A processor uses a 32-bit instruction format and supports byte-addressable memory access. The ISA of the processor has 150 distinct instructions. The instructions are equally divided into two types, namely R-type and I-type, whose formats are shown below. R-type Instruction Format: I-type Instruction Format: In the OPCODE, 1 bit is used to distinguish between I-type and R-type instructions and the remaining bits indicate the operation. The processor has 50 architectural registers, and all register fields in the instructions are of equal size. Let $X$ be the number of bits used to encode the UNUSED field, $Y$ be the number of bits used to encode the OPCODE field, and $Z$ be the number of bits used to encode the immediate value/address field. The value of $X + 2Y + Z$ is _________

1. OPCODE Field ($Y$):

   • There are 150 instructions, equally divided into R-type (75 instructions) and I-type (75 instructions).

   • 1 bit distinguishes between the two types.

   • The remaining bits must uniquely identify 75 operations: $\lceil \log_2(75) \rceil = 7$ bits.

   • Total bits in the OPCODE field: $Y = 1 + 7 = 8$ bits.

2. Register Fields:

   • The processor has 50 architectural registers.

   • Each register field requires: $\lceil \log_2(50) \rceil = 6$ bits.

3. UNUSED Field ($X$) in R-type Instruction:

   • Total instruction size = 32 bits.

   • R-type contains: $\text{OPCODE} + \text{UNUSED} + 3 \times (\text{Register fields}) = 32$

   • $8 + X + 3 \times 6 = 32 \implies 8 + X + 18 = 32 \implies X = 6$ bits.

4. Immediate/Address Field ($Z$) in I-type Instruction:

   • I-type contains: $\text{OPCODE} + 2 \times (\text{Register fields}) + \text{Immediate/Address} = 32$

   • $8 + 2 \times 6 + Z = 32 \implies 8 + 12 + Z = 32 \implies Z = 12$ bits.

5. Calculate $X + 2Y + Z$:

   $$X + 2Y + Z = 6 + 2(8) + 12 = 6 + 16 + 12 = 34$$

Thus, the correct answer is 34.