Recent questions and answers in Operating System - GATE CSE Doubts

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os timothy williams memory protection
Memory protection is normally done by A. the processor and the associated hardware B. the operating system C. the compiler D. the user program answer is given as A is it not B?

asked 5 days ago in Operating System by eyeamgj (21 points) | 8 views

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os disk timothy willliams
A certain moving arm disk storage, with one head, has the following specifications: Number of tracks/recording surface =200=200 Disk rotation speed =2400=2400 rpm Track storage capacity =62,500=62,500 bits the average latency time(assume that the head can move from one track ... entire track) is answer given as 2.5 sec is it correct ? i am getting 12.5 ms please see .

asked 5 days ago in Operating System by eyeamgj (21 points) | 7 views

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Applied mock 1 msq

answered Oct 19 in Operating System by spike500 (17 points) | 28 views

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Gate Applied Course Mock Test
Given a single paging system with a TLB cache, where main memory access time is 800 ns and TLB access time is 100 ns. The minimum hit rate needed to reduce the effective memory overhead to 300 ns is _____. (Round off upto 2 decimal places)

answered Oct 19 in Operating System by spike500 (17 points) | 24 views

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Made Easy Full Length Demo Test 2021
I am confused! Is bounded waiting satisfied here? And isnt it the definition of progress that the admin is speaking? Am I missing something?

asked Oct 18 in Operating System by nvs16 (9 points) | 24 views

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Ace Academy Bits and Bytes. Semaphores in operating system

asked Oct 14 in Operating System by CSHuB (13 points) | 17 views

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ME practice sheet
How to solve this question?

asked Oct 10 in Operating System by Kindaichi (15 points) | 23 views

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Self Doubt (PYQ)
https://gateoverflow.in/538/gate1991-11-a In this question is bounded waiting satisfied? According to me it should not be satisfied as if turn = k and i wants to enter then it will be stuck in the “while turn != i do skip;” and j can keep on entering and leaving the CS infinite times.

asked Oct 9 in Operating System by Mellophi (333 points) | 14 views

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Unacademy Gate Challenge
The above processes are scheduled using FCFS, note that there is 1 unit of extra CPU scheduling overhead for every process. What is the fraction of CPU scheduling overhead in the schedule length of the above chart? My actual doubt is, what is the schedule length here?? Schedule length starts from 0 or from the min(Arrival time) of any process?

asked Oct 5 in Operating System by kawinm (5 points) | 29 views

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GATE2014_set2_33
Gate2014 set 2 question 33 In this question demand paging is not mentioned,but the answer used it.In MRU non referenced frames are replaced first and then MRU works???

answered Oct 3 in Operating System by Siddhartha1997 (5 points) | 26 views

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#applied gate test series
Below we have to do a performance comparison of a FAT file system to a more primitive system that uses linked allocation (where file blocks are linked together via pointers in each block) with respect to the number of disk accesses. Consider the ... the linked list approach for finding block one by one.Then How Random access is easier in FAT than Linked List alloaction.

asked Oct 2 in Operating System by mrinmoyh (21 points) | 8 views

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Round Robin doubt source: N B Venkateswarlu
Consider a system running ten I/O-bound tasks and one CPU-bound task. Assume that the I/O-bound tasks issue an I/O operation one for every millisecond of CPU computing and that each I/O operation takes 10 milliseconds to complete. Also, assume ... for a Round-Robin scheduler when: (a) The time quantum is 1 msec. (b) The time quantum is 10 msec.

asked Oct 2 in Operating System by ghostman23111 (5 points) | 12 views

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Gate 1993 2 marks question
The details of an interrupt cycle are shown in figure.

asked Oct 2 in Operating System by learner 2102 (5 points) | 9 views

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Syncronization
How to check whether the given any block having bounded waiting or not?

answered Sep 29 in Operating System by iarnav (231 points) | 31 views

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Made easy workbook os memory management
Consider a computer system with logicala addresss space and physicaladdress space of 216 Bytes. Page size is 512 Bytes. Page table entry size is 2 bytes. If page table entry contain besides frame some other information like one bit for ... many bits are still availble in the page table entry to store aging information of the page? Memory is byte addressable

asked Sep 27 in Operating System by Scion_of_fire (47 points) | 8 views

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Self Doubt (Operating System)
If a process is running in the CPU and a new process arrives in the ready queue, is the working of the current process stopped (context switched) then compared and then again rescheduled in the CPU or no context switching happens?

