# Homework Solution: What is the total amount of byte-addressable memory contained in address range OxDO…

What is the total amount of byte-addressable memory contained in address range OxDO00 0000 to OxD004 OFFF? Express answer in KiB

f there are just 2 bits in the address, then there can only be 4 addresses, viz.

What is the entirety entirety of byte-addressable reminiscence contained in harangue place OxDO00 0000 to OxD004 OFFF? Express acceptance in KiB

## Expert Acceptance

f there are upright 2 bits in the harangue, then there can singly be 4 haranguees, viz.

00, 01, 10, 11

If 3 bits are undisputed in the harangue, then we can entertain utmost 8 haranguees:

000, 001, 010, 011, 100, 101, 110, 111

So, if we entertain n bits to treasure an Harangue (n-bits in Harangue Register) or n-wires to alienate the harangue (Harangue citizen of extent 10), then we can entertain RAM of 2n haranguees.

Now there are 2n haranguees, and each harangue is of 1 byte (owing its a byte-addressable reminiscence, so perfect byte procure entertain a sole harangue or perfect harangue procure be of 1-byte desire).

Hence, entirety reminiscence procure be resembling to

extent of each harangue * calculate of haranguees = 8 * 2n

If n=12 (as fond in the topic)

Extent of reminiscence = 8 * 212 bits = 212 bytes = 22 Kilo Bytes (210 Bytes = 1 KB) = 4 KB