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Inside the Optimizer: Plan Costing

Summary: A detailed look at costing, and more undocumented optimizer fun.

The SQL Server query optimizer generates a number of physical plan alternatives from a logical requirement expressed in T-SQL.  If full cost-based optimization is required, a cost is assigned to each iterator in each alternative plan, and the plan with the lowest overall cost is ultimately selected for execution.

Costing Details

Graphical execution plans show the total estimated cost of an iterator, and the I/O and CPU cost components:


In the simplest cases (like that shown above) the total estimated cost of an iterator is a simple sum of the I/O and CPU components.  In other circumstances, the final cost may be modified by other considerations – perhaps as a result of a row goal.  The I/O and CPU costs are calculated based on cardinality estimates, data sizes, statistical information, and the type of iterator.  Iterators may also be costed differently depending on context.  For example, a Merge Join iterator has a zero estimated I/O cost when performing a one-to-many join.  When running in many-to-many mode, a worktable in tempdb is required, and an associated I/O cost appears:


The cost values represent expected execution times (in seconds) on a particular hardware configuration.  The interested reader can find an in-depth look at some of the ‘magic numbers’ used by the costing component in an excellent article by Joe Chang, who also blogs on this site.  It is important not to take the costing numbers literally.  They are obtained from a mathematical model used to compare candidate plans in a consistent fashion.  It is a practical certainty that your server has very different performance characteristics from those assumed by the model.  The great thing is that it does not typically matter, since it is relative plan cost that drives the final plan choice.  Although very detailed, the costing model does make a number of simplifying assumptions to prioritise speed and consistency over a more accurate reflection of reality.  One good example of this is that costing assumes that all query executions start with a cold cache.

Why Use a Cost-Based Model?

Consider the task of choosing between possible physical join implementations.  Most people are aware that Nested Loops is a good choice where both inputs are relatively small, Merge works well on suitably-ordered medium to large inputs, and Hash join is particularly suited  to very large joins, particularly in a parallel execution plan.  One optimization approach is to code a detailed set of rules based on that sort of reasoning.  This is known as rule-based optimization, and has featured in a number of database products over the years.  Experience has shown, however, that rule-based schemes lack flexibility and extensibility: introducing a new iterator or database feature, for example, might require a large number of new rules, and many specific changes to existing ones.

An alternative approach is cost-based optimization, based on a common, relatively abstract framework.  Adding a new iterator becomes as easy as implementing the required interfaces to the existing model, reusing features which provide base cost information (such as the cost to retrieve a page from disk) and contextual data (like cardinality estimation and average row size).  Using this more object-oriented approach, relatively simple costing rules can model complex behaviours.  For example, Nested Loops can be modelled with no start-up cost, and a relatively high cost per row.  Hash join can be given a significant start-up cost (to build its hash table) but a lower cost per row after that.  In each case, the exact cost may further depend on row size and distribution information – the key point is that the cost is context-sensitive and dynamic.

Costing Example

Let’s look at a query written to return the quantity of products in stock at the AdventureWorks warehouse, which have names starting with the first five letters of the alphabet:

    total_qty = SUM(inv.Quantity)
FROM Production.Product AS p WITH (INDEX(AK_Product_Name))
JOIN Production.ProductInventory AS inv ON
    inv.ProductID = p.ProductID
    p.Name LIKE '[A-E]%'

We get the following plan, where SQL Server has chosen to compute SUM(Quantity) for all products, and merge-join those results with the products that have names starting A-E (I used an index hint just to make sure you get the same plan I did):


I have deliberately chosen to feature a plan where each iterator’s total cost (‘Operator Cost’) is close to being a simple sum of the I/O and CPU cost components.  Just the Sort and Merge Join iterators add a small amount for overheads: the Sort adds 0.000038, and the Merge Join adds 0.0000031.  One final thing to notice is that each iterator knows the cumulative cost of all iterators below it (its subtree cost).  Let’s now force a nested loops join plan with the same query, and compare the costs:


Looking at the highlighted top-level subtree cost, we can see that this plan is very slightly more expensive than the Merge equivalent overall, which explains the optimizer’s choice.  We can also see that the Stream Aggregate and Nonclustered Index Seek have Operator Costs equal to the sum of their I/O and CPU costs.  The Nested Loops iterator adds a small overhead of 0.000038 to its stated CPU cost.  The costing of the Clustered Index Seek is somewhat more complex:

Costing the Clustered Index Seek

CPU Cost

The CPU cost is calculated as 0.0001581 for the first row, and 0.000011 for subsequent rows.  In this case, SQL Server expects each execution of the Clustered Index Seek to produce 2.47454 rows, so the CPU cost is 0.0001581 (for the first row) plus 1.47454 * 0.000011 (for the remainder).  This calculation produces exactly the 0.0001597 CPU cost seen in the plan.

