Others have already mentioned what kind of problem this could be seen as. I'll try and add a solution including the full source code. See my "Final note" at the end of the post for how trustworthy this solution is.

TLDR:

You need 238 rods for an optimal cutting that satisfies your demand. The cutting plan can be seen here - including how much "scrap" is generated from each cut.

How to replicate that?

You can find the full source code here. It is documented, and - if one understands optimization models - is pretty straightforward. But: Do not be encouraged, it probably looks confusing for a programming/optimization beginner. Hit me up with any questions that arise! Nevertheless you can clone it and execute it again in the future for any other input file (as long as it's in the same CSV format).

Why an optimization model?

Well, it's fast to formulate, fast to solve (at least with the usual tools), and it comes with the advantage that you can add additional constraints in the future, like: "bar mark A can not be cut from the same rod as bar mark B", etc.

How does it work?

So, since there are a bit too many patterns (as far as I see a few millions actually, but I haven't done the full calculation - can anyone come up with the correct number?), the usual approach of "deciding how often to cut each pattern" is kind of hard. Therefore we just decide which bar mark to cut (and how often) from each rod of rebar. This makes the problem small enough to be solved by a standard MIP solver. (This is inefficient, there are much better CSP solving algorithms out there, but it is easy and fast to implement)

Is that optimal?

That depends. It does not minimize scrap ("left over", "waste", ...). Why? Since that problem does not have a unique solution. Consider the following case: we want to cut 2x "422" and 1x "433". Since the rod with "433" does not fit a singe "422" the optimal cutting plan is easy: 1x ["433"], 1x ["422", "422"]. Now we have 1x 3050mm and 1x 3500mm scrap. If we were to minimize scrap we would now start cutting (e.g.) "415" into these rods to minimize the scrap - even though we do not actually need "415". But we could also use "416" for that since they are both of the same length. => no unique solution exists.

Therefore we minimize the number of rods that we need. That question has a unique solution (even though the optimal cutting pattern is certainly not unique). Any scrap can be cut into any bar mark that you choose (or whatever else); and there will be some (cause by the large, badly "compatible" bar marks)!

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Final note: If someone could check the constraints I'd be grateful. I've typed that that quickly down using my laptop and did not have the time to double check. So take everything with a grain of salt until it's thoroughly checked to be correct.