GATE CS 2014 Set 1 Q9 — Machine Instructions and Addressing Modes

To find the maximum value of the immediate operand, we need to determine the maximum number of bits that can be allocated to it within the 32-bit instruction format.

Let's break down the instruction fields:

1. Opcode Field:
   The machine needs to support $45$ distinct instructions.
   $$\text{Opcode bits} = \lceil \log_2(45) \rceil = 6 \text{ bits}$$
2. Register Operands:
   There are $64$ registers, so each register operand requires:
   $$\text{Register bits} = \log_2(64) = 6 \text{ bits}$$
   Since the instruction contains two register operands, they require a total of:
   $$2 \times 6 = 12 \text{ bits}$$
3. Immediate Operand Field:
   The total instruction length is $1 \text{ word} = 32 \text{ bits}$.
   The bits remaining for the immediate operand ($I$) are:
   $$I = 32 - (\text{Opcode bits} + \text{Register bits})$$
   $$I = 32 - (6 + 12) = 14 \text{ bits}$$

Since the immediate operand is an unsigned integer, the maximum value it can represent with $14$ bits is: