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Algorithm to round two numbers to the nearest evenly divisible ones

The title is not really well phrased, I'm aware - can't think of a better way of writing it though.
Here's the scenario - I have two input boxes, both representing integer quantities. One is represented in our units, the other in the vendor's units. There is a multiplier defining how to convert from ours to theirs. In the below example, I'm saying that two of our units is equal to five of theirs. So, for example,
decimal multiplier = 0.4; // Two of our units equals five of theirs
int requestedQuantity = 11; // Our units
int suppliedQuantity = 37; // Their units
// Should return 12, since that is the next highest whole number that results in both of us having whole numbers (12 of ours = 30 of theirs)
int correctedFromRequestedQuantity = GetCorrectedRequestedQuantity(requestedQuantity, null, multiplier);
// Should return 16, since that is the next highest whole number that results in both of us having whole numbers (16 of ours = 40 of theirs);
int correctedFromSuppliedQuantity = GetCorrectedRequestedQuantity(suppliedQuantity, multiplier, null);
Here's the function I've written to handle this. I'm not doing a divide by zero check on the multiplier / rounder since I've already checked for that elsewhere. It seems crazy to do all that converting, but is there a better way of doing it?
public int GetCorrectedRequestedQuantity(int? input, decimal? multiplier, decimal? rounder)
{
if (multiplier == null)
{
if (rounder == null)
return input.GetValueOrDefault();
else
return (int)Math.Ceiling((decimal)((decimal)Math.Ceiling(input.GetValueOrDefault() / rounder.Value) * rounder.Value));
}
else if (input.HasValue)
{
// This is insane...
return (int)Math.Ceiling((decimal)((decimal)Math.Ceiling((int)Math.Ceiling((decimal)input * multiplier.Value) / multiplier.Value) * multiplier.Value));
}
else
return 0;
}

Represent the multiplier as a fraction in lowest terms. I don't know if .NET has a fractions class but if not you can probably find a C# implementation, or just write your own. So assume the multiplier is given by two integers a / b in lowest terms, with a ≠ 0 and b ≠ 0. That also means that conversion in the other direction is given by multiplying by b / a. In your example, a = 2 and b = 5, and a / b = 0.4.
Now suppose you want to convert an integer X. If you think about it a bit you'll see what you really want is to nudge X up until b divides X. The number you need to add to X is simply (b - (X%b)) % b. So to convert on one direction is just
return (a * (X + (b - (X % b) % b))) / b;
and to convert Y going in the other direction is just
return (b * (Y + (a - (X % a) % a))) / a;

My best idea of my head is to semi brute-force it. It does sound like it is basically Fraction Mathematics. So there might be a way easier solution for this.
First we need to find in what sort of "Batch" the multiplier becomes whole. That way, we can stop working with floats/doubles altogether. Ideally this should be supplied with the multiplier (as float math is messy).
double currentMultiple=multiplier;
int currentCount=0;
//This is the best check for "is an integer" could think off.
while(currentMultiple % 1 = 0){
//The Framework can detect Arithmetic Overflow. Let us turn that one on
//If we ever get there, likely the math is non-solveable
checked{
currentMultiple+= multiplier;
currentCount += 1;
}
}
//You get here either via exception or because you got a multiple that solves it.
//Store the value of currentCount into a variable "OurBatchSize"
//Also store the value of currentMultiple in "TheirBatchSize"
Getting the closest Multiple of OurBatchSize:
int requestedQuantity = 11; // Our units
int result = OurBatchSize;
int batchCount = 0;
while(temp < requestedQuantity){
result += OurBatchSize;
batchCount++
}
//result contains the answer here. Return it
//batchCount * TheirBatchSize will also tell you how much they get.
Edit: Credit for this goes mostly to James Reinstate Monica Polk. He had the math idea to use Modulo for this. Here is what I got with explanation:
int result;
int rest = requestedAmout % BatchSize;
if (rest != 0){
//Correct upwards to the next multiple
int DistanceToNextMultiple = BatchSize - Rest;
result = requestedAmout + DistanceToMultiple;
}
else{
//It already is right
result = requestedAmout;
}
For the BatchSize of 4, you will get:
13; 13%4=1; 4-1=3; 13+3=16;
14; 14%4=2; 4-2=2; 14+2=16;
15; 15%4=3; 4-3=1; 15+1=16;
16; 16%4=0; Else is used. 16 is already right.

C# Round up and down

I'm counting the results of my database.
If it's lower then 50, i want to divide them by 2.
Example:
if(CountResults < 50)
{
//CountResults = 39
int divided = CountResults / 2; //Results in 19
}
What i want:
if(CountResults < 50)
{
//CountResults = 39
int divided = CountResults / 2; //Results in 19,5
Math.Round(divided, 0);
}
I want to be able to round it upwards and down.
So i get the result 19.5 twice. Once i want it to be 19, and once to be 20..
How do i achieve this?

