Recently I came across the statement that
In SJF IO bound jobs get priority over CPU bound jobs.
I found this statement in page 4 of this slide and also in page 3 of this slide. I decide to attach the corresponding pictures below, if in case the link breaks in the future.
But I am having difficulty to understand the above and it seems rather counter intuitive to me. My argument is as follows:
I assume a CPU bound process is one which uses has higher CPU burst:(CPU Burst+IO Burst) and I assume a process as IO bound which has higher IO Burst:(CPU Burst+IO Burst). I assumed it from the knowledge I have received after reading the textbook "Operating Concepts" by Galvin et. al and the excerpt is below:
An I/O-bound process is one that spends more of its time doing I/O than it spends doing computations. A CPU-bound process, in contrast, generates I/O requests infrequently, using more of its time doing computations.
Which I guess agrees with what the professor says here.
Based on this I came up with the following examples:
Suppose I have two jobs
JOB 1: CPU BURST = 10 units; IO BURST =100 units
JOB 2: CPU BURST= 100 units; IO BURST=10 units...
SJF shall schedule JOB1 first which is IO Bound...
———————————————————————————
suppose I have two other jobs
JOB 3: CPU BURST = 10 units; IO BURST =1 units
JOB 4: CPU BURST= 100 units; IO BURST=200 units...
SJF shall schedule JOB3 first which is CPU bound...
From the above example I do not find any such correlation that SJF gives priority to IO bound jobs.
Related
I'm trying to understand how to compute the CPU utilisation for audio and video use cases.
In real time audio applications, this is what I typically do:
if an application takes 4ms to process 28ms of audio data, I say that the CPU utilisation is 14.28% (4/28).
How should this be done for applications like resize/crop? let's say I'm resizing an image from 162*122 to 128*128 size image at 1FPS, and it takes 11ms.. What would be the CPU utilisation?
CPU utilization is quite complicated, and strongly depends on stuff like:
The CPU itself
The algorithms utilized for the task
Other tasks running alongside the CPU
CPU utilization is also strongly related to the process scheduling of your PC, hence the operating system used, so most operating systems will expose some kind of API for CPU utilization diagnostics, but such API is highly platform-dependent.
But how does CPU utilization calculations work anyway?
The most simple way in which CPU utilization is calculated is taking a (for example) 1 second period, in which you observe how long the CPU has been idling (not executing any processes), and divide that by the time interval you selected. For example, if the CPU did useful calculations for 10 milliseconds, and you were observing for 500ms, this would mean that the CPU utilization is 2%.
Answering your question / TL; DR
You can apply this principle in your program. For the case you provided (processing video), this could be done in more or less the same way: you calculate how long it takes to calculate one frame, and divide that by the length of a frame (1 / FPS). Of course, this could be done for a longer period of time, to get a more accurate reading, in the following way: you track how much time it takes to process, for example, 2 seconds of video, and divide that by 2. Then, you'll have your CPU utilization.
NOTE: if you aren't able to process the frame in time, for example, your video is 10FPS (0.1ms), and processing one frame takes 0.5ms, then your CPU utilization will be seemingly 500%, but obviously you can't utilize more than 100% of your CPU, so you should just cap the CPU utilization at 100%.
Let's take this processor as an example: a CPU with 2 cores and 4 threads (2 threads per core).
From what I've read, such a CPU has 2 physical cores but can process 4 threads simultaneously through hyper threading. But, in reality, one physical core can only truly run one thread at a time, but using hyper threading, the CPU exploits the idle stages in the pipeline to process another thread.
Now, here is Kubernetes with Prometheus and Grafana and their CPU resource units measurement - millicore/millicpu. So, they virtually slice a core to 1000 millicores.
Taking into account the hyper threading, I can't understand how they calculate those millicores under the hood.
How can a process, for example, use 100millicore (10th part of the core)? How is this technically possible?
PS: accidentally, found a really descriptive explanation here: Multi threading with Millicores in Kubernetes
This gets very complicated. So k8s doesn't actually manage this it just provides a layer on top of the underlying container runtime (docker, containerd etc). When you configure a container to use 100 millicore k8's hands that down to the underlying container runtime and the runtime deals with it. Now once you start going to this level you have to start looking at the Linux kernel and how it does cpu scheduling / rate with cgroups. Which becomes incredibly interesting and complicated. In a nutshell though: The linux CFS Bandwidth Control is the thing that manages how much cpu a process (container) can use. By setting the quota and period params to the schedular you can control how much CPU is used by controlling how long a process can run before being paused and how often it runs. as you correctly identify you cant only use a 10th of a core. But you can use a 10th of the time and by doing that you can only use a 10th of the core over time.
For example
if I set quota to 250ms and period to 250ms. That tells the kernel that this cgroup can use 250ms of CPU cycle time every 250ms. Which means it can use 100% of the CPU.
if I set quota to 500ms and keep the period to 250ms. That tells the kernel that this cgroup can use 500ms of CPU cycle time every 250ms. Which means it can use 200% of the CPU. (2 cores)
if I set quota to 125ms and keep the period to 250ms. That tells the kernel that this cgroup can use 125ms of CPU cycle time every 250ms. Which means it can use 50% of the CPU.
