6: Costs in File Handling Flashcards
How are files stored on a disk?
Similar to a relational table, they also have indexes associated with them, and are stored on the disk to improve efficiency of retrieval.
What are the steps for storing a file or index on a disk ?
- Formatting disk
- Dividing disk into pages (equal size blocks, between 512 to 4096 bytes)
- The block forming the unit of transfer between disk (secondary storage) and main memory (primary storage in buffers). Buffer size = 1 block.
Can hold more from secondary storage - ‘cluster’ - several contiguous blocks on disk which can be transferred as a single unit to be held in one buffer. This can be done by adjusting the byte-size of the buffer to match the byte-size of the cluster. (magic)
What is spanned organisation ?
When part of a record is stored in one block and a pointer is used within the block to point to the rest of the record in another block.
Used for variable length records.
What’s an unspanned organisation ?
When a record must fit entirely within one block ie record-size cannot exceed block-size.
Used for fixed-length records, ie when B>=R. (Can also be used for variable-length records).
How is data held in a block on a disk found ?
By using an address, that specifies the:
- Surface number
- Track number (within the surface)
- Block number (within the track)
How long does it take to retrieve a block and store it in a buffer ?
It takes the sum time of the following actions:
- Seek time: taken to move read/write head over the track required in the address
- Rotational Delay (latency): time taken for the beginning of the block required in the address to rotate into position under the read/write head.
- Block Transfer time: taken to transfer the data block to the buffer
Total retrieval time = seek time + latency + block transfer time.
Usually seek time and latency are much larger than block transfer time.
What are the factors to consider when calculating the costs of storing records?
- Block Size B [bytes]
- Record Size R [bytes]
- Blocking Factor B/R [bfr] ie the number of records in a block, rounded down
How many blocks are needed to store the file on disk? Given:
- Block Size (B) = 512 bytes
- Record Size (R) (fixed length) = 128 bytes
- Relation (‘File’) Size (F) = 50 records
50*128 = 6400 bytes in total required.
6400/512 = 12.5 blocks required => 13 in total, 12 full blocks holding 4 records per block, plus 1 block extra to hold the remaining 2 records.
512/128 = 4 records in one block 50/4 = 12.5 = 13 blocks
How many blocks are required to store an index on disk for the relation given?
- Block Size (B) = 512 bytes
- Record Size (R) (fixed length) = 128 bytes
- Relation (‘File’) Size (F) = 50 records
There is one index entry per block:
Key-Field Size = 20 bytes
Pointer-Field Size = 10 bytes
20 + 10 = 30 extra bytes per block.
We have 13 blocks: 30*13 = 390 bytes.
390/512 = 0.762 => 1 block
128+30 = 158 bytes
158*50 = 7900
7900/512 = 15.5 bytes/block
Therefore, we need 15.5-12-5 = 3 records to store an index of 30 bytes per block.
3/4 = 0.75 => 1 block.
What is the cost (in terms of blocks) to store the following file on disk:
Block Size = 512 bytes Record Size (fixed length) = 75 bytes Relation Size = 325 records
75*325 = 24375 bytes
24375/512 = 47.6 => 48 blocks required.
512/75 = 6.826 records per block => 6 rounded DOWN 325/6 = 54.1 => 55 blocks rounded UP
What is the cost (in terms of blocks) to store the following file on disk:
Block Size = 512 bytes Record Size (fixed length) = 75 bytes Relation Size = 325 records
Index Entry = 1 per block Key-Field size = 25 bytes Pointer-Field Size = 10 bytes
55 * 35 = 1925
1925/512 = 3.7597 => 4 blocks
