How scalable is my Nutanix cluster really?

In a previous post I showed a chart which plots concurrency [X-axis] against throughput (IOPS) on the Y-Axis.  Here is that plot again:

Experienced performance chart ogglers will notice the familiar pattern of Littles Law, whereby throughput (X) rises quickly as concurrency (N) is increased.  As we follow the chart to the right, the slope flattens out and we achieve a lower increase in throughput, even as we increase concurrency by the same amount at each stage.  The flattening of the curve is best understood as Amdahls Law.

Anyone who follows Dr. Neil Gunther and his Universal Scalability Law, will also recognize this curve.

The USL states that taking the values of concurrency and throughput as inputs, we can in fact calculate the scalability of the system.  Specifically we are able to calculate the key factors of contention and crosstalk – which limit absolute linear scalability and eventually result in less throughput as additional load is submitted even as the capacity of the system is saturated.

I was fortunate to find both a very useful tool, and an easy-to-read summary of the USL from the Vivid Cortex site.  Both were written by Baron Schwartz.  I encourage anyone interested in scalability to check out his paper.

Using his Excel spreadsheet, I was able to input the numbers from my test and derive values that determine scalability.

Taking the largest number (0.074%)  the “contention value” (i.e the impact we expect due to Amdahls law) as the limit to linear scaling – we can say that for this particular cluster, running this particular (simplistic/synthetic) workload the Nutanix cluster scales 99.926% linear.  Although I did not crank up the concurrency beyond 576, the model shows us that this cluster will start to degrade performance if we try to push concurrency beyond 600 or so.  Again, the USL model is for this particular workload – on this particular cluster.  Doubling the concurrency of the offered load to 1200 will only net us 500,000 IOPS according to the model.

The high linearity (99.926%) is expected. The workload is 100% read, and with the data-locality feature of Nutanix filesystem – we expect close to 100% scalability.

We will return to these measures of scalability in the future to look at more realistic workloads.

Here is the Excel Sheet with my data : VividCortex_USL_Worksheet_v1 You are here

Cache behavior – How long will it take to fill my cache?

When benchmarking filesystems or storage, we need to understand the caching effects. Most often this involves filling the cache and reaching steady state. But how long will it take to fill a cache of a given size? The answer depends of course on the size of the cache, the IO size and the IO rate. So, to simpify let’s just say that a cache consists of some number of entries. For instance a 4GB cache would have 1 million 4KB entries. In my example this is simply a 1M entry cache.

In terms of time to fill the cache, it’s simpler to think about how many entries will need to be read before the cache is filled.

For a random workload, it will be more than 1M “reads”. Let’s see why.

The first read will be inserted into the cache, the second read will probably be inserted into the cache, but there is a small (1/1000000) chance that the second read will actually be already in the cache since it’s random. As the cache gets fuller – the chances of a given read already being present in cache increases. As a result it will take a lot more than 1 million reads to populate the entire cache with a random read workload.

The question is this. Is is possible to predict, how many “reads” it will take to fill the cahe?

The experiment.

In this experiment, we create an array to represent the cache. It has 1M entries. Then using a random number generator, simulate the workload and measure how long it takes to populate the cahche.

Results

After 1,000,000 “reads” there are 633,000 positive entries (entries that have data in them). So what happened to the other 367,000? The 367,000 represent cache “hits” on an existing entry. Since the read “workload” is 100% random, there is some chance that a subsequent read will be for an entry that is already cached. Over the life of 1,000,000 reads around 37% are for an entry that is already cached.

After 2,0000,000 reads the cache contains 864,000 entries. Another 1,000,000 reads yields 950,000.

The fuller that the cache becomes, the fewer new entries are added. Intuitively this makes sense because as the cache becomes more full, more of the “random reads” are satisfied by an existing cache entry.

In my experiments it takes about 17,000,000 “reads” to ensure that every cache entry is filled in a 1M entry cache. Here are the data for 19 runs.

 Iteration Positive Entries Empty Slots 1 631998 368002 2 864334 135666 3 950184 49816 4 981630 18370 5 993266 6734 6 997577 2423 7 999080 920 8 999660 340 9 999879 121 10 999951 49 11 999985 15 12 999996 4 13 999998 2 14 999998 2 15 999999 1 16 999999 1 17 1000000 0 18 1000000 0
• For 500,000 Entries it takes 15 iterations to fill all the entries.
• For 2,000,000 Entries it takes 19 iterations to fill all the entries.

Interestingly, the ratio of positive to empty entries after one iteration is always about 0.632:0.368

• 0.368 is roughly 1/e
• .632 is roughly 1-(1/e).