Question

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: \begin{center} \setlength{\tabcolsep}{15pt} % Adjust column spacing \renewcommand{\arraystretch}{1.5} % Adjust row spacing \begin{tabular}{|c|c|c|c|c|} \hline OPCODE & UNUSED & DST Register & SRC Register1 & SRC Register2
\hline \end{tabular} \end{center} \vspace{0.5cm} I-type Instruction Format: \begin{center} \setlength{\tabcolsep}{15pt} % Adjust column spacing \renewcommand{\arraystretch}{1.5} % Adjust row spacing \begin{tabular}{|c|c|c|c|} \hline OPCODE & DST Register & SRC Register & \# Immediate Value/Address
\hline \end{tabular} \end{center} 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 ............

Hint:

The OPCODE size is determined by the total number of instructions. Ensure register and immediate field sizes fit within the 32-bit instruction format.

Answer 1

Step 1: Instruction format analysis. Each instruction is \(32\) bits in total. Based on the formats: R-type instruction fields: \[ \text{OPCODE} + \text{UNUSED} + \text{DST Register} + \text{SRC Register1} + \text{SRC Register2} = 32 \, \text{bits}. \] I-type instruction fields: \[ \text{OPCODE} + \text{DST Register} + \text{SRC Register} + \text{Immediate Value/Address} = 32 \, \text{bits}. \] Step 2: Determine the number of bits for the registers. The processor has \(50\) architectural registers. The number of bits required to uniquely identify a register is: \[ \text{Register bits} = \lceil \log_2(50) \rceil = 6 \, \text{bits}. \] Thus, each register field (DST Register, SRC Register1, SRC Register2) uses \(6\) bits. Step 3: Determine the number of bits for the OPCODE field. The ISA contains \(150\) distinct instructions, equally divided into R-type and I-type. Since \(1\) bit in the OPCODE field is used to distinguish between instruction types, the remaining bits encode the operations: \[ \text{OPCODE bits} = \lceil \log_2(150) \rceil = 8 \, \text{bits}. \] Step 4: Calculate the number of bits for the UNUSED field (R-type) and Immediate Value/Address field (I-type). R-type: From the total \(32\) bits, the UNUSED field is: \[ X = 32 - (\text{OPCODE} + 3 \times \text{Register bits}) = 32 - (8 + 3 \times 6) = 6 \, \text{bits}. \] I-type: The Immediate Value/Address field uses: \[ Z = 32 - (\text{OPCODE} + 2 \times \text{Register bits}) = 32 - (8 + 2 \times 6) = 12 \, \text{bits}. \] Step 5: Compute \(X + 2Y + Z\). Substituting the values of \(X\), \(Y\), and \(Z\): \[ X + 2Y + Z = 6 + 2 \times 8 + 12 = 34. \] Final Answer: \[ \boxed{34} \]