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I have a combinations problem that's bothering me. I'd like someone to give me their thoughts and point out if I'm missing some obvious solution that I may have overlooked.
Let's say that there is a shop that buys all of its supplies from one supplier. The supplier has a list of items for sale. Each item has the following attributes:
size, cost, quantity, m, b
m and b are constants in the following equation:
sales = m * (price) + b
This line slopes downward. The equation tells me how many of that item I will be able to sell if I charge that particular price. Each item has its own m and b values.
Let's say that the shop has limited storage space, and limited funds. The shop wants to fill its warehouse with the most profit-dense items possible.
(By the way, profit density = profit/size. I'm defining that profit density be only with regard to the items size. I could work with the density with regard to size and cost, but to do that I'd have to know the cost of warehouse space. That's not a number I know currently, so I'm just going to use size.)
The profit density of items drops the more you buy (see below.)
If I flip the line equation, I can see what price I'd have to charge to sell some given amount of the item in some given period of time.
price = (sales-b)/m
So if I buy n items and wanted to sell all of them, I'd have to charge
price = (n-b)/m
The revenue from this would be
price*n = n*(n-b)/m
The profit would be
price*n-n*cost = n*(n-b)/m - n*cost
and the profit-density would be
(n*(n-b)/m - n*cost)/(n*size)
or, equivalently
((n-b)/m - cost)/size
So let's say I have a table containing every available item, and each item's profit-density.
The question is, how many of each item do I buy in order to maximise the amount of money that the shop makes?
One possibility is to generate every possible combination of items within the bounds of cost and space, and choose the combo with the highest profitability. In a list of 1000 items, this takes too long. (I tried this and it took 17 seconds for a list of 1000. Horrible.)
Another option I tried (on paper) was to take the top two most profitable items on the list. Let's call the most profitable item A, the 2nd-most profitable item B, and the 3rd-most profitable item C. I buy as many of item A as I can until it's less profitable than item B. Then I repeat this process using B and C, for every item in the list.
It might be the case however, that after buying item B, item A is again the most profitable item, more so than C. So this would involve hopping from the current most profitable item to the next until the resources are exhausted. I could do this, but it seems like an ugly way to do it.
I considered dynamic programming, but since the profit-densities of the items change depending on the amount you buy, I couldn't come up with a resolution for this.
I've considered multiple-linear regression, and by 'consider' I mean I've said to myself "is multi-linear regression an option?" and then done nothing with it.
My spidey-sense tells me that there's a far more obvious method staring me in the face, but I'm not seeing it. Please help me kick myself and facepalm at the same time.
If you treat this as a simple exercise in multivariate optimization, where the controllable variables are the quantities bought, then you are optimizing a quadratic function subject to a linear constraint.
If you use a Lagrange multiplier and differentiate then you get a linear equation for each quantity variable involving itself and the Lagrange multiplier as the only unknowns, and the constraint gives you a single linear equation involving all of the quantities. So write each quantity as a linear function of the Lagrange multiplier and substitute into the constraint equation to get a linear equation in the Lagrange multiplier. Solve this and then plug the Lagrange multiplier into the simpler equations to get the quantities.
This gives you a solution if you are allowed to buy fractional and negative quantities of things if required. Clearly you are not, but you might hope that nothing is very negative and you can round the non-integer quantities to get a reasonable answer. If this isn't good enough for you, you could use it as a basis for branch and bound. If you make an assumption on the value of one of the quantities and solve for the others in this way, you get an upper bound on the possible best answer - the profit predicted neglecting real world constraints on non-negativity and integer values will always be at least the profit earned if you have to comply with these constraints.
You can treat this as a dynamic programming exercise, to make the best use of a limited resource.
As a simple example, consider just satisfying the constraint on space and ignoring that on cost. Then you want to find the items that generate the most profit for the available space. Choose units so that expressing the space used as an integer is reasonable, and then, for i = 1 to number of items, work out, for each integer value of space up to the limit, the selection of the first i items that gives the most return for that amount of space. As usual, you can work out the answers for i+1 from the answers for i: for each value from 0 up to the limit on space just consider all possible quantities of the i+1th item up to that amount of space, and work out the combined return from using that quantity of the item and then using the remaining space according to the answers you have already worked out for the first i items. When i reaches the total number of items you will be working out the best possible return for the problem you actually want to solve.
