Azure Rate Card API returns a MeterRates field (see documentation).
Azure UsageAggregate gives a quantity (see documentation).
According to the azure support page. This is the forum to ask questions.
So, how are meter rates applied?
Example meter rates:
{"0":20, "100":15, "200":10}
If I have a quantity of 175 is the amount 100*20 + 75*15 or 175*15?
Why specify an included quantity?
Example: rates:{"0":23} with included quantitiy 10 could be expressed as rates:
{"0":0, "10":23}
example meter rates: {"0":20, "100":15, "200":10}
if I have a quantity of 175 is the amount 100*20 + 75*15 or 175*15 ?
The pricing is tiered pricing. So when you get the rates it essentially tells you that:
from 0 - 99 units, the unit rate is 20
from 100 - 199 units, the unit rate is 15
from 200 units and above, the unit rate is 10
Based on this logic, your calculation should be:
99 * 20 + 75 * 15 = 3105
One thing which confuses me though is the upper limit. Above calculation is based on the information I received from Azure Billing team. What confused me is what would happen if the consumption is say 99.5 units? For the first 99 units it is fine but I am not sure how the additional 0.5 units will be calculated.
Guarav gets at the core of the issue and is why I marked it as the answer. Based on that I devised the following code to implement the logic. It falls in two parts:
Creating a bucket list from the meter rates
Processing a quantity with the bucket list to determine an amount
The following function creates a list of buckets (each bucket object is a simple POCO with Min, Max and Rate properties). The list is attached to a meter object that has the other properties from the rate card api.
private Dictionary<int, RateBucket> ParseRateBuckets(string rates)
{
dynamic dRates = JsonConvert.DeserializeObject(rates);
var rateContainer = (JContainer)dRates;
var buckets = new Dictionary<int, RateBucket>();
var bucketNumber = 0;
foreach (var jToken in rateContainer.Children())
{
var jProperty = jToken as JProperty;
if (jProperty != null)
{
var bucket = new RateBucket
{
Min = Convert.ToDouble(jProperty.Name),
Rate = Convert.ToDouble(jProperty.Value.ToString())
};
if (bucketNumber > 0)
buckets[bucketNumber - 1].Max = bucket.Min;
buckets.Add(bucketNumber, bucket);
}
bucketNumber++;
}
return buckets;
}
The second function uses the meter object with two useful properties: the bucket list, and the included quantity. According to the rate card documentation (as I read it) you don't start counting billable quantity until after you surpass the included quantity. I'm sure there's some refactoring that could be done here, but the ordered processing of the buckets is the key point.
I think I've addressed issue on the quantity by recognizing that it's a double and not an integer. Therefore the quantity associated with any single bucket is the difference between the bucket max and the bucket min (unless we've only filled a partial bucket).
private double CalculateUsageCost(RateCardMeter meter, double quantity)
{
var amount = 0.0;
quantity -= meter.IncludedQuantity;
if (quantity > 0)
{
for (var i = 0; i < meter.RateBuckets.Count; i++)
{
var bucket = meter.RateBuckets[i];
if (quantity > bucket.Min)
{
if (bucket.Max.HasValue && quantity > bucket.Max)
amount += (bucket.Max.Value - bucket.Min)*bucket.Rate;
else
amount += (quantity - bucket.Min)*bucket.Rate;
}
}
}
return amount;
}
Finally, the documentation is unclear about the time scope for the tiers. If I get a discounted price based on quantity, over what time scope do I aggregate quantity? The usage api allows me to pull data either daily or hourly. I want to pull my data hourly so I can correlate my costs by time of day. But when is it appropriate to actually calculate the bill? Seems like hourly is wrong, daily may work, but it might only be appropriate over the entire month.
Recently I just did this similar task. Following is my example (I think you can use regex to remove those characters rather than like me using replace). The first function parse the rate info string to generate a key-value pair collection, and the second function is used to calculate the total price.
private Dictionary<float, double> GetRatesDetail(string r)
{
Dictionary<float, double> pairs = null;
if(string.IsNullOrEmpty(r) || r.Length <=2)
{
pairs = new Dictionary<float, double>();
pairs.Add(0, 0);
}
else
{
pairs = r.Replace("{", "").Replace("}", "").Split(',')
.Select(value => value.Split(':'))
.ToDictionary(pair => float.Parse(pair[0].Replace("\"", "")), pair => double.Parse(pair[1]));
}
return pairs;
}
public decimal Process(Dictionary<float, double> rates, decimal quantity)
{
double ret = 0;
foreach (int key in rates.Keys.OrderByDescending(k => k))
{
if (quantity >= key)
{
ret += ((double)quantity - key) * rates[key];
quantity = key;
}
}
return (decimal)ret;
}
Related
The home gas-meter is providing the actual gas count. With the subtraction (last minus before) I get 'Delta_Gas'. This is working.
