My goal for the year is more time learning math and less time running MySQL benchmarks. I haven't done serious benchmarks for more than 12 months. It was a great experience but I want to learn new things. MySQL 8.0.14 has been released with fixes for a serious bug I found via the insert benchmark. I won't confirm whether it has been fixed. I hope someone else does.
My tests and methodology are described in posts for sysbench, linkbench and the insert benchmark. I hope the upstream distros (MySQL, MariaDB, Percona) repeat my tests and methodology and I am happy to answer questions about that. I even have inscrutable shell scripts that make it easy to run the tests. Despite being a lousy example of how to use Bash, they are portable enough to run on my home and work hardware.
Tuesday, January 22, 2019
Monday, January 21, 2019
Optimal configurations for an LSM and more
I have been trying to solve the problem of finding an optimal LSM configuration for a given workload. The real problem is larger than that, which is to find the right index structure and the right configuration for a given workload. But my focus is RocksDB so I will start by solving for an LSM.
This link is to slides that summarizes my effort. I have expressed the problem to be solved using differentiable functions to express the cost that is to be minimized. The cost functions have a mix of real and integer valued parameters for which values must be determine to minimize the cost. I have yet to solve the functions, but I am making progress and learning more math. This might be a constrained optimization problem and Lagrange Multipliers might be useful. The slides are from a talk I am about to present at the MongoDB office in Sydney where several WiredTiger developers are based. I appreciate that Henrik Ingo set this up.
My work has things in common with the excellent work by Harvard DASlab lead by Stratos Idreos. I have years of production experience on my side, they have many smart and ambitious people on their side. There will be progress. I look forward to more results from their Data Calculator effort. And I have learned a lot from the Monkey and Dostoevsky papers by Niv Dayan et al.
This link is to slides that summarizes my effort. I have expressed the problem to be solved using differentiable functions to express the cost that is to be minimized. The cost functions have a mix of real and integer valued parameters for which values must be determine to minimize the cost. I have yet to solve the functions, but I am making progress and learning more math. This might be a constrained optimization problem and Lagrange Multipliers might be useful. The slides are from a talk I am about to present at the MongoDB office in Sydney where several WiredTiger developers are based. I appreciate that Henrik Ingo set this up.
My work has things in common with the excellent work by Harvard DASlab lead by Stratos Idreos. I have years of production experience on my side, they have many smart and ambitious people on their side. There will be progress. I look forward to more results from their Data Calculator effort. And I have learned a lot from the Monkey and Dostoevsky papers by Niv Dayan et al.
Sunday, January 20, 2019
Bugs in Windows 10 parental controls
I use Windows 10 parental controls with my two children. Sometimes I am surprised at the bugs I encounter, but I can't rant too much because of glass houses and stones. My old favorite was that a hard reset before the time limit reached zero allowed my clever child to get more time. Apparently Microsoft takes storage efficiency very seriously and didn't want to waste a disk write and/or fsync on persisting the usage counter every few minutes. I haven't tried to reproduce this recently but never heard back after filing a bug report.
Now I have a new favorite bug. I am 5 hours behind their timezone and granted another hour to my daughter. It is 4pm here and 9pm there. The landing page after granting the time tells me my child can use the computer until 5pm (my timezone). Child tries to login and immediately encounters the timeout dialog. Apparently timezones are a hard problem. But less screen time is a good thing.
Now I have a new favorite bug. I am 5 hours behind their timezone and granted another hour to my daughter. It is 4pm here and 9pm there. The landing page after granting the time tells me my child can use the computer until 5pm (my timezone). Child tries to login and immediately encounters the timeout dialog. Apparently timezones are a hard problem. But less screen time is a good thing.
Tuesday, January 15, 2019
Geek code for LSM trees
This is a link to slides from my 5-minute talk at the CIDR 2019 Gong Show. The slides are a brief overview of the geek code for LSM trees. If you click on the settings icon in the slide show you can view the speaker notes which have links to blog posts that have more details. I also pasted the links below. Given time I might add to this post, but most of the content is in my past blog posts. Regardless I think there is more to be discovered about performant, efficient and manageable LSM trees.
The key points are there are more compaction algorithms to discover, we need to make it easier to describe them and compaction is a property of a level, not of the LSM tree.
