Suppose a relation is stored in a B+-tree file organization. Suppose secondaryindices store record identifiers that are pointers to records on disk. What would be the effect on the secondary indices if a node split happened in the file organization?
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Suppose a relation is stored in a B+-tree file organization. Suppose secondary
indices store record identifiers that are pointers to records on disk. What would be the effect on the secondary indices if a node split happened in the file organization?
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- A B+-tree is to be stored on disk whose block size is 2048 bytes. to be stored are 64 bytes, and their key is 24 bytes. The data records Determine the values for M and L for the B+-tree. Assume pointers are 4 bytes each. oints For the problem above, in the worst case, how many levels are needed to store 16,000,000 records?Suppose a relation is stored in a B+-tree file organization. Suppose secondaryindices store record identifiers that are pointers to records on disk. a. What would be the effect on the secondary indices if a node split happenedin the file organization?b. What would be the cost of updating all affected records in a secondaryindex?c. How does using the search key of the file organization as a logical recordidentifier solve this problem?d. What is the extra cost due to the use of such logical record identifiers?Suppose a relation is stored in a B+-tree file organization. Suppose secondaryindices store record identifiers that are pointers to records on disk. What would be the cost of updating all affected records in a secondary index?
- Suppose a relation is stored in a B+-tree file organization. Suppose secondaryindices store record identifiers that are pointers to records on disk. What is the extra cost due to the use of such logical record identifiers?Consider a table T in a relational database with a key field K. A B-tree of order p is used as an access structure on K, wherep denotes the maximum number of tree pointers in a B- tree index node. Assume that K is 10 bytes long; disk block size is 512 bytes; each data pointer PD is 8 bytes long and each block pointer Pg is 5 bytes long. In order for each B- tree node to fit in a single disk block, the maximum value of p is_?Suppose a relation is stored in a B+-tree file organization. Suppose secondaryindices store record identifiers that are pointers to records on disk. How does using the search key of the file organization as a logical record identifier solve this problem?
- The statement "using no index at all is probably a better idea then using a tree-based index" is true if: Select one: a. The indexed relation is fairly static O b. The indexed relation is so small it can fit in the main memory buffer c. The statement is incorrect, i.e., it is always beneficial to use an index O d. Alternative (a) is correct only if one assumes no bufferingIf a B+tree file header record is 256 bytes, and sequence set blocks are 100 bytes, what is the byte offset for a RBN of {N} (assuming the first sequence set block has an RBN of 1)?Assume one file has r = 10^6 records. Each record takes R = 100 bytes, of which 10 bytes are for the key of record. Suppose the key balues range from 1 to 1,000,000, inclusive. Assume the block size B is 1000 bytes for all files, and that an address (block pointer, tree node poiinter, or data record pointer) takes 10 bytes. What is the total number of blocks required by the B+-tree?
- Computer Science Propose a relation schema for storing the access rights associated with user groups in a distributed database catalog, and give a fragmentation scheme for that relation, assuming that all members of a group are at the same site.Compute total no of blocks} Consider a file of 8192 records. Each record is 16 bytes long and its key field Is of size 6 bytes. The file is ordered on a key field, and the file organization is unspanned. The file is stored in a file system with block size 512 bytes, and the size of a block pointer is 10 bytes. If the primary index is built on the key field of the file, and a multilevel index scheme is used to store the primary index, number of first-level and second level blocks in the multilovel index are the respectively 16 and 1 b. 32 and 1 16 and 2 8 and 1 c. d.Consider a file of 16384 records. Each record is 32 bytes long and its key field is of size 6 bytes. The file is ordered on a non-key field, and the file organization is unspanned. The file is stored in a file system with block size 1024 bytes, and the size of a block pointer is 10 bytes. If the secondary index is built on the key field of the file, and a multilevel index scheme is used to store the secondary index, the number of first-level and second-level blocks in the multilevel index are respectively