🔥 UGC NET Rapid Revision Series | DAY-12 OPERATING SYSTEM

[00:00:42]Hello all. Good morning.

[00:00:45]How are you all? I'm really sorry that I'm late.

[00:00:49]Okay, I'm I'm late for again 10 minute I think 7 minutes I'm late, right?

[00:00:55]Okay. Sorry.

[00:01:01]One second.

[00:01:02]Let me read your messages.

[00:01:07]How about operating system revision?

[00:01:09]Easy, right?

[00:01:15]Can you able to recall the concepts of operating system?

[00:01:22]Uh let me um check the messages. Just wait a bit slow. One second.

[00:01:31]Okay. Harani kaishrishna.

[00:01:45]Okay. Good morning all of you.

[00:01:50]Okay.

[00:01:53]August 2024.

[00:02:01]Then we have to do gate and set. Okay.

[00:02:05]Main focus August 2024 and set and gate. Okay. And evening we will do more gate questions.

[00:02:17]Okay. So let us start without wasting time. So tomorrow we have discrete structures. Okay. Maths.

[00:02:26]Okay. Good morning.

[00:02:29]Yes. So, let us start. This is your first question. Easy question. First, tell me the answer first.

[00:03:06]Yes, you are right.

[00:03:09]Process management, memory management and file management are the functions of operating system but not post. What is post? Post is a power on self test.

[00:03:17]Right? System booting power on self test. test for booting test. Okay. Uh that is performed by BIOS. It's not a function of operating system. And this is the second question from deadlock.

[00:03:34]Okay.

[00:03:39]Not needed Tell me the answer.

[00:04:11]Answer C.

[00:04:15]Any anybody else got any other answer?

[00:04:19]Hrishna.

[00:04:23]Answer two. I mean option B that means answer two.

[00:04:30]Option four. Kavanga 2.

[00:04:42]Good morning. Kbarisha also three.

[00:04:47]Okay. Let us see.

[00:04:57]Okay. Easy. Total number of resources of five. Okay.

[00:05:09]Okay.

[00:05:17]Right?

[00:05:20]need needed uh resource to one process and for the remaining process minus one.

[00:05:26]Do you can you recall that?

[00:05:29]Yes. So let us start we have only five out of five we are giving to one process one uh I'm just writing P1 P2 P3 likewise I'm writing because I don't know how many process is going to process so what I'm going to do I'm just write like this and five instances of resource we have so for one process I'm giving full what is the need three the remaining process minus one that means 3us Okay, that's all my resources is over.

[00:06:03]Now I don't have I don't have any other resources for allocating to process three. That means only two process can give these uh resources so that catalog can never be occurring.

[00:06:44]Why you are going to use the formula?

[00:06:47]Okay. Now moving to the next question that is I asked you August 2024.

[00:07:46]Answer Mahasendul Shajida.

[00:07:55]Yes, you are right. Correct answer is option three. That is AC D. Process process control block. Process control blockcess information process control block. It will reside in the main memory that you know. Okay.

[00:08:15]cell in process complete information state whether it is running waiting or terminate I'll show you the details okay see this is this is our class slide okay it's residing memory over process over process control block okay for each process one process control block will be there and throughout the lifetime of the process it will be there and it's deleted once the process deleted okay and uh this information I took from this website. Okay. So, information uh process state, current state of the process, whether it is ready, running, waiting or whatever.

[00:08:58]Contains information about the process state such as whether it is running, waiting or terminated. And then processes process ID program counter. Program counter.

[00:09:21]Yes.

[00:09:25]All right. And what else? Registers.

[00:09:28]Yes. That is in the option four I think.

[00:09:29]Yes. CPU registers. So scheduling information and CPU registers.

[00:09:36]Okay.

[00:09:38]Program code and data segments. Okay. It stores a process memory management information. See see this memory limits.

[00:09:46]Memory management information. Memory management information means what about uh what are the things in memory management? That is memory limits. Page table or information. Okay. Segment table. All these details will be there in this memory management information.

[00:10:03]Okay. So, okay. Program code and data segments of the process will not be there.

[00:10:14]Next question about mutual exclusion.

