maxElementsOnDisk when using ehcache with terracotta - ehcache

I am using ehcache with terracotta, and I wonder what is the meaning of the attribute maxElementsOnDisk in such cotext.
Does it mean the max elemnts allowed on the terracotta layer?
If not, what is the attribute controlling the number of elements allowed on the tarracotta layer?
If I understand correctly maxEntriesLocalHeap represents the number of entries allowed on the local heap tier, and it can overflow to terracotta layer which can overflow to disk (please correct me if I am wrong) but I am not sure what is the name of the element controlling each element.
BTW, in my use case I will never want entries to be written to disk, If there is no more room in the local heap or terracotta layer the entry should be evicted.
Yosi

There are 2 stores and related options:
L1: MemoryStore (local JVM of node) -> maxElementsInMemory
L2: DiskStore -> maxElementsOnDisk
The L2 size represents the maximum cache size. Elements can overflow from L1 to L2.
When using Terracotta, the maxElementsOnDisk value is overridden to provide the L2 size. Also, the DiskStore is never used because the elements overflow to Terracotta (L2 Store).
You can read about it in the official FAQs and a related bug report.

Related

Minimum associativity for a PIPT L1 cache to also be VIPT, accessing a set without translating the index to physical

This question comes in context of a section on virtual memory in an undergraduate computer architecture course. Neither the teaching assistants nor the professor were able to answer it sufficiently, and online resources are limited.
Question:
Suppose a processor with the following specifications:
8KB pages
32-bit virtual addresses
28-bit physical addresses
a two-level page table, with a 1KB page table at the first level, and 8KB page tables at the
second level
4-byte page table entries
a 16-entry 8-way set associative TLB
in addition to the physical frame (page) number, page table entries contain a valid bit, a
readable bit, a writeable bit, an executable bit, and a kernel-only bit.
Now suppose this processor has a 32KB L1 cache whose tags are computed based on physical addresses. What is the minimum associativity that cache must have to allow the appropriate cache set to be accessed before computing the physical address that corresponds to a virtual address?
Intuition:
My intuition is that if the number of indices in the cache and the number of virtual pages (aka page table entries) is evenly divisible by each other, then we could retrieve the bytes contained within the physical page directly from the cache without ever computing that physical page, thus providing a small speed-up. However, I am unsure if this is the correct intuition and definitely don't know how to follow through with it. Could someone please explain this?
Note: I have computed the number of page table entries to be 2^19, if that helps anyone.
What is the minimum associativity that cache must have to allow the appropriate cache set to be accessed before computing the physical address that corresponds to a virtual address?
They're only specified that the cache is physically tagged.
You can always build a virtually indexed cache, no minimum associativity. Even direct-mapped (1 way per set) works. See Cache Addressing Methods Confusion for details on VIPT vs. PIPT (and VIVT, and even the unusual PIVT).
For this question not to be trivial, I assume they also meant "without creating aliasing problems", so VIPT is just a speedup over PIPT (physically indexed, phyiscally tagged). You get the benefit of allowing TLB lookup in parallel with fetching tags (and data) for the ways of the indexed set without any downsides.
My intuition is that if the number of indices in the cache and the number of virtual pages (aka page table entries) is evenly divisible by each other, then we could retrieve the bytes contained within the physical page directly from the cache without ever computing that physical page
You need the physical address to check against the tags; remember your cache is physically tagged. (Virtually tagged caches do exist, but typically have to get flushed on context switches to a process with different page tables = different virtual address space. This used to be used for small L1 caches on old CPUs.)
Having both numbers be a power of 2 is normally assumed, so they're always evenly divisible.
Page sizes are always a power of 2 so you can split an address into page number and offset-within-page by just taking different ranges of bits in the address.
Small/fast cache sizes also always have a power of 2 number of sets so the index "function" is just taking a range of bits from the address. For a virtually-indexed cache: from the virtual address. For a physically-indexed cache: from the physical address. (Outer caches like a big shared L3 cache may have a fancier indexing function, like a hash of more address bits, to avoid aliasing for addresses offset from each other by a large power of 2.)
The cache size might not be a power of 2, but you'd do that by having a non-power-of-2 associativity (e.g. 10 or 12 ways is not rare) rather than a non-power-of-2 line size or number of sets. After indexing a set, the cache fetches the tags for all the ways of that set and compare them in parallel. (And for fast L1 caches, often fetch the data selected by the line-offset bits in parallel, too, then the comparators just mux that data into the output, or raise a flag for no match.)
Requirements for VIPT without aliasing (like PIPT)
For that case, you need all index bits to come from below the page offset. They translate "for free" from virtual to physical so a VIPT cache (that indexes a set before TLB lookup) has no homonym/synonym problems. Other than performance, it's PIPT.
My detailed answer on Why is the size of L1 cache smaller than that of the L2 cache in most of the processors? includes a section on that speed hack.
Virtually indexed physically tagged cache Synonym shows a case where the cache does not have that property, and needs page coloring by the OS to let avoid synonym problems.
How to compute cache bit widths for tags, indices and offsets in a set-associative cache and TLB has some more notes about cache size / associativity that give that property.
Formula:
min associativity = cache size / page size
e.g. a system with 8kiB pages needs a 32kiB L1 cache to be at least 4-way associative so that index bits only come from the low 13.
A direct-mapped cache (1 way per set) can only be as large as 1 page: byte-within-line and index bits total up to the byte-within-page offset. Every byte within a direct-mapped (1-way) cache must have a unique index:offset address, and those bits come from contiguous low bits of the full address.
To put it another way, 2^(idx_bits + within_line_bits) is the total cache size with only one way per set. 2^N is the page size, for a page offset of N (the number of byte-within-page address bits that translate for free).
The actual number of sets (in this case = lines) depends on the line size and page size. Using smaller / larger lines would just shift the divide between offset and index bits.
From there, the only way to make the cache bigger without indexing from higher address bits is to add more ways per set, not more ways.

