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Although we know that this cardinality of 16 is actually wrong, the plan seems to have followed the standard cardinality formula for single table access quite correctly. To reprise the working for selectivity: Selectivity of ind_pad = 1 Selectivity of n1 = 1 / 25 Selectivity of n2 = 1 / 25 Total selectivity = 1 / 625 Hence cardinality = 10,000 / 625 = 16 Similarly, we can check view user_indexes for the values for the blevel (2), leaf_blocks (1,107), and clustering_factor (6,153), and put them into the standard cost formula: Cost = blevel + ceiling(selectivity * leaf_blocks) + ceiling(selectivity * clustering_factor) 2 + ceiling(1,107 / 625) + ceiling(6,153 / 625) 2 + 2 + 10 4 + 10 -- note the cost of the index (line 2 of the execution plan) 14 -- the total cost of the table access (line 1 of the plan)

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// Calculate the score. int scoreForPlay = CalculateScore(row, col, playType, workingBoard); PrintBoard(workingBoard); if (scoreForPlay == -1) { Console::WriteLine("The move is not a legal move."); if (moveNum == 0) { Console::WriteLine("The first play must use the center square."); } else { Console::WriteLine( "You must use at least one existing tile on the board."); } Console::WriteLine(); continue; } if (!ConfirmPlay(scoreForPlay)) continue; RecordPlay(workingTiles, workingBoard); // If more tiles are in the bag, draw tiles to replace those played. if (bag->Count > 0) { ReplacePlayedTiles(); } // // // if { The game ends when a player "goes out" -- she uses up all the tiles in her hand and there are none in the bag. The player is eligible for the end game bonus. (bag->Count == 0 && players[playerNum]->tiles->Count == 0) endBonus = true; gameOver = true; } UpdateScore(playerNum, scoreForPlay); moveComplete = true; } catch(Exception^) { moveComplete = false; } } return gameOver; }

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But look what happens when we generate the execution plan of the join. We know that we are going to pick up just one row from the driver table, so the cardinality of the final result should not be changed from the 16 that we got from the first query, and the cost ought to be Cost of acquiring the one row from the driver + 1 (cardinality from driver) * 14 (cost as for single-table query) Instead we see Execution Plan (8.1.7.4 or 9.2.0.6, or 10.1.0.4 autotrace) ---------------------------------------------------------0 SELECT STATEMENT Optimizer=ALL_ROWS (Cost=292 Card=16 Bytes=1712) 1 0 NESTED LOOPS (Cost=292 Card=16 Bytes=1712) 2 1 TABLE ACCESS (BY INDEX ROWID) OF 'DRIVER' (Cost=1 Card=1 Bytes=49) 3 2 INDEX (UNIQUE SCAN) OF 'D_PK' (UNIQUE) 4 1 TABLE ACCESS (BY INDEX ROWID) OF 'T1' (Cost=291 Card=16 Bytes=928) 5 4 INDEX (RANGE SCAN) OF 'T1_I1' (NON-UNIQUE) (Cost=45 Card=16) You will notice that this execution plan uses the traditional nested loop structure even in 9i and 10g remember that one of the conditions I mentioned originally for the use of the new structure was that the driving table should return more than one row. The cardinality for the execution plan is indeed unchanged at 16, but look what s happened to the cost. The cost of the index line has jumped from 4 to 45, and the incremental cost of visiting the table has jumped from 10 to (291 45 =) 246. Where did these numbers come from A quick trip back to the view user_indexes tells us the answer:

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select blevel, avg_leaf_blocks_per_key, avg_data_blocks_per_key from user_indexes where and ; table_name = 'T1' index_name = 'T1_I1'