question regarding ME networks and item conversion...

[CreeperGoBoom]

im having a small issue regarding storage, ok so i have a 64k storage cell, it is 50 percent full, but there are no item counts above 300, i have moved the mega numbers to deep storage units

so does anyone know a list of item conversion rates? that is, how much storage one particular item takes up per unit?

because for the life of me i cant figure out whats taking up so much space!!

[Gorian]

Here: http://wiki.technicpack.net/ME_64K_Storage

[CreeperGoBoom]

ah so each item uses the same space regardless of what it is.. thanks :D

[Gorian]

Yup. That was just for the 64k storage, which you said you were using. It's different for each one I believe. Just read the wiki, these guys are doing a good job of getting it all up to date (it's the "official" wiki for tekkit btw)

[jakalth]

basicly, each new item type takes up a set amount of memory in "formatting" the storage cell. the larger the cell, the more memory is used per item type. how much memory each individual item of a given type takes up(once the storage cell is accepting those items) is far smaller then the formatting size. When storing small amounts of a large number of different items, it is more cost effective to make several smaller storage cells to spread out the storage. But when storing a larger amount of a few items(like metal ingots for example). pre-formatting a single large cell to only accept those items proves to be a better way to go. Or just output the large stacks of a single item into DSU's like you are already doing. Storage bus's are your friend.

[Lethosos]

In a wierd way, that's counter-intuitive. Real-world data compression rates favor small implementations for large single-data storage, where you only hold one data item in memory and keep a list of pointers to duplicates on the disk.

I'm guessing this is more of a balance implementation than a logical progression.

[jakalth]

well, it is strait forward, once you understand it. If you have 10 different items on a single storage cell, and each item has a single stack of 64. It will take up less space then if you had 20 different items on a single storage cell, and each item has a single stack of 32. both cells would have the same total number of items stored on them, but the total memory used would be less for the cell that only has 10 different types of items. this is due to the higher cost of adding another type of item, vs the much lower cost of increasing how many items are in each stack.

[jakalth]

lower size storage cells cost less to add a new type of item to their memory then larger cells. This makes it more efficient to store smaller amounts of a larger variety of items in them, in terms of total memory used. For example; a 64K cell uses 512bytes per new type of item added, while a 4K cell only used 32bytes per new type of item added. BUT, they both cost the same amount to increase the size of an item stack.

[Gorian]

In a wierd way, that's counter-intuitive. Real-world data compression rates favor small implementations for large single-data storage, where you only hold one data item in memory and keep a list of pointers to duplicates on the disk.

I'm guessing this is more of a balance implementation than a logical progression.

deduplication is awesome! :D

[phazeonphoenix]

I'd have to say that the "files" being stored aka the digital representation of that block would be very complex and non-compressable like a video file. The storage math used in AE is loosely based on the actual file storage techniques. There is a logic to it but it has been "fudged" here and there.

[CharlieChop]

think of it as space resrved for item storage.

think of each disk as a cluster of 63 dsu. so if you place one cobble the entire dsu is just for cobble the only reason that 64k disks loose more space is because the dsu inside are bigger.

now that I think of it he prolly just did it like that so the other disks didnt become obsolete late game.