How does Direct mapped cache implement spatial locality?

An array is declared as floatA[2048] . Each array element is 4Bytes in size.This program is run on a computer that has a
direct mapped data cache of size 8Kbytes, with block (line) size of 16Bytes.
Which elements of the array conflict with element A[0] in the data cache?
Ultimately A[0],A[512],A[1024],A[1536] map to cache block 0
As per my understanding, when A[0] is required for the first time, A[0],A[1],A[2],A[3](since one cache block can hold 4 elements) are brought into the cache and placed in cache blocks 0, 1, 2,and 3 respectively.
Other approach would be to bring only A[0] and place it in cache block 0. (Spatial locality not used here)
What is the general practice in such a scenario?
All four elements A[3:0] are stored in cache block 0 - since these 4 elements together form 16B. Depending on how the hardware system is set up, the next 16B are then stored on cache block 1 (the decision of which cache line (16B contiguous data granule) maps into which set is made while designing hardware and is based on certain bits of the address.

What values should be provided for DataNucleus' L2 cache size

With DataNucleus' L2 cache configuration, the "datanucleus.cache.level2.maxSize" is used to configure the cache.
The docs say that -1 is the default, and imply that this is turned on.
I'm guessing that -1 therefore means provide a cache up until some sort of -Xmx heap limit; but is that right?
If some other value is used, what are the units? Is it number of cached objects, bytes used, Mbytes used, something else?
number of objects, for the L2 cache providers that support that.

