You specify the desired K-safety value as part of the cluster configuration when you initialize the database root
directory. By default, VoltDB uses a K-safety value of zero (no duplicate partitions). You can specify a larger K-safety
value using the kfactor property. For example, in the following configuration file, the K-safety value is
set to 2:
Documentationdeployment:
cluster:
kfactor: 2When you start the database specifying a K-safety value greater than zero, the appropriate number of partitions out of the cluster will be assigned as duplicates. For example, if you start a cluster with 3 nodes and the default partitions per node of 8, there are a total of 24 partitions. With K=1, half of those partitions (12) will be assigned as duplicates of the other half. If K is increased to 2, the cluster would be divided into 3 copies consisting of 8 partitions each.
The important point to note when setting the K value is that, if you do not change the hardware configuration, you are dividing the available partitions among the duplicate copies. Therefore performance (and capacity) will be proportionally decreased as K-safety is increased. So running K=1 on a 6-node cluster will be approximately equivalent to running a 3-node cluster with K=0.
If you wish to increase reliability without impacting performance, you must increase the cluster size to provide the appropriate capacity to accommodate for K-safety.
Of course, to ensure a system failure does not impact the database, not only do the partitions need to be duplicated, but VoltDB must ensure that the duplicates are kept on separate nodes of the cluster. To achieve this, VoltDB calculates the maximum number of unique partitions that can be created, given the number of nodes, partitions per node, and the desired K-safety value.
When the number of nodes is an integral multiple of the duplicates needed, this is easy to calculate. For example, if you have a six node cluster and choose K=1, VoltDB will create two instances of three nodes each. If you choose K=2, VoltDB will create three instances of two nodes each. And so on.
If the number of nodes is not a multiple of the number of duplicates, VoltDB does its best to distribute the partitions evenly. For example, if you have a three node cluster with two partitions per node, when you ask for K=1 (in other words, two of every partition), VoltDB will duplicate three partitions, distributing the six total partitions across the three nodes.
By now it should be clear that there is a correlation between the K value and the number of nodes and partitions in the cluster. Ideally, the number of nodes is a multiple of the number of copies needed (in other words, the K value plus one). This is both the easiest configuration to understand and manage.
However, if the number of nodes is not an exact multiple, VoltDB distributes the duplicated partitions across the cluster using the largest number of unique partitions possible. This is the highest whole integer where the number of unique partitions is equal to the total number of partitions divided by the needed number of copies:
Unique partitions = (nodes * partitions/node) / (K + 1)
Therefore, when you specify a cluster size that is not a multiple of K+1, but where the total number of partitions is, VoltDB will use all of the partitions to achieve the required K-safety value.
Note that the total number of partitions must be a whole multiple of the number of copies (that is, K+1). If neither the number of nodes nor the total number of partitions is divisible by K+1, then VoltDB will not let the cluster start and will display an appropriate error message. For example, if the configuration specifies 3 sites per host and a K-safety value of 1 but the voltdb start command specifies a server count of 3, the cluster cannot start because the total number of partitions (3X3=9) is not a multiple of the number of copies (K+1=2). To start the cluster, you must either change the configuration to increase the K-safety value to 2 (so the number of copies is 3) or change the sites per host to 2 or 4 so the total number of partitions is divisible by 2.
Finally, if the configuration specifies a K value higher than the available number of nodes, it is not possible to achieve the requested K-safety. Even if there are enough partitions to create the requested duplicates, VoltDB cannot distribute the duplicates to distinct nodes. For example, if you start a 3 node cluster when the configuration specifies 4 partitions per node (12 total partitions) and a K-safety value of 3, the number of total partitions (12) is divisible by K+1 (4) but not without some duplicates residing on the same node. In this situation, VoltDB issues an error message. You must either reduce the K-safety or increase the number of nodes.