Links to posts with more details:
The key points are there are more compaction algorithms to discover, we need to make it easier to describe them and compaction is a property of a level, not of the LSM tree.
Links to posts with more details:
- Describing tiered and leveled compaction
- Number of levels that minimized write amplification
- Combining tiered and leveled compaction
- Tiered vs leveled, why not both
- Name that compaction algorithm
- Original LSM paper that got this started
- Review of SlimDB with references to the first tiered compaction, Stepped Merge
Thursday, January 10, 2019
LSM math: fixing mistakes in my last post
My last post explained the number of levels in an LSM that minimizes write amplification using 3 different estimates for the per-level write-amp. Assuming the per-level growth factor is w then the 3 estimates were approximately w, w+1 and w-1 and named LWA-1, LWA-2 and LWA-3 in the post.
I realized there was a mistake in that post for the analysis of LWA-3. The problem is that the per-level write-amp must be >= 1 (and really should be > 1) but the value of w-1 is <= 1 when the per-level growth factor is <= 2. By allowing the per-level write-amp to be < 1 it easy to incorrectly show that a huge number of levels reduces write-amp as I do for curve #3 in this graph. While I don't claim that (w-1) or (w-1)/2 can't be a useful estimate for per-level write-amp in some cases, it must be used with care.
Explaining LWA-3
The next challenge is to explain how LWA-3 is derived. That comes from equation 12 on page 9 of the Dostoevsky paper. Start with the (T-1)/(K+1) term and with K=1 then this is (T-1)/2. T in the paper is the per-level growth factor so this is the same as (w-1)/2. The paper mentions that this is derived using an arithmetic series but does not show the work. I show my work but was not able to reproduce that result.
Assume that the per-level growth factor is w, all-to-all compaction is used and the LSM tree has at least 3 levels. When full L1 has size 1, L2 has size w and L3 has size w*w. There are four derivations below - v1, v2, v3, v4. The results are either w/2 or (w+1)/2 which doesn't match (w-1)/2 from the paper. Fortunately, my previous post shows how to minimize total write-amp assuming the per-level write-amp is w/2 or (w+1)/2. I will contact the author to figure out what I am missing.
The analysis below is for merges from L1 to L2, but it holds for merges from Ln to Ln+1. I think that v1 and v2 are correct and their estimate for per-level write-amp is (w+1)/2. As explained below I don't think that v3 or v4 are correct, their estimate for per-level write-amp is w/2.
I have yet to explain how to get (w-1)/2.
v1
Assume that merges are triggered from Ln to Ln+1 when a level is full -- L1 has size 1, L2 has size w, L3 has size w*w. A level is empty immediately after it is merged into the next level. So L2 gets full, then is merged into L3 and becomes empty, then slowly gets larger as L1 is merged into it w times. The per-level write-amp from this is (w+1)/2.
* merges into L2 write output of size 1, 2, ..., w
* then L2 is full
* sum of that sequence -> w*(w+1)/2
* average value is sum/w -> (w+1)/2
1) Moving data of size 1 from L1 to L2 writes (w+1)/2 on average
2) Therefore per-level write-amp for L1 -> L2 is (w+1)/2
Note that per-level write-amp is (avg merge output to Ln / size of Ln-1)
* avg merge output to L2 is (w+1)/2
* size of Ln-1 is 1
v2
Assume that merges are triggered from Ln to Ln+1 when a level is almost full -- L1 has size 1 * (w-1)/w, L2 has size w * (w-1)/w, L3 has size (w*w) * (w-1)/w. The trigger conditions can be reduced to L1 has size (w-1)/w, L2 has size (w-1) and L3 has size w*(w-1).
This assumes that w merges are done from L1 to L2 for L2 to go from empty to full. Each merge adds data of size (w-1)/w because L1:L2 merge is triggered when L1 has that much data. Thus L2 has size (w-1) after w merges into it at which point L2:L3 merge can be done. The per-level write-amp from this is the same as it was for v1.