This is a very brief explanation. Here is some further reading:
https://blog.krybot.com/a?ID=00750-cfae57ed-c7dd-45a2-9dfa-09d42b7bd2d7
https://www.kernel.org/doc/html/latest/scheduler/sched-bwc.html
it's my first post here.
I'm currently learning Modern Operating Systems and I'm stuck at this question : A computer system has enough room to hold five programs in its main memory. These programs are idle waiting for I/O half of the time. What fraction of the CPU time is wasted?
The answer is 1/32, but why ?
The answer is 1/32, but why ?
The sentence "These programs are idle waiting for I/O half of the time" is ambiguous. Let's look at a few different ways of interpreting this sentence and see if they match the expected answer:
a) "Each of the 5 programs spends 50% of the total time waiting for IO". In this case, while one program is waiting for IO the CPU could be being used by other programs; and all programs combined could use 100% of CPU time with no time wasted. In fact, you'd be able to use 100% of CPU time with only 2 programs (the 1st program uses the CPU while the 2nd program waits for IO, then the 2nd program uses the CPU while the 1st task waits for IO, then ...). This can't be the intended meaning of "These programs are idle waiting for I/O half of the time" because the answer (possibly zero CPU time wasted) doesn't match the expected answer.
b) "All of the programs are idle waiting for I/O at the same time, for half the time". This can't be the intended meaning of the question because the answer would obviously be "50% of CPU time is wasted" and doesn't match the expected answer.
c) "Each program spends half of the time available to it waiting for IO". In this case, the first program has 100% of CPU time available to it but spends 50% of the time using the CPU and waits for IO for the other 50% of the time, leaving 50% of CPU time available for the next program; then the 2nd program uses 50% of the remaining CPU time (25% of total time) using the CPU and 50% of the remaining CPU time (25% of total time) waiting for IO, leaving 25% of CPU time available for the next program; then the third program uses 50% of the remaining CPU time (12.5% of total time) using the CPU and 50% of the remaining CPU time (12.5% of total time) waiting for IO, leaving 12.5% of CPU time available to the next programs, then...
In this case, the remaining time is halved by each program, so you get a "negative power of 2" sequence (1/2, 1/4, 1/8, 1/16, 1/32) that arrives at an answer that matches the expected answer.
Because we get the right answer for this interpretation, we can assume that this is what "These programs are idle waiting for I/O half of the time" was supposed to mean.
Hello I'm undergraduate student of computer programming
I take operating system class and I have question about MLFQ scheduling
Suppose that MLFQ has two ready queues, and they use both round-robin scheduling method which of time quantum is 3 seconds and 5 seconds each
Of course priority of Q1 is higher.
Then suppose that there are no ready process in Q1, so process in Q2 can be allocated on CPU.
However, a process that finished IO bound burst returns to Q1, and process in CPU which belonged to Q2 is not done(just 3 seconds gone)
In this situation, What happen?
Is the process in Q1 preempt the CPU right now?
Or process in CPU now(which is come from Q2) is allocated on CPU until time quantum of it(still 2 seconds needed) is reached?
Thank you for reading.
The process that arrives after completing I/O burst in Q1, will preempte the process in Q2. Quoting from the Operating Systems principles by Galvin,Gagne and Silberchatz:
A process that arrives in Q1, will preempt a process in Q2. A process
in Q1 will in turn be preempted by a process arriving in Q0(if Q0
exists).
SO,Any process that arrives in a higher priority Q, will preempt the process of a lower priority Q, even if the time quantum of lower process is not reached. A process gets to have the CPU as long as the ready Q of higher priority is empty.
How does a long term scheduler decide which job is I/O bound and which one is CPU bound?
I heard that by using cpu burst we can distinguish between I/O bound and CPU bound jobs, but how is the CPU burst calculated without processing the program?
Generally, the CPU scheduler assigns time slices to processes/threads and switches between them whenever a) the time slice has run out or b) the process/thread blocks for I/O.
An I/O-bound job will be blocking for I/O very often, while a process/thread that always makes use of his full time slice can be assumed to be CPU-bound. So by distinguishing whether a process/thread blocks at the end of the time slice or by calling some wait_for_io_completion() function, you can effectively characterize those types of processes.
Note, that in real life, things get more complicated, because most of the time applications are not either I/O-bound or CPU-bound but switch roles all the time. This is why scheduling is about heuristics and not about correct solutions, because you cannot (always) predict the future.
CPU bound uses more of its time doing computations than I/O bound.
answered by tumaini kami david
Answers. Generally, the CPU scheduler assigns time slices to processes/threads and switches between them whenever a) the time slice has run out or b) the process/thread blocks for I/O. ... CPU bound uses more of its time doing computations than I/O bound.strong text
IO BOUND PROCESS :
Io bound process spends more time doing io than computations,many short cpu burst.
COU BOUND PROCESS :
process spends more time doing computations;few very long cpu bursts.