If you have constraints for both space and cost, then the state of the dynamic program is not the single variable (space) but a pair of variables (space, cost) but you can still solve it, although with more work. Consider all possible values of (space, cost) from (0, 0) up to the actual constraints - you have a 2-dimensional table of returns to compute instead of a single set of values from 0 to max-space. But you can still work from i=1 to N, computing the highest possible return for the first i items for each limit of (space, cost) and using the answers for i to compute the answers for i+1.
I have a field that calculates an average per row (Avg_Amt), it gives a correct output. My problem is how can I calculate the total summary of the average field (Avg Amt) then place it to the report footer.
Thanks & Best Regards.
Create 2 formula fields one for average and one for sum and insert this formula the first one
formula = Average ({Feild Name})
in the second formula from the report fields drag and drop the average formula and add the sum operation like :
formula = SUM({#Average})
If it is a straight average, use the sum wizard to add a sum for each field, then insert a formula field see below, this will actually get you to the accurate basis points average
(Sum ({#Field_1})/Sum ({#Field_2}))*100
I'd like to sum up moving averages for a number of different categories when storing log records. Imagine a service that saves web server logs one entry at a time. Let's further imagine, we don't have access to the logged records. So we see them once but don't have access to them later on.
For different pages, I'd like to know
the total number of hits (easy)
a "recent" average (like one month or so)
a "long term" average (over a year)
Is there any clever algorithm/data model that allows to save such moving averages without having to recalculate them by summing up huge quantities of data?
I don't need an exact average (exactly 30 days or so) but just trend indicators. So some fuzziness is not a problem at all. It should just make sure that newer entries are weighted higher than older ones.
One solution probably would be to auto-create statistics records for each month. However, I don't even need past month statistics, so this seems like overkill. And it wouldn't give me a moving average but rather swap to new values from month to month.
An easy solution would be to keep an exponentially decaying total.
It can be calculated using the following formula:
newX = oldX * (p ^ (newT - oldT)) + delta
where oldX is the old value of your total (at time oldT), newX is the new value of your total (at time newT); delta is the contribution of new events to the total (for example the number of hits today); p is less or equal to 1 and is the decay factor. If we take p = 1, then we have the total number of hits. By decreasing p, we effectively decrease the interval our total describes.
If all you really want is a smoothed value with a given time constant then the easiest thing is to use a single pole recursive IIR filter (aka AR or auto-regressive filter in time series analysis). This takes the form:
Xnew = k * X_old + (1 - k) * x
where X_old is the previous smoothed value, X_new is the new smoothed value, x is the current data point and k is a factor which determines the time constant (usually a small value, < 0.1). You may need to determine the two k values (one value for "recent" and a smaller value for "long term") empirically, based on your sample rate, which ideally should be reasonably constant, e.g. one update per day.
It may be solution for you.
You can aggregate data to intermediate storage grouped by hour or day. Than grouping function will work very fast, because you will need to group small amount of records and inserts will be fast as well. Precision decisions up to you.
It can be better than auto-correlated exponential algorithms because you can understand what you calculate easier and it doesn't require math each step.
For last term data you can use capped collections with limited amount of records. They supported natively by some DBs for example MongoDB.
Here's the scenario.
I have one hundred car objects. Each car has a property for speed, and a property for price. I want to arrange images of the cars in a grid so that the fastest and most expensive car is at the top right, and the slowest and cheapest car is at the bottom left, and all other cars are in an appropriate spot in the grid.
What kind of sorting algorithm do I need to use for this, and do you have any tips?
EDIT: the results don't need to be exact - in reality I'm dealing with a much bigger grid, so it would be sufficient if the cars were clustered roughly in the right place.