It's not clear how you are going to use your code twice, but if you want to divide integer into two integer parts just subtract first result from totals:
if(CountResults < 50)
{
//CountResults = 39
int divided1 = CountResults / 2; // 19
int divided2 = CountResults - divided1; // 20
}
First result will use integer division and it will give you result rounded towards zero (19 in your case). Further reading: C# Specification 7.7.2 Division Operator
Second result will give you rest which will be either equal to first result (if there was no rounding), or it will be equal to division rounded from zero (20 in your case).

The rounding part can be accomplished using these 2 nice methods:
Math.Floor brings it to the floor
Math.Celing lifts it to the celing ;)
The calculation part is a little more tricky. This statement:
int divided = CountResults / 2; //Results in 19,5
cannot really be true, or let's say it does not matter what is behind the comma because when it is assigned to the variable int devided it will loose this information and no rounding is anymore required.
When you want a result of type double (meaning e.g. 19,5 ) and you want to round that result, you need at least one of the parameters of the calculation to be of type double double!
Example
double var1 = 39;
int res_low = (int)Math.Floor(var1 / 2);
int res_high = (int)Math.Ceiling(var1 / 2);
note that writing 2 is implicitly seen by the compiler as int and writing 2.0 is implicitly seen as double. So this would yield the same result:
int var2 = 39;
int res_low2 = (int)Math.Floor(var2 / 2.0);
int res_high2 = (int)Math.Ceiling(var2 / 2.0);

Time to Temperature Calculation

This might not be the correct place for this, so apologies in advance if it isn't.
My situation - I need to come up with a simple formula/method of giving it an hour E.g. 13, 15, 01 etc, and based on that number, the method will return the 'approx' temperature for that particular time.
This is very approximate and it will not use weather data or anything like that, it will just take the hour of the day and return a value between say -6 deg C > 35 deg C. (very extreme weather, but you get the idea.)
This is the sort of examples I would like to know how to do:
Just as a note, I COULD use an ugly array of 24 items, each referencing the temp for that hour, but this needs to be float based - e.g. 19.76 should return 9.25 deg...
Another note: I don't want a complete solution - I'm a confident programmer in various languages, but the maths have really stumped me on this. I've tried various methods on paper using TimeToPeak (the peak hour being 1pm or around there) but to no avail. Any help would be appreciated at this point.

EDIT
Following your comment, here is a function that provides a sinusoidal distribution with various useful optional parameters.
private static double SinDistribution(
double value,
double lowToHighMeanPoint = 0.0,
double length = 10.0,
double low = -1.0,
double high = 1.0)
{
var amplitude = (high - low) / 2;
var mean = low + amplitude;
return mean + (amplitude * Math.Sin(
(((value - lowToHighMeanPoint) % length) / length) * 2 * Math.PI));
}
You could use it like this, to get the results you desired.
for (double i = 0.0; i < 24.0; i++)
{
Console.WriteLine("{0}: {1}", i, SinDistribution(i, 6.5, 24.0, -6.0, 35.0));
}
This obviously discounts environmental factors and assumes the day is an equinox but I think it answers the question.
So,
double EstimatedTemperature(double hour, double[] distribution)
{
var low = Math.Floor(hour);
var lowIndex = (int)low;
var highIndex = (int)Math.Ceiling(hour);
if (highIndex > distribution.Count - 1)
{
highIndex = 0;
}
if (lowIndex < 0)
{
lowIndex = distribution.Count - 1;
}
var lowValue = distribution.ElementAt(lowIndex);
var highValue = distribution.ElementAt(highIndex);
return lowValue + ((hour - low) * (highValue - lowValue));
}
assuming a rather simplistic linear transition between each point in the distibution. You'll get erroneous results if the hour is mapped to elements that are not present in the distribution.

For arbitrary data points, I would go with one of the other linear interpolation solutions that have been provided.
However, this particular set of data is generated by a triangle wave:
temp = 45*Math.Abs(2*((t-1)/24-Math.Floor((t-1)/24+.5)))-10;

The data in your table is linear up and down from a peak at hour 13 and a minimum at hour 1. If that is the type of model that you want then this is really easy to put into a formulaic solution. You would just simply perform linear interpolation between the two extremes of the temperature based upon the hour value. You would have two data points:
(xmin, ymin) as (hour-min, temp-min)
(xmax, ymax) as (hour-max, temp-max)
You would have two equations of the form:
The two equations would use the (x0, y0) and (x1, y1) values as the above two data points but apply them the opposite assignment (ie peak would be (x0, y0) on one and (x1, y1) in the other equation.
You would then select which equation to use based upon the hour value, insert the X value as the hour and compute as Y for the temperature value.
You will want to offset the X values used in the equations so that you take care of the offset between when Hour 0 and where the minimum temperature peak happens.

