## Wednesday, October 3, 2018

### Minimizing write amplification in an LSM

Write-amplification for an LSM with leveled compaction is minimized when the per-level growth factor (fanout) is the same between all levels. This is a result for an LSM tree using a given number of levels. To find the minimal write-amplification for any number of levels this result can be repeated for 2, 3, 4, ... up to a large value. You might find that a large number of levels is needed to get the least write-amp and that comes at price of more read-amp, as the RUM Conjecture predicts.

In all cases below I assume that compaction into the smallest level (from a write buffer flush) has no write-amp. This is done to reduce the size of this blog post.

tl;dr - for an LSM with L1, L2, L3 and L4 what values for per-level fanout minimizes write-amp when the total fanout is 1000?
• (10, 10, 10) for leveled
• (6.3, 12.6, 12.6) for leveled-N assuming two of the levels have 2 sorted runs
• (>1, >1, >1) for tiered

Minimizing write-amp for leveled compaction

For an LSM with 4 levels (L1, L2, L3, L4) there is a per-level fanout between L1:L2, L2:L3 and L3:L4. Assume this uses classic leveled compaction so the total fanout is size(L4) / size(L1). The product of the per-level fanouts must equal the total fanout. The total write-amp is the sum of the per-level write-amp. I assume that the per-level write amp is the same as the per-level fanout although in practice and in theory it isn't that simple. Lets use a, b and c as the variables for the per-level fanout (write-amp) then the math problem is:
1. minimize a+b+c
2. such that a*b*c=k and a, b, c > 1
While I have been working on my math skills this year they aren't great and corrections are welcome. This is a constrained optimization problem that can be solved using Lagrange Multipliers. From above #1 is the sum of per-level write-amp and #2 means that the product of per-level fanout must equal the total fanout. The last constraint is that a, b and c must (or should) all be > 1.

This result uses Lagrange Multipliers for an LSM tree with 4 levels do there are 3 variables: a, b, c. But the math holds for an LSM tree with fewer levels or with more levels. If there are N levels then there are N-1 variables.

L(a, b, c) = a + b + c - lambda * (a*b*c - k)
dL/da = 1 - lambda * bc
dL/db = 1 - lambda * ac
dL/dc = 1 - lambda * ab
then
lambda = 1/bc = 1/ac = 1/ab
bc == ac == ab
and a == b == c to minimize the sum in #1

I wrote a Python script to discover the (almost) best values and the results match the math above.

Minimizing write-amp for tiered compaction

Assuming you can reason about tiered compaction using the notion of levels then the math changes a bit because the per-level write-amp with tiered equals 1 regardless of the per-level fanout. For tiered with 4 levels and 3 variables the problem is:
1. minimize 1+1+1
2. such that a*b*c = k and a, b, c > 1
Any values for a, b and c are sufficient as long they satisfy the constraints in #2. But it still helps to minimize a+b+c if that is predicts read-amp because a, b and c are also the number of sorted runs in L2, L3 and L4. So my advice is to use a == b == c in most cases.

Minimizing write-amp for leveled-N compaction
I explain leveled-N compaction here and here. It is like leveled compaction but allows a level to have more than one sorted run. This reduces the per-level write-amp at the cost of more read-amp. Sometimes that is a good trade.

The math above can also be used to determine how to configure per-level fanout to minimize write-amp for leveled-N. Assume an LSM tree with 4 levels (L1, L2, L3, L4) and 2 sorted runs in L2 and L3. The problem is:
1. minimize a + b/2 + c/2
2. such that a*b*c = k and a, b, c > 1
For leveled compaction I assume that per-level write-amp is all-size(Ln+1) / all-size(Ln) for compaction from Ln into Ln+1. For leveled-N I assume it is run-size(Ln+1) / all-size(Ln) where all-size is the size of all sorted runs on that level and run-size is the size of one sorted run. The astute reader might notice that all-size(Ln) == run-size(Ln) for traditional leveled. For leveled-N I assume that fanout continues to be run-size(Ln+1) / run-size(Ln).

Therefore with leveled-N the per-level write-amp is b/2 for L2 to L3 and c/2 for L3 to L4 because there are 2 sorted runs in the compaction input (twice as much data) in those cases. Were there 3 sorted runs then the values would be b/3 and c/3.

Lagrange Multipliers can be used to solve this assuming we want to minimize a + b/2 + c/2.

L(a, b, c) = a + b/2 + c/2 - lambda * (a*b*c - k)
dL/da = 1   - lambda * bc
dL/db = 1/2 - lambda * ac
dL/dc = 1/2 - lambda * ab
then
lambda = 1/bc = 1/2ac = 1/2ab
bc == 2ac -> b == 2a
bc == 2ab -> c == 2a
2ac == 2ab -> c == b
and 2a == b == c to minimize the sum

If the total fanout is 1000 then the per-level fanout values that minimize write-amp are 10, 10, 10 for leveled and 6.30, 12.60, 12.60 for this example with leveled-N and can be computed by "bc -l"
# for leveled-N
e(l(1000/4)/3)
6.29960524947436582381

e(l(1000/4)/3) * 2
12.59921049894873164762

# and for leveled
e(l(1000)/3)

9.99999999999999999992

One way to think of this result is that with leveled compaction the goal is to use the same per-level fanout between levels. This also uses the same per-level write-amp between levels because per-level write-amp == the per-level fanout for leveled.

But with leveled-N compaction we need to adjust the per-level fanout for levels to continue to get the same per-level write-amp between levels.