Question 54

Instruction format details:

• Total instruction length = 16 bits.
• Number of registers = 64 $\implies$ register address field = $\log_2 64 = 6$ bits.
• I-type instruction format: Opcode, 1 register (6 bits), 4-bit immediate.
  • Total operand bits = $6 + 4 = 10$ bits.
  • Opcode bits for I-type = $16 - 10 = 6$ bits.
• R-type instruction format: Opcode, 2 registers ($6 + 6 = 12$ bits).
  • Opcode bits for R-type = $16 - 12 = 4$ bits.

Expanding opcode relationship:

• The R-type opcode is 4 bits long, supporting up to $2^4 = 16$ bit patterns.
• Let $x$ be the number of patterns allocated to R-type instructions.
• The remaining $16 - x$ 4-bit patterns can be expanded by 2 additional bits ($6 - 4 = 2$) to form 6-bit I-type opcodes.
• Total possible I-type opcodes = $(16 - x) \times 2^2 = (16 - x) \times 4$.

Solve for $x$:

Therefore, the maximum number of distinct R-type opcodes is 14.