* merges into L2 write output of size (w-1)/w * [1, 2, ..., w]
* then L2 is full
* sum of that sequence -> (w-1)/w * w*(w+1)/2 = (w-1)(w+1)/2
* average value is sum/w -> (w-1)(w+1)/(2*w)
As from v1, per-level write-amp is (avg merge output to Ln / size of Ln-1)
* avg merge output to L2 = (w-1)(w+1)/(2*w)
* size of L1 = (w-1)/w
start with: ( (w-1)(w+1)/(2*w) ) / ( (w-1)/w )
simplify to: (w+1)/2
v3
Merges are triggered the same as for v1 but I assume that only w-1 merges are done from Ln to Ln+1 rather than w. Ln+1 won't be full at the end of that, for example L2 would have size w-1 rather than the expected size w. But I was curious about the math. The per-level write-amp is w/2.
* merges into L2 write output of size 1, 2, ..., w-1
* sum of that sequence -> (w-1)*w/2
* average value is sum/(w-1) -> w/2
1) Moving data of size 1 from L1 to L2 writes w/2 on average
2) Therefore per-level write-amp for L1 -> L2 is w/2
v4
Merges are triggered the same as for v2. But as with v3, only w-1 merges are done into a level. Again I don't think this is correct because a level won't have enough data to trigger compaction at that point. The per-level write-amp here is the same as for v3.
* merges into L2 write output of size (w-1)/w * [1, 2, ..., w-1]
* sum of that sequence -> (w-1)/w * (w-1)*w/2 = (w-1)(w-1)/2
* average value is sum/(w-1) -> (w-1)/2
As from v1, per-level write-amp is (avg merge output to Ln / size of Ln-1)
* avg merge output to L2 = (w-1)/2
* size of L1 = (w-1)/w
start with: ( (w-1)/2 ) / ( (w-1)/w )
simplify to: w/2
I realized there was a mistake in that post for the analysis of LWA-3. The problem is that the per-level write-amp must be >= 1 (and really should be > 1) but the value of w-1 is <= 1 when the per-level growth factor is <= 2. By allowing the per-level write-amp to be < 1 it easy to incorrectly show that a huge number of levels reduces write-amp as I do for curve #3 in this graph. While I don't claim that (w-1) or (w-1)/2 can't be a useful estimate for per-level write-amp in some cases, it must be used with care.
Explaining LWA-3
The next challenge is to explain how LWA-3 is derived. That comes from equation 12 on page 9 of the Dostoevsky paper. Start with the (T-1)/(K+1) term and with K=1 then this is (T-1)/2. T in the paper is the per-level growth factor so this is the same as (w-1)/2. The paper mentions that this is derived using an arithmetic series but does not show the work. I show my work but was not able to reproduce that result.
Assume that the per-level growth factor is w, all-to-all compaction is used and the LSM tree has at least 3 levels. When full L1 has size 1, L2 has size w and L3 has size w*w. There are four derivations below - v1, v2, v3, v4. The results are either w/2 or (w+1)/2 which doesn't match (w-1)/2 from the paper. Fortunately, my previous post shows how to minimize total write-amp assuming the per-level write-amp is w/2 or (w+1)/2. I will contact the author to figure out what I am missing.
The analysis below is for merges from L1 to L2, but it holds for merges from Ln to Ln+1. I think that v1 and v2 are correct and their estimate for per-level write-amp is (w+1)/2. As explained below I don't think that v3 or v4 are correct, their estimate for per-level write-amp is w/2.
I have yet to explain how to get (w-1)/2.
v1
Assume that merges are triggered from Ln to Ln+1 when a level is full -- L1 has size 1, L2 has size w, L3 has size w*w. A level is empty immediately after it is merged into the next level. So L2 gets full, then is merged into L3 and becomes empty, then slowly gets larger as L1 is merged into it w times. The per-level write-amp from this is (w+1)/2.
* merges into L2 write output of size 1, 2, ..., w
* then L2 is full
* sum of that sequence -> w*(w+1)/2
* average value is sum/w -> (w+1)/2
1) Moving data of size 1 from L1 to L2 writes (w+1)/2 on average
2) Therefore per-level write-amp for L1 -> L2 is (w+1)/2
Note that per-level write-amp is (avg merge output to Ln / size of Ln-1)
* avg merge output to L2 is (w+1)/2
* size of Ln-1 is 1
v2
Assume that merges are triggered from Ln to Ln+1 when a level is almost full -- L1 has size 1 * (w-1)/w, L2 has size w * (w-1)/w, L3 has size (w*w) * (w-1)/w. The trigger conditions can be reduced to L1 has size (w-1)/w, L2 has size (w-1) and L3 has size w*(w-1).
