The order of an internal node in a B* tree index is the maximum number of children it can have. Suppose that a child pointer takes 6 bytes, the search field value takes 14 bytes, and the block size is 512 bytes. What is the order of the internal node?
| A. | 24 |
| B. | 25 |
| C. | 26 |
| D. | 27 |
Answer : C Explanation : sunitha joshi said: (12:39pm on Thursday 4th April 2013)
how is it calculated? any formula plz suggest
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Option: A Explanation : Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. |
