Question

Assume a byte-addressable memory consisting of 1G (230) bytes. A Direct Mapped Cache contains 256 (28) blocks of 2048 (211) bytes each.

Therefore the 30 bits address is partitioned into three areas of the following sizes (in bits):

offset	bits
block	bits
tag	bits

Answer 1

• The size of physical address is 30 bits.

• Each block contains 2¹¹ bytes, so the lowest 11 bits of physical address, denote offset.
• There are 2⁸ blocks, so the middle 8 bits of physical address, denote the index/block .
• Remaining (30 - 11 - 8 = ) 11 highest bits of physical address, denote the tag.

In summary :

• offset => 11 bits
• block => 8 bits
• tag => 11 bits

Following is the physical-address fabrication :

Tag	Block	Offset
A29 - A19	A18 - A11	A10 - A0