This assumes that w merges are done from L1 to L2 for L2 to go from empty to full. Each merge adds data of size (w-1)/w because L1:L2 merge is triggered when L1 has that much data. Thus L2 has size (w-1) after w merges into it at which point L2:L3 merge can be done. The per-level write-amp from this is the same as it was for v1.
* merges into L2 write output of size (w-1)/w * [1, 2, ..., w]
* then L2 is full
* sum of that sequence -> (w-1)/w * w*(w+1)/2 = (w-1)(w+1)/2
* average value is sum/w -> (w-1)(w+1)/(2*w)
As from v1, per-level write-amp is (avg merge output to Ln / size of Ln-1)
* avg merge output to L2 = (w-1)(w+1)/(2*w)
* size of L1 = (w-1)/w
start with: ( (w-1)(w+1)/(2*w) ) / ( (w-1)/w )
simplify to: (w+1)/2
v3
Merges are triggered the same as for v1 but I assume that only w-1 merges are done from Ln to Ln+1 rather than w. Ln+1 won't be full at the end of that, for example L2 would have size w-1 rather than the expected size w. But I was curious about the math. The per-level write-amp is w/2.
* merges into L2 write output of size 1, 2, ..., w-1
* sum of that sequence -> (w-1)*w/2
* average value is sum/(w-1) -> w/2
1) Moving data of size 1 from L1 to L2 writes w/2 on average
2) Therefore per-level write-amp for L1 -> L2 is w/2
v4
Merges are triggered the same as for v2. But as with v3, only w-1 merges are done into a level. Again I don't think this is correct because a level won't have enough data to trigger compaction at that point. The per-level write-amp here is the same as for v3.
* merges into L2 write output of size (w-1)/w * [1, 2, ..., w-1]
* sum of that sequence -> (w-1)/w * (w-1)*w/2 = (w-1)(w-1)/2
* average value is sum/(w-1) -> (w-1)/2
As from v1, per-level write-amp is (avg merge output to Ln / size of Ln-1)
* avg merge output to L2 = (w-1)/2
* size of L1 = (w-1)/w
start with: ( (w-1)/2 ) / ( (w-1)/w )
simplify to: w/2
Wednesday, January 9, 2019
LSM math: revisiting the number of levels that minimizes write amplification
I previously used math to explain the number of levels that minimizes write amplification for an LSM tree with leveled compaction. My answer was one of ceil(ln(T)) or floor(ln(T)) assuming the LSM tree has total fanout = T where T is size(database) / size(memtable).
Then I heard from a coworker that the real answer is less than floor(ln(T)). Then I heard from Niv Dayan, first author of the Dostoevsky paper, that the real answer is larger than ceil(ln(T)) and the optimal per-level growth factor is ~2 rather than ~e.
All of our answers are correct. We have different answers because we use different functions to estimate the per-level write-amp. The graph of the functions for total write-amp using the different cost functions is here and you can see that the knee in the curve occurs at a different x value for two of the curves and the third curve doesn't appear to have a minimum.
While working on this I learned to love the Lambert W function. But I wonder whether I made the math below for LWA-2 harder than necessary. I am happy to be corrected. I appreciate the excellent advice on Quora: here, here and here. The online graphing calculator Desmos is another great resource.
Math
I use differentiable functions to express the total write-amp as a function of the number of levels, then determine the value (number of levels) at which the first derivative is zero as that might be the global minimum. Constants, variables and functions below include:
I wrote above that the 3 different answers came from using 3 different estimates for the per-level write-amp and I label these LWA-1, LWA-2 and LWA-3. When w is the per-level growth factor then the per-level write-amp functions are:
First I explain LWA-1 and LWA-2. Compacting 1 SST from Ln to Ln+1 requires merging 1 SST from Ln with ~w SSTs from Ln+1 where w=10 by default with RocksDB. The output will be between w and w+1 SSTs. If the output is closer to w then LWA-1 is correct. If the output is closer to w+1 then LWA-2 is correct. This paper explains why the per level write-amp is likely to be less than w. Were I to use f*w where f < 1 for LWA-1 then the math still holds. Maybe that is a future blog post.
