in best fit there will be external fragmentation right 1 votes 1 votes rahul sharma 5 commented Dec 11, 2017 reply Follow Share Please explain option c!!! 1 votes 1 votes Rishabh Gupta 2 commented Jan 21, 2018 reply Follow Share B should be TRUE. 0 votes 0 votes Please log in or register to add a comment.