Secondary B + -tree: Secondary B + -tree is constructed on a relation r with n r 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×(n r /f). So, the total cost is maximum of either of these two values; (2×n r ) random disk access, or (n r +2×(n r /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×n r ) 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...

Database System Concepts

7th Edition

ISBN: 9789332901384

Author: Henry F. Korth & S. Sudarshan Abraham Silberschatz

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 14, Problem 8PE

Explanation of Solution

Secondary B⁺-tree:

Secondary B⁺-tree is constructed on a relation r with n_r 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×(n_r/f).

So, the total cost is maximum of either of these two values;

(2×n_r) random disk access, or

(n_r+2×(n_r/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×n_r) 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, Problem 7PE

Chapter 14, Problem 9PE

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