i have an task to calculate CPU utilization, I have 4 proccess
P1 wait for I/O 30% of his time.
P2 wait for I/O 40% of his time.
P3 wait for I/0 20% of his time.
P4 wait for I/0 50% of his time.
my result is 0.99999993...it seems to me unreasonable
The probability that all processes are waiting for I/O (and therefore the CPU is idle) is:
0.3 * 0.4 * 0.2 * 0.5 = 0.012
The CPU is therefore busy with a probability of: (1 - 0.012) = 0.988, i.e. CPU utilization = 98.8%.
Related
The high cpu usage rate threshold looks to have been lowered on iOS 15. Possibly from 80% in 60sec to 15% in 60sec? I have noticed that my app does NOT run correctly on IOS 15 and the background operations seem to stop like location, and some timers... I have a lot of #Published property's being updated while in the background, would this affect the background operations terminating after ~40 seconds in the background? If So how would I go about Updating my UI and keeping the constant updates to the published property.
I am getting the message:
Event: cpu usage
Action taken: none
CPU: 9 seconds cpu time over 36 seconds (25% cpu average), exceeding limit of 15% cpu over 60 seconds
CPU limit: 9s
Limit duration: 60s
CPU used: 9s
CPU duration: 36s
Duration: 35.85s
Duration Sampled: 25.92s
Steps: 5
I've read in many places that a simple and decent way to get the % of CPU utilization is by this formula:
CPU utilization = 1 - p^n
where:
p - blocked time
n - number of processes
But i can't find an explanation for it. Seems it has to do with statistics, but i can't wrap my head around it.
My starting point is: if i have 2 processes with 50% wait time, then the formula would yield 1 - 1/4 = 75% CPU utilization. But my broken logic begs the question: if one process is blocked on I/O and the other is swapped in to run when the first is blocked(whatever the burst is), that means that while one waits, the second runs and their wait time overlap. Isn't that 100% CPU utilization? I think this is true only when the first half of the programs is guaranteed to run without IO need.
Question is: How is that formula taking into account every other possibility?
You need to think in terms of probabilities. If the probability of each of the cores to be idle (waiting for IO) is 0.5 then the probability of the CPU to be in idle is the probability of all of the cores to be in idle at the same time. That is 0.5 * 0.5 = 0.25 and so the probability the CPU is doing work is 1 - 0.25 = 0.75 = 75%
CPU utilisation is given as 1 - probability of CPU to being in the idle state
and CPU remain in the idle state when all the process loaded in the main memory is blocked time(I/O)
So if n process has wait time of 50% the probability that all the process are in
block(I/O) state
I missed a class and now I'm confused.
Im trying to solve this task:
On a server with 2 CPU are 3 processes running
They are waiting 10% of their time on I/O
How high is the CPU load
The only formula i got is
CPU-load of a 1 CPU system = 1 - p^n
p = %of time idle
n = number of processes
I have no clue how to account for the second CPU in the formula.
Or can i say a CPU runs 2 processes and the other only 1.
One processor is idle if 2 processes cannot run. The probability of one processor being idle is
.1 x .1 + .1 x .1 + .1 x .1 - .1 x .1 x .1
Both processors are idle if 3 processes cannot run. The probability of both processors being idle is:
.1 x .1 x .1
Is the question then whether one processor is idle or both processors is idle? If the former and one processor is running, to you count this as being half idle?
I am amazed at the useless busywork they put students through.
I'm facing difficulty with the following question :
Consider a disk drive with the following specifications .
16 surfaces, 512 tracks/surface, 512 sectors/track, 1 KB/sector, rotation speed 3000 rpm. The disk is operated in cycle stealing mode whereby whenever 1 byte word is ready it is sent to memory; similarly for writing, the disk interface reads a 4 byte word from the memory in each DMA cycle. Memory Cycle time is 40 ns. The maximum percentage of time that the CPU gets blocked during DMA operation is?
the solution to this question provided on the only site is :
Revolutions Per Min = 3000 RPM
or 3000/60 = 50 RPS
In 1 Round it can read = 512 KB
No. of tracks read per second = (2^19/2^2)*50
= 6553600 ............. (1)
Interrupt = 6553600 takes 0.2621 sec
Percentage Gain = (0.2621/1)*100
= 26 %
I have understood till (1).