LWA-3 assumes that all-to-all compaction is used rather than some-to-some. I explain the difference here. RocksDB/LevelDB leveled uses some-to-some but all-to-all is interesting. With all-to-all when compaction from Ln to Ln+1 finishes then Ln is empty and slowly gets full after each merge into it. Assume the per-level growth factor is w and Ln-1, Ln and Ln+1 are full at sizes 1, w and w*w. Then Ln becomes full after w merges from Ln-1 and those write output of size 1, 2, ..., w-1, w. The sum of the first w integers is w(w+1)/2. Divide this by w to get the averge -- (w+1)/2. However above LWA-3 is (w-1)/2 not (w+1)/2. I will explain that in another blog post. Note that in LWA-3 the numerator, w-1, is more interesting than the denominator, 2. Dividing by any constant doesn't change where the minimum occurs assuming there is a minimum and that is visible on this graph that shows the impact of dividing by 2 on the total write-amp.
Read on to understand the impact of using w-1, w or w+1 as the function for per-level write-amp. The difference might be more significant than you expect. It surprised me.
Minimizing TWA
This graph shows the total write-amp for LWA-1, LWA-2 and LWA-3. I call the total write-amp TWA-1, TWA-2 and TWA-3. Two of the curves, for TWA-1 and TWA-2, appear to have a minimum. One occurs for x between 4 and 6, the other for x between 6 and 8. The third curve, for TWA-3, doesn't appear to have a minimum and is decreasing as x (number of levels) grows.
The next graph uses the first derivative for the total write-amp functions, so it is for TWA-1', TWA-2' and TWA-3'. A global minimum for TWA-x can occur when TWA-x' = 0 and from the graph TWA-1'=0 when x=6.931 and TWA-2'=0 when x=5.422 which matches the estimate from the previous paragraph. From the graph it appears that TWA-3' approaches zero as x gets large but is never equal to zero.
The next step is to use math to confirm what is visible on the graphs.
Min write-amp for LWA-1
See my previous post where I show that n = ln(T) minimizes total write-amp if n isn't limited to an integer and then the per-level growth factor is e. Since the number of levels must be an integer then one of ceil(ln(T)) or floor(ln(T)) minimized total write-amp.
Min write-amp for LWA-2
Update-1 - I think I made a few mistakes here. So you can stop reading until update-2 arrives.
Update-2 - this post explains my mistake and uses math to estimate that per-level write-amp = (w+1)/2 when all-to-all compaction is used. I am still unable to derive (w-1)/2.
Then I heard from a coworker that the real answer is less than floor(ln(T)). Then I heard from Niv Dayan, first author of the Dostoevsky paper, that the real answer is larger than ceil(ln(T)) and the optimal per-level growth factor is ~2 rather than ~e.
All of our answers are correct. We have different answers because we use different functions to estimate the per-level write-amp. The graph of the functions for total write-amp using the different cost functions is here and you can see that the knee in the curve occurs at a different x value for two of the curves and the third curve doesn't appear to have a minimum.
While working on this I learned to love the Lambert W function. But I wonder whether I made the math below for LWA-2 harder than necessary. I am happy to be corrected. I appreciate the excellent advice on Quora: here, here and here. The online graphing calculator Desmos is another great resource.
Math
I use differentiable functions to express the total write-amp as a function of the number of levels, then determine the value (number of levels) at which the first derivative is zero as that might be the global minimum. Constants, variables and functions below include:
- T - total fanout, = size(database) / size(memtable)
- n - number of levels in the LSM tree
- LWA, LWA-x - function for the per-level write-amp
- TWA, TWA-x - function for the total write-amp, = n * LWA
- w - per-level growth factor, = T^(1/n) for all levels to minimize write-amp
I wrote above that the 3 different answers came from using 3 different estimates for the per-level write-amp and I label these LWA-1, LWA-2 and LWA-3. When w is the per-level growth factor then the per-level write-amp functions are:
- LWA-1 = w -- I used this to find that the best n = ceil(ln(T)) or floor(ln(T))
- LWA-2 = w + 1 -- with this the best n is less than that found with LWA-1
- LWA-3 = (w - 1) / 2 -- with this the best n is greater than that found with LWA-1
I can also state the per-level write-amp functions directly with T and n. I didn't above to make it easier to see the differences.