Now I am trying to sum up gas these consumption values greater 0. In a typical cycle after gas burner ON max.10 values need to be summed up every minute. With burner OFF gas consumption should be set to zero(0).
Finally I want to see the gas consumption values for each single burner ON/OFF cycle.
Who can support my approach as I am a little bit lost?
var v1 = dp('/User Registers/Delta_Gas');
var sum = dp('/User Registers/Sum_Gas')
var s1;
AddValues(s1);
setInterval(AddValues, 3600000, s1); //1h Laufzeit
function AddValues(s1)
const add = add + v1
const sum = add
{
if(v1 > 0) { for (var i = 0; i < 10; i++) {
console.log(sum)
}else{
console.log(0)
}
Example 1:
``
Given S="300.01" and B-["300.00", "200.00*,*100.00"].
R[0]="150.00" (=300.01 300.00/600.00) R[1]="100.00" (=150.01* 200.00/300.00)
R[2]="50.01" (=50.01*100.00/100.00)
Example 2 (Pay careful attention to this one).
Given S="1.00" and B=["0.05","1.00"]. 1. First we consider 1.00 because it is the largest,
a. 1.00*1.00/1.05~0.95238...
b. Round 0.95238... to "0.95". Rounding down to carry pennies to smaller departments. c. Set R[1]=0.95. Notice, this is in the same place as 1.00. It is the 2nd value in the result! 2. Now we have 0.05 left
Next we look at the smaller B[0]=0.05 department
a. 0.05 0.05/0.05 = 0.05 b. No rounding required
c. Set R[0]=0.05. R=["0.05", "0.95"]
`
Write a function:
class Solution { public String[] solution(String 5, String[] B); }
that, given a string S representing the total excess billables and an array B consisting of K strings representing the undiscounted bills for each customer. The return value should be an array of strings R (length M) in the same order as B representing the amount of the discount to each customer.
Notes:
The total S should be completely refunded. Neither more nor less than S should be
returned. Don't lose or gain a penny!
Be careful with the types you choose to represent currencies. Floating points numbers are notoriously error prone for precise calculations with currencies.
Test Output
Amounts should be rounded down to the nearest $0.01. By design, fractional pennies are pushed to groups with smaller unadjusted invoices.
Results should be represented with 2 decimal places of precision as strings, even if these are both zeroes. ex. "100.00" 5. You may assume sensible inputs. The total to be discounted will never exceed the total of the
unadjusted invoices.
Please do pay attention to returning the discounts in the same order as the incoming invoices.
Answer:::
def solution(A):
answer = 0
current_sum = 0
#Currently there is one empty subarray with sum 0
prefixSumCount = {0:1}
for list_element in A:
current_sum = current_sum + list_element
if current_sum in prefixSumCount:
answer = answer + prefixSumCount[current_sum]
if current_sum not in prefixSumCount:
prefixSumCount[current_sum] = 1
else:
prefixSumCount[current_sum] = prefixSumCount[current_sum] + 1
if answer > 1000000000:
return -1
else:
return answer
#Sample run
A = [2,-2,3,0,4,-7]
print(solution(A))
strong text
Find my solution for JavaScript
You can avoid function sumArray and use the sum funciton with reducer within solution function.
enter code here
function solution(S, B) {
// write your code in JavaScript (Node.js 8.9.4)
let copyArray=[...B];
let solutionObj={};
//ordered array to consider last first
copyArray.sort();
//calculating sum of values within array
let sumArray=B.reduce((acc,value)=> acc+Number(value),0);
//calculating values of array
//loop for ading on to solvin array
let initial=S;
for(i=copyArray.length-1;i>=0;i--){
//obtaining index of value addded to solution array
let index=B.indexOf(copyArray[i]);
let value=initial*copyArray[i]/sumArray;
value=i==0?Math.ceil(value*100)/100:Math.floor(value*100)/100;
solutionObj[index]=value.toFixed(2);
}
return Object.values(solutionObj) ;
}
console.log(solution(300.01,["300.00","200.00","100.00"]))
console.log(solution(1.00,["0.05","1.00"]))
These are the resulting entries
Solution in java for the same coding exercise
public String[] solution(String S, String[] B) {
List<Double> list = Arrays.stream(B).sorted(Comparator.comparingDouble(v->Double.valueOf((String) v)).reversed()).map(Double::parseDouble).collect(Collectors.toList());
Double S1 = Double.valueOf(S);
String R[] = new String[B.length];
Double total = 0.00;
for (int i = 0; i< list.size(); i++){
Double individualValue = list.get(i);
Double sumTotal = 0.0;
for(int j = i+1;j < list.size(); j++){
sumTotal+=list.get(j);
}
BigDecimal data = new BigDecimal(S1 * (individualValue / (individualValue + sumTotal)));
data = data.setScale(2, RoundingMode.FLOOR);
total+=data.doubleValue();
R[i] = String.valueOf(data);
S1 = S1 - Double.valueOf(R[i]);
}
Double diff = new BigDecimal(Double.valueOf(S) - total).setScale(2, RoundingMode.HALF_DOWN).doubleValue();
if (diff > 0) {
R[B.length - 1] = String.valueOf(Double.valueOf(R[B.length - 1]) + diff);
}
return R;
}
Sorry if the text is confusing, I don't speak English.
