Concept

Write amplification — where it appears

How many elements' worth of storage a structure writes for each element it stores. It separates an in-place tree from a log-structured one by a factor of tens, it appears in no operation count, and it is what decides which structure a storage engine is built from.

Named by 5 essays across one field — each of them below, with the objects they name alongside it.

elements of block written per key insertedB-tree, in place49.3Log-structured, T = 23.0 · 16× less than the treeLog-structured, T = 42.0 · 25× less than the treeLog-structured, T = 81.0 · 49× less than the treeLog-structured, T = 161.0 · 49× less than the treeB = 64, M = 4,096 (M/B = 64)25× between the two structures at T = 4

The writes nobody counted

Sixteen thousand keys inserted into a B-tree write 49.3 elements' worth of blocks for every key stored. The same keys into a log-structured store write 2.0. Every operation counter reports the two as the same work — the same insertions, the same comparisons, the same number of updates — and the factor of 24 decides which structure a storage engine is built from.

applied · Transfer
1100ε — the exponent the fanout is B toelements written per key33445679ε = 1 — the B-tree ·fanout 256 · 3transfers a queryε = 0.5 · fanout 16 · 5a querythe number above eachpoint is what a querycostsB = 256, M = 16,384 (M/B = 64)131,072 random keys

One dial between two structures

A B-tree writes 226 elements of block for every key stored and a log-structured store writes two. They are presented as rival designs. They are one design at two settings of an exponent that nothing in either description mentions, and every setting between them is available.

applied · Transfer
0.020.050.10.20.5121235810filter bits a keywasted reads per absent lookupno filters: all 3 levelssame rate every levelrates sized to levelsn = 65,536, size ratio 4, first run 1,024 keysthe same memory at every point

The filter each run carries

A log-structured store turns every lookup for a missing key into a read of every level, and a Bloom filter on each run buys those reads back with memory. Spread five bits a key evenly across three levels and a missing key still wastes 0.279 reads. Give the small levels more bits and the large one fewer — the same memory — and it wastes 0.201. At four levels the gap is 0.382 against 0.209, because sized filters stop the waste growing with the number of levels.

applied · Transfer
234681016110size ratioblock transfers · levelsa range of 100 keyslevelsan absent point lookup1,048,576 keys, 5 bits a keyfilters answer one of these two

The read a filter has no key for

A Bloom filter on every run of a log-structured store turns a lookup for a missing key from a read of every level into a fraction of one — 0.72 transfers across eight levels at five bits a key, and 0.00027 at twenty. A range query over the same store reads nine transfers at five bits and nine at twenty, because a filter answers whether one named key is in a run and a range has no key to name.

applied · Transfer
6080100how far a full leaf looks for room, in siblingsmean leaf fill, per cent12481664randomascendingdescendingascending, 1% late131,072 keys · leaves of 64 · fan-out 64a reach of 1 is the adjacent-sibling rule

A key passed along the row

A full B+-tree leaf that offers a key to its immediate neighbours before splitting leaves an ascending stream 67.2% full. Let it look one sibling further and the same stream fills its leaves completely, 2,048 leaves where there were 3,048, for about the same key moves. On random keys each doubling of the reach adds a few points of fill and more writes than it saves, and at the whole parent a key travels sixteen leaves on average. On a stream that is mostly in order the same reach costs almost nothing.

applied · Transfer

Named alongside it

The objects these essays reach for when they reach for this one.

Block transferLsm treeTrade offB-treeBloom filterDesign parameterExternal-memory modelMemory hierarchySpace accountingAmortised analysisBlock sizeBuffered tree

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