[00:10:27]Yeah. Yeah.

[00:11:00]Okay. Fourth one. What is the answer?

[00:11:14]What is the answer? Good morning.

[00:11:17]Okay.

[00:11:20]Fourth one. Fourth one is option A because mutual exclusion ensure that only one process can be in critical section. Yes, that is the definition.

[00:11:31]Mutual exclusion is designed to prevent conflicts. Yes, mutual exclusion can use various algorithm. Yes, various algorithm.

[00:11:42]But last option mutual exclusion allows multiple process to access the critical section.

[00:11:51]First statement contradictory statement option D that's why it is wrong. Next question.

[00:12:11]I know I remember you day and I you were in PGT. I know I think we all mentioned Google meet also right one time during the PTB section.

[00:12:28]This is question number five option.

[00:12:42]What is the answer?

[00:12:47]Option B. Some students telling option C.

[00:12:53]Answer is of course option C.

[00:13:00]Threat virtual address space file system signal signal path that is interrupt.

[00:13:11]Signal is matched with interrupt. No doubt in that four is matched with P. So these options gone. Okay. Then uh file system of course this file system disk only three 3s. Yes. So that we can choose this. Okay. Then other other than I is thread thread and uh what what else? Virtual address space. Okay.

[00:13:36]Virtual address space is memory related, right? And thread the CPU.

[00:13:40]What is the concept? Thread is represent unit of execution on the CPU. Right?

[00:13:47]Thread represents a unit of execution that runs on the CPU. Virtual address space provides an abstraction of physical memory.

[00:13:55]File system organize and manages data stored on disk. Signal is a software level mechanism that is analogous to hardware inter.

[00:14:09]Next question. August 2024 simple question.

[00:14:31]Answer sixth one.

[00:14:42]Yes. Simple. No explanation required.

[00:14:44]But why I choose chosen this question?

[00:14:47]Because yes.

[00:14:59]Now this is your seventh question.

[00:15:01]Asking gauge expected question. Okay.

[00:16:30]One second.

[00:16:47]Okay. What is the answer you got?

[00:16:52]Okay. Praadi mamaya kavanga mahasindil rani krishna all of you given wrong answer.

[00:17:03]Ghami and da da and ja prascila anita gave the correct answer.

[00:17:14]Okay. What is the doubt in this?

[00:17:18]What is critical section means critical section is the area where process can use the shared resources. Shared resources area only one process can enter that is called mutual exclusion and progress means whatcess I mean one process need to access Okay. Then that process we should allow that process to enter into the critical section if it is free means that is called what progress. Okay. So mutual exclusion progress.

[00:18:02]So the answer is mutual exclusion only.

[00:18:04]Why? Because here the initial values random assign. Okay. So we can uh check the program. See here while S1 S2 this is the program for P1 first process then second P2 okay first P1 condition S1= to S2 that means there is variables S1 S2 equal it is not allowed P1 to enter into the critical section. Okay S1 S2 both same then you are not allowed to access the critical section. You just wait, keep on wait till this condition got wrong.

[00:18:44]S1 and S2 not you can see a semicolon here. This is called infinite loop.

[00:19:00]I mean okay. So it cannot allow it cannot go.

[00:19:10]If suppose the values s1=0 and s2=0 what will happen? The p1 will keep on checking but it is not satisfied. So it will wait. But P2 if P2 wants to enter P2 is checking S1 not equal to S_sub_2 S1 not equal to S2.

[00:19:30]Okay.

[00:19:36]But this is the favorable condition for P2.

[00:19:44]P2 will go to the critical section.

[00:19:46]Enter into the critical section. It will do it will access the resource after finishing. Okay. Resource it will change the value S2= S1 S2= S1 so that the P1 can access the critical section. Okay. That means only one process is allowing to enter into the critical process.

[00:20:12]Okay. That means mutual exclusion is okay. Mutual exclusion.

[00:20:17]Okay. But what about progress?

[00:20:24]Mutual exclusion I mean critical area.

[00:20:27]Nobody is there. Okay. And the values are S1=0 and S2=0.

[00:20:33]Okay.

