Word addressable memory

Question

If the memory size would be 512KB then should we devide by 32?

Answer 1

32 bits instruction size

1024 different operations so 10 bits to represent it

512k to address word so 9+10=19 bits to address words

1word = 32 bits = 4 bytes----so 19+2 =21 to address byte

so left bits are 32-10-21 = 1 bit to represent the registers

Answer 2

Given that 512 k words memory i.e 2^19 words (locations) thus 19 bits required to address a word.

Also given that 1024 operations thus 10 bits for opcode

Therefore, number of bits used for register equals 32 - 19 - 10 = 3 bits