EBK DATABASE SYSTEM CONCEPTS
7th Edition
ISBN: 9781260049268
Author: SILBERSCHATZ
Publisher: MCG COURSE
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Chapter 14, Problem 8PE
Explanation of Solution
Secondary B+-tree:
- Secondary B+-tree is constructed on a relation r with nr tuples.
- The cost of building the tree index when one record is inserted at a time with the given assumptions.
- The cost of insertion is the sum of below costs:
- The cost to locate the page number of the requisite leaf node. In case of insertions this cost may vary as the non-leaf nodes are already in memory.
- The cost to read one random disk access after reaching the leaf level.
- The cost to update is also one random disk access.
- The cost to write one page.
- Not that if a node split occurs due to insertion the cost increases by an extra page write.
- Now, in the worst scenario, each leaf node is just half filled. So, the number of splits is given by 2×(nr/f).
- So, the total cost is maximum of either of these two values;
- (2×nr) random disk access, or
- (nr+2×(nr/f)) page writes.
- Here, the cost of page writes is assumed to be negligible. So, the cost of random disk access is more.
- So, substitute the given values in above in the above formula to get;
- (2×nr) random disk access
- =2×10000000×10×(1/1000)sec
- =200000sec
- 55.5hours
- first sort the file in ascending order using the function which is assumed to be available.
- Now, for each pair of value and pointer from the file call the insert-in-leaf function.
- In insert-in-leaf function;
- Check if the tree is empty, if yes, this is the first root node to be inserted. If it is considered as “L”.
- Otherwise, since the values are sorted so insertion will occur in the last leaf node. So, transverse the leaf node to get the last leaf node.
- Check if this node “L” is full or not. If not, insert the given value and pointer pair at the first available location is node “L”.
- If this node is full, split it. In this case,
- Create a leaf node and assume it as “L1”.
- Set the pointer in node “L” to this node “L1”.
- Initialize “K1” to the last value from nodes chain “L”.
- Call the insert_in_parent function with desired values to insert this node in the parent...
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Chapter 14 Solutions
EBK DATABASE SYSTEM CONCEPTS
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