Let's assume that the following indices are now set on the schema from the previous task: - Level 2 B+ index on the attribute Vehicle.year Hash index on the attribute Rental.date Our DBMS has the following algorithms to perform the join: Block nested loops join Index nested loops join Translate the following SQL query into relational algebra and find the plan that reads the fewest pages or blocks: SELECT Vehicle.brand, Vehicle.model FROM Vehicle JOIN Rental ON (Vehicle.vid = Rental.vid) WHERE Vehicle.year = 2015 AND Rental.date > '31.12.2022' A schema of the car rental information system is given: Vehicle (vid, brand, model, year, no_doors, transmission); Customer (cid, name, surname, address, telephone); Rental (rid, vid, cid, eid, date); Employee (eid, name, surname, address, telephone); Additional information is given: Page on the disk = 8000 | Vehicle | = 5.000 records, 240 bytes, 20 records/page, 250 pages | Customer | = 10.000 records, 200 bytes, 40 records/page, 250 pages | Rental | = 300.000 records, 40 bytes, 200 records/page, 1500 pages | Employee | = 100 records, 200 bytes, 40 records/page, 3 pages Buffer can store 2500 pages.
Let's assume that the following indices are now set on the schema from the previous task: - Level 2 B+ index on the attribute Vehicle.year Hash index on the attribute Rental.date Our DBMS has the following algorithms to perform the join: Block nested loops join Index nested loops join Translate the following SQL query into relational algebra and find the plan that reads the fewest pages or blocks: SELECT Vehicle.brand, Vehicle.model FROM Vehicle JOIN Rental ON (Vehicle.vid = Rental.vid) WHERE Vehicle.year = 2015 AND Rental.date > '31.12.2022' A schema of the car rental information system is given: Vehicle (vid, brand, model, year, no_doors, transmission); Customer (cid, name, surname, address, telephone); Rental (rid, vid, cid, eid, date); Employee (eid, name, surname, address, telephone); Additional information is given: Page on the disk = 8000 | Vehicle | = 5.000 records, 240 bytes, 20 records/page, 250 pages | Customer | = 10.000 records, 200 bytes, 40 records/page, 250 pages | Rental | = 300.000 records, 40 bytes, 200 records/page, 1500 pages | Employee | = 100 records, 200 bytes, 40 records/page, 3 pages Buffer can store 2500 pages.
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Transcribed Image Text:Let's assume that the following indices are now set on the schema from the previous
task:
-
Level 2 B+ index on the attribute Vehicle.year
Hash index on the attribute Rental.date
Our DBMS has the following algorithms to perform the join:
Block nested loops join
Index nested loops join
Translate the following SQL query into relational algebra and find the plan that reads
the fewest pages or blocks:
SELECT Vehicle.brand, Vehicle.model
FROM Vehicle JOIN Rental ON (Vehicle.vid = Rental.vid)
WHERE Vehicle.year = 2015 AND Rental.date > '31.12.2022'
Transcribed Image Text:A schema of the car rental information system is given:
Vehicle (vid, brand, model, year, no_doors, transmission);
Customer (cid, name, surname, address, telephone);
Rental (rid, vid, cid, eid, date);
Employee (eid, name, surname, address, telephone);
Additional information is given:
Page on the disk = 8000
| Vehicle | = 5.000 records, 240 bytes, 20 records/page, 250 pages
| Customer | = 10.000 records, 200 bytes, 40 records/page, 250 pages
| Rental | = 300.000 records, 40 bytes, 200 records/page, 1500 pages
| Employee | = 100 records, 200 bytes, 40 records/page, 3 pages
Buffer can store 2500 pages.
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