https://github.com/open-telemetry/opentelemetry-collector-contrib/blob/main/receiver/hostmetricsreceiver/internal/scraper/processscraper/documentation.md
I have been using this library which gives me 3 values for a single process :
user time, system time & wait time
One example value is : 0.05, 0.01, 0.00
How can I calculate CPU percent of the particular process ?
To calculate the total CPU load/utilization percent of the system, we need to calculate "total system cpu time (during the period)" + "total user cpu time (during the period)" / "period"
In your case, suppose you take sample every 2 seconds, then for every sample, you need to calculate:
= ( (process.cpu.time.sys - previous_process.cpu.time.sys) + (process.cpu.time.user - previous_process.cpu.time.user) ) / 2
Related
I'm wokring on project with Omron PLC, I need to show on HMI elapsed time after I started my system but problem is that, I can see my time just in seconds but I need in hour and min type. How to display elapsed time in hour and minute type from Omron Nx1 PLC to HMI?
One solution is to calculate the total hours and minutes from the total secconds and display on the HMI. You could concatenate the values into a string, but since I don't know what your purpose is, it's easier to use the values directly in integers with two different variables.
As you didn't define the language, here's an example in Structured-text. The Time given in Seconds in TotalSeconds will be separated into Hours and Minutes (and also Seconds as a bonus!).
Note: I'm putting the variable declaration as text, but I know that in Sysmac it is possible to declare it as a table...
Declaration
VAR
TotalSeconds : DINT;
Seconds : DINT;
Minutes : DINT;
Hours : DINT;
rest : DINT;
END_VAR
Program
rest := TotalSeconds MOD 3600;
Seconds := rest MOD 60;
Minutes := (rest - seconds) / 60;
Hours := (TotalSeconds - rest) / 3600;
Example
242 s >>> 0 h / 4 min / 2 s
33868 s >>> 9 h / 24 min / 28 s
So I was told to ask this on here instead of StackExchage:
If I have a program P, which runs on a 2GHz machine M in 30seconds and is optimized by replacing all instances of 'raise to the power 4' with 3 instructions of multiplying x by. This optimized program will be P'. The CPI of multiplication is 2 and CPI of power is 12. If there are 10^9 such operations optimized, what is the percent of total execution time improved?
Here is what I've deduced so far.
For P, we have:
time (30s)
CPI: 12
Frequency (2GHz)
For P', we have:
CPI (6) [2*3]
Frequency (2GHz)
So I need to figure our how to calculate the time of P' in order to compare the times. But I have no idea how to achieve this. Could someone please help me out?
Program P, which runs on a 2GHz machine M in 30 seconds and is optimized by replacing all instances of 'raise to the power 4' with 3 instructions of multiplying x by. This optimized program will be P'. The CPI of multiplication is 2 and CPI of power is 12. If there are 10^9 such operations optimized,
From this information we can compute time needed to execute all POWER4 ("raise to the power 4) instructions, we have total count of such instructions (all POWER4 was replaced, count is 10^9 or 1 G). Every POWER4 instruction needs 12 clock cycles (CPI = clock per instruction), so all POWER4 were executed in 1G * 12 = 12G cycles.
2GHz machine has 2G cycles per second, and there are 30 seconds of execution. Total P program execution is 2G*30 = 60 G cycles (60 * 10^9). We can conclude that P program has some other instructions. We don't know what instructions, how many executions they have and there is no information about their mean CPI. But we know that time needed to execute other instructions is 60 G - 12 G = 48 G (total program running time minus POWER4 running time - true for simple processors). There is some X executed instructions with Y mean CPI, so X*Y = 48 G.
So, total cycles executed for the program P is
Freq * seconds = POWER4_count * POWER4_CPI + OTHER_count * OTHER_mean_CPI
2G * 30 = 1G * 12 + X*Y
Or total running time for P:
30s = (1G * 12 + X*Y) / 2GHz
what is the percent of total execution time improved?
After replacing 1G POWER4 operations with 3 times more MUL instructions (multiply by) we have 3G MUL operations, and cycles needed for them is now CPI * count, where MUL CPI is 2: 2*3G = 6G cycles. X*Y part of P' was unchanged, and we can solve the problem.
