CS486586 Introduction To Databases Fall 2017 Quarter Assignm

Cs486586 Introduction To Databasesfall 2017 Quarterassignment 6 Disks

Cs486586 Introduction To Databasesfall 2017 Quarterassignment 6 Disks

Analyze disk access times for different data arrangements and evaluate the ability of indexes to answer various queries without accessing actual data records. Also, explore the implementation adaptations required for hash-join algorithms to perform outer joins effectively when one of the tables does not fit entirely into memory.

Paper For Above instruction

Database systems often need to estimate disk access times to optimize query performance, especially when reading large datasets. Disk I/O can be a significant bottleneck, and understanding the factors influencing access time is essential for designing efficient storage and retrieval strategies. Additionally, the use of indexes can sometimes eliminate the need to access raw data records, thus reducing I/O and improving query response times. Finally, understanding how to modify join algorithms like hash joins to handle outer joins is crucial when implementing efficient query processing, especially in cases where datasets are too large to fit entirely into memory.

Part I: Disk Access Considerations

The characteristics of a hard disk—such as average seek time, rotational delay, and the number of surfaces—directly influence the time required to read data. The seek time (5 ms) is the average time to position the read/write head over the correct track, while the rotational delay (3 ms) is the average time for the desired sector to rotate under the head.

Question 1

Estimate the time to read a list of 150 pages randomly distributed across the disk.

When pages are randomly distributed, each page may require a separate seek and rotational delay. For each page, the total time is approximate to the sum of seek time and rotational delay, in addition to the transfer time. Assuming the transfer time per page is negligible compared to seek and rotational delays, the estimated time is:

  • Seek time per page: 5 ms
  • Rotational delay per page: 3 ms
  • Total per page: 8 ms

For 150 pages, the total estimated disk time is:

150 pages * 8 ms = 1,200 ms or approximately 1.2 seconds.

Question 2

Estimate the time to read 150 pages randomly distributed within tracks of a single cylinder.

If pages are located within the same track (cylinder), the seek time is minimized—often negligible—as the head already positions over the track. The rotational delay dominates, but since pages are within the same track, the head does not need to move between tracks. The average rotational delay remains 3 ms, as the desired sector can be expected to pass under the head within this time.

Assuming sequential reading within a track and minimal seek overhead, the total time approximates to:

  • Seek time: negligible (close to 0 ms)
  • Rotational delay: 3 ms (average)
  • Data transfer time: depends on the number of pages and track size; assuming negligible transfer time or that it scales linearly with number of pages, the main delays are seek and rotational. For approximation, total time is about 150 pages * 3 ms = 450 ms.

Thus, the estimated time is approximately 450 ms, significantly faster than entirely random distributions.

Question 3

Estimate the time to read 150 pages arranged consecutively in tracks of a single cylinder.

If pages are arranged consecutively in tracks, the access can be sequential, drastically reducing rotational delays. The head would need to reposition only once, and subsequent pages can be read with minimal seek time.

The total time in this case involves:

  • Initial seek time to position the head: ~5 ms
  • Rotational delay: negligible once aligned, as sequential reading minimizes waiting
  • Data transfer: proportional to the number of pages, assuming a high transfer rate that makes transfer time negligible compared to seek and rotational delays

Considering only the seek time and minimal rotational delay, the total estimated time is approximately 5 ms plus the transfer time, which might be around 150 pages * negligible transfer per page, say less than 1 ms per page, summing to about 150 ms. Overall, the total is roughly 155 ms.

Question 4

To speed up reading times for questions 1 and 2 when order does not matter, the following strategies are advised:

  • Batch reading: Read large chunks of data, minimizing head movements and rotational delays by sequentially accessing data.
  • Pre-fetching: Use anticipatory reading strategies to load subsequent pages into cache before they are requested.
  • Optimized data layout: Arrange data sequentially or within minimal seek zones, such as within the same track or cylinder.
  • Reduced seek operations: Group pages that are spatially close, decreasing the number of seeks needed.

Part II: Evaluating Queries with Just Indexes

Indexes are crucial for optimizing query execution, especially in large databases. Properly designed indexes can sometimes allow query evaluation without accessing the full data records, significantly reducing I/O. Here, three unclustered indexes are considered, each with specific search keys associated with the tables Pokemon and Captured.

Question 5

SELECT MAX(difficulty) FROM Captured;

This query seeks the maximum difficulty value across all entries. Since max aggregate can be computed by scanning the index on difficulty, this query can be answered solely using information from the index with . By scanning this index, we identify the maximum difficulty value, without accessing the full table.

