solution(with explanation) Question 1 [4 marks] Which of the following is a fals
ID: 3744303 • Letter: S
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solution(with explanation)
Question 1 [4 marks] Which of the following is a false statement about B+ trees? A. B+-trees are balanced B. non-leaf nodes include direct pointers to data records C. insertion of a key can lead to node splitting D. deletion of a key can lead to node coalescing Question 2 [4 marks] Which of the following factors determines the size of a bitmap index on an attribute "X" in relation "R"? A. The number of distinct values in "X" B. The number of tuples in "R" C. The data type for attribute "X" D. Answers A & B above E. Answers A &B&C Question 3 [4 marks] Which of the following is a false statement about cost- based query optimization? A. It selects a query plan in a shorter time than heuristic-based optimization B. It requires estimating the execution cost of a query plarn C. It selects the query plan with the minimum execution cost D. All of the aboveExplanation / Answer
1)D
a)True,in B++ trees length from root to leafs are equal so B++ tree is balanced
b)True,In B++ trees data nodes are stored at last level of the tree,So nonleafs contains pointers to data records
c)True,in all the cases the nodes dont spilt but when the no of nodes is greater than the order it leads to spliting
d)False,Because delelting of a node dont causes other nodes to merge
2)A
In bitmap index choosing choosing indexing on unique column is better because underling index must be larger than a corresponding B-tree,So here we will index on "No of distinct values in X".
3)A
Because a cost based optimizer will find out all of the possible ways in which a query can be execute and each way will be assigned a ‘cost’, which indicates how efficiently that query can be run. Then, the cost based optimizer will select the way with least cost and execute the query using that way, because that is the most efficient way to run the query.
4)A
In Bitmap Indexing bits are used to represent the unique values,as we have four rows in the above table,Here the price column has only two unique values we use two bits for representation
5)A,
Here we have two values so we use 1100 bitmap to represent HONDA
6)To get records that are greater than 30 and less than 75 first we have to visit N1 as it is a root and also N3 as it is a first nonleaf node at level 1 and next N7 followed by N8 and N9 these are only nodes which contains the elements that are greater than 30 and less than 75.
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