OK guys I have a real world problem, and I need some algorithm to figure it out.
We have a bunch of orders waiting to be shipped, each order will have a volume (in cubic feet), let's say, V1, V2, V3, ..., Vn
The shipping carrier can provide us four types of containers, and the volume/price of the containers are listed below:
Container Type 1: 2700 CuFt / $2500;
Container Type 2: 2350 CuFt / $2200;
Container Type 3: 2050 CuFt / $2170;
Container Type 4: 1000 CuFt / $1700;
No single order will exceed 2700 CuFt but likely to pass 1000 CuFt.
Now we need a program to get an optimized solution on freight charges, that is, minimum prices.
I appreciate any suggestions/ideas.
EDIT:
My current implementation is using biggest container at first, apply the first fit decreasing algorithm for bin packing to get result, then parse through all containers and adjust container sizes according to the content volume...
I wrote a similar program when I was working for a logistics company. This is a 3-dimensional bin-packing problem, which is a bit trickier than a classic 1-dimensional bin-packing problem - the person at my job who wrote the old box-packing program that I was replacing made the mistake of reducing everything to a 1-dimensional bin-packing problem (volumes of boxes and volumes of packages), but this doesn't work: this problem formulation states that three 8x8x8 packages would fit into a 12x12x12 box, but this would leave you with overlapping packages.
My solution was to use what's called a guillotine cut heuristic: when you put a package into the shipping container then this produces three new empty sub-containers: assuming that you placed the package in the back bottom left of the container, then you would have a new empty sub-container in the space in front of the package, a new empty sub-container in the space to the right of the package, and a new empty sub-container on top of the package. Be certain not to assign the same empty space to multiple sub-containers, e.g. if you're not careful then you'll assign the section in the front-right of the container to the front sub-container and to the right sub-container, you'll need to pick just one to which to assign it. This heuristic will rule out some optimal solutions, but it's fast. (As a concrete example, say you have a 12x12x12 box and you put an 8x8x8 package into it - this would leave you with a 4x12x12 empty sub-container, a 4x8x12 empty sub-container, and a 4x8x8 empty sub-container. Note that the wrong way to divide up the free space is to have three 4x12x12 empty sub-containers - this is going to result in overlapping packages. If the box or package weren't cubes then you'd have more than one way to divide up the free space - you'd need to decide whether to maximize the size of one or two sub-containers or to instead try to create three more or less equal sub-containers.) You need to use a reasonable criteria for ordering/selecting the sub-containers or else the number of sub-containers will grow exponentially; I solved this problem by filling the smallest sub-containers first and removing any sub-container that was too small to contain a package, which kept the quantity of sub-containers to a reasonable number.
There are several options you have: what containers to use, how to rotate the packages going into the container (there are usually six ways to rotate a package, but not all rotations are legal for some packages e.g. a "this end up" package will only have two rotations), how to partition the sub-containers (e.g. do you assign the overlapping space to the right sub-container or to the front sub-container), and in what order you pack the container. I used a randomized algorithm that approximated a best-fit decreasing heuristic (using volume for the heuristic) and that favored creating one large sub-container and two small sub-containers rather than three medium-sized sub-containers, but I used a random number generator to mix things up (so the greatest probability is that I'd select the largest package first, but there would be a lesser probability that I'd select the second-largest package first, and so on, with the lowest probability being that I'd select the smallest package first; likewise, there was a chance that I'd favor creating three medium-sized sub-containers instead of one large and two small, there was a chance that I'd use three medium-sized boxes instead of two large boxes, etc). I then ran this in parallel several dozen times and selected the result that cost the least.
There are other heuristics I considered, for example the extreme point heuristic is slower (while still running in polynomial time - IIRC it's a cubic time solution, whereas the guillotine cut heuristic is linear time, and at the other extreme the branch and bound algorithm finds the optimal solution and runs in exponential time) but more accurate (specifically, it finds some optimal solutions that are ruled out by the guillotine cut heuristic); however, my use case was that I was supposed to produce a fast shipping estimate, and so the extreme point heuristic wasn't appropriate (it was too slow and it was "too accurate" - I would have needed to add 10% or 20% to its results to account for the fact that the people actually packing the boxes would inevitably make sub-optimal choices).
I don't know the name of a program offhand, but there's probably some commercial software that would solve this for you, depending on how much a good solution is worth to you.
Zim Zam's answer is good for big boxes, but assuming relatively small boxes you can use a much simpler algorithm that amounts to solving an integral linear equation with a constraint:
Where a, b, c and d are integers being the number of each type of container used:
Given,
2700a + 2350b + 2050c + 1000d <= V (where V is the total volume of the orders)
You want to find a, b, c, and d such that the following function is minimized:
Total Cost C = 2500a + 2200b + 2170c + 1700d
It would seem you can brute force this problem (which is NP hard). Calculate every possible viable combination of a, b, c and d, and calculate the total cost for each combination. Note that no solution will ever use more than 1 container of type d, so that cuts down the number of possible combinations.
I am assuming orders can be split between containers.
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