= Evaluating and Refining Data Models
:stem: latexmath

Once you’ve created a physical model, there are some steps you’ll want
to take to evaluate and refine table designs to help ensure optimal
performance.

== Calculating Partition Size

The first thing that you want to look for is whether your tables will
have partitions that will be overly large, or to put it another way, too
wide. Partition size is measured by the number of cells (values) that
are stored in the partition. Cassandra’s hard limit is 2 billion cells
per partition, but you’ll likely run into performance issues before
reaching that limit.

In order to calculate the size of partitions, use the following formula:

[latexmath]
++++
\[N_v = N_r (N_c - N_{pk} - N_s) + N_s\]
++++

The number of values (or cells) in the partition (N~v~) is equal to the
number of static columns (N~s~) plus the product of the number of rows
(N~r~) and the number of of values per row. The number of values per row
is defined as the number of columns (N~c~) minus the number of primary
key columns (N~pk~) and static columns (N~s~).

The number of columns tends to be relatively static, although it is
possible to alter tables at runtime. For this reason, a primary driver
of partition size is the number of rows in the partition. This is a key
factor that you must consider in determining whether a partition has the
potential to get too large. Two billion values sounds like a lot, but in
a sensor system where tens or hundreds of values are measured every
millisecond, the number of values starts to add up pretty fast.

Let’s take a look at one of the tables to analyze the partition size.
Because it has a wide partition design with one partition per hotel,
look at the `available_rooms_by_hotel_date` table. The table has four
columns total (N~c~ = 4), including three primary key columns (N~pk~ =
3) and no static columns (N~s~ = 0). Plugging these values into the
formula, the result is:

[latexmath]
++++
\[N_v = N_r (4 - 3 - 0) + 0 = 1N_r\]
++++

Therefore the number of values for this table is equal to the number of
rows. You still need to determine a number of rows. To do this, make
estimates based on the application design. The table is storing a record
for each room, in each of hotel, for every night. Let's assume the
system will be used to store two years of inventory at a time, and there
are 5,000 hotels in the system, with an average of 100 rooms in each
hotel.

Since there is a partition for each hotel, the estimated number of rows
per partition is as follows:

[latexmath]
++++
\[N_r = 100 rooms/hotel \times 730 days = 73,000 rows\]
++++

This relatively small number of rows per partition is not going to get
you in too much trouble, but if you start storing more dates of
inventory, or don’t manage the size of the inventory well using TTL, you
could start having issues. You still might want to look at breaking up
this large partition, which you'll see how to do shortly.

When performing sizing calculations, it is tempting to assume the
nominal or average case for variables such as the number of rows.
Consider calculating the worst case as well, as these sorts of
predictions have a way of coming true in successful systems.

== Calculating Size on Disk

In addition to calculating the size of a partition, it is also an
excellent idea to estimate the amount of disk space that will be
required for each table you plan to store in the cluster. In order to
determine the size, use the following formula to determine the size S~t~
of a partition:

[latexmath]
++++
\[S_t = \displaystyle\sum_i sizeOf\big (c_{k_i}\big) + \displaystyle\sum_j sizeOf\big(c_{s_j}\big) + N_r\times \bigg(\displaystyle\sum_k sizeOf\big(c_{r_k}\big) + \displaystyle\sum_l sizeOf\big(c_{c_l}\big)\bigg) +\]
++++

[latexmath]
++++
\[N_v\times sizeOf\big(t_{avg}\big)\]
++++

This is a bit more complex than the previous formula, but let's break it
down a bit at a time. Let’s take a look at the notation first:

* In this formula, c~k~ refers to partition key columns, c~s~ to static
columns, c~r~ to regular columns, and c~c~ to clustering columns.
* The term t~avg~ refers to the average number of bytes of metadata
stored per cell, such as timestamps. It is typical to use an estimate of
8 bytes for this value.
* You'll recognize the number of rows N~r~ and number of values N~v~
from previous calculations.
* The *sizeOf()* function refers to the size in bytes of the CQL data
type of each referenced column.