My problem is:
1. I have the number of pages that was printed.
2. The duration of printing (start time and finish)
3. I want to plot a chart by hour whith the number of pages per hour
Example:
900 pages
1:30 hous
I want this Array of hour: [600, 300]
I think this is more a mathematical problem, but I don't had a good idea to do this. The are a lot of data and i need to do a algorithm fast and optimized.
Obs: I am more interested in the logic, not in the programming language.
Ok, not sure that works for everything but I think is a good start.
I have made the assumption that your duration is in minutes or else I believe you can transform it to minutes.
function something(pages,totalDutation){
// pages = 900
// totalDutation = 90
var printsPerMinute = pages / totalDutation //get the prints per minute!
var printsPerHour = Math.floor(printsPerMinute * 60) //calculate the prints made in one hour
var countOfHours = parseInt(totalDutation / 60) //divide the total duration by 60 to get the count of hours
var remainingPrints = (totalDutation % 60) * printsPerMinute //add the extra prints that didn't complete a whole hour.
var result = Array(countOfHours).fill(printsPerHour) //create an array and fill it.
if(remainingPrints) result.push(remainingPrints)
return result
}
Take a look at the NodeJS example below to get an idea of how it can be done.
Might need a bit of fine-tuning, though.
Calculate the number of prints per hour
Make as many iterations as hours taken
Do (total prints - prints per hour) as long as a > b,
otherwise return the remaining total prints
const getPagesPerHour = function (totalPrints, totalTime) {
const printsPerHour = (totalPrints / totalTime) * 60;
return Array.apply(null, { length: Math.ceil(totalPrints/printsPerHour) }).map(function (val, key) {
if (totalPrints > printsPerHour) {
totalPrints = (totalPrints - printsPerHour);
return printsPerHour;
} else {
return totalPrints;
}
});
}
console.log(getPagesPerHour(900, 90)); // [600, 300]
I've got spectrum from a Fourier transformation. It looks like this:
Police was just passing nearby
Color represents intensity.
X axis is time.
Y axis is frequency - where 0 is at top.
While whistling or a police siren leave only one trace, many other tones seem to contain a lot of harmonic frequencies.
Electric guitar plugged directly into microphone (standard tuning)
The really bad thing is, that as you can see there is no major intensity - there are 2-3 frequencies that are almost equal.
I have written a peak detection algorithm to highlight the most sigificant peak:
function findPeaks(data, look_range, minimal_val) {
if(look_range==null)
look_range = 10;
if(minimal_val == null)
minimal_val = 20;
//Array of peaks
var peaks = [];
//Currently the max value (that might or might not end up in peaks array)
var max_value = 0;
var max_value_pos = 0;
//How many values did we check without changing the max value
var smaller_values = 0;
//Tmp variable for performance
var val;
var lastval=Math.round(data.averageValues(0,4));
//console.log(lastval);
for(var i=0, l=data.length; i<l; i++) {
//Remember the value for performance and readibility
val = data[i];
//If last max value is larger then the current one, proceed and remember
if(max_value>val) {
//iterate the ammount of values that are smaller than our champion
smaller_values++;
//If there has been enough smaller values we take this one for confirmed peak
if(smaller_values > look_range) {
//Remember peak
peaks.push(max_value_pos);
//Reset other variables
max_value = 0;
max_value_pos = 0;
smaller_values = 0;
}
}
//Only take values when the difference is positive (next value is larger)
//Also aonly take values that are larger than minimum thresold
else if(val>lastval && val>minimal_val) {
//Remeber this as our new champion
max_value = val;
max_value_pos = i;
smaller_values = 0;
//console.log("Max value: ", max_value);
}
//Remember this value for next iteration
lastval = val;
}
//Sort peaks so that the largest one is first
peaks.sort(function(a, b) {return -data[a]+data[b];});
//if(peaks.length>0)
// console.log(peaks);
//Return array
return peaks;
}
The idea is, that I walk through the data and remember a value that is larger than thresold minimal_val. If the next look_range values are smaller than the chosen value, it's considered peak. This algorithm is not very smart but it's very easy to implement.
