Sql-server – SQL Server – Overlapping NC Indexes, Drop Question

It all depends on the definitions and the key (and non-key) columns defined in the nonclustered index. The clustered index is the actual table data. Therefore it contains all of the data in the data pages, whereas the nonclustered index is only containing columns' data as defined in the index creation DDL.

Let's set up a test scenario:

use testdb;
go

if exists (select 1 from sys.tables where name = 'TestTable')
begin
    drop table TestTable;
end
create table dbo.TestTable
(
    id int identity(1, 1) not null
        constraint PK_TestTable_Id primary key clustered,
    some_int int not null,
    some_string char(128) not null,
    some_bigint bigint not null
);
go

create unique index IX_TestTable_SomeInt
on dbo.TestTable(some_int);
go

declare @i int;
set @i = 0;

while @i < 1000
begin
    insert into dbo.TestTable(some_int, some_string, some_bigint)
    values(@i, 'hello', @i * 1000);

    set @i = @i + 1;
end

So we've got a table loaded with 1000 rows, and a clustered index (PK_TestTable_Id) and a nonclustered index (IX_TestTable_SomeInt). As you've seen in your testing, but just for thoroughness:

set statistics io on;
set statistics time on;

select some_int
from dbo.TestTable -- with(index(PK_TestTable_Id));

set statistics io off;
set statistics time off;
-- nonclustered index scan (IX_TestTable_SomeInt)
-- logical reads: 4

Here we have a nonclustered index scan on the IX_TestTable_SomeInt index. We have 4 logical reads for this operation. Now let's force the clustered index to be used.

set statistics io on;
set statistics time on;

select some_int
from dbo.TestTable with(index(PK_TestTable_Id));

set statistics io off;
set statistics time off;
-- clustered index scan (PK_TestTable_Id)
-- logical reads: 22

Here with the clustered index scan we have 22 logical reads. Why? Here's why. It all matters on how many pages that SQL Server has to read in order to grab the entire result set. Get the average row count per page:

select
    object_name(i.object_id) as object_name,
    i.name as index_name,
    i.type_desc,
    ips.page_count,
    ips.record_count,
    ips.record_count / ips.page_count as avg_rows_per_page
from sys.dm_db_index_physical_stats(db_id(), object_id('dbo.TestTable'), null, null, 'detailed') ips
inner join sys.indexes i
on ips.object_id = i.object_id
and ips.index_id = i.index_id
where ips.index_level = 0;

Take a look at my result set of the above query:

As we can see here, there are an average of 50 rows per page on the leaf pages for the clustered index, and an average of 500 rows per page on the leaf pages for the nonclustered index. Therefore, in order to satisfy the query more pages need to be read from the clustered index.

Under the covers clustered and nonclustered indexes are the same. The clustered index just has the additional property that is is guaranteed to INCLUDE all columns. Therefore the data does not need to be maintained somewhere else. So, a clustered index and a nonclustered index that INCLUDEs all columns are virtually the same from an update cost perspective.

However, every index needs to be maintained if it contains a column that was changed during an updated. That means, the more indexes you have, the more expensive updates get.

So in your situation, I would try to keep the number of indexes to a minimum. That will help update performance more than worrying about if a particular index is better clustered or covering.

That all being said, your updates still need to find the row(s) to update as quickly as possible. Because you have two orders of magnitude more updates then select, updates should be looked at first when designing the indexing strategy. After they are taken care of, look at providing the minimal number of appropriate indexes for the read queries.