asked Sep 22 in Operating System by Mellophi (333 points) | 41 views

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#Applied Course Test Series #Operating Systems #Synchronization

answered Sep 22 in Operating System by k.saurav (5 points) | 33 views

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Workbook made-easy os virtual memory
Consider a system having a LA = 28 bits. The address space is divided into page with page size be power of 2. PTE size is 4 bytes. What must be the page size in a system (in bytes) such that the page table of process exactly fits in 4 page frame of memory.

asked Sep 22 in Operating System by Scion_of_fire (47 points) | 11 views

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Made easy workbook os virtual memory
Virtual memory fetch strategies determine when a page or segment should be moved from.....to... (a) main memory, the TLB (b) secondary storage, main memory (c) main memory, secondary storage (d) the TLB, registers Ans: C Please explain approach of this ans

asked Sep 22 in Operating System by Scion_of_fire (47 points) | 8 views

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Self Doubt in Binary Semaphore
Suppose binary Semaphore S=1 P0: Up(S) Critical Section. So now what will be the value of S after performing up operation and will the process P0 enter critical section?

asked Sep 22 in Operating System by Avi100 (5 points) | 35 views

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Process synchronization (interested variable mechanism)

asked Sep 21 in Operating System by Kshyamano (5 points) | 11 views

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Made easy os workbook
Consider the following statements: S1: Throughout always increases with increase in degree of multiprogramming S2: interrupt can be handled in between the execution an instruction Which of the above we true? a) Only S1 B)only s2 C) both S1 and S2 d) neither s1 Nor S2 is true

asked Sep 21 in Operating System by Scion_of_fire (47 points) | 19 views

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Made easy os workbook
Match the pair List I (a) system call (b) command interpreter (c)swapper (d) dispatcher List II (1) context switching (2) medium term schedule (3) interrupt (4) user mode

asked Sep 21 in Operating System by Scion_of_fire (47 points) | 14 views

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Made Easy Operating system doubt
Multilevel feedback queue scheduling suffers from starvation [True or False]. I think it is false, Multilevel queue scheduling suffers starvation, while in Multilevel feedback queue scheduling we move the process from lower priority queue to higher priority ... starvation, and therefore it should be false. [As per madeeasy key it is true, Someone please explain]

answered Sep 18 in Operating System by Ehraz Hasan (366 points) | 23 views

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2 answers

ace test series
Consider a system with VAS of 8MB; Page table entry is 32-bits.The optimal Page Size in order to minimize internal fragmentation and page table size overhead is________ KB

answered Sep 16 in Operating System by Mellophi (333 points) | 37 views

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MadeEasy Subject Wise OS
Consider the following code comprising of a swap function which is atomic. void swap(bool *a, bool *b){ bool temp; temp = *a; *a = *b; *b = temp; } bool lock = false void process(int i){ bool key; while(1){ // ... Bounded Waiting The code snippet solves the synchronization problem [According to me it should be bounded waiting but MadeEasy answer key says option (D)]

answered Sep 15 in Operating System by Ehraz Hasan (366 points) | 28 views

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Self Doubt (OS)
https://gateoverflow.in/74241/two-process-critical-section-software-solution In this question people are saying bounded waiting is satisfied. But in parbegin there are only 2 processes and once a process is complete it will not be scheduled again as parbegin statement is not inside any loop. So how can bounded waiting be not satisfied?

asked Sep 14 in Operating System by Mellophi (333 points) | 35 views

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MadeEasy Subject Wise Operating System
Consider a system where 2-level paging is used with TLB support. The access time for TLB is 20 ns while main memory access time is 80 ns. 130 pages references found in TLB and total 400 reference is generated. What is the effective memory access time?

answered Sep 13 in Operating System by Ehraz Hasan (366 points) | 16 views

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MadeEasy Subject Wise OS
Consider a system using 2-level paging and the virtual address is 38 bits. The most significant 10 bits are used to index the page directory and next 16 bits index the page table. Each entry in both levels is 4 bytes. What is the maximum number of page ... table only, it can have multilevel page table but it is one only. why a process will have more than one page table?]

