GATE CS 2025 Set 1 Q27 — Machine Instructions and Addressing Modes

To find the maximum number of bits available for the immediate operand, we analyze the fields of the instruction format based on the given example ADD R1, #25:

• Opcode Field: There are $50$ distinct instruction types. The number of bits required for the opcode is: $$\lceil \log_2(50) \rceil = 6\text{ bits}$$
• Register Field: There are $64$ general-purpose registers. The number of bits required to address one register is: $$\log_2(64) = 6\text{ bits}$$
• Immediate Operand Field: Let this require $x$ bits.

The total instruction length is $32$ bits. Since the instruction ADD R1, #25 references only one register ($R1$ acts as both source and destination) and one immediate value, we have:

Thus, the maximum number of bits that can be used for the immediate operand is $20$.