A processor has 40 distinct instructions and 24 general purpose registers. A…

Key idea: count how many bits are needed for the opcode and for the two register operands, then subtract from 32 to get the immediate field size.

• Opcode bits: need enough bits to encode 40 instructions, so bits = ceil(log2(40)) = 6 bits (since 2^5 = 32 < 40 ≤ 64 = 2^6).

• Register operand bits: need bits to encode 24 registers, so bits per register = ceil(log2(24)) = 5 bits (2^4 = 16 < 24 ≤ 32 = 2^5). Two register operands use 2 × 5 = 10 bits.

• Immediate bits: 32 (total instruction length) − 6 (opcode) − 10 (two registers) = 16 bits.

Answer: The immediate operand field is 16 bits long.