Just an idea inspired by Mr Cantor:
calculate max(speed) and max(price)
normalize all speed and price data into range 0..1
for each car, calculate the "distance" to the possible maximum
based on a²+b²=c², distance could be something like
sqrt( (speed(car[i])/maxspeed)^2 + (price(car[i])/maxprice)^2 )
apply weighting as (visually) necessary
sort cars by distance
place "best" car in "best" square (upper right in your case)
walk the grid in zigzag and fill with next car in sorted list
Result (mirrored, top left is best):
1 - 2 6 - 7
/ / /
3 5 8
| /
4
Treat this as two problems:
1: Produce a sorted list
2: Place members of the sorted list into the grid
The sorting is just a matter of you defining your rules more precisely. "Fastest and most expensive first" doesn't work. Which comes first my £100,000 Rolls Royce, top speed 120, or my souped-up Mini, cost £50,000, top speed 180?
Having got your list how will you fill it? First and last is easy, but where does number two go? Along the top or down? Then where next, along rows, along the columns, zig-zag? You've got to decide. After that coding should be easy.
I guess what you want is to have cars that have "similar" characteristics to be clustered nearby, and additionally that the cost in general increases rightwards, and speed in general increases upwards.
I would try to following approach. Suppose you have N cars and you want to put them in an X * Y grid. Assume N == X * Y.
Put all the N cars in the grid at random locations.
Define a metric that calculates the total misordering in the grid; for example, count the number of car pairs C1=(x,y) and C2=(x',y') such that C1.speed > C2.speed but y < y' plus car pairs C1=(x,y) and C2=(x',y') such that C1.price > C2.price but x < x'.
Run the following algorithm:
Calculate current misordering metric M
Enumerate through all pairs of cars in the grid and calculate the misordering metric M' you obtain if you swapt the cars
Swap the pair of cars that reduces the metric most, if any such pair was found
If you swapped two cars, repeat from step 1
Finish
This is a standard "local search" approach to an optimization problem. What you have here is basically a simple combinatorial optimization problem. Another approaches to try might be using a self-organizing map (SOM) with preseeded gradient of speed and cost in the matrix.
Basically you have to take one of speed or price as primary and then get the cars with the same value of this primary and sort those values in ascending/descending order and primaries are also taken in the ascending/descending order as needed.
Example:
c1(20,1000) c2(30,5000) c3(20, 500) c4(10, 3000) c5(35, 1000)
Lets Assume Car(speed, price) as the measure in the above list and the primary is speed.
1 Get the car with minimum speed
2 Then get all the cars with the same speed value
3 Arrange these values in ascending order of car price
4 Get the next car with the next minimum speed value and repeat the above process
c4(10, 3000)
c3(20, 500)
c1(20, 1000)
c2(30, 5000)
c5(35, 1000)
If you post what language you are using them it would we helpful as some language constructs make this easier to implement. For example LINQ makes your life very easy in this situation.
cars.OrderBy(x => x.Speed).ThenBy(p => p.Price);
Edit:
Now you got the list, as per placing this cars items into the grid unless you know that there will be this many number of predetermined cars with these values, you can't do anything expect for going with some fixed grid size as you are doing now.
One option would be to go with a nonuniform grid, If you prefer, with each row having car items of a specific speed, but this is only applicable when you know that there will be considerable number of cars which has same speed value.
So each row will have cars of same speed shown in the grid.
Thanks
Is the 10x10 constraint necessary? If it is, you must have ten speeds and ten prices, or else the diagram won't make very much sense. For instance, what happens if the fastest car isn't the most expensive?
I would rather recommend you make the grid size equal to
(number of distinct speeds) x (number of distinct prices),
then it would be a (rather) simple case of ordering by two axes.
If the data originates in a database, then you should order them as you fetch them from the database. This should only mean adding ORDER BY speed, price near the end of your query, but before the LIMIT part (where 'speed' and 'price' are the names of the appropriate fields).
As others have said, "fastest and most expensive" is a difficult thing to do, you ought to just pick one to sort by first. However, it would be possible to make an approximation using this algorithm:
Find the highest price and fastest speed.
Normalize all prices and speeds to e.g. a fraction out of 1. You do this by dividing the price by the highest price you found in step 1.
Multiply the normalized price and speed together to create one "price & speed" number.
Sort by this number.
This ensures that is car A is faster and more expensive than car B, it gets put ahead on the list. Cars where one value is higher but the other is lower get roughly sorted. I'd recommend storing these values in the database and sorting as you select.
