I am recently taking a course called The principle of operating system, and I learned CPU scheduling. I am confused about Round robin scheduling, for I/O-bound process, for example, the process will use CPU for 2ms and does I/O for 8ms. Will scheduler still assign a quantum to this process when it is doing I/O? Also, when this process is doing I/O, will the scheduler wait for the I/O to complete even when the quantum expires or it will just start to execute next process? Any help would be appreciated!
Will scheduler still assign a quantum to this process when it is doing I/O?
Typically each task has a state, maybe one of:
running, currently using CPU
ready to run, waiting to use CPU
blocked, waiting for something (disk IO, a mutex, a time delay, a network packet to arrive, ...)
The scheduler only cares about tasks that are running and ready to run - e.g. it might have a (circular singly linked?) list of tasks that want CPU time, and when a task blocks the task is removed from that list (and then later when whatever the task was waiting for happens and the task is unblocked, the task is put back on the list).
Traditionally; when a task is put back on the list it's put back on the end of the list, so that a task can't repeatedly block briefly to get a new time slice and hog the CPU.
This means that if there are 2 tasks and one blocks, a round robin scheduler might do "task A, task B, task A, task B" while task A is running/ready to run; then switch to "task B, task B, task B, task B" while task A is blocked; then after task A unblocks it'd go back to "task B, task A, task B, task A, ..." (starting with task B because task A was put on the end of the list and not the start of the list).
The other thing is that tasks literally can't decide to do something that would cause them to block unless they're currently running; which means that whenever a task blocks it doesn't use its whole time slice. For example, if the scheduler is giving out 1 ms time slices then a task may block after using 0.3 ms of its time slice, leaving a remainder of 0.7 ms. For this reason the scheduler needs a timer with higher precision and the length of time slices will be rounded to the precision of the timer IRQ (e.g. if the scheduler is using a timer IRQ that occurs every 0.2 ms; then that remaining 0.7 ms left after one task blocks might be rounded to 0.8 ms leaving a spare 0.1 ms due to rounding, and the next task might actually get 1.1 ms of CPU time instead of 1.0 ms of CPU time because of that "rounding to the timer's precision").
Also; when all tasks are blocked the scheduler's timer can be suppressed/disabled (and the CPUs put into a power saving state) to reduce power consumption by preventing pointless timer IRQs from waking the CPU out of a power saving state; and when only one task can run the scheduler's timer can be also be suppressed/disabled (and the task given an "effectively infinite" time slice) to prevent the overhead of pointless timer IRQs from decreasing the performance of the task.
Note 1: Almost all universities ignore reality; starting with the extremely dodgy assumption that its possible to know how long a task will use CPU time and when it will block or use IO (followed by the assumption that everything happens in nicely "aligned to time slice duration" boundaries).
Note 2: Almost all universities assume that "IO" means the initiating task is blocked; either because the disk controller does the IO while the CPU does other things, or because one or more different task/s use the CPU to do the IO while the initiating task blocks (e.g. your task calls "read()", your task is blocked and a file system task is unblocked, the file system task asks the disk controller's driver to fetch some data, the file system task is blocked and the disk controller driver's task is unblocked, then ...). This isn't strictly true in all cases (but may be true in all cases for some operating systems).
In general time critical I/O will typically be handled by an interrupt handler rather than a round-robin scheduled process.
For example, say you have a UART with no hardware FIFO, a character arriving in its data-register, must be read before it is overwritten by the next received character. In this case the character might be placed in a software FIFO buffer (a pipe or queue). That buffer would need to be large enough to capture all data received while the receiving process is not running. When the receiving process is scheduled it will receive all buffered data at once.
In other cases, I/O may use DMA operations, which occur in parallel to CPU operations. There is still often an interrupt handler involved but it would be a DMA controller interrupt rather than an interrupt from the I/O device.
Non time-critical I/O may simply be polled or asserted in a round-robin process when no precise timing is required.
If an application has a great deal of time-critical I/O and also time critical data processing. Round-robin scheduling may not be appropriate. Real-time operating systems generally use priority based premptive scheduling, with round-robin for tasks if equal priority.