Explaining LWA- LWA-1 = T^(1/n)
- LWA-2 = T^(1/n) + 1
- LWA-3 = (T^(1/n) - 1) / 2
First I explain LWA-1 and LWA-2. Compacting 1 SST from Ln to Ln+1 requires merging 1 SST from Ln with ~w SSTs from Ln+1 where w=10 by default with RocksDB. The output will be between w and w+1 SSTs. If the output is closer to w then LWA-1 is correct. If the output is closer to w+1 then LWA-2 is correct. This paper explains why the per level write-amp is likely to be less than w. Were I to use f*w where f < 1 for LWA-1 then the math still holds. Maybe that is a future blog post.
LWA-3 assumes that all-to-all compaction is used rather than some-to-some. I explain the difference here. RocksDB/LevelDB leveled uses some-to-some but all-to-all is interesting. With all-to-all when compaction from Ln to Ln+1 finishes then Ln is empty and slowly gets full after each merge into it. Assume the per-level growth factor is w and Ln-1, Ln and Ln+1 are full at sizes 1, w and w*w. Then Ln becomes full after w merges from Ln-1 and those write output of size 1, 2, ..., w-1, w. The sum of the first w integers is w(w+1)/2. Divide this by w to get the averge -- (w+1)/2. However above LWA-3 is (w-1)/2 not (w+1)/2. I will explain that in another blog post. Note that in LWA-3 the numerator, w-1, is more interesting than the denominator, 2. Dividing by any constant doesn't change where the minimum occurs assuming there is a minimum and that is visible on this graph that shows the impact of dividing by 2 on the total write-amp.
Read on to understand the impact of using w-1, w or w+1 as the function for per-level write-amp. The difference might be more significant than you expect. It surprised me.
Minimizing TWA
This graph shows the total write-amp for LWA-1, LWA-2 and LWA-3. I call the total write-amp TWA-1, TWA-2 and TWA-3. Two of the curves, for TWA-1 and TWA-2, appear to have a minimum. One occurs for x between 4 and 6, the other for x between 6 and 8. The third curve, for TWA-3, doesn't appear to have a minimum and is decreasing as x (number of levels) grows.
The next graph uses the first derivative for the total write-amp functions, so it is for TWA-1', TWA-2' and TWA-3'. A global minimum for TWA-x can occur when TWA-x' = 0 and from the graph TWA-1'=0 when x=6.931 and TWA-2'=0 when x=5.422 which matches the estimate from the previous paragraph. From the graph it appears that TWA-3' approaches zero as x gets large but is never equal to zero.
The next step is to use math to confirm what is visible on the graphs.
Min write-amp for LWA-1
See my previous post where I show that n = ln(T) minimizes total write-amp if n isn't limited to an integer and then the per-level growth factor is e. Since the number of levels must be an integer then one of ceil(ln(T)) or floor(ln(T)) minimized total write-amp.
Min write-amp for LWA-2
I can reuse some of the math from my previous post. But this one is harder to solve.
# wa is the total write-amp
# n is the number of levels
# t is the total fanout
wa = n * ( t^(1/n) + 1 )
wa = n*t^(1/n) + n
# the difference between this and the previous post is '+1'
wa' = t^(1/n) + n * ln(t) * t^(1/n) * (-1) * (1/n^2) + 1
wa' = t^(1/n) - (1/n) * ln(t) * t^(1/n) + 1
At this point the difference between this and the previous post is '+1'. But wait this starts to get interesting.
# critical point for this occurs when wa' = 0
t^(1/n) - (1/n) * ln(t) * t^(1/n) + 1 = 0
# multiply by t^(-1/n)
1 - (1/n) * ln(t) + t^(-1/n) = 0
# move some terms to RHS
t^(-1/n) = (1/n) ln(t) - 1
# use ln on LHS and RHS to get rid of '^(1/n)'
ln ( t^(-1/n) ) = ln( (1/n) * ln(t) - 1 )
(-1/n) ln(t) = ln( (1/n) * ln(t) - 1
I got stuck here but eventually made progress.
# let a = (1/n) ln(t) and rewrite
-a = ln(a - 1)
# let x=a-1, a=x+1 and rewrite
-(x+1) = ln(x)
# do e^LHS = e^RHS
e^-(x+1) = e^ln(x)
e^-x * e^-1 = x
# multiply LHS and RHS by e^x
e^-1 = e^x * x
# e^-1 -> (1/e)
(1/e) = e^x * x
At last I can use Lambert W function!
