How to solve memory address problems

Question

A byte is 8 bits. If it's byteaddressable, you can't reference an address by anything other than the start of some 8 bits. That is, in a 2^2 byte memory, you have 4 bytes. The lowest address starts at 0 bytes, and the highest address starts at 3 bytes. (0, 1, 2, 3 = 4 bytes total)

If the bytes are contiguous (they are juxtaposed- touching each other rather than spread out) then you can fit all 4 bytes into a 4 byte memory perfectly.

a)

If you have 2^24 bytes then you have 2^(24 + 3) bits because you're doing (2^24 * 2^3) = 2^(24+3). Thus you have 134,217,728 total bits.

The highest address would be one byte before the end, so the address at 2^24 - 1. Note that it's 2^24 - 1 and not 2^27 - 1 because you are addressing it by bytes and not bits. Lowest address would be 0.

Lowest address = 0

Highest address = 2^24 - 1

b)

A word just means a grouping of bytes. A 1-byte word is literally the same thing as a byte, it just implies that the word is some meaningful piece of data, whereas a byte is not necessarily a meaningful piece of data.

A 16-bit word == a 2-byte word because 8 bits are in a byte, thus if you have 2^24 bytes available, you only have a total of 2^23 words.

Lowest address = 0

Highest address = max number of words - 1 = 2^23 - 1.

c)

Same thing as with a 4-byte word instead of 2. Thus:

2^22 bytes available to store words.

Lowest address = 0

Highest address = max number of words - 1 = 2^22 - 1.

Feel free to correct me if you see any errors. Hope I helped.