The concept that a process either uses the CPU or does I/O however makes no sense, a process runs on the CPU, whether it is performing I/O or data processing. In fact for memory mapped I/O the CPU makes no real distinction.
Related
If I am running a set of processes and they all want these burst times: 3, 5, 2 respectively, with the total expected time of execution being 10 time units.
Is it possible for one of the processes to take up more that what they ask for? For example even though it asked for 3 it took 11 instead because it was waiting on the user to enter some input. So the total execution time turns out to be 18.
This was all done in a non-preemptive cpu scheduler.
The reality is that software has no idea how long anything will take - my CPU runs at a different "nominal speed" to your CPU, both our CPUs keep changing their speed for power management reasons, and the speed of software executed by both our CPUs is effected by things like what other CPUs are doing (especially for SMT/hyper-threading) and what other devices happen to be doing at the time (their effect on caches, shared RAM bandwidth, etc); and software can't predict the future (e.g. guess when an IRQ will occur and take some time and upset the cache contents, guess when a read from memory will take 10 times longer because there was a single bit error that ECC needed to correct, guess when the CPU will get hot and reduce its speed to avoid melting, etc). It is possible to record things like "start time, burst time and end time" as it happens (to generate historical data from the past that can be analysed) but typically these things are only seen in fabricated academic exercises that have nothing to do with reality.
Note: I'm not saying fabricated academic exercises are bad - it's a useful tool to help learn basic theory before moving on to more advanced (and more realistic) theory.
Instead; for a non-preemptive scheduler, tasks don't try to tell the scheduler how much time they think they might take - the task can't know this information and the scheduler can't do anything with that information (e.g. a non-preemptive scheduler can't preempt the task when it takes longer than it guessed it might take). For a non-preemptive scheduler; a task simply runs until it calls a kernel function that waits for something (e.g. read() that waits for data from disk or network, sleep() that waits for time to pass, etc) and when that happens the kernel function that was called ends up telling the scheduler that the task is waiting and doesn't need the CPU, and the scheduler finds a different task to run that can use the CPU; and if the task never calls a kernel function that waits for something then the task runs "forever".
Of course "the task runs forever" can be bad (not just for malicious code that deliberately hogs all CPU time as a denial of service attack, but also for normal tasks that have bugs), which is why (almost?) nobody uses non-preemptive schedulers. For example; if one (lower priority) task is doing a lot of heavy processing (e.g. spending hours generating a photo-realistic picture using ray tracing techniques) and another (higher priority) task stops waiting (e.g. because it was waiting for the user to press a key and the user did press a key) then you want the higher priority task to preempt the lower priority task "immediately" (e.g. because most users don't like it when it takes hours for software to respond to their actions).
Since idle tasks are generally used to safely consume CPU time that is not required by other software, what would happen if there was no idle task? Would the RTOS just automatically create one? Also, what other purpose do idle tasks serve other than consuming time?
what would happen if there was no idle task? Would the RTOS just automatically create one?
I doubt there is any RTOS that would do that. If there would be no idle task, then the list of runnable tasks would be empty and the scheduler would probably crash. Generally the single most important reason for idle thread's existence is to make the list of runnable tasks "never empty". This simplifies the code of scheduler.
Also, what other purpose do idle tasks serve other than consuming time?
In some systems idle task can perform some low priority activities (for example some garbage collection). It can also switch the core to low-power mode, especially on embedded devices. In that case when the idle task is run it means that there is nothing more to do, so the core can be stopped and wait for the next event (hardware interrupt or timeout) without using too much power. When the next event arrives the core is awakened by hardware and the event is processed. Either some "normal" thread will start running, or - if there is still nothing more to do - idle thread will resume and again switch to low-power mode.
If the CPU clock is running, instructions must be executed; if there were no idle task, then your OS is broken. The idle loop is an intrinsic part of the RTOS, not a user task, so the RTOS does not need to "create one automatically".
A low priority user task that never yields will prevent the idle loop from running; which is not necessarily a good thing. Such a task is not the same thing as the idle loop. For one thing any CPU usage tools the RTOS supports would report 100% usage all the time if such a task eusted - execution of the idle loop is not included is CPU usage because the CPU is always ready to respond to any interrupt event when idle - the loop does not ever cause any ready task to be delayed.
