Question

A machine has a 32-bit architecture with 1-word long instructions. It has 24 registers and supports an instruction set of size 40. Each instruction has five distinct fields, namely opcode, two source register identifiers, one destination register identifier, and an immediate value. Assuming that the immediate operand is an unsigned integer, its maximum value is \(\_\_\_\_\).

Hint:

To design instruction formats, allocate bits logically for each field and calculate remaining bits for operands like immediate values. Use \(\lceil \log_2(n) \rceil\) to determine the required bits.

Answer 1

Step 1: Determine the bit allocation for instruction components.
The instruction is 32 bits long, divided among the following fields: Opcode: With 40 supported instructions, the number of bits required is: \[ \lceil \log_2 40 \rceil = 6 \, {bits}. \] Source Registers: Each register requires \(\lceil \log_2 24 \rceil = 5\) bits. For two source registers: \[ 2 \times 5 = 10 \, {bits}. \] Destination Register: Also requires \(\lceil \log_2 24 \rceil = 5 \, {bits}. \] Immediate Value: Occupies the remaining bits in the instruction. Step 2: Compute the bits available for the immediate value.
The total bits allocated to the opcode and registers are: \[ 6 + 10 + 5 = 21 \, {bits}. \] The remaining bits for the immediate value are: \[ 32 - 21 = 11 \, {bits}. \] Step 3: Determine the maximum value of the immediate operand.
An 11-bit unsigned integer can represent values up to: \[ 2^{11} - 1 = 2047. \] Final Answer: \[ \boxed{{2047}} \]