Can anybody explain me how has 0.2621 come ? How is the interrupt time calculated? Please help .
Reversing form the numbers you've given, that's 6553600 * 40ns that gives 0.2621 sec.
One quite obvious problem is that the comments in the calculations are somewhat wrong. It's not
Revolutions Per Min = 3000 RPM ~ or 3000/60 = 50 RPS
In 1 Round it can read = 512 KB
No. of tracks read per second = (2^19/2^2)*50 <- WRONG
The numbers are 512K / 4 * 50. So, it's in bytes. How that could be called 'number of tracks'? Reading the full track is 1 full rotation, so the number of tracks readable in 1 second is 50, as there are 50 RPS.
However, the total bytes readable in 1s is then just 512K * 50 since 512K is the amount of data on the track.
But then it is further divided by 4..
So, I guess, the actual comments should be:
Revolutions Per Min = 3000 RPM ~ or 3000/60 = 50 RPS
In 1 Round it can read = 512 KB
Interrupts per second = (2^19/2^2) * 50 = 6553600 (*)
Interrupt triggers one memory op, so then:
total wasted: 6553600 * 40ns = 0.2621 sec.
However, I don't really like how the 'number of interrupts per second' is calculated. I currently don't see/fell/guess how/why it's just Bytes/4.
The only VAGUE explanation of that "divide it by 4" I can think of is:
At each byte written to the controller's memory, an event is triggered. However the DMA controller can read only PACKETS of 4 bytes. So, the hardware DMA controller must WAIT until there are at least 4 bytes ready to be read. Only then the DMA kicks in and halts the bus (or part of) for a duration of one memory cycle needed to copy the data. As bus is frozen, the processor MAY have to wait. It doesn't NEED to, it can be doing its own ops and work on cache, but if it tries touching the memory, it will need to wait until DMA finishes.
However, I don't like a few things in this "explanation". I cannot guarantee you that it is valid. It really depends on what architecture you are analyzing and how the DMA/CPU/BUS are organized.
The only mistake is its not
no. of tracks read
Its actually no. of interrupts occured (no. of times DMA came up with its data, these many times CPU will be blocked)
But again I don't know why 50 has been multiplied,probably because of 1 second, but I wish to solve this without multiplying by 50
My Solution:-
Here, in 1 rotation interface can read 512 KB data. 1 rotation time = 0.02 sec. So, one byte data preparation time = 39.1 nsec ----> for 4B it takes 156.4 nsec. Memory Cycle time = 40ns. So, the % of time the CPU get blocked = 40/(40+156.4) = 0.2036 ~= 20 %. But in the answer booklet options are given as A) 10 B)25 C)40 D)50. Tell me if I'm doing wrong ?
This was an exam question I could not solve, even after searching about response time.
I thought that answer should be 220, 120
Effectiveness of RR scheduling depends on two factors: choice of q, the time quantum, and the scheduling overhead s. If a system contains n processes and each request by a process consumes exactly q seconds, the response time (rt) for a request is rt= n(q+s) . This means that response is generated after spending the whole CPU burst and being scheduled to the next process. (after q+s)
Assume that an OS contains 10 identical processes that were initiated at the same time. Each process contains 15 identical requests, and each request consumes 20msec of CPU time. A request is followed by an I/O operation that consumes 10 sec. The system consumses 2msec in CPU scheduling. Calculate the average reponse time of the fisrt requests issued by each process for the following two cases:
(i) the time quantum is 20msec.
(ii) the time quantum is 10 msec.
Note that I'm assuming you meant 10ms instead of 10s for the I/O wait time, and that nothing can run on-CPU while an I/O is in progress. In real operating systems, the latter assumption is not true.
Each process should take time 15 requests * (20ms CPU + 10ms I/O)/request = 450ms.
Then, divide by the time quantum to get the number of scheduling delays, and add that to 450ms:
450ms / 20ms = 22.5 but actually it should be 23 because you can't get a partial reschedule. This gives the answer 450ms + 2ms/reschedule * 23 reschedules = 496ms.
450ms / 10ms = 45. This gives the answer 450ms + 2ms/reschedule * 45 reschedules = 540ms.