The idle task, or "idle loop" is typically just that, and empty loop that the program counter is set to when there is nothing else to do. In some architectures the loop may include a "wait-for-interrupt" instruction that stops core execution (stops clocking the core) to reduce power consumption. Since any context switch necessarily requires an interrupt to occur, the processor can if WFI is supported just stop in this loop.
Some RTOS support user hooks for the idle loop; low-priority run-to-completion functions that can operate in the background in the idle loop context.
what other purpose do idle tasks serve other than consuming time?
Most commonly, it does two things:
1. Garbage(resource) collection or cleaning
2. Initiate steps to reduce power consumption
I'm studying operating system on this book and my prof's slides. I'm arrived at the "Process scheduling algorithms" chapter. Talking about the RoundRobin (RR) algorithm i found some inconsistencies. I understand that is a preemptive version of the FCFS algorithm with a time-slice (quatum).
From now on, I will use the following notation:
#1 = prof's version
#2 = book's version
#3 = other version
Here's the inconsistency (suppose a quantum of 100ms):
#1 The RR uses two queue (Q1, Q2):
Q1: queue for processes that did not end their quantum;
Q2: queue for processes that did end their quantum;
The scheduler takes the process from the head of Q1;
If the process ends before the quantum expires, the process release the CPU on purpose and the scheduler takes the next process from Q1
If the process doesn't end before the quantum expires, is preempted and placed at the end of Q2;
When a process is ready it's placed at the end of Q1;
When Q1 is empty, Q1 and Q2 are swapped;
So when a process is blocked for an I/O request (e.g. after 30ms) and its quantum is not expired yet, is placed at the end of Q1 (I guess) and when it will be scheduled again it will use the CPU for his remaining time (70ms in this case).
#2 (The book did not talk about multiple queues, so I assume it will use just one queue)
The scheduler takes the process from the head of the ready queue;
If the process ends before the quantum expires, the process release the CPU on purpose and the scheduler takes the next process from the ready queue
If the process doesn't end before the quantum expires, is preempted and placed at the end of the ready queue;
#3 Source
The scheduler takes the first process in the ready queue;
If the process ends before the quantum expires, the process release the CPU on purpose and the scheduler takes the next process from the ready queue
If the process doesn't end before the quantum expires, is preempted and placed at the end of the ready queue;
If the process is blocked by a I/O request, it's placed in a waiting queue and when it will became ready will be placed again in the ready queue;
To me, these are 3 different implementations of the RR scheduling algorithm. I think that the most valuable is the #3 because the #1 can cause a starvation (if the process is placed in Q2 and new processes keep coming in Q1, then the process will never be scheduled again) and #2 will waste CPU time when a process is blocked for an I/O request. So, my question is: which one is the right one?
Round robin in theory
Round Robin scheduling can be quite good visualized when thinking about an analog clock: The hand is turning around at constant speed, so it's in the slice of a single digit for 1/12 of the time it takes for one complete run.
A single digit thus has some slice of the total available amount of resource. And, most importantly, there's a fixed order in which the digits get served: After the hand just passed some digit, it'll only get visited again after the hand passed all the other digits.
Looking at the variants you presented, number #2 the version from the book matches this exactly: After a task has been served it is put at the end of a (often so called) ready queue, and thus only gets served again after all of the other tasks have been served once.
Round robin is, as a theoretical scheduling algorithm, only considering the scheduling of multiple consumers (tasks) to a single resource (CPU).
Variants
Some common variations of the basic round robin scheduling are to either use different slice sizes for different tasks, or to dynamically adjust the slice of a task based on some metric, or even to provide more than a single slice to some tasks.
In practice
When you are scheduling tasks, you have to schedule them to more than the CPU as single resource, there will be other resources that need to be managed, like IO devices.
Very simple schedulers just ignore that fact, and leave tasks that are currently waiting for some other resource in the task queue for the CPU.
So when such a task gets its time slice, all it'll do is find that it still needs to wait for that other resource and hands back the CPU, just to get put back into the task queue by the scheduler. Starting the task, checking that the task still needs to wait for the other resource, and stopping the task takes some time that could better be spend for a task that actually can use the CPU.