# Given: e^x * x = K, then x = W(K)
x = W(e^-1) ~= 0.27846
# because a=x+1
a ~= 1.27846
# a = (1/n) ln(t) -> n = (1/a) ln(t), t=1024
n = 1/1.27846 * ln(1024)
# The value for n that minimizes total write-amp
# from the graph I claimed that n=5.422. this is close
n = 5.4217
# wa is the total write-amp
# n is the number of levels
# t is the total fanout
wa = n * ( t^(1/n) + 1 )
wa = n*t^(1/n) + n
# the difference between this and the previous post is '+1'
wa' = t^(1/n) + n * ln(t) * t^(1/n) * (-1) * (1/n^2) + 1
wa' = t^(1/n) - (1/n) * ln(t) * t^(1/n) + 1
At this point the difference between this and the previous post is '+1'. But wait this starts to get interesting.
# critical point for this occurs when wa' = 0
t^(1/n) - (1/n) * ln(t) * t^(1/n) + 1 = 0
# multiply by t^(-1/n)
1 - (1/n) * ln(t) + t^(-1/n) = 0
# move some terms to RHS
t^(-1/n) = (1/n) ln(t) - 1
# use ln on LHS and RHS to get rid of '^(1/n)'
ln ( t^(-1/n) ) = ln( (1/n) * ln(t) - 1 )
(-1/n) ln(t) = ln( (1/n) * ln(t) - 1
I got stuck here but eventually made progress.
# let a = (1/n) ln(t) and rewrite
-a = ln(a - 1)
# let x=a-1, a=x+1 and rewrite
-(x+1) = ln(x)
# do e^LHS = e^RHS
e^-(x+1) = e^ln(x)
e^-x * e^-1 = x
# multiply LHS and RHS by e^x
e^-1 = e^x * x
# e^-1 -> (1/e)
(1/e) = e^x * x
At last I can use Lambert W function!
# Given: e^x * x = K, then x = W(K)
x = W(e^-1) ~= 0.27846
# because a=x+1
a ~= 1.27846
# a = (1/n) ln(t) -> n = (1/a) ln(t), t=1024
n = 1/1.27846 * ln(1024)
# The value for n that minimizes total write-amp
# from the graph I claimed that n=5.422. this is close
n = 5.4217
Min write-amp for LWA-3
Update-1 - I think I made a few mistakes here. So you can stop reading until update-2 arrives.
Update-2 - this post explains my mistake and uses math to estimate that per-level write-amp = (w+1)/2 when all-to-all compaction is used. I am still unable to derive (w-1)/2.
I started to work on this without paying attention to the curve for LWA-3'. From the graph it appears to converge to 0 but is always less than 0, TWA-3 is decreasing as x, number of levels, gets large. Therefore make the number of levels as large as possible, 2M or 2B, to minimize total write-amp as visible in this graph.
But more levels in the LSM tree comes at a cost -- more read-amp. And the reduction in write-amp is small when the number of levels increases from 20 to 200 to 2000 to 2M. Again, this is visible in the graph. Besides, if you really want less write-amp then use tiered compaction rather than leveled with too many levels.
The other consideration is the minimal per-level growth factor that should be allowed. If the min per-level growth factor is 2. Then then that occurs when the number of levels, n, is:
# assume total fanout is 1024
2^n = 1024
log2(2^n) = log2(1024)
n = log2(1024) = 10
Alas the total fanout isn't always a power of 2. Given that the number of levels must be an integer then the goal is to use the smallest number of levels such that the per-level growth factor >= 2. Therefore when x isn't limited to an integer there is no answer -- just make x as large as possible (1M, 1B, etc) in which case the per-level growth factor converges to 1 but is always greater than 1.
The above can be repeated where the constraint is either the max number of levels or a different value for the min per-level growth factor (either <2 or >2). Regardless, if LWA-3 is the cost function then total write-amp is minimized by using as many levels as possible subject to these constraints.
Below is some math for LWA-3 and LWA-3'.