The final CPU cost contribution is obtained by multiplying this per-execution CPU cost by the expected number of executions.  In this case, SQL Server expects the Index Seek on the Product table to produce 43.2 rows, so the final CPU cost of the Clustered Index Seek is 43.2 * 0.0001597, which comes to 0.00689904.

I/O Cost

The I/O cost of 0.003125 is exactly 1/320 – reflecting the model’s assumption that the disk subsystem can perform 320 random I/O operations per second (each fetching a single 8KB page).  Again, this estimate needs to be modified to account for the expected 43.2 executions.  However, the costing component is smart enough to recognise that the total number of pages that need to be brought in from disk can never exceed the number of pages required to store the whole table.  The Product Inventory table uses only 7 pages, so the modelled total I/O cost is approximately 7 * 0.003125, which comes to 0.021875.

Total Operator Cost

Adding the CPU and I/O subtotals, we get a total operator cost of 0.02877404.  This is not quite the 0.00287423 cost shown in the plan, because I have simplified the calculation slightly to more clearly illustrate the process.

Undocumented Features

So far, we’ve seen how the I/O and CPU costs contribute to plan selection.  In this last section, I want to show you a couple of undocumented features which allow you to change the I/O and CPU cost calculations.  As usual, this is presented for entertainment and educational reasons only – never use these tools on anything other than a personal test system.

There are two undocumented DBCC commands, SETCPUWEIGHT and SETIOWEIGHT, which apply a multiplier to the cost values normally produced by the costing component.  By default, both are set to 1, which produces the unmodified I/O and CPU costs you normally see.  Each command takes a single parameter of the real floating-point type.  The easiest way to enter a literal floating-point value is to express it in scientific notation.  A third undocumented DBCC command, SHOWWEIGHTS, can be used to display the effective settings for the current session.  As usual, we also need to enable trace flag 3604 to see the output from these DBCC commands (which appears in the SSMS messages window).  Let’s say we want to see the effect of making I/Os cheaper.  We can adjust the base I/O costing to 60% of its default value by executing this command:

DBCC    TRACEON (3604);     -- Show DBCC output
DBCC SETCPUWEIGHT(1E0); -- Default CPU weight
DBCC SETIOWEIGHT(0.6E0); -- I/O multiplier = 0.6
DBCC SHOWWEIGHTS; -- Show the settings

If we now run the query used in the costing example, we find that the optimizer now chooses the nested loops plan, because it now has a lower overall cost than the Merge join plan.  We can achieve the same affect by raising the CPU weight to 1.7, while resetting the I/O cost multiplier to 1.  We can produce a plan based on a Hash join by lowering the CPU cost multiplier below 0.8, or raising the I/O cost multiplier to 1.3:


There is a great deal to be learned about the workings of the costing component and the optimizer as a whole, by playing around with these settings.  One particularly interesting combination is to set both cost multipliers to zero.  This can cause the optimizer to reveal some of its internal workings, as in the following example:

    cnt = COUNT_BIG(*)
FROM Production.ProductSubcategory AS ps
JOIN Production.Product AS p ON
    p.ProductSubcategoryID = ps.ProductSubcategoryID
        SELECT SQRT(PI())
        FROM Production.ProductInventory AS inv
            inv.ProductID = p.ProductID
            AND inv.Shelf = N'A'

With both multipliers set to zero, the optimizer produces a plan containing three consecutive sorts:


Notice that all three sort in the same way, and all operator costs are zero.  The extra sorts are the result of the application of enforcer rules, and are normally removed from the final plan by other rules.  Setting the costing multipliers to zero removes the advantage in doing this.  Have fun with this, and don’t forget to reset everything to the default when you have finished experimenting.

