GATE CS 2018 Q51 — Machine Instructions and Addressing Modes

To find the maximum value of $N$, we use the concept of expanding opcodes. The instruction size is $2\text{ bytes} = 16\text{ bits}$.

• Register addressing:
  • Integer registers ($16$ registers): $\log_2(16) = 4\text{ bits}$
  • Floating-point registers ($64$ registers): $\log_2(64) = 6\text{ bits}$
• Type-1 (3Rs): Uses $3 \times 4 = 12\text{ bits}$ for operands, leaving $4\text{ bits}$ for the opcode.
  • Total possible opcodes = $2^4 = 16$
  • Used = $4$
  • Remaining 4-bit opcodes = $16 - 4 = 12$
• Type-2 (2Fs): Uses $2 \times 6 = 12\text{ bits}$ for operands, leaving $4\text{ bits}$ for the opcode.
  • Available 4-bit opcodes = $12$
  • Used = $8$
  • Remaining 4-bit opcodes = $12 - 8 = 4$
• Type