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 Arjun commented Nov 20, 2015 reply Follow Share c should be false.. https://en.wikipedia.org/wiki/Monitor_(synchronization) 3 votes 3 votes Akash Kanase commented Nov 20, 2015 reply Follow Share Updated. 2 votes 2 votes Madhab commented Jun 2, 2016 reply Follow Share why b is false? 1 votes 1 votes vamsi2376 commented Jun 22, 2016 reply Follow Share option b should be true. 2 votes 2 votes Arjun commented Jun 22, 2016 reply Follow Share B is false- reason already given in answer. 4 votes 4 votes Harsh181996 commented Dec 30, 2016 reply Follow Share Arjun Sir , In Option B) This is false. Example :- Peterson's solution is purely software based solution without use of hardware. Sir but the context here is if we need indivisible machine instructions or not ? So both {test and set (which uses hardware) } and {peterson's solution (which uses Software Implementation) } will fall in the same category according to the context.So it should be true. Where am I going wrong ? 4 votes 4 votes bhuv commented Sep 26, 2017 reply Follow Share More specifically for option B https://en.wikipedia.org/wiki/Peterson's_algorithm#Note. 1 votes 1 votes rahul sharma 5 commented Dec 11, 2017 reply Follow Share Can you explain option E.It asks about best fit and your are saying first fit.It can be a typeo.Please verify 1 votes 1 votes rahul sharma 5 commented Dec 11, 2017 reply Follow Share @Arjun sir. How can monitor causes deadlock? 3 votes 3 votes bhuv commented Dec 11, 2017 reply Follow Share Whatever be the case "External Fragmentation" can occur is all three types i.e best fit, first fit, wrost fit. That might be an typeo in the answer for part E. But answer is correct the statement is false. 9 votes 9 votes Rishabh Gupta 2 commented Jan 21, 2018 reply Follow Share Option B should be TRUE. It's written clearly in Galvin. Also see the wikipedia page: https://en.wikipedia.org/wiki/Peterson's_algorithm#Note 3 votes 3 votes MiNiPanda commented Sep 8, 2018 reply Follow Share I didn't get C from the reference given.. At most one thread can be active inside the monitor at any given time. If others try to enter we don't allow them and they are kept waiting. Once the process inside has done executing, then one of the waiting processes is notified. In this way how can deadlock occur? 0 votes 0 votes Asim Siddiqui 4 commented Oct 15, 2018 reply Follow Share http://courses.cs.vt.edu/~cs5204/fall00/monitor.html navigate to 'LIMITATIONS OF MONITORS' this might help. 3 votes 3 votes srestha commented Nov 5, 2018 reply Follow Share https://gateoverflow.in/247579/%23os-usergate-process-synchronisation 0 votes 0 votes zeeshanmohnavi commented Dec 8, 2018 reply Follow Share @Asim Siddiqui 4's link really explains the rationale behind option (C) 0 votes 0 votes zeeshanmohnavi commented Dec 8, 2018 reply Follow Share @Arjun @Akash Kanase The Wikipedia link given in the answer for option (B) contains the following text. The algorithm satisfies the three essential criteria to solve the critical section problem, provided that changes to the variables turn, flag[0], and flag[1] propagate immediately and atomically. The while condition works even with preemption. Doesn't this imply that atomic instructions are essential to implement Peterson's algorithm as well? 1 votes 1 votes Raj Singh 1 commented Jan 3, 2019 reply Follow Share Point E is: The best fit techniques for memory allocation ensures that memory will never be fragmented. Your explanation is: False. Memory can get fragmented with First fit. Question asks about best fit, you explained first fit. I guess two are different, even though I feel both can cause fragmentation. 0 votes 0 votes Raj Singh 1 commented Jan 3, 2019 reply Follow Share @Asim Siddiqui 4 @zeeshanmohnavi I doubt about monitor point. Monitor is a construct just like read-write instructions (test-and-set, compare-and-swap, exchange etc) and these are used to develop code which will satisfy critical section problem criteria (bounded wait, mutual exclusion and progress) and as such it does not guarantees against or for deadlock on itself. Thats why @Asim Siddiqui 4 link describes scenario where deadlock can occur while using monitor, whereas page 21 of this ppt says "guarantees against deadlock". Am I correct with this? 0 votes 0 votes 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.