GATE2025_CS1_Q37

A processor has 64 general-purpose registers and 50 distinct instruction types. An instruction is encoded in 32-bits. What is the maximum number of bits that can be used to store the immediate operand for the given instruction?

ADD R1, #25      // R1 = R1 + 25

✅ Final Answer:

The correct answer is option (B) 20.

🔹 Short Answer:

The 32-bit instruction must be partitioned to encode the opcode, register, and immediate value. To maximize bits for the immediate value, we minimize bits for other fields.
• Bits for 50 opcodes = ⌈log₂(50)⌉ = 6 bits.
• Bits for 64 registers = ⌈log₂(64)⌉ = 6 bits.
• Remaining bits for immediate operand = 32 - 6 - 6 = 20 bits.

🔸 Long Answer:

This problem requires us to understand how an instruction is encoded into a fixed-size binary word. The total instruction size is 32 bits. These 32 bits must be divided among all the necessary fields of the instruction. For the given instruction ADD R1, #25, the fields are:

1. Opcode: Represents the operation (e.g., `ADD`).
2. Register Operand: Specifies the destination/source register (e.g., `R1`).
3. Immediate Operand: The constant value used in the operation (e.g., `#25`).

To find the maximum possible size for the immediate operand, we must allocate the minimum necessary number of bits for the other fields.

Step 1: Calculate Bits for Opcodes

There are 50 distinct instruction types. To uniquely represent each one, we need to find the smallest integer `k` such that 2^k ≥ 50.

• 2⁵ = 32 (Not enough)
• 2⁶ = 64 (Sufficient)

Therefore, we need a minimum of 6 bits to encode the opcode.

Step 2: Calculate Bits for Registers

The processor has 64 general-purpose registers. To uniquely address each one, we need to find the smallest integer `k` such that 2^k ≥ 64.

• 2⁶ = 64 (Exactly enough)

So, we need exactly 6 bits to encode the register operand.

Step 3: Calculate Remaining Bits for the Immediate Operand

We subtract the bits allocated for the opcode and the register from the total instruction size.

Available bits = Total Bits - Opcode Bits - Register Bits

Available bits = 32 - 6 - 6 = 20 bits.

Instruction Format Visualization

The 32-bit instruction can be visualized as being partitioned like this:

The maximum number of bits that can be reserved for the immediate operand is 20.

Bibliography:
- Patterson, D. A., & Hennessy, J. L. (2016). Computer Organization and Design: The Hardware/Software Interface. Morgan Kaufmann. (Chapter 2: Instructions: Language of the Computer).
- Stallings, W. (2018). Computer Organization and Architecture: Designing for Performance. Pearson.

Ask your doubts in comment form our GURUJEE will reply ASAP, click notify me button so that you get notified on reply.