# wa is the total write-amp
# n is the number of levels
# t is the total fanoutwa = n * ( t^(1/n) - 1 ) / 2
wa = (n*t^(1/n) - n ) / 2
# the big difference between this and the previous post is '+1'
wa' = [ t^(1/n) + n * ln(t) * t^(1/n) * (-1) * (1/n^2) - 1 ] / 2
wa' = [ t^(1/n) - (1/n) * ln(t) * t^(1/n) - 1 ] / 2
# determine when wa' = 0[ t^(1/n) - (1/n) * ln(t) * t^(1/n) - 1 ] / 2 = 0
# multiply LHS and RHS by 2t^(1/n) - (1/n) * ln(t) * t^(1/n) - 1 = 0# multiply LHS and RHS by t^(-1/n)
1 - (1/n) * ln(t) - t^(-1/n) = 0
# move last term to RHS
1 - (1/n) * ln(t) = t^(-1/n)
# probably a good idea to stop here
# LHS is likely to be <0 so can't use ln(LHS) = ln(RHS)
But more levels in the LSM tree comes at a cost -- more read-amp. And the reduction in write-amp is small when the number of levels increases from 20 to 200 to 2000 to 2M. Again, this is visible in the graph. Besides, if you really want less write-amp then use tiered compaction rather than leveled with too many levels.
The other consideration is the minimal per-level growth factor that should be allowed. If the min per-level growth factor is 2. Then then that occurs when the number of levels, n, is:
# assume total fanout is 1024
2^n = 1024
log2(2^n) = log2(1024)
n = log2(1024) = 10
Alas the total fanout isn't always a power of 2. Given that the number of levels must be an integer then the goal is to use the smallest number of levels such that the per-level growth factor >= 2. Therefore when x isn't limited to an integer there is no answer -- just make x as large as possible (1M, 1B, etc) in which case the per-level growth factor converges to 1 but is always greater than 1.
The above can be repeated where the constraint is either the max number of levels or a different value for the min per-level growth factor (either <2 or >2). Regardless, if LWA-3 is the cost function then total write-amp is minimized by using as many levels as possible subject to these constraints.
Below is some math for LWA-3 and LWA-3'.
# wa is the total write-amp
# n is the number of levels
# t is the total fanoutwa = n * ( t^(1/n) - 1 ) / 2
wa = (n*t^(1/n) - n ) / 2
# the big difference between this and the previous post is '+1'
wa' = [ t^(1/n) + n * ln(t) * t^(1/n) * (-1) * (1/n^2) - 1 ] / 2
wa' = [ t^(1/n) - (1/n) * ln(t) * t^(1/n) - 1 ] / 2
# determine when wa' = 0[ t^(1/n) - (1/n) * ln(t) * t^(1/n) - 1 ] / 2 = 0
# multiply LHS and RHS by 2t^(1/n) - (1/n) * ln(t) * t^(1/n) - 1 = 0# multiply LHS and RHS by t^(-1/n)
1 - (1/n) * ln(t) - t^(-1/n) = 0
# move last term to RHS
1 - (1/n) * ln(t) = t^(-1/n)
# probably a good idea to stop here
# LHS is likely to be <0 so can't use ln(LHS) = ln(RHS)
Monday, January 7, 2019
Define "better"
Welcome to my first rant of 2019, although I have written about this before. While I enjoy benchmarketing from a distance it is not much fun to be in the middle of it. The RocksDB project has been successful and thus becomes the base case for products and research claiming that something else is better. While I have no doubt that other things can be better I am wary about the definition of better.
There are at least 3 ways to define better when evaluating database performance. The first, faster is better, ignores efficiency, the last two do not. I'd rather not ignore efficiency. The marginal return of X more QPS eventually becomes zero while the benefit of using less hardware is usually greater than zero.
There are at least 3 ways to define better when evaluating database performance. The first, faster is better, ignores efficiency, the last two do not. I'd rather not ignore efficiency. The marginal return of X more QPS eventually becomes zero while the benefit of using less hardware is usually greater than zero.
- Optimize for throughput and ignore efficiency (faster is better)
- Get good enough performance and then optimize for efficiency
- Get good enough efficiency and then optimize for throughput
Call to action
I forgot to include this before publishing. Whether #1, #2 or #3 is followed I hope that more performance results include details on the HW consumed to create that performance. How much memory and disk space were used? What was the CPU utilization? How many bytes were read from and written to storage? How much random IO was used? I try to report both absolute and relative values where relative values are normalized by the transaction rate.
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