[00:20:35]Right. Critical section three. P2. P2 chance because P2 can enter but P2 doesn't want to enter. P2 doesn't want to enter but P1 want to enter. P1 wants to enter. Then what will what will happen can P1 get the chance?

[00:20:54]Tell me using this code the value is S1=0 and S2=0. P2 is not entering into the critical section but P1 needs the critical section. Then what will happen?

[00:21:08]Is the P1 get a chance? P1 chance.

[00:21:14]Tell me.

[00:21:31]Tell me S1 S2 P2.

[00:21:41]Okay.

[00:21:43]But P1 wants wants a critical session.

[00:21:45]Then what will happen?

[00:21:48]Tell me. That is called progress.

[00:21:51]Progress. No. Progress is not satisfied.

[00:21:54]Even though P even though this critical section is free free P1 wants to enter but P2 enter I P2 in the condition change progress is not implemented by this code. Is it clear?

[00:22:19]Yes. If no process is in the critical section and some process want to enter one of them should eventually be allowed. If it is not allowed means progress is not satisfied. I hope it is clear.

[00:22:34]Okay. And in the boundary waiting another one condition is there. You remember three conditions are there now for the for the critical section problem solution. Okay. Initial solution progress and bounded waiting. Bounded waiting also not here. Bounded waiting means limited.

[00:23:01]Okay. Because of this progress.

[00:23:05]So here solution.

[00:23:11]Okay.

[00:23:15]Is it clear or not?

[00:23:18]Tell me.

[00:23:29]Okay, next one.

[00:23:42]Okay, I will take more question from this critical section. Okay, I think you need more practice.

[00:23:57]Grade level 11 question.

[00:24:30]What is the answer?

[00:24:34]84 also wrong. The correct answer is option two. Raid level two. Raid level two.

[00:24:44]Error correcting hamming code.

[00:24:48]See this. This is our class slide.

[00:24:50]See rate two bit level striping with dedicated parity used error correction code named hamming code. See error correcting rate level one mirroring.

[00:25:06]Okay zero striping.

[00:25:10]Next rate level four block level striping with dedicated parity.

[00:25:18]Raid level four. See block blocked is level five.

[00:25:33]Then level three is bit this one.

[00:25:44]So it's match with three see by level. Sorry.

[00:25:58]Okay. And rate six is there. It can handle double failure. Two dise I think December 2025 P plus Q redundant scheme. Okay. Raid 1 Z that is 1 and zero. Good combination.

[00:26:18]Okay, right combination.

[00:26:22]Next question.

[00:27:07]10th one it's related to privilege and instruction simple question that is option one only one.

[00:27:30]Okay. What is privilege instruction?

[00:27:32]Means privilege instruction.

[00:27:37]It can only be executed by OS kernel.

[00:27:40]Okay. And it is designed to perform operation that can directly affect the hardware. Yes.riv privilege instruction.

[00:27:50]Okay. And third option, user application can execute privileged instruction if they have the correct permission. No, if they got the permission also user privilege instruction access only with the help of system.

[00:28:10]See they must request the OS through a system call and the colonel execute the privilege instruction. Okay. system but execution not user. Okay. And the last option it usually executed in user mode to en ensure the safety and security. No, it is uh this privilege instruction will be executed in kernel mode only. Simple question. Next one. This types of question you can expect from paper one.

[00:28:42]Okay. ICT paper.

[00:28:47]This type of fastest to slowest.

[00:29:12]Yes.

[00:29:16]Uh the answer of course then slow.

[00:29:33]Okay. So this is the last which one is faster cashier or register.

[00:29:51]Suppose in this question if they have included this option register.

[00:29:56]Yes. Before cash. Very good. Register to the cash. Register the fastest.

[00:30:02]Yes. Good.

[00:30:05]Next question.

[00:30:14]Message passing explain based on this this question.

[00:30:20]Okay.

[00:30:31]Anyone can recall this trend was yes.

[00:30:52]Okay. Kbarisha. Kbarisha. Correct.

[00:30:54]Brahma. It's wrong. It's not D. The answer is option A.

[00:31:00]See this is our class.

