I have a table with 368 blocks, no free blocks, and the clustering factor of the primary key is 345. how is it possible? I thought that CF will give you the amount of table blocks that an index will need to access for a given rowset. Something like smallest( table blocks, ceil( (CF/tot table rows)*rowset/avg rows per block)) (real formula must be much complicated. However if I have an avg ro Here are the brute facts: select EMPTY_BLOCKS,AVG_SPACE_FREELIST_BLOCKS ,NUM_FREELIST_BLOCKS from dba_tables where table_name = 'CUSTOMERS' and owner = 'SOE' ; EMPTY_BLOCKS AVG_SPACE_FREELIST_BLOCKS NUM_FREELIST_BLOCKS ------------ ------------------------- ------------------- 0 0 0 The table was analysed with dbms_stat, percent = 100: exec dbms_stats.gather_table_stats( ownname=>'SOE', tabname=> 'CUSTOMERS', Degree=> 4, estimate_percent=> 100, granularity=>'ALL', cascade=>TRUE, stattab=>'sm_stattab', statid=>'CUSTOMERS09270906', statown=>'SOE') Table name Owner NUM_ROWS BLOCKS Size (m) LAST_ANALYSED ------------------------------ ---------------- ---------- ---------- -------- ------------------- CUSTOMERS SOE 36364 368 3 09/27/2006 08:39:50 But still the clust factor reported by dba_indexes for CUSTOMERS_PK is inferiors to CUSTOMERS block count. Table Index Column Clust Distinct Name Name U Name Factor Keys NUM_ROWS -------------------- ------------------------------ - -------------------- ---------- ---------- ---------- CUSTOMERS CUSTOMERS_PK U CUSTOMER_ID 345 36112 36112 This is oracle 10.1.0.4 B. Polarski