Nested IF statement to check between values - if-statement

I'm struggling with an IF statement in Google Docs.
I'm using it financially to find out interest accrued on a value.
Interest is either 1%, 2% or 3% depending on the amount.
If the amount is between £0 and £1999 then 1% is given
If it is between £2000 and £2999 then 2% is given
And anything over £3000 comes to 3%
Here is my attempt:
=IF(M8>=3000,(M8/100)*3,IF(M8<2999),(M8/100)*2,IF(M8<1999),(M8/100)*1)
My thought process here was that it would first check for the >=3000 and work out 3%. If that statement wasn't true it would continue on to the 2% calc and so on.
But this doesn't work. I get "wrong number of arguments for IF" error

You are not closing the IF statements properly and you only need two.
The logic should be: if number is more than or equal to 3000, multiply by 3, if number is less than 3000, but more than 1999(more than or equal to 2000) multiply by 2, in any other case (when number is less than or equal to 1999) multiply by 1.
Here is a working function
=IF(M8>=3000,(M8/100)*3,IF(AND(M8>1999,M8<3000),(M8/100)*2,(M8/100)*1))

Related

gcp monitoring "Any time series violates" vs "All time series violate"

What's the difference between the two options "Any time series violates" and "All time series violate"? I can imagine what would the former one do easily, but I have no idea what would the latter one do.
All time series? how long is its range? and why does it have a for option?
What's the difference between the two options "Any time series violates" and "All time series violate"? I can imagine what would the former one do easily, but I have no idea what would the latter one do.
First, what is "time series violates" - its when CURRENT VALUE of metric is outside of expected range, e.g: above the threshold specified.
Second, "any/all/percent/number" - let's say you have 5 time series, e.g.: cpu usage on 5 instances, then per dropdown options the whole alert condition will violate when:
"any time series": any 1 of the time series is in violation
"all time series": all 5 of the time series are in violation
"percent of time series" (40%): 2 out of 5 of the time series are in violation, and yes, selecting 39% or 41% on small numbers will give you different results, so
"number of time series" (3): 3 out of 5 of the time series are in violation
Third, for aka Duration box, - it looks like "if my time series violates FOR 5 minutes, then violate the condition". And for some simpler alerts this can even work, but once you try to combine it with say, "metric is absent" or other complicated configuration, you will see that what actually happens is "wait for 5 minutes after the problem is there, and only then trigger the violation".
In practice, the use of for field is discouraged and its better to keep it on default "Most recent value".
If you do need the "cpu usage is above 90% for 5 minutes", then correct way of doing it is by denoizing/smoothing your data:
set alignment period to 5 minutes (or whatever is the sliding window that you want)
then choose reasonable aligner (like, mean which will average the values)
and then while the chart will have less datapoints, they would be less noisy and you can act upon the latest value.
Any time will trigger if there is a violation of any time series, inside the window chosen on "for".
Let's say there is 5 time series, it will trigger if there is a violation in one of them.
For the all time series, it will trigger if it happens 5 out of 5.

