[SOLVED] adressability of memory system of a computer is 2 bytes.I need 18 bits to access a location in memory.What is the total size of the memory in bytes?

adressability of memory system of a computer is 2 bytes.I need 18 bits to access a location in memory.What is the total size of the memory in bytes?

Title is a question for my Computer Architecture homework. My problem is i can't figure out the correct units for the formula.

I know the formula for max memory size should be:

TotalSizeOfMemory = Addressability x 2^NumberOfBits

My question is, should the addressability be expressed in bytes or bits? And is the result expressed in bits or bytes? thanks

Solution

My question is, should the addressability be expressed in bytes or bits?

It is reasonable to specify the unit of addressability in bytes or bits (or both), though when the unit of addressability is 2 bytes aka 16 bits, many would commonly call that a "Word" of memory (to differentiate from the standard 8-bit Byte).

And is the result expressed in bits or bytes?

1 Word per Location × # Locations = # of Words
2 Bytes per Location × # Locations = # of Bytes
16 Bits per Location × # Locations = # of Bits

Notice how the unit Location cancels out, because it is in the denominator for the first factor (units per Location) and in the numerator for the second factor — thus, the unit that is left for the whole formula is simply the number of words, bytes, or bits, depending on how you choose to express the unit of addressability.