Question

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?

• A. 16
• B. 20
• C. 22
• D. 24

Hint:

When calculating the number of bits for the immediate operand, subtract the bits required for the instruction type and the registers from the total instruction size.

Answer 1

The Correct Option is B

Given Data:
Number of general-purpose registers = 64
Number of instruction types = 50
Instruction format size = 32 bits

Step 1: Compute Bits for Register Specification
Since there are 64 registers, the number of bits required to uniquely identify a register is: \[ \text{Register bits} = \log_2(64) = 6 \]
Step 2: Compute Bits for Instruction Type
Since there are 50 instruction types, the number of bits required to identify an instruction type is: \[ \text{Instruction bits} = \log_2(50) \approx 6 \]
Step 3: Compute Bits for Immediate Operand
The total instruction format size is 32 bits. The remaining bits in the instruction word will be used for the immediate operand: \[ \text{Immediate operand bits} = 32 - 6 (\text{register bits}) - 6 (\text{instruction bits}) \] \[ = 32 - 12 = 20 \]
Final Answer: The maximum number of bits that can be used to store the immediate operand is \( 20 \).