[00:31:05]I think you didn't do the properly correct since synchronization message passing can be blocking. Blocking or non-blocking blocking means synchronous.

[00:31:18]Nonblocking means options blocking sent non-blocking send.

[00:31:23]Blocking received. Non-blocking receive.

[00:31:27]Recall the concept. If both send and received are blocked.

[00:31:32]And receiver both are blocking means it is called renus. It is used in a programming language. Okay. Random sender and receiver both blocking it is called renas.

[00:31:46]Yes.

[00:31:49]Next question. August 2024.

[00:31:57]must go through the notes once that okay before the exam notes full and full go through pano PDF or the your class notes Okay. So correct answes and circular access.

[00:33:14]Then finally virtual memory that is extension of physical memory illusion right illusion providing that secondary storage secondary memory for the execution purpose that is extense physical memory.

[00:33:32]13th question.

[00:34:17]Yes and correcting the correct answer is option three only.

[00:34:25]What is multi-processor system means?

[00:34:27]Multi-processor system is controlled by one operating system. Yes, because multiprocessor system multiple CPUs will be there. Okay. Multiple CPUs.

[00:34:40]Multiple CPUs controlled by one operating system.

[00:34:46]Second option multiple computers connected. No, multiprocessor means multiple CPUs connected by control connected and it is controlled by one object. It is not like multiple computers. That's why it is wrong. And the last option multiprocess system is classified as multiple instruction stream and multiple data stream. Yes, ma right multiple instruction stream and multiple data stream system and according to flying taxonomy this is called MIMD which display how multiprocess system operate.

[00:35:25]Okay, next question.

[00:36:16]Which one takes more time? Turka wrong and wrong.

[00:36:25]Question number 14. Let the time taken to switch between user and kernel mod t1 and the time for contact switching that is t2 of course for t2 t1 is less than t2. Why? Because user to kernel mode switches involves only privilege mode either user 0 or exiting theel butcess.

[00:36:58]These are the steps involved in the contact switch. First of all in the current state then next process state load. Right?

[00:37:08]Then DCB information process control block information update. Then finally memory management.

[00:37:18]Okay. So all these process are all these steps are involved in the context switch. That's why it takes more time than switching from user to kernel mode.

[00:37:30]Next question.

[00:37:46]Expected questions continuous allocation linked allocation index multi Yes.

[00:38:43]What is the answer?

[00:38:46]Okay. Everyone say option C. Yes.

[00:38:49]Continuous contigious memory allocation.

[00:38:52]That's why minimal access number of disk access is minimal.

[00:39:13]random access required. No, right. So lingual allocation supports only sequential access. So it is matched with this. Then index allocation level index allocation.

[00:39:28]So wastage of space then multi-le supports very large files large multi-level allocation.

[00:39:43]Okay. So multi-le allocation uses a hierarchy of index block single double triple likewise. So it support very large files.

[00:39:52]Next one 16th It's very very important.

[00:40:37]Which of the following is true?

[00:41:02]important.

[00:41:14]Okay, correct answer is uh I think only Krishna gave the correct answer. A thread is usually defined as a lightweight process because an operating system maintains smaller data structures for a thread than for a process. In relation to this, which of the following is true? on per thread basis always maintains only CPU register. Thread basis register CPU OS maintains only CPU register state OS does not maintain separate ST for each thread on per thread basis. OS does not maintain virtual memory state which is a correct answer. This is the true statement. See a thread has its own CPU register state, program counter, stack and scheduling information, but it's thread belonging to the same process chair, core segment, data segment, open file, virtual address, virtual memory.

[00:42:28]Okay, you already know segment and data segment and virtual address also.

[00:42:33]Virtual memory also it is sharing.

[00:42:38]That's why option C is the correct answer. Virtual memory state is maintained per process not per thread maintain.

[00:42:51]Next question numerical directly formula.

[00:43:34]Why in the previous question the option one is wrong?

[00:43:38]Maintains only CPU register state.

[00:43:52]Sorry. Yes, you got the answer.

[00:44:22]Okay. So what is the heat ratio here?

[00:44:24]35. So just multiply 0.35 into primary access time that is 10 right. Then what is the um uh what miss ratio? So 1 minus.35 that is 65 0.65 into memory access time that is 100 ncond that's all when you do the calculation you will get 68.5.

