Final answer:
The maximum value of the immediate operand in a machine with a 32-bit architecture, 1-word instructions, and 64 registers supporting 45 instructions is 16383.
Step-by-step explanation:
The student asked about the maximum value of the immediate operand in a 32-bit architecture machine which supports 1-word long instructions with 64 registers. Each register is 32 bits long, and there are 45 instructions that require an immediate operand along with two register operands. To calculate the maximum value of the immediate operand, we need to divide the instruction word into different parts: one part for the instruction code, two parts for the registers, and one part for the immediate operand.
Since there are 64 registers, we need 6 bits to uniquely identify each register (since 2^6 = 64), so for two registers we need 2 * 6 = 12 bits. To identify 45 instructions, we need at least 6 bits (since 2^5 = 32 and 2^6 = 64, which is the smallest power of 2 greater than 45). With 6 bits for the opcode and 12 bits for the registers, we've used 18 bits, leaving 14 bits for the immediate operand (32 - 18 = 14).
Therefore, the maximum value of the immediate operand is 2^14 - 1, which is 16383, as the operand is an unsigned integer.