Putting them in a 10x10 grid is easy. Start outputting items, and when you get to a multiple of 10, start a new row.
Another option is to apply a score 0 .. 200% to each car, and sort by that score.
Example:
score_i = speed_percent(min_speed, max_speed, speed_i) + price_percent(min_price, max_price, price_i)
Hmmm... kind of bubble sort could be simple algorithm here.
Make a random 10x10 array.
Find two neighbours (horizontal or vertical) that are in "wrong order", and exchange them.
Repeat (2) until no such neighbours can be found.
Two neighbour elements are in "wrong order" when:
a) they're horizontal neighbours and left one is slower than right one,
b) they're vertical neighbours and top one is cheaper than bottom one.
But I'm not actually sure if this algorithm stops for every data. I'm almost sure it is very slow :-). It should be easy to implement and after some finite number of iterations the partial result might be good enough for your purposes though. You can also start by generating the array using one of other methods mentioned here. Also it will maintain your condition on array shape.
Edit: It is too late here to prove anything, but I made some experiments in python. It looks like a random array of 100x100 can be sorted this way in few seconds and I always managed to get full 2d ordering (that is: at the end I got wrongly-ordered neighbours). Assuming that OP can precalculate this array, he can put any reasonable number of cars into the array and get sensible results. Experimental code: http://pastebin.com/f2bae9a79 (you need matplotlib, and I recommend ipython too). iterchange is the sorting method there.
I'm trying to calculate the median of a set of values, but I don't want to store all the values as that could blow memory requirements. Is there a way of calculating or approximating the median without storing and sorting all the individual values?
Ideally I would like to write my code a bit like the following
var medianCalculator = new MedianCalculator();
foreach (var value in SourceData)
{
medianCalculator.Add(value);
}
Console.WriteLine("The median is: {0}", medianCalculator.Median);
All I need is the actual MedianCalculator code!
Update: Some people have asked if the values I'm trying to calculate the median for have known properties. The answer is yes. One value is in 0.5 increments from about -25 to -0.5. The other is also in 0.5 increments from -120 to -60. I guess this means I can use some form of histogram for each value.
Thanks
Nick
If the values are discrete and the number of distinct values isn't too high, you could just accumulate the number of times each value occurs in a histogram, then find the median from the histogram counts (just add up counts from the top and bottom of the histogram until you reach the middle). Or if they're continuous values, you could distribute them into bins - that wouldn't tell you the exact median but it would give you a range, and if you need to know more precisely you could iterate over the list again, examining only the elements in the central bin.
There is the 'remedian' statistic. It works by first setting up k arrays, each of length b. Data values are fed in to the first array and, when this is full, the median is calculated and stored in the first pos of the next array, after which the first array is re-used. When the second array is full the median of its values is stored in the first pos of the third array, etc. etc. You get the idea :)
It's simple and pretty robust. The reference is here...
http://web.ipac.caltech.edu/staff/fmasci/home/astro_refs/Remedian.pdf
Hope this helps
Michael
I use these incremental/recursive mean and median estimators, which both use constant storage:
mean += eta * (sample - mean)
median += eta * sgn(sample - median)
where eta is a small learning rate parameter (e.g. 0.001), and sgn() is the signum function which returns one of {-1, 0, 1}. (Use a constant eta if the data is non-stationary and you want to track changes over time; otherwise, for stationary sources you can use something like eta=1/n for the mean estimator, where n is the number of samples seen so far... unfortunately, this does not appear to work for the median estimator.)
This type of incremental mean estimator seems to be used all over the place, e.g. in unsupervised neural network learning rules, but the median version seems much less common, despite its benefits (robustness to outliers). It seems that the median version could be used as a replacement for the mean estimator in many applications.
Also, I modified the incremental median estimator to estimate arbitrary quantiles. In general, a quantile function tells you the value that divides the data into two fractions: p and 1-p. The following estimates this value incrementally:
quantile += eta * (sgn(sample - quantile) + 2.0 * p - 1.0)
The value p should be within [0,1]. This essentially shifts the sgn() function's symmetrical output {-1,0,1} to lean toward one side, partitioning the data samples into two unequally-sized bins (fractions p and 1-p of the data are less than/greater than the quantile estimate, respectively). Note that for p=0.5, this reduces to the median estimator.
