Exercise 5: Space Utilisation
Consider the following page/record information:
- page size = 1KB = 1024 bytes = 210 bytes
- records:
(a:int,b:varchar(20),c:char(10),d:int)
- records are all aligned on 4-byte boundaries
-
c field padded to ensure d starts on 4-byte boundary
- each records has 4 field-offsets at start of record (each 1 byte)
-
char(10) field rounded up to 12-bytes to preserve alignment
- maximum size of
b values = 20 bytes; average size = 16 bytes
- page has 32-bytes of header information, starting at byte 0
- only insertions, no deletions or updates
Calculate C = average number of records per page.
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