To solve this, one usually has a task queue for each resource that is managed, i.e. one for the CPU, one for each IO device, etc. When a task is doing a blocking IO call, it is then removed from the queue for the CPU and put into the queue of the device it is accessing. This way, tasks that are waiting for a resource other than the CPU don't sit in the CPU task queue (and thus waste no time getting started and immediately stopped again).
This is what #3 is talking about (when you look again you see that they're talking about "waiting queues")
Another kind of "waiting queue"
Often there's also a "waiting queue" when you're talking about multi level schedulers: In that case, the ready queue is the queue of tasks that are ready and get scheduled by the primary scheduler (using round robin, for example). If a task blocks because of an IO operation, it is - as described above - put into the queue of the corresponding resource. When that resource gets available again, the task is first put into the waiting queue, from which a secondary scheduler (which is just a task for the primary scheduler) eventually takes it and puts it into the ready queue of the primary scheduler.
What your prof is about
The version #1 is probably better to understand if you rename the queues into something like "queue with tasks that did not yet run in this turn" and "queue with tasks that did already run in this turn". This is basically just a "workaround" if you don't want to have circular lists. So it's round robin, too, but a bit obscured.
[..] can cause a starvation (if the process is placed in Q2 and new processes keep coming in Q1, then the process will never be scheduled again) [..]
This is a very good observation. This could be solved if new, ready tasks get inserted at the end of Q2 instead of Q1. If suitable, this is a very good start for a discussion when your prof is asking for questions.
The first implementation can not only cause starvation, it can also cause deadlocks. If only we modify the step 5 of first implementation, the method can be made just fine.
The second approach is round-robin in its purest form.
The third approach is not round-robin but is in fact a smarter version which understands that I/O bound process should not be given another chance too soon as it will probably not be ready yet.
If you will continue reading that book you will read the next scheduling algorithm called Multi-level Queue. That is actually the best of all implementations of different variations of Round-robin. In Multi-level queue we put all incoming processes in one queue and then depending upon whether they finished in their first time slice or not, put them in another queue. This new queue has higher time slice and ends up holding CPU bound processes.
Using Multi-level queues (like 4-5 of them) the CPU combs out all the incoming processes into various classes and then picks optimum number of processes from each queue so that it is neither over-subscribed nor under-subscribed.
Any process which arrives in the system is queue at the end of the ready queue. A process which is at the head of the ready queue is selected and allowed to execute on the CPU for a time quantum q.At the expiry of q, the process is queued at the tail of the ready queue.The next process scheduled is the one at the head of ready queue.This is know Round Robin Scheduling.
OR
In round robin scheduling, processes are dispatched FIFO but are given a limited amount of CPU time called a time-slice or a quantum.If a process does not complete before its CPU times expires,the CPU is preempted and given to the next waiting process,the preempted process is than placed at the back of the ready queue.
It sounds reasonable that the os/rtos would schedule an "Idle task". In that case, wouldn't it be power consuming? (it sounds reasonable that the idle task will execute: while (true) {} )
This depends on the OS and the CPU architecture. On x86 (Intel compatible) the operating system might execute HLT instructions, making the CPU wait until something interesting happens, such as a hardware interrupt. This supposedly consumes very little power. Operating systems report the time spent doing this as "idle" and may even assign it to a fictional "idle" process.
So, when in Windows task manager you see that the System Idle Process is consuming 90% CPU what it really means is that the CPU does not have an actual a program to run 90% of the time.
Here's a good article on the subject: What does an idle CPU do?
Historically it's been a lot of different schemes, especially before reducing power consumption in idle was an issue.
Generally there is an "idle" process/task that runs at the lowest priority and hence always gets control when there's nothing else to do. Many older systems would simply have this process run a "do forever" loop with nothing of consequence in the loop body. One OS I heard of would run machine diagnostics in the idle process. A number of early PCs would run a memory refresh routine (since memory needed to be cycled regularly or it would "evaporate").
(A benefit of this scheme is that 100% minus the % CPU used by the idle process gives you the % CPU utilization -- a feature that was appreciated by OS designers.)
But the norm on most modern systems is to either run a "halt" or "wait" instruction or have a special flag in the process control block that even more directly tells the processor to simply stop running and go into power-saving mode.