I would love to see an incremental mode estimator of a similar form...
(Note: I also posted this to a similar topic here: "On-line" (iterator) algorithms for estimating statistical median, mode, skewness, kurtosis?)
Here is a crazy approach that you might try. This is a classical problem in streaming algorithms. The rules are
You have limited memory, say O(log n) where n is the number of items you want
You can look at each item once and make a decision then and there what to do with it, if you store it, it costs memory, if you throw it away it is gone forever.
The idea for the finding a median is simple. Sample O(1 / a^2 * log(1 / p)) * log(n) elements from the list at random, you can do this via reservoir sampling (see a previous question). Now simply return the median from your sampled elements, using a classical method.
The guarantee is that the index of the item returned will be (1 +/- a) / 2 with probability at least 1-p. So there is a probability p of failing, you can choose it by sampling more elements. And it wont return the median or guarantee that the value of the item returned is anywhere close to the median, just that when you sort the list the item returned will be close to the half of the list.
This algorithm uses O(log n) additional space and runs in Linear time.
This is tricky to get right in general, especially to handle degenerate series that are already sorted, or have a bunch of values at the "start" of the list but the end of the list has values in a different range.
The basic idea of making a histogram is most promising. This lets you accumulate distribution information and answer queries (like median) from it. The median will be approximate since you obviously don't store all values. The storage space is fixed so it will work with whatever length sequence you have.
But you can't just build a histogram from say the first 100 values and use that histogram continually.. the changing data may make that histogram invalid. So you need a dynamic histogram that can change its range and bins on the fly.
Make a structure which has N bins. You'll store the X value of each slot transition (N+1 values total) as well as the population of the bin.
Stream in your data. Record the first N+1 values. If the stream ends before this, great, you have all the values loaded and you can find the exact median and return it. Else use the values to define your first histogram. Just sort the values and use those as bin definitions, each bin having a population of 1. It's OK to have dupes (0 width bins).
Now stream in new values. For each one, binary search to find the bin it belongs to.
In the common case, you just increment the population of that bin and continue.
If your sample is beyond the histogram's edges (highest or lowest), just extend the end bin's range to include it.
When your stream is done, you find the median sample value by finding the bin which has equal population on both sides of it, and linearly interpolating the remaining bin-width.
But that's not enough.. you still need to ADAPT the histogram to the data as it's being streamed in. When a bin gets over-full, you're losing information about that bin's sub distribution.
You can fix this by adapting based on some heuristic... The easiest and most robust one is if a bin reaches some certain threshold population (something like 10*v/N where v=# of values seen so far in the stream, and N is the number of bins), you SPLIT that overfull bin. Add a new value at the midpoint of the bin, give each side half of the original bin's population. But now you have too many bins, so you need to DELETE a bin. A good heuristic for that is to find the bin with the smallest product of population and width. Delete it and merge it with its left or right neighbor (whichever one of the neighbors itself has the smallest product of width and population.). Done!
Note that merging or splitting bins loses information, but that's unavoidable.. you only have fixed storage.
This algorithm is nice in that it will deal with all types of input streams and give good results. If you have the luxury of choosing sample order, a random sample is best, since that minimizes splits and merges.
The algorithm also allows you to query any percentile, not just median, since you have a complete distribution estimate.
I use this method in my own code in many places, mostly for debugging logs.. where some stats that you're recording have unknown distribution. With this algorithm you don't need to guess ahead of time.
The downside is the unequal bin widths means you have to do a binary search for each sample, so your net algorithm is O(NlogN).
David's suggestion seems like the most sensible approach for approximating the median.
A running mean for the same problem is a much easier to calculate:
Mn = Mn-1 + ((Vn - Mn-1) / n)
Where Mn is the mean of n values, Mn-1 is the previous mean, and Vn is the new value.
In other words, the new mean is the existing mean plus the difference between the new value and the mean, divided by the number of values.