[00:44:51]Okay, simple.

[00:44:54]But you can expect this this type of question.

[00:44:58]Next question.

[00:45:17]Whatever new information you are getting from question now itself you have to write. Okay.

[00:45:40]18th one.

[00:45:46]17th.

[00:45:47]18th one.

[00:45:51]Option BT.

[00:46:07]Okay. So I will directly the option Ri and Anita.

[00:46:17]Correct answer is option A.

[00:46:22]Okay. For ls for ls list.

[00:46:27]Okay. That is a that is a external commain.

[00:46:32]Okay.

[00:46:36]Internal command. It's built into the shell itself. Separate executable files they not required. Okay. Separate comm library function. Print is a library function. It is used to display output on the screen.

[00:47:03]And the last one is ls that is list.

[00:47:06]Okay. It's used to display the files and directories present in a uh in a directory. Okay.

[00:47:23]separate executable file.

[00:47:26]Okay. So, echo is a internal command and ls is external and is a library function that you might know right. Next question.

[00:47:41]If the page size is increase, what will happen?

[00:48:14]Recall the paging concept.

[00:48:46]Yes, everyone is saying correct answer I'm wrong. Answer is becomes small larger page size more unused space will be there likely in the last therefore internal fragmentation increases for example page size 4 KB okay and process needs uh maybe 3 3.4 Four four page size is four and process may need 3 maybe 3.5 3.5 Kess okay 8 km I mean again more space that means page size increase will get increased.

[00:50:03]Okay that's why option B is the answer.

[00:50:11]Okay, because in paging memory is divided into fixer size pages, right? That's why Anita also wrong.

[00:50:28]Which of the following statement is false?

[00:50:56]Long-term scheduleuler selects process from disk and loads them into memory for execution. 28th is concept of switching is regard as option D. Answer what is the answer? False statement.

[00:51:15]False statement is option B only. Degree of multirogramming is controlled by whom? Who is controlling the degree of multirogramming?

[00:51:24]H tell me who is controlling the degree of multirogramming.

[00:51:38]Which scheduleuler is controlling the yes long-term scheduleuler is controlling the degree of multirogramming because long-term scheduleuler select process from disk and load them into memory.

[00:51:55]So that is degree of multirogramming.

[00:51:57]Medium-term scheduleuler is responsible for swapping to reduce the degree of multirogramming. This is a correct statement and concept of cont switching is regarded as overhead. Yes, this is over. Okay. And this is our class slide.

[00:52:11]Long-term cellular controls the degree of multirogramming. Medium-term cellular reduces the degree of multirogramming.

[00:52:18]Next question.

[00:52:23]Consider the following program for Yes. 213 zombie process. Why? Because okay greater than Zcess equal 100 parent.

[00:53:54]I meancess in the process table. Then what will happen? it will become a zombie process till the parent wakes up and exit the weight code. Okay, that's why the answer is zombie process.

[00:54:32]Next question.

[00:54:35]Computer network questioning RPC remote procedure connected 22.

[00:55:14]So connection between two network client and server client server communication.

[00:55:21]Yes, correct answer is client correct format to sending into the network. You need to convert this parameters into a format.

[00:55:44]Okay. And uh that is okay. Then other side process of converting procedure parameters into a format that can be transmitted over the network that is client step skeleton is at the this is wrong. Skeleton is at the server side but it ending.

[00:56:14]Okay. So this is the correct answer.

[00:56:21]And this is the um steps. Okay.

[00:56:27]So just read it.

[00:56:31]Next one.

[00:56:54]23. Yes, very good. 23 is it prevents crashing. Which of the following statement is true about working set model? Working set model.

[00:57:05]What is working set model?

[00:57:12]working.

[00:57:22]Okay. Working set means frequently accessing memory set. Okay.

[00:57:37]Thrashing. What is mean by thrashing?

[00:57:42]Yes, thrashing.

[00:57:44]Thrashing is swapping and swapping out.

[00:57:47]Swapping in and swapping out frequent that is that is called trashing. Okay.