Here is an example of how you could do this using a simple set of values in the function, if you wish, add these as parameters;
public double GetTemp(double hour)
{
int min = 1;
int max = min + 12;
double lowest = -10;
double highest = 35;
double change = 3.75;
return (hour > max) ? ((max - hour) * change) + highest : (hour < min) ? ((min - hour)*change) + lowest : ((hour - max) * change) + highest;
}
I have tested this according to your example and it is working with 19.75 = 9.6875.
There is no check to see whether the value entered is within 0-24, but that you can probably manage yourself :)

You can use simple 2 point linear approximation. Try somthing like this:
function double hourTemp(double hour)
{
idx1 = round(hour);
idx2 = idx1 + 1;
return (data[idx2] - data[idx1]) * (hour - idx1) + data[idx1];
}
Or use 3,5 or more points to get polynom cofficients with Ordinary Least Squares method.
Your sample data similar to the sin function so you can make sin function approximation.

How to initialize an array with numbers separated by a specific interval in C#

I want to create an array containing values from 0 to 1 with interval of 0.1. I can use:
float[] myArray = new float[10];
float increment = 0.1;
for(i = 0; i < 10; i++)
{
myArray[i] = increment;
increment += 0.1;
}
I was wondering whether there is a function like Enumerable.Range that permits to specify also the increment interval.

An interesting fact is that every answer posted so far has fixed the bug in your proposed code, but only one has called out that they've done so.
Binary floating point numbers have representation error when dealing with any quantity that is not a fraction of an exact power of two. ("3.0/4.0" is a representable fraction because the bottom is a power of two; "1.0/10.0" is not.)
Therefore, when you say:
for(i = 0; i < 10; i++)
{
myArray[i] = increment;
increment += 0.1;
}
You are not actually incrementing "increment" by 1.0/10.0. You are incrementing it by the closest representable fraction that has an exact power of two on the bottom. So in fact this is equivalent to:
for(i = 0; i < 10; i++)
{
myArray[i] = increment;
increment += (exactly_one_tenth + small_representation_error);
}
So, what is the value of the tenth increment? Clearly it is 10 * (exactly_one_tenth + small_representation_error) which is obviously equal to exactly_one + 10 * small_representation_error. You have multiplied the size of the representation error by ten.
Any time you repeatedly add together two floating point numbers, each subsequent addition increases the total representation error of the sum slightly and that adds up, literally, to a potentially large error. In some cases where you are summing thousands or millions of small numbers the error can become far larger than the actual total.
The far better solution is to do what everyone else has done. Recompute the fraction from integers every time. That way each result gets its own small representation error; it does not accumulate the representation errors of previously computed results.

Ugly, but...
Enumerable.Range(0,10).Select(i => i/10.0).ToArray();

No, there's no enumerable range that allows you to do that, but you could always divide by 10:
foreach (int i in Enumerable.Range(0, 10))
array[i] = (i + 1) / 10.0f
Note that this avoids the error that will accumulate if you repeatedly sum 0.1f. For example, if you sum the 10 elements in the myArray in your sample code, you get a value that's closer to 5.50000048 than 5.5.

Here is one way:
Enumerable.Range(1,10).Select(i => i /10.0)

Well you could use this:
Enumerable.Range(1,10).Select(x => x / 10.0).ToArray()
Not sure if that's better though.

Division to the nearest 1 decimal place without floating point math?

I am having some speed issues with my C# program and identified that this percentage calculation is causing a slow down. The calculation is simply n/d * 100. Both the numerator and denominator can be any integer number. The numerator can never be greater than the denominator and is never negative. Therefore, the result is always from 0-100. Right now, this is done by simply using floating point math and is somewhat slow, since it's being calculated tens of millions of times. I really don't need anything more accurate than to the nearest 0.1 percent. And, I just use this calculated value to see if it's bigger than a fixed constant value. I am thinking that everything should be kept as an integer, so the range with 0.1 accuracy would be 0-1000. Is there some way to calculate this percentage without floating point math?
Here is the loop that I am using with calculation:
for (int i = 0; i < simulationList.Count; i++)
{
for (int j = i + 1; j < simulationList.Count; j++)
{
int matches = GetMatchCount(simulationList[i], simulationList[j]);
if ((float)matches / (float)simulationList[j].Catchments.Count > thresPercent)
{
simulationList[j].IsOverThreshold = true;
}
}
}

Instead of n/d > c, you can use n > d * c (supposing that d > 0).
(c is the constant value you are comparing to.)
This way you don't need division at all.
However, watch out for the overflows.

If your units are in tenths instead of ones, then you can get your 0.1 accuracy using integer arithmetic:
Instead of:
for (...)
{
float n = ...;
float d = ...;
if (n / d > 1.4) // greater than 140% ?
...do something like:
for (...)
{
int n = 10 * ...;
int d = ...;
if (n / d > 14) // greater than 140% ?

Instead of writing
if ((float)matches / (float)simulationList[j].Catchments.Count > thresPercent)
write this:
if (matches * theresPercent_Denominator > simulationList[j].Catchments.Count * thresPercent_Numerator)
In this way, you get rid of the floating points.
Note: thresPercent can be expressed as thresPercent_Numerator / theresPercent_Denominator, as long as the number is a rational number.) I think this is the optimal way on PC. For some other platform, you may further optimize it by left-shift or right-shift, if theresPercent_Denominator and/or thresPercent_Numerator are 2's power. (Normally left-shift is enough, but may need use right-shift by rearrange the equation to division, to prevent from overflow)