Paul White
twitter: @SQL_Kiwi

Published Wednesday, September 1, 2010 8:34 PM by Paul White

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jchang said:

interesting that we can adjust the relative CPU-IO weights,

but we really want to be able to

1) adjust the ratio for "random" IO and sequential IO (currently 1/320 = 0.003125 and 1/1350 = 0.000740741).

On a disk storage system configured for sequential, we might like to dial sequential to 13500 (100MB/sec). On SSD, perhap random and sequential should be closer?

2) decide whether IO costs should be reduced in a parallel execution plan. On a weak storage system, a parallel plan will not get anymore IO, so IO cost should stay the same as non-parallel. On a powerful storage system, we would like IO costs reduce, because IO throughtput might only be limited by the ability of SQL Server (the the processor cores) to consume data.

September 2, 2010 3:37 PM

Paul White said:

Hi Joe,

Sorry for the slow reply - I didn't see the comment notification email for some reason.

I agree it would be wonderful to be able to modify the random:sequential cost ratio.  That was the goal I was pursuing when I came across the techniques illustrated in my post.

As you point out in your superb article, the costing model has been changed before, so we can at least hope for further changes in future.

I would be extremely surprised if work had not already been done to see how SQL Server should adapt to the increasing use of SSDs and very parallel systems.  Perhaps the future changes will be of the sort used to 'optimize for ad-hoc workloads' rather than modifying base costs directly, but who knows ;c)

Thanks again for reading this entry and taking the time to comment!


September 7, 2010 6:08 AM

Doug said:

I am curious in the article you say the optimizer assumes that the cache is cold? The link for Joe Chang says

"The formula for IO represents the probability that an IO will already be in memory after a number of pages have already been accessed"

Seems like the query plans will be off for databases that reside mostly in cache id assumes its cold?


March 15, 2011 3:56 PM

Paul White said:

Hi Doug,

The costing model does assume the query starts on a server with nothing in the data cache.  It does avoid costing more pages than exist in the structure, and makes some adjustments to reflect the fact that later page reads might be satisfied from cache due to an earlier fetch for this same query...and so on.

So yes, query plans are optimized for worst-case performance (reading every page from disk).


March 15, 2011 4:18 PM

Rich said:

If I execute the first query (no query hint), the execution plan I get has no combination Index Seek & Sort for Product.Name; instead, I get a Clustered Index Scan of PK_Product_ProductID fed to the Merge Join.

SQL Server 2005 SP 3 Enterprise

Difference in costing model between 2005 and 2008?


May 13, 2011 10:08 AM

Paul White said:

Hi Rich,

Yes.  In 2005, the estimated cost for the scan plan is 0.0295338, versus  0.03146 for the seek + sort.  In 2008 R2, the cost for the seek + sort is 0.0322007 against 0.0384227 for the scan.


May 13, 2011 3:54 PM

Rodrigo Ribeiro Gomes said:

Hi Paul. Nice content!

Just I notice a little thing on calculation on the clustered index operator.

You says that for the first row the cost is 0.0001581 and for all subsequent is 0.000011.

Using the same estimated number of rows, 2.47454, I seen that 1.47454 * 0.000011 result in 0.00001621994 and adding it to 0.0001581 result is 0.00017431994 that is considerably different that 0.0001597.

The correct value would would be 0.0000011 (with five decimal zeros, instead of four?)? with this value, the calculations is correct for me.

Another questions is: this values of 0.0001581 and supposed 0.0000011 are fixed for Clustered Index Seek? Each operator can have values like this?


February 19, 2015 2:25 PM


HI paul,

Thanx for the in depth  article , I have a question here

you wrote

"The cost values represent expected execution times (in seconds) on a particular hardware configuration."

and in one MSDN Article i read

"units for these costs: they're not in seconds, MB or anything."

refrence :

Did i interpret something wrong.

September 7, 2015 2:48 AM

Paul White said:

The cost values were originally the number of seconds the query was expected to run for on the SQL Server developer's desktop PC ("Nick's machine). As a practical matter, the numbers are best regarded these days as unitless.

September 7, 2015 4:53 AM

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