Question

1. a) How many instructions could fit into a 256 byte memory unit, given a 32-bit architecture?

b) How many address bits are needed to specify each byte in a 512 byte memory unit?

c) How many minterms could you have in a circuit with 3 inputs and two flip flops?

it would be nice if you draw and explain! :D

Answer 1

a)  32 bit architecture is given. There fore, 32 bits = 32/8 = 4Bytes and we have memory unit of 256 Bytes.
So number of instructions possible : 256B/4B = 64

b) Memory unit is of 512 Byte, we are asked here to give the number of bits required to address each byte.
So number of bytes = 512. Number of bits required to represent the number 512 = log₂(512) = 9 bits.
Therefore, number of address bits are needed to specify each byte in a 512 byte memory unit = 9 bits.

c) I have attached the image for the solution to this question, if doubt please revert back in the comments.