Redis 10x more memory usage than data

I am trying to store a wordlist in redis. The performance is great.
My approach is of making a set called "words" and adding each new word via 'sadd'.
When adding a file thats 15.9 MB and contains about a million words, the redis-server process consumes 160 MB of ram. How come I am using 10x the memory, is there any better way of approaching this problem?
Well this is expected of any efficient data storage: the words have to be indexed in memory in a dynamic data structure of cells linked by pointers. Size of the structure metadata, pointers and memory allocator internal fragmentation is the reason why the data take much more memory than a corresponding flat file.
A Redis set is implemented as a hash table. This includes:
an array of pointers growing geometrically (powers of two)
a second array may be required when incremental rehashing is active
single-linked list cells representing the entries in the hash table (3 pointers, 24 bytes per entry)
Redis object wrappers (one per value) (16 bytes per entry)
actual data themselves (each of them prefixed by 8 bytes for size and capacity)
All the above sizes are given for the 64 bits implementation. Accounting for the memory allocator overhead, it results in Redis taking at least 64 bytes per set item (on top of the data) for a recent version of Redis using the jemalloc allocator (>= 2.4)
Redis provides memory optimizations for some data types, but they do not cover sets of strings. If you really need to optimize memory consumption of sets, there are tricks you can use though. I would not do this for just 160 MB of RAM, but should you have larger data, here is what you can do.
If you do not need the union, intersection, difference capabilities of sets, then you may store your words in hash objects. The benefit is hash objects can be optimized automatically by Redis using zipmap if they are small enough. The zipmap mechanism has been replaced by ziplist in Redis >= 2.6, but the idea is the same: using a serialized data structure which can fit in the CPU caches to get both performance and a compact memory footprint.
To guarantee the hash objects are small enough, the data could be distributed according to some hashing mechanism. Assuming you need to store 1M items, adding a word could be implemented in the following way:
hash it modulo 10000 (done on client side)
HMSET words:[hashnum] [word] 1
Instead of storing:
words => set{ hi, hello, greetings, howdy, bonjour, salut, ... }
you can store:
words:H1 => map{ hi:1, greetings:1, bonjour:1, ... }
words:H2 => map{ hello:1, howdy:1, salut:1, ... }
...
To retrieve or check the existence of a word, it is the same (hash it and use HGET or HEXISTS).
With this strategy, significant memory saving can be done provided the modulo of the hash is
chosen according to the zipmap configuration (or ziplist for Redis >= 2.6):
# Hashes are encoded in a special way (much more memory efficient) when they
# have at max a given number of elements, and the biggest element does not
# exceed a given threshold. You can configure this limits with the following
# configuration directives.
hash-max-zipmap-entries 512
hash-max-zipmap-value 64
Beware: the name of these parameters have changed with Redis >= 2.6.
Here, modulo 10000 for 1M items means 100 items per hash objects, which will guarantee that all of them are stored as zipmaps/ziplists.
As for my experiments, It is better to store your data inside a hash table/dictionary . the best ever case I reached after a lot of benchmarking is to store inside your hashtable data entries that are not exceeding 500 keys.
I tried standard string set/get, for 1 million keys/values, the size was 79 MB. It is very huge in case if you have big numbers like 100 millions which will use around 8 GB.
I tried hashes to store the same data, for the same million keys/values, the size was increasingly small 16 MB.
Have a try in case if anybody needs the benchmarking code, drop me a mail
Did you try persisting the database (BGSAVE for example), shutting the server down and getting it back up? Due to fragmentation behavior, when it comes back up and populates its data from the saved RDB file, it might take less memory.
Also: What version of Redis to you work with? Have a look at this blog post - it says that fragmentation has partially solved as of version 2.4.

How does lookup the L1 and L2 cache?

Recently I was reading some material on cpu cache. I am wondering how does the cpu lookup the L1 and L2 cache and in what format is the data in the cpu cache stored?
I think a linear scan of the cache would be inefficient, are there any better solutions?
Thanks.
It uses index bits and tags extracted from the address it is looking up.
Say you are accessing some 32 bit address ADDR
ADDR will have bits: 31--------------------------0, [------tag|index|offset]
Then depending on the size of your cache:
Let's say you have a 32K, Direct Mapped cache with 32bytes per block.
Offset bits are used to find the data within each line because 8bytes is a minimum data size to be brought into the cache (well you always get the full 32bytes, but within the 32bytes you will have your data.)
This accounts for a cache with 1024 lines or sets, again each line with 32bytes. In order to index the 1024 sets you need 10bits. Thus the 10 bits from your address are used as an index into the cache. The offset bits are used to see where inside that line your data is , and the tag bits are used to match the address that you are looking up since two or more addresses will map into the same line of the cache.
Makes sense?
I do not know your answer, but I can recommend a good book that might lead you to one - The Essentials Of Computer Organization and Architecture

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