Imho the simplification, that seq scan startup cost is 0.0 is
only valid when we expect to return most of the rows.
When expecting only 1 row, imho the costs need to be 50 % for
startup and 50 % rest. Assuming, that on average the row will be
in the middle of that relation file.
When returning 10% of the rows startup would be 10 % ...
The reasoning beeing, that you need to read a few pages before you
find the first match.
Andreas
