Estimate RAM/memory requirements SQL server- SharePoint 2010

The memory required for SharePoint Server 2010 is directly related to the size of the content databases that you are hosting on a server that is running SQL Server.
As you add service applications and features, your requirements are likely to increase. The following table gives guidelines for the amount of memory we recommend.

Combined size of content databases	RAM recommended for computer running SQL Server
Minimum for small production deployments	8 GB
Minimum for medium production deployments	16 GB
Recommendation for up to 2 terabytes	32 GB
Recommendation for the range of 2 terabytes to 5 terabytes	64 GB
Recommendation for more than 5 terabytes	Additional RAM over 64 GB can improve SQL Server caching speed

Note:
These values are higher than those recommended as the minimum values for SQL Server because of the distribution of data required for a SharePoint Server 2010 environment. For more information about SQL Server system requirements, see Hardware and Software Requirements for Installing SQL Server 2008 (http://go.microsoft.com/fwlink/p/?LinkId=129377).

Other factors that may influence the memory required include the following:

• The use of SQL Server mirroring.
• The frequent use of files larger than 15 megabytes (MB).

How to estimate content database size

Estimate content database storage

The following process describes how to approximately estimate the storage required for content databases, without considering log files:

1. Calculate the expected number of documents. This value is referred to as D in the formula.
   
   How you calculate the number of documents will be determined by the features that you are using. For example, for My Sites or collaboration sites, we recommend that you calculate the expected number of documents per user and multiply by the number of users. For records management or content publishing sites, you may calculate the number of documents that are managed and generated by a process.
   
   If you are migrating from a current system, it may be easier to extrapolate your current growth rate and usage. If you are creating a new system, review your existing file shares or other repositories and estimate based on that usage rate.
2. Estimate the average size of the documents that you will be storing. This value is referred to as S in the formula. It may be worthwhile to estimate averages for different types or groups of sites. The average file size for My Sites, media repositories, and different department portals can vary significantly.
3. Estimate the number of list items in the environment. This value is referred to as L in the formula.
   
   List items are more difficult to estimate than documents. We generally use an estimate of three times the number of documents (D), but this will vary based on how you expect to use your sites.
4. Determine the approximate number of versions. Estimate the average number of versions any document in a library will have (this value will usually be much lower than the maximum allowed number of versions). This value is referred to as V in the formula.
   
   The value of V must be above zero.
5. Use the following formula to estimate the size of your content databases:
   
   Database size = ((D × V) × S) + (10 KB × (L + (V × D)))
   
   The value of 10 KB in the formula is a constant that roughly estimates the amount of metadata required by SharePoint Server 2010. If your system requires significant use of metadata, you may want to increase this constant.

As an example, if you were to use the formula to estimate the amount of storage space required for the data files for a content database in a collaboration environment with the following characteristics, you would need approximately 105 GB.

 

Input	Value
Number of documents (D)	200,000 Calculated by assuming 10,000 users times 20 documents
Average size of documents (S)	250 KB
List items (L)	600,000
Number of non-current versions (V)	2 Assuming that the maximum versions allowed is 10

Database size = (((200,000 x 2)) × 250) + ((10 KB × (600,000 + (200,000 x 2))) = 110,000,000 KB or 105 GB