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In the 8i-style calculations, the optimizer allows 10% (rounded to an exact multiple of the block size) of the sort_area_size for write buffers This still leaves a lot missing which according to my theory is the allowance built into the model for the binary insertion tree Subtract 10% (rounded to the next complete block) from the initial value, and then subtract 25% of the rest for the sort tree: 1,048,576 bytes = 128 blocks (at 8KB); subtract 13 blocks (10%) and you get 115 blocks; subtract 25%, and you get 8625 blocks rounding up to the nearest block this is 87 blocks 712,704 bytes, which is what we see here Note, however, that this 25% allowance does not change as the row size changes (which it probably ought to if my theory were correct); nor does it change as you move from 32-bit Oracle to 64-bit Oracle. .net code 128 reader C# Code 128 Reader SDK to read, scan Code 128 in C#.NET class ...
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