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:

- minimize a+b+c
- such that a*b*c=k and a, b, c > 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)

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

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.

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:

**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:

- minimize 1+1+1
- 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.

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:

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.

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"

# 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.

**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:

- minimize a + b/2 + c/2
- such that a*b*c = k and a, b, c > 1

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

bc == 2ac -> b == 2a

bc == 2ab -> c == 2a

2ac == 2ab -> c == b

and a == 2b == 2c 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)

e(l(1000/4)/3)

6.29960524947436582381

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

12.59921049894873164762

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.