In code this would look something like:
new_mean = prev_mean + ((value - prev_mean) / count)
though obviously you may want to consider language-specific stuff like floating-point rounding errors etc.
I don't think it is possible to do without having the list in memory. You can obviously approximate with
average if you know that the data is symmetrically distributed
or calculate a proper median of a small subset of data (that fits in memory) - if you know that your data has the same distribution across the sample (e.g. that the first item has the same distribution as the last one)
Find Min and Max of the list containing N items through linear search and name them as HighValue and LowValue
Let MedianIndex = (N+1)/2
1st Order Binary Search:
Repeat the following 4 steps until LowValue < HighValue.
Get MedianValue approximately = ( HighValue + LowValue ) / 2
Get NumberOfItemsWhichAreLessThanorEqualToMedianValue = K
is K = MedianIndex, then return MedianValue
is K > MedianIndex ? then HighValue = MedianValue Else LowValue = MedianValue
It will be faster without consuming memory
2nd Order Binary Search:
LowIndex=1
HighIndex=N
Repeat Following 5 Steps until (LowIndex < HighIndex)
Get Approximate DistrbutionPerUnit=(HighValue-LowValue)/(HighIndex-LowIndex)
Get Approximate MedianValue = LowValue + (MedianIndex-LowIndex) * DistributionPerUnit
Get NumberOfItemsWhichAreLessThanorEqualToMedianValue = K
is (K=MedianIndex) ? return MedianValue
is (K > MedianIndex) ? then HighIndex=K and HighValue=MedianValue Else LowIndex=K and LowValue=MedianValue
It will be faster than 1st order without consuming memory
We can also think of fitting HighValue, LowValue and MedianValue with HighIndex, LowIndex and MedianIndex to a Parabola, and can get ThirdOrder Binary Search which will be faster than 2nd order without consuming memory and so on...
Usually if the input is within a certain range, say 1 to 1 million, it's easy to create an array of counts: read the code for "quantile" and "ibucket" here: http://code.google.com/p/ea-utils/source/browse/trunk/clipper/sam-stats.cpp
This solution can be generalized as an approximation by coercing the input into an integer within some range using a function that you then reverse on the way out: IE: foo.push((int) input/1000000) and quantile(foo)*1000000.
If your input is an arbitrary double precision number, then you've got to autoscale your histogram as values come in that are out of range (see above).
Or you can use the median-triplets method described in this paper: http://web.cs.wpi.edu/~hofri/medsel.pdf
I picked up the idea of iterative quantile calculation. It is important to have a good value for starting point and eta, these may come from mean and sigma. So I programmed this:
Function QuantileIterative(Var x : Array of Double; n : Integer; p, mean, sigma : Double) : Double;
Var eta, quantile,q1, dq : Double;
i : Integer;
Begin
quantile:= mean + 1.25*sigma*(p-0.5);
q1:=quantile;
eta:=0.2*sigma/xy(1+n,0.75); // should not be too large! sets accuracy
For i:=1 to n Do
quantile := quantile + eta * (signum_smooth(x[i] - quantile,eta) + 2*p - 1);
dq:=abs(q1-quantile);
If dq>eta
then Begin
If dq<3*eta then eta:=eta/4;
For i:=1 to n Do
quantile := quantile + eta * (signum_smooth(x[i] - quantile,eta) + 2*p - 1);
end;
QuantileIterative:=quantile
end;
As the median for two elements would be the mean, I used a smoothed signum function, and xy() is x^y. Are there ideas to make it better? Of course if we have some more a-priori knowledge we can add code using min and max of the array, skew, etc. For big data you would not use an array perhaps, but for testing it is easier.
On homogeneous random ordered and for big enough list, this pseudo code can work:
# find min on the fly
if minDataPoint > dataPoint:
minDataPoint = dataPoint
# find max on the fly
if maxDataPoint < dataPoint:
maxDataPoint = dataPoint
# estimate median base on the current data
estimate_mid = (maxDataPoint + minDataPoint) / 2
#if **new** dataPoint is closer to the mid? stor it
if abs(midDataPoint - estimate_mid) > abs(dataPoint - estimate_mid):
midDataPoint = dataPoint
Inspired by #lakshmanaraj