There's always code to run, the idle task is the code if there's nothing else. It may execute a special CPU instruction to power down the CPU until a hardware interrupt arrives. On x86 CPUs it's hlt (halt).
This answer is specific to Windows NT-based OS.
Idle thread functioality
Tasks may vary between architectures, but generally these are the tasks performed by idle threads:
Enable interrupts to allow pending interrupts be delivered
Disable interrupts (using STI or CLI instructions, more on wiki)
On the DEBUG (or checked) builds, query if a kernel debugger is attached and allow breakpoints if been requested
Handle deferred procedure calls
Check if there are any runnable threads ready for execution. If there is one, update the idle processor control block with a pointer to the thread
Check the queues of other processors, if possible schedule thread awaiting execution on the idle processor
Call a power management routine, which may halt a processor or downgrade CPU tick rate and do other similar power saving activities
Additional info
When there are no runnable threads for a logical processor, Windows executes a kernel-mode idle thread. There is only 1 Idle process that has as many idle threads as there are logical processors. So on a Quad core machine with 4 logical/physical processors, there will be 1 Idle process and 4 idle threads.
In Windows, Idle process has ID = 0, so do all the Idle threads. These objects are represented by standard EPROCESS/KPROCESS and ETHREAD/KTHREAD data structures. But they are not executive manager processes and threads objects. There are no user-land address space and no user-land code is executed..
Idle process is statically allocated at system boot time before the process manager and object manager are set up. Idle thread structures are allocated dynamically as logical processors are brought live.
Idle thread priority is set to 0. However, this value doesn't actually matter as this thread only gets executed when there are no other threads available to run. Idle thread priority is never compared with priority of any other threads.
Idle threads are also special cases for preemption. The idle thread main routine KiIdleLoop (implementation from reactos) performs several tasks that are not interrupted by other threads. When there are no runnable threads available to run on a processor, that processor is marked as idle in a processor control block. Then if a runnable threads arrives to the queue scheduled for execution, that thread's address pointer is stored in the NextThread pointer of the idle processor control block. During the run of an idle thread, this pointer address gets checked on every iteration inside a while loop.
Source: Windows Internals. M. Russinovich. 6-th edition. Part 1, p.453 - 456.
I understand that CPU scheduling algorithms are classified into
Interactive - Round Robin, Priority scheduling
Batch Scheduling - FCFS,SJF
But I cant understand the reason behind the naming Interactive and Batch Scheduling..??
Why are algorithms like RR called interactive and those like FCFS called batch scheduling??
Thanks in advance...
The idea of Batch Scheduling is that there will be no change in the schedule during runtime: a process is scheduled to do an operation on data, and it runs until the process is finished. In 'interactive' scheduling, a new process could be launched while another process is running, and so time would be allocated for that process as well as the other. In batch scheduling the schedule is determined at the beginning of the operation.
Example of priority (interactive) scheduling:
Process A has a high priority, and process B has a low priority. Process A runs until it requires some input from the user. While A is waiting, the CPU gives some time to process B. Once the input for A has been gathered, process B is swapped out and process A is given the CPU, due to its higher priority.
Example of batch (FCFS) scheduling:
Process A and process B are processes to be scheduled. Process A is given to the CPU first, so B will not receive any time until A finishes running. Even if A pauses for user input, B will not run (and the CPU time while waiting for input is effectively wasted).
Of course, as with everything this low-level, it's not entirely that simple: to gain the illusion of multi-tasking, time is generally divided up between processes even when nothing is waiting for I/O. In priority scheduling, this may mean that more time slices are given to A than B while both are running so that A executes quicker. Both interactive and batch scheduling have their pros and cons: while interactive scheduling gives a quicker response time to the user and divides time up more 'fairly', an overhead is incurred due to how long a 'context switch' takes, which is the time taken for the processor to switch from working on process A to process B.
Interactive scheduling policies assign a time-slice to each process. Once the time-slice is over, the process is swapped even if not yet terminated. It can also be said that scheduling of this kind are preemptive.
Batch scheduling policies, instead, are non-preemptive. Once a Process is in the Running-status, it will not change status until it terminates.