asked Sep 13 in Operating System by Ollie (453 points) | 13 views

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Made easy os workbook
Consider a virtual memory of 256 tera Bytes. The page size is 4KB. This logical address space is mapped into a physical memory of 256 Mega bytes. Given the address: OX 304F 48A C5345. What is the page offset in decimal? 837 709 345 563

answered Sep 13 in Operating System by Ehraz Hasan (366 points) | 21 views

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INODE GATE ACADEMY
is the answer 580KB?

answered Sep 9 in Operating System by RavGopal (145 points) | 39 views

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GATE 2020 question doubt
0.95( 20+100 ) + 0.05[20 + 0.9( ( 1+1 )100 ) + 0.1( 0.8( 5000+100+20 ) + 0.2( 5000+5000+100+20 ) ) ] Will this formula work if TLB update time is not negligible? Question came in gate 2020 https://www.geeksforgeeks.org/gate-gate-cs-2020-question-63/

asked Sep 9 in Operating System by Shashank Rustagi (513 points) | 25 views

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GATE ACADEMY TEST SERIES SUBJECT WISE OPERATING SYSTEMS 2

asked Sep 8 in Operating System by Shashank Rustagi (513 points) | 31 views

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Made Easy Test Series - OS: Paging
I think the answer should be 22 MB. We get Page Table Entry size = (22/8) B Page table size = 2^23 * 22/8 B = 22 MB. Why have they rounded 22/8 as 3B? Answer should be 22 MB and not 24 MB. Please clarify. Thanks!

answered Sep 8 in Operating System by Shashank Rustagi (513 points) | 32 views

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Made Easy Test Series - OS: Memory Management
I think the answer should be B). S1 is incorrect. This is because TLB needs to be saved on a context switch. TLB stores frame numbers of frequently referred pages and therefore it has to be updated on a context switch between the process. We cannot simply flush it for a process going out. Please clarify. Thanks!

answered Sep 8 in Operating System by Shashank Rustagi (513 points) | 30 views

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#doubt from a question mentioned
as per my knowledge of information in multicore systems we use spinlocks so that while one thread is busy waiting on while loop other thread can work on Critical section. my question is in gate question when busy waiting is being used using while ... process is busy waiting no other process can work on CS. like in this qsn https://gateoverflow.in/1839/gate2006-61

answered Sep 7 in Operating System by Shashank Rustagi (513 points) | 17 views

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Made Easy Test Series - OS: Process Scheduling
I think the answer should be 65%. They say 8 processes arrive in 60 seconds. That means the first process will arrive at the time instant t = 7.5s. Now at the time t = 60s, 8th process will be arriving. However till this mark, 7 processes are ... 39s % of time CPU is idle = (39/60) * 100 = 65% Please clarify once if I am correct or not. Thanks!

answered Sep 7 in Operating System by Ehraz Hasan (366 points) | 27 views

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OS NPTEL ASSIGNMENT QUESTION
Degree of concurrency in threads can be arranged in the manner One-to-one > many-to-one > many-to-many One-to-one > many-to-many > many-to-one Many-to-many > many-to-one > one-to-one None of the above answer given is A but according to me answer is B. Someone please confirm

asked Sep 7 in Operating System by Akanksha Agrawal (309 points) | 54 views

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Effective access time in OS
Can someone share the notes related to calculating effective access time Because I'm not able to solve those questions which is asked with cache miss,page fault and TLB miss.

asked Sep 7 in Operating System by kaleen bhaiya (9 points) | 20 views

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Made Easy Test Series: OS - Process Scheduling
How is the answer LIFO? If I take the following example: Process Arrival Time Burst Time Priority P0 0 3 1 P1 0 5 1 P2 3 3 1 P3 6 4 1 Now we know that the priority is dynamic and so if we draw the Gantt chart: P0 P1 P0 ... (first out) but it is the first to arrive. LIFO is clearly not being followed. Can someone clairify how the answer is LIFO? Thanks!

asked Sep 7 in Operating System by sankalpmittal (33 points) | 19 views

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