Normal index in oracle.

Question

i have a city_tbl with columns ( city_cd, city_name,state_cd) in oracle 11g release2
city_cd is a primary key.
i ran the query
select city_name from city_tbl where state_cd = 'AL'
expalin plan shows full table scan
now i create normal index on state_cd column
after running the query still explain plan shows full table scan.
after creating the index why there is full table scan ?

Answer 1

Hi,
it looks like the optimizer is doing the right thing by ignoring your index and doing the full table scan. Full table scan is a perfectly valid way of retrieving data and in many cases is more efficient than an index access, because a) it uses multiblock reads b) doesn't have to go through index blocks first before reading the data.
In your case, the optimizer estimates that the cost of reading the table via a FTS is equivalent to 34 single-block reads. Index access in your case is more expensive, because while it's pretty cheap to acquire rowids of desired rows (only 2 reads are needed), the desired table rows are scattered around 115 blocks, so Oracle would have to make 117 reads to retrieve the data via an index.
Of course, this is only true if the optimizer is right in its assumptions about table size, index structure, predicate selectivity and clustering factor. If it's not, then you need to find where exactly the optimizer is wrong and correct it (e.g. by collecting a histogram on the column in WHERE clause).
Best regards,
Nikolay