The first term asks you to sum the size of the partition key columns.
For this example, the `available_rooms_by_hotel_date` table has a single
partition key column, the `hotel_id`, which is of type `text`. Assuming
that hotel identifiers are simple 5-character codes, you have a 5-byte
value, so the sum of the partition key column sizes is 5 bytes.

The second term asks you to sum the size of the static columns. This
table has no static columns, so the size is 0 bytes.

The third term is the most involved, and for good reason—it is
calculating the size of the cells in the partition. Sum the size of the
clustering columns and regular columns. The two clustering columns are
the `date`, which is 4 bytes, and the `room_number`, which is a 2-byte
short integer, giving a sum of 6 bytes. There is only a single regular
column, the boolean `is_available`, which is 1 byte in size. Summing the
regular column size (1 byte) plus the clustering column size (6 bytes)
gives a total of 7 bytes. To finish up the term, multiply this value by
the number of rows (73,000), giving a result of 511,000 bytes (0.51 MB).

The fourth term is simply counting the metadata that that Cassandra
stores for each cell. In the storage format used by Cassandra 3.0 and
later, the amount of metadata for a given cell varies based on the type
of data being stored, and whether or not custom timestamp or TTL values
are specified for individual cells. For this table, reuse the number of
values from the previous calculation (73,000) and multiply by 8, which
gives 0.58 MB.

Adding these terms together, you get a final estimate:

[latexmath]
++++
\[Partition size = 16 bytes + 0 bytes + 0.51 MB + 0.58 MB = 1.1 MB\]
++++

This formula is an approximation of the actual size of a partition on
disk, but is accurate enough to be quite useful. Remembering that the
partition must be able to fit on a single node, it looks like the table
design will not put a lot of strain on disk storage.

Cassandra’s storage engine was re-implemented for the 3.0 release,
including a new format for SSTable files. The previous format stored a
separate copy of the clustering columns as part of the record for each
cell. The newer format eliminates this duplication, which reduces the
size of stored data and simplifies the formula for computing that size.

Keep in mind also that this estimate only counts a single replica of
data. You will need to multiply the value obtained here by the number of
partitions and the number of replicas specified by the keyspace’s
replication strategy in order to determine the total required total
capacity for each table. This will come in handy when you plan your
cluster.

== Breaking Up Large Partitions

As discussed previously, the goal is to design tables that can provide
the data you need with queries that touch a single partition, or failing
that, the minimum possible number of partitions. However, as shown in
the examples, it is quite possible to design wide partition-style tables
that approach Cassandra’s built-in limits. Performing sizing analysis on
tables may reveal partitions that are potentially too large, either in
number of values, size on disk, or both.

The technique for splitting a large partition is straightforward: add an
additional column to the partition key. In most cases, moving one of the
existing columns into the partition key will be sufficient. Another
option is to introduce an additional column to the table to act as a
sharding key, but this requires additional application logic.

Continuing to examine the available rooms example, if you add the `date`
column to the partition key for the `available_rooms_by_hotel_date`
table, each partition would then represent the availability of rooms at
a specific hotel on a specific date. This will certainly yield
partitions that are significantly smaller, perhaps too small, as the
data for consecutive days will likely be on separate nodes.

Another technique known as *bucketing* is often used to break the data
into moderate-size partitions. For example, you could bucketize the
`available_rooms_by_hotel_date` table by adding a `month` column to the
partition key, perhaps represented as an integer. The comparision with
the original design is shown in the figure below. While the `month`
column is partially duplicative of the `date`, it provides a nice way of
grouping related data in a partition that will not get too large.

image::data_modeling_hotel_bucketing.png[image]

If you really felt strongly about preserving a wide partition design,
you could instead add the `room_id` to the partition key, so that each
partition would represent the availability of the room across all dates.
Because there was no query identified that involves searching
availability of a specific room, the first or second design approach is
most suitable to the application needs.

_Material adapted from Cassandra, The Definitive Guide. Published by
O'Reilly Media, Inc. Copyright © 2020 Jeff Carpenter, Eben Hewitt. All
rights reserved. Used with permission._
