Question

The memory unit of a computer has 256k words of 32 bits each. The computer has an instruction format with 4 fields: an opcode field; a mode field to specify 1 of 7 addressing modes; a register address field to specify 1 0f 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:
A: How large must the mode field be?

B: How large must the register field be?

C: How large must the address field be?

D: How large is the opcode field?

Answer 1

Answer:

A. 3 bits required

B. 6 bits are required

c. 8 bits are required

d. 15 bits are required

Explanation:

A. In the address mode selector, we need to specify 1 out of 7 addressing modes. Therefore, there must be 3 bits (2³ = 8)

B. There are 60 address registers. Therefore the bits to represent 60 numbers are 6 bits. (2^6 = 64)

C. As indicated, the memory bit word required is 8 bits.

D. Total Bits = Opcode + Address mode + Register Add + Memory ADD

32 bits = Opcode+ 3 bits + 6 bits + 8 bits

Opcode =  32 bits - 17 bits = 15 bits.

Therefore 15 bits are required for Opcode field.