The order of an internal node in a…


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

Ravi Garg edited question
    Ganesh Nimbolkar (anonymous)

    Madhav Purohit but they are asking order so m order tree have m-1 keys in a node so as you are getting 25 which is number of keys in node hence 26 will be order of tree ????

    Ganesh Nimbolkar answered
      Ganesh Nimbolkar (anonymous)

      C) 26

      Ganesh Nimbolkar answered
        Madhav Purohit (anonymous)

        Actually number of child pointer is 1 more than the search field so
        let x be number of search key and x+1 be child ptrs

        Now we get14x+(x+1)6=512

        Solve ths u get x=25

        Madhav Purohit answered
          Ravi Garg (anonymous)

          The correct answer is c ๐Ÿ™‚

          Ganesh Nimbolkar.. you are right

          Ravi Garg answered
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