RRDTool data values (e.g. max value) are different in different time resolutions

currently I'm experimenting a bit with RRDTool. I'm aware that the accuracy gets lower the longer the time periods are selected. But I thought I could bypass this with my datasource settings.
For example temperature and humidity from my house, resoultion 1h:
And now with the resolution of 1d:
As you could see, there is a great difference for the max. value of the blue line.
I created my datasources and archives with this values:
"rrdtool create temp.rrd --step 30",
"DS:temp:GAUGE:60:U:U",
"DS:humidity:GAUGE:60:U:U",
"RRA:AVERAGE:0.5:1:1051200",
"RRA:MAX:0.5:1:1051200",
"RRA:MIN:0.5:1:1051200",
I thought that 1051200 (1 year = 31536000 / 30 s (resoulution) = 1051200) is correct for saving every value for a year and that there should be no need for interpolating.
Is it possible to get the exact values displayed even if the resolution changes (for example the max humidity (Luftfeuchtigkeit) at 99.9%)?
Here are my values for image creation:
"--start" => "-1h", (-1d etc-)
"--title" => "Haustemperatur",
"--vertical-label" => "°C / % RLF",
"--width" => 800,
"--height" => 600,
"--lower-limit" => "-5",
"DEF:temperatur=$rrdFile:temperatur:LAST",
"DEF:humidity=$rrdFile:humidity:LAST",
"LINE1:temperatur#33CC33:Temperatur",
"GPRINT:temperatur:LAST:\t\tAktuell\: %4.2lf °C",
"GPRINT:temperatur:AVERAGE:Schnitt\: %4.2lf °C",
"GPRINT:temperatur:MAX:Maximum\: %4.2lf °C\j",
"LINE1:humidity#0000FF:Relative Luftfeuchtigkeit",
"GPRINT:humidity:LAST:Aktuell\: %4.2lf %%",
"GPRINT:humidity:AVERAGE:Schnitt\: %4.2lf %%",
"GPRINT:humidity:MAX:Maximum\: %4.2lf %%\j",
Thanks for your help and any suggestions.
P.S. I'm using a library to generate the graphs and the database, please do not be surprised about possible syntax errors.
Your problem is that you are causing the values to be rolled-up on the fly at graph time, but have not correctly specified which rollup function to use. Your second graph is showing the MAXIMUM of the LAST in the interval, not the true Maximum.
There are a few issues to explain with this configuration:
Firstly, your RRD is defined using 3 RRAs with 1cdp=1pdp and different consolidation functions (AVG, MIN, MAX). This means they are functionally identical, but they do not save you any time at graphing as they have not done any pre-rollup for you! You should definitely consider having just one of these (probably AVG) and adding others at lower resolution to help speed up graphing when you have a bigger time window.
Secondly, you need to specify the on-the-fly rollup function. When graphing, RRDTool will work out the best RRA to use based on your DEF lines, and will perform any additional consolidation required on the fly. This can take a long time if the only available RRA is too high-granularity.
Your graph request uses DEF:temperatur=$rrdFile:temperatur:LAST but you do not actually have a LAST type RRA, so RRDTool will grab the last average. Your RRA data points are at 30s interval, but your second graph has (approx) 5min per pixel, meaning that RRDTool needs to grab the 10 entries from the RRA, and print the last. Looking at the data in the top graph, it seems that the last in that interval was the 66 value, though previous ones were 100.
So you have a choice. Do you want the graph to show the average for the time period, the maximum, or both? Do you want the figures at the bottom to show the maximum of the average, or the maximum of everything?
For example
"DEF:temperatur=$rrdFile:temperatur:AVERAGE",
"DEF:humidity=$rrdFile:humidity:AVERAGE",
"DEF:temperaturmax=$rrdFile:temperatur:MAX;reduce=MAX",
"DEF:humiditymax=$rrdFile:humidity:MAX;reduce=MAX",
"LINE1:temperatur#33CC33:Temperatur",
"LINE1:temperaturmax#66EE66:Maximum Temperatur",
"GPRINT:temperatur:LAST:\t\tAktuell\: %4.2lf °C",
"GPRINT:temperatur:AVERAGE:Schnitt\: %4.2lf °C",
"GPRINT:temperaturmax:MAX:Maximum\: %4.2lf °C\j",
"LINE1:humidity#0000FF:Relative Luftfeuchtigkeit",
"LINE1:humiditymax#3333FF:Maximum Luftfeuchtigkeit",
"GPRINT:humidity:LAST:Aktuell\: %4.2lf %%",
"GPRINT:humidity:AVERAGE:Schnitt\: %4.2lf %%",
"GPRINT:humiditymax:MAX:Maximum\: %4.2lf %%\j",
In this case, we define a separate DEF for the maximum data set, so that we can always obtain the highest value even after consolidation. This is also used in the GPRINT so that we get the MAX of the MAX rather than the MAX of the AVERAGE. The Maximum line is now drawn separately to the average line, so that we can see the effect of any rollup of data - the lines will be together at high-resolution but get further apart as the time window widens and resolution decreases.
TheDEF is set to force any rollup function used for the maxima to be MAX rather than AVG, so we can be sure to get the maximum rather than average of maxima.
We are also using AVERAGE rather than LAST in order to get more meaningful data after rollup. Note that we could also use a separate DEF for the LAST as well if we wanted to though it is of less usefulness.
Note that, if you ever expect to be generating graphs over more than a few days, you should definitely consider adding some lower-resolution RRAs for AVERAGE and MAX or else the graphs will generate very slowly. RRDTool is designed with the intention that data will be rolled up over time, rather than (as in a traditional database) every sample kept as-is. So, unless you really need to have 30s resolution data kept for an entire year, you may prefer to keep this high resolution data for only a week, and then have separate RRAs that roll up to 1 hour resolution and keep for longer. Many people keep the 30s for 2 days, then 30min-summary for 2 weeks, 2h-summary for 2 months, and then 1day-summary for 2 years.
For more information, see the RRDTool manual pages.