Question 6

SELECT AVERAGE(difficulty) FROM Captured GROUP BY player;

This aggregate with a grouping operation depends on access to the difficulty values linked to each player. The index on can facilitate grouping by retrieving difficulty for each charName, then aggregating it per player. Since player attribute is in Captured, retrieving the index and then joining with the table to map charNames to players is necessary. However, with suitable index and minimal data access, this can be estimated or computed directly from the index by scanning all entries, then associating each with the corresponding player.

Question 7

SELECT charName, MIN(difficulty) FROM Captured GROUP BY charName;

The index on is appropriate here. By scanning the index, for each charName, the minimum difficulty can be found efficiently, with no need to access the full table.

Question 8

SELECT COUNT(*) FROM Pokemon WHERE stamina = 85 AND cType = 'Fire';

Since indexes are on and , and both conditions are on different attributes, the index on helps filter entries with stamina=85, and then filtering these on cType using the index on cType can further narrow the result set. Combining indexes (via bitmap or other techniques) allows evaluation without accessing full records.

Question 9

SELECT cType, COUNT(charName) FROM Pokemon GROUP BY cType;

The index on allows grouping by cType directly, enabling counting of charName entries per cType by scanning the index and aggregating counts, avoiding full table scans.

Question 10

SELECT pType, difficulty FROM Pokemon, Captured WHERE stamina = 90 AND Pokemon.charName = Captured.charName;

This join condition can utilize the indexes on (from Pokemon) and (from Captured). The process involves filtering Pokemon entries with stamina=90 using the index, then joining with Captured based on charName, which is present in the index on . This approach can be achieved using index lookups and join operations without scanning entire tables.

Question 11

SELECT COUNT(DISTINCT stamina) FROM Pokemon WHERE cType = 'Fire';

The index on can facilitate extracting distinct stamina values for Fire-type Pokémon by scanning the index entries where cType='Fire' (assuming a composite index on cType and an attribute like stamina, or using two indexes combined). This provides the count without accessing full records.

Question 12

SELECT charName, AVERAGE(difficulty) FROM Captured GROUP BY charName HAVING COUNT(DISTINCT player) > 1;

This involves grouping by charName, calculating average difficulty, and filtering on the count of distinct players. Using the index on , one can gather difficulty values per charName efficiently. The count of distinct players may require additional data access, as the index doesn't contain player data; hence, full data access is necessary or auxiliary data structures to track distinct players per charName.

Part III: Operator Implementation

Outer joins extend inner joins by including all records from one table and matching records from the other, with nulls where no match exists. Implementing a RIGHT OUTER JOIN using hash join techniques requires modifications to the typical inner join algorithm, especially when the inner table does not fit entirely into memory.

Question 13

In a hash-join where Captured is the inner relation fitting in memory, the standard approach constructs a hash table on Captured's join attribute, charName. To adapt this to produce the additional rows for a right outer join, after processing each matching inner record, the algorithm must identify and include all outer table records (Pokemon) that do not have matches in Captured.

  • Initially, build a hash table on Captured’s charName values.
  • For each Pokemon record, probe the hash table; if a match exists, output the combined row, indicating a match.
  • After processing all Pokemon records, create a complete set of Pokemon records which may lack matches in Captured.
  • For each unmatched Pokemon record, generate a row with NULLs for Captured attributes, ensuring all outer records are included.

This approach requires maintaining a separate list of unmatched Pokemon records or marking which Pokemon do not have a matching Captured record during the probe phase, then outputting these after the main hash join completes.

Question 14

When the outer relation (Pokemon) fits in memory but the inner relation (Captured) does not, the hash join can be adapted similarly by building a hash table on the inner relation (Captured), as in standard hash join algorithms. To produce the full outer join result, the algorithm must also include unmatched outer records.

  • Partition the Captured relation into chunks that fit into memory, building a hash table on each chunk sequentially.
  • For each chunk, create a hash table keyed on charName.
  • Then scan the entire outer relation (Pokemon); for each Pokémon, probe the hash table. If an inner match is found, output a combined row; if not, output a row with NULLs for Captured attributes.
  • To handle unmatched Pokémon records, after processing all chunks of the inner relation, keep track of Pokémon records not matched and output them with NULLs for the inner relation's data.

This process involves maintaining a list or marking flags on Pokémon records during the join process to identify whether they have matching Captured records. After processing all inner chunks, output the unmatched Pokémon records with nulls, ensuring the outer join property.

References

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