Answer:
Offset bits: 3-bits
Set number of cache: 12-bits
Tag bits: 7-bits
22-bit physical address
Explanation:
Since the system is 32K so,
=2⁵.2¹⁰
=2¹⁵
As we know that it is 8-way set associative so,
=2¹⁵/2³
=2¹⁵⁻³
=2¹²
2¹² are cache blocks
22-bit physical address
Off-set bits are 3 as they are calulated from 8-way set associative information.
Set number of cache : 12-bits
For tag-bits:
Add off-set bits and cache bits and subtract from the total bits of physical address.
=22 - (12+3)
=22 - 15
=7
I need the options to help you out, thanks. :)
Answer:
The correct option is;
A. The file will not take up as much space on the database
Explanation:
Changing the field size of a field that does not contain data limits the size of taken up by the data values added to the field. The field size for text field determines the disc space that each value of the field allowed to take up by Access
When the field size is changed for fields containing data, the data values in the field that have values higher than the maximum field size are truncated while the field size of subsequent data are limited as stated above.
