Consider a hypothetical 32-bit microprocessor having 32-bit instructions composed of two fields the first byte contains the opcode and the remainder the immediate operand or an operand address What?

Answer 1

• What is the maximum directly addressable memory capacity (in bytes)? 2(32-8) = 16,777,216 bytes = 16 MB • Discuss the impact on the system speed if the microprocessor data bus has • a 32-bit local address bus and a 16-bit local data bus. Instruction and data transfers would take three bus cycles each - one for the address and two for the data. • a 16-bit local address bus and a 16-bit local data bus. Instruction and data transfers would take four bus cycles each - two for the address and two for the data. • How many bits are needed for the program counter and the instruction register? 24 bits for the PC (24-bit addresses), 32 bits for the IR (32-bit addresses)