[00:58:01]again transferring and swapping out thrashing. Thrashing is not a good good property. Okay. System spends more time swapping pages in and out.

[00:58:17]Why? because it's ensuring that each process get in a frame for its current working set. Okay, next question.

[00:58:33]Yes, Sha correct.

[00:59:04]memory management unit.

[00:59:15]Yes, it's a hardware unit, right? It's a hardware unit that maps virtual address to physical address. That is the definition of memory management. It's a hardware unit. It's a chip to map virtual address to physical address.

[00:59:28]Okay.

[00:59:30]This is essential system that use virtual memory where each process has its own address space.

[00:59:37]Okay. Sorry. Next one question so that you can learn it now. Full form of POSIX in Pthread concept full form of POSIX is portable operating system interface. What is this? It is a set of ITE standard designed to maintain compatibility between operating system operating system between operating system.

[01:00:04]compatibility sets portable operating system interface.

[01:00:16]Okay, it's a part of the standard and provide a standardized C programming interface for creating and managing threats. Okay. So, P concept. Okay.

[01:00:35]Next one.

[01:00:45]Either you can expect non preemptive or preempt What is the answer? 26th one.

[01:01:19]Oh, actually 25th question. 26th.

[01:01:31]Yes Krishna correct grandma correct shaj wrong nonpreemptive CPU scheduling nonp prereemption of CPUcess it is terminated or it is going for the voluntarily input output operation So when it is going to the terminated state or it is going to the waiting state a process switch from running state to waiting state for the input output input output processing I mean input output waiting.

[01:02:29]Okay. and terminating also it is overpreting.

[01:02:36]Okay.

[01:02:38]But running to ready preeemption and waiting to ready also preemption.

[01:02:46]Okay.

[01:02:50]27th What is the answer?

[01:03:20]Yes.

[01:03:23]Opensource software. Software that is distributed with its source code allowing users to use, modify and redistribute it with its original rights. Opensource software. Okay.

[01:03:34]Example Linux and Firefox opensource software.

[01:03:50]Okay.

[01:03:57]Proprietary software.

[01:04:01]Okay. Open we can modify and redistribute it.

[01:04:21]Linux rm command prompts the user for confirmation before removing each file. rm for removing.

[01:04:30]Okay.

[01:04:32]Something we will add here for getting a confirmation. Are you want to delete or not?

[01:04:51]interactive mode that is ask for confirmation before deleting each file.

[01:04:56]And next question.

[01:05:07]Okay.

[01:05:12]Wrong.

[01:05:18]Change the access permission. discuss access permission.

[01:05:27]Yes, ch mode. Yes, very good. CH mode is change mode. Choose to change the file and directory per set.

[01:05:39]Yes. And what is pwd? Do you remember what is pwd?

[01:05:44]H pwd is print working directory print working directory and it displays the current directory path and cat for concatenation. Okay, file concatenate combined cq and ss means secure shell.

[01:06:05]And 30th question.

[01:06:20]So the following is active first during the normal.

[01:06:33]What is the answer? Yes. Option C ROM BIOS RAM BIOS BIOS BIOS it's ROM BIOS exe first then performs the power on self test after that read the SMOS boot configuration then finally boot load from the hard sequence question and this is your last question virtual memory system can be combined by which of the following technique Demand paging techncept.

[01:07:51]Paging and segmentation technique both segmentation both combined virtual memory paging means break the memory into small equaliz pages. Okay.

[01:08:05]Only needed pages are loaded into RAM from the disk making memory usage efficient and allowing the program to be larger.

[01:08:12]Memory segment division based on logic it is dividing. Okay.

[01:08:30]Then whenever it is needed it is it can be stored on this when not in use and loaded into RAM as needed. That is virtual memory. Right. virtual memory concept this paging and segmentation combin okay so that's all 31 question discuss and uh we will do more uh gate question in the evening section okay and tomorrow max discrete mathematics discrete structures we will do okay so thank you all And this is our ninth subject revision. Okay. And tomorrow is the last subject.

[01:09:16]So late.

[01:09:19]Sorry for that. You can continue your reision. Okay. So bye all of you. Thank you so much.