Combinatorics optimization case

Lobotomy

I hope im posting this in the right place, I don't know under which category combinatorics goes. Anyway:

If we have
Y number of categories of building blocks.

Each category has X number of variants.

Each variant is associated with a unique cost.

One variant from each category is put together into an assembly.

Z number of assemblies are put together, arbitrarily using variants from each category.

Hypothesis: If total cost of each category is minimized, with respect to the number of each variant used in all assemblies – then total cost of all assemblies is also minimized. Is this true? Can it be proven mathematically? How?

Example:
Categories of building blocks: Tetraeders Cubes Oktaeders Spheres
4 Variants in each category: Red Blue Green Yellow

Total of 16 parts, we have one red, one blue one green and one yellow tetraeder etc....

Combine these together arbitrarily in 100 different assemblies with 4 parts each (one from each category).

Cost of each variant is randomly distributed between 1 and 100.

chiro

MHF Helper
Hey Lobotomy.

Think about how you would define cost arbitrarily as a function of the other variables.

Also think about other variables are related - is category a function of variant (or reverse)? What about blocks in terms of categories? What about assembly in terms of blocks?

You don't need to make it absolutely specific if you don't have the information but even getting something simple out that is useful will help you understand this.

With a simple model you can look at how derivatives of different functions translate to the derivatives of other functional relations.

Lobotomy

I am not really following what you mean here. I am assuming all variables are independent. Categories and variants are independent.

Reconsidering cost, I would say that cost can be increased or decreased for a variant. You can decrease cost by making it specific to a unique assembly, and you increase cost by making it a general reusable variant, thus highering the quantity of that variant used. So lets say that you can add a new color for teraeders that is used in a specific assembly, and that unique variant will have a lower cost. However there is a cost benefit of reusing variants, cost declines when volume go up (like economy of scales)... umm i think I have oversimplified the problem in my original post.

The issue would then be to whether assemblies should be optimised on cost (gives alot of variants with low volume) or if categories should be optimized (adding together the cost of each variant, with respect to the quantity used in all variants). My hypothesis is that for the total cost of all assemblies to be minimized, it is better not to suboptimize each assembly cost, but to optimize category cost. Do you get what i mean?

So I'm still not following what you meant, but somehow the problem formulation became a bit clearer for me =)

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chiro

MHF Helper
Lets say we have the cost of an assembly: you could write

Cost(A) = Cost(Category1) + Cost(Category2) + Cost(Category3) + ... + Cost(CategoryN).

If Cost functions for categories are always increasing then minimizing the derivative of Cost(A) [Cost function for an assembly] is done by minimizing each of the individual category cost functions.

This is what I was trying to say in the first response.

You don't need to have specific definitions right away - but if you are able to represent the basic structure of the functions then you also can get an idea in answering your question.

Your cost function for assemblies are functions of categories which are a function of variants. If all categories are completely independent of each other then you get the description I mentioned above.

So for getting cost of a category your category cost function is that of a variant. So you have Cost(Variant) = function of variant and each value of variant has some cost. You could organize it so that the function increases or if you have a function definition (like f(x) = x^2) then you use that.

Once you outline these things then it will be a lot easier for you to answer your question.

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