However, it can't tell which is the major frequency of the string, much like I anticipated:
The red dots highlight the strongest peak
Here's a jsFiddle to see how it really works (or rather doesn't work).
What you see in the spectrum of a string tone is the set of harmonics at
f0, 2*f0, 3*f0, ...
with f0 being the fundamental frequency or pitch of your string tone.
To estimate f0 from the spectrum (Output of FFT, abs value, probably logarithmic) you should not look for the strongest component, but the distance between all these harmonics.
One very nice method to do so is a second (inverse) FFT of the (abs, real) spectrum. This produces a strong line at t0 == 1/f0.
The sequence fft -> abs() -> fft-1 is equivalent to calculating the auto-correlation function (ACF) thanks to the Wiener–Khinchin theorem.
The precission of this approach depends on the length of the FFT (or ACF) and your sampling rate. You can improve precission a lot if you interpolate the "real" max between the sampling points of the result using a sinc function.
For even better results you could correct the intermediate spectrum: Most sounds have an average pink spectrum. If you amplify the higher frequencies (according an inverse pink spectrum) before the inverse FFT the ACF will be "better" (It takes the higher harmonics more into account, improving acuracy).
I have these 2 values:
decimal totalAmountDue = 1332.29m;
short installmentCount = 3;
I want to create 3 installments that have an even amount based on the totalAmountDue (extra pennies apply starting with the lowest installment number going to the highest installment number) using this class:
public class Installment
{
public Installment( short installmentNumber, decimal amount )
{
InstallmentNumber = installmentNumber;
Amount = amount;
}
public short InstallmentNumber { get; private set; }
public decimal Amount { get; private set; }
}
The installments should be as follows:
{ InstallmentNumber = 1, Amount = 444.10m }
{ InstallmentNumber = 2, Amount = 444.10m }
{ InstallmentNumber = 3, Amount = 444.09m }
I am looking for an interesting way to create my 3 installments. Using a simple LINQ to objects method would be nice. I have been trying to understand more about functional programming lately and this seems like it could be a fairly good exercise in recursion. The only decent way I can think of doing this is with a traditional while or for loop at the moment...
There's not a whole lot here that is "functional". I would approach the problem like this:
var pennies = (totalAmountDue * 100) % installmentCount;
var monthlyPayment = totalAmountDue / installmentCount;
var installments = from installment in Enumerable.Range(1, installmentCount)
let amount = monthlyPayment + (Math.Max(pennies--, 0m) / 100)
select new Installment(installment, amount);
You might be able to work something out where you constantly subtract the previous payment from the total amount and do the division rounding up to the nearest penny. In F# (C# is too wordy for this) it might be something like:
let calculatePayments totalAmountDue installmentCount =
let rec getPayments l (amountLeft:decimal) = function
| 0 -> l
| count -> let paymentAmount =
(truncate (amountLeft / (decimal)count * 100m)) / 100m
getPayments (new Installment(count, paymentAmount)::l)
(amountLeft - paymentAmount)
(count - 1)
getPayments [] totalAmountDue installmentCount
For those unfamiliar with F#, what that code is doing is setting up a recursive function (getPayments) and bootstrapping it with some initial values (empty list, starting values). Using match expressions it sets up a terminator (if installmentCount is 0) returning the list so far. Otherwise it calculates the payment amount and calls the recursive method adding the new installment to the front of the list, subtracting the payment amount from the amount left, and subtracting the count.
This is actually building the list in reverse (adding on to the front each time), so we throw away the extra pennies (the truncate) and eventually it catches up with us so the penny rounding works as expected. This is obviously more math intensive than the add/subtract code above since we divide and multiply in every iteration. But it is fully recursive and takes advantage of tail recursion so we'll never run out of stack.
The trouble with C# here is that you want a sequence of installments and recursion and there's no idiomatic built-in structure for doing that in C#. Here I'm using F#'s list which is immutable and O(1) operation to prepend.
You could possibly build something using the Scan() method in the Reactive Extensions to pass state from once instance to another.
Talljoe,
I think you are pushing me in the right direction. This code below seems to work. I had to switch out how the penny math was working but this looks pretty good (I think)
decimal totalAmountDue = 1332.29m;
short installmentCount = 8;
var pennies = (totalAmountDue * 100) % installmentCount;
var monthlyPayment = Math.Floor(totalAmountDue / installmentCount * 100);
var installments = from installmentNumber in Enumerable.Range(1, installmentCount)
let extraPenny = pennies-- > 0 ? 1 : 0
let amount = (monthlyPayment + extraPenny) / 100
select new Installment(installmentNumber, amount);