P' time in seconds = ( MUL_count * MUL_CPI + OTHER_count * OTHER_mean_CPI ) / Frequency
P' time = (3G*2 + X*Y) / 2GHz
Improvement is not so big as can be excepted, because POWER4 instructions in P takes only some part of running time: 12G/60G; and optimization converted 12G to 6G, without changing remaining 48 G cycles part. By halving only some part of time we get not half of time.
In an e-commerce system how do I calculate what tax rate was used (e.g. 20%, 17.5%, etc) based on the total order value, and the total tax value.
For example, if I have total order value of £60 and a total tax value of £10, how do I calculate that the tax rate is 20%? (Going forward, the pre-vat total is £50 x 1.2 = £60)
What about doing the forward calculation in reverse?
60 / (60 - 10) - 1 = 0.2 (i.e. 20%)
Given a set of tasks:
T1(20,100) T2(30,250) T3(100,400) (execution time, deadline=peroid)
Now I want to constrict the deadlines as Di = f * Pi where Di is new deadline for ith task, Pi is the original period for ith task and f is the factor I want to figure out. What is the smallest value of f that the tasks will continue to meet their deadlines using rate monotonic scheduler?
This schema will repeat (synchronize) every 2000 time units. During this period
T1 must run 20 times, requiring 400 time units.
T2 must run 8 times, requiring 240 time units.
T3 must run 5 times, requiring 500 time units.
Total is 1140 time units per 2000 time unit interval.
f = 1140 / 2000 = 0.57
This assumes long-running tasks can be interrupted and resumed, to allow shorter-running tasks to run in between. Otherwise there will be no way for T1 to meet it's deadline once T3 has started.
The updated deadlines are:
T1(20,57)
T2(30,142.5)
T3(100,228)
These will repeat every 1851930 time units, and require the same time to complete.
A small simplification: When calculating factor, the period-time cancels out. This means you don't really need to calculate the period to get the factor:
Period = 2000
Required time = (Period / 100) * 20 + (Period / 250) * 30 + (Period / 400) * 100
f = Required time / Period = 20 / 100 + 30 / 250 + 100 / 400 = 0.57
f = Sum(Duration[i] / Period[i])
To calculate the period, you could do this:
Period(T1,T2) = lcm(100, 250) = 500
Period(T1,T2,T3) = lcm(500, 400) = 2000
where lcm(x,y) is the Least Common Multiple.
Here is a probability problem: you observe .5 cars on average passing in front of you every 5 minutes on a road. What is the probability of seeing at least 1 car in 10 minutes?
I'm trying to solve this in 2 ways. The first way is to say: P(no car in 5 minutes) = 1 - .5 = .5. P(no car in first 5 minutes and no car in second 5 minutes) = P(no car in first 5 minutes) * P(no car in second 5 minutes) by independence. Therefore P(at least 1 car in 10 minutes) = 1 - .5*.5 = .75.
However, if I try the same, with a Poisson distribution with rate lambda = .5 per unit of time, for 2 units of time, I get: P(at least 1 car in 2 units of time) = 1 - exp(-2*lambda) = .63.
Am I doing something wrong? If not, what explains the discrepancy?
Thanks!
Your first calculation is incorrect. An average .5 cars / 5 minutes does not imply P(no car in 5 minutes) = 0.5. Consider for instance a process where every five minute, you see either no car with probability 90%, or 5 cars with probability 10%. On average you will see 0.5 cars every five minute, but the probability you see 0 cars in the next 5 minutes is clearly not 50%.
I haven't checked the computations for your second example; the calculation logic is looks correct, but the conclusion is incorrect: you are making an assumption about the distribution (Poisson) which is plausible but not implied by the problem statement.
If you take again my example, which is consistent with your problem description, the probability to see 0 cars in 10 minutes is 0.9 x 0.9 = 0.81, which gives you 19% of seeing one car or more. We could arbitrarily change my example to give you a wide variety of probabilities.
From your problem statement, the only thing you can say is that "in the long run, you'll see 0.5 cars every 5 minutes". Beyond that you can't make a statement on what should be expected within 10 minutes, unless you make some assumptions about the distribution of the cars arrivals.