Divide Large Number into Multiple Smaller Units

I am new to Stack overflow but I am having logic issues with Google sheets and need some advice on how to proceed.
My goal is I have 3 cup sizes, small medium and large. Example: Small holds 50ML, Medium 100ML,Large 200ML,etc
I want to take a large number and evenly distribute it among the cups to show me how many cups I will need with the least amount of cups used. Example : 170ML = 1 Large, 0 Medium, 0 Small. I only need 1 large to hold everything. Also, 240ML would suggest 1 Large, 0 medium, 1 small. Since small can hold the remaining 40 and medium would be too big of a container.
Problem is I don't understand how to break down my original large number into the smaller number as I have to check and compare if it will fit, I also have to be able to add more if there's a remainder and as far as I know the Google sheet functions only run and represent numbers once.
I've already tried breaking it down to my large container first then in my second row with medium cups I take the first result and subtract from my large number to see if anything is left. If there is, all I can do is add 1 or set the number, I can't seem to scale it up if it requires more than 1 cup which isn't what I want.
I've been going crazy trying to find an easy solution to this but it seems to get more complex as if I need IF statements of some kind.
If anyone has any ideas I'd be happy to hear them out.
you could use IF statement for example like this:
=ARRAYFORMULA(IF(A2:A<>"",
IF(A2:A>=B1, QUOTIENT(A2:A, B1),
IF(A2:A> C1+D1, QUOTIENT(B1, A2:A), 0)), ))

RRDTool Counter increment lower than time

I create a standard RRDTool database with a default step of 5mn (300s).
I have different types of values in it, some gauges which are easily processed, but I have other values I would have in COUNTER but here is my problem :
I read the data in a program, and get the difference between values over two steps is good but the counter increment less than time (It can increment by less than 300 during a step), so my out value is wrong.
Is it possible to change the COUNTER for not be a number by second but by step or something like that, if it's not I suppose I have to calculate the difference in my program ?
Thank you for helping.
RRDTool is capable of handling fractional values, so there is no problem if the counter increments by less than the seconds interval since the last update.
RRDTool stores everything as a Rate. If your DS is of type GAUGE, then RRDTool assumes that the incoming value is alreayd a rate, and only applies Data Normalisation (more on this later). If the type is COUNTER or DERIVE, then the value/timepoint you are updating with is compared to the previous value/timepoint to obtain a rate thus: r=(x2 - x1)/(t2 - t1). The rate obtained is then Normalised. The other DS type is ABSOLUTE, which assumes the counter was reset on the last read, giving r=x2/(t2 - t1).
The Normalisation step adjusts the data point based on assuming a linear progression from the last data point so that it lies exactly on an interval boundary. For example, if your step is 5min, and you update at 12:06, the data point is adjusted back to what it would have been at 12:05, and stored against 12:05. However the last unadjusted DP is still preserved for use at the next update, so that overall rates are correct.
So, if you have a 300s (5min) interval, and the value increased by 150, the rate stored will be 0.5.
If the value you are graphing is something small, e.g. 'number of pages printed', this might seem counterintuitive, but it works well for large rates such as network traffic counters (which is what RRDTool was designed for).
If you really do not want to display fractional values in the generated graphs or output, then you can use a format string such as %.0f to enforce no decimal places and the displayed number will be rounded to the nearest integer.

How can I round money values to the nearest $5.00 interval?

I have an Informix-SQL based Pawnshop app which calculates an estimate of how much money should be loaned to a customer, based on the weight and purity of gold. The minimum the pawnshop lends is $5.00. The pawnshop employee will typically lend amounts which either ends with a 5 or 0. examples: 10, 15, 20, 100, 110, 125, etc. They do this so as to not run into shortage problems with $1.00 bills. So, if for example my system calculates the loan should be: $12.49, then round it to $10, $12.50 to $15.00, $13.00 to $15.00, $17.50 to $20.00, and so on!..The employee can always override the rounded amount if necessary. Is it possible to accomplish this within the instructions section of a perform screen or would I have to write a cfunc and call it from within perform?.. Are there any C library functions which perform interval rounding of money values?.. On another note, I think the U.S. Government should discontinue the use of pennies so that businesses can round amounts to the nearest nickel, it would save so much time and weight in our pockets!
I'd just divide by 5, round to the appropriate integer, and multiply by 5.
There might be an easier way but that would work.
The C function round and relatives would do what you ask.
float in_fives = roundf( exact_change / 5. ) * 5.;
There is also rint, which you might avoid because it is not guaranteed to round $12.50 properly.