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A machine has a $32\text{-bit}$ architecture, with $1\text{-word}$ long instructions. It has $64$ registers, each of which is $32$ bits long. It needs to support $45$ instructions, which have an immediate operand in addition to two register operands. Assuming that the immediate operand is an unsigned integer, the maximum value of the immediate operand is ____________
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16,383.

32 bit architecture means word size of CPU is 32 bit, which means instruction size = 32 bit.

It is given that register memory size is 64 register, to access 1 register 6 bits are required(log 64 base2)

The size of ISA(instruction set architecture) = 45 which require 6 bit for opcode(as 5 won’t be sufficient)

So, now the feilds of instruction will have bifurcation like this,

opcode+register1+register2+Immediate value = 32  bit

Immediate value = 32 – (6+6+6) = 14

Now, Immediate value is given to be unsigned and the range of unsigned is 0 to 2^(n) – 1

So maximum unsigned value that can be represented in this instruction = 2^(14) – 1 = 16,383.

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  1. Bit Allocation:
  • Opcode: To support 45 instructions, we need 6 bits for the opcode ($2^6 = 64,$ which is greater than $45$).
  • Register Operands: To address 64 registers, we need 6 bits for each register operand ($2^6 = 64$). So, two register operands require 12 bits.
  • Immediate Operand: The remaining bits in the 32-bit instruction word are available for the immediate operand. This leaves $32 - 6 - 12 = 14$ bits for the immediate operand.
  1. Maximum Value:
  • With 14 bits, the maximum unsigned integer value that can be represented is 2^14 - 1 =  16383

Therefore, the maximum value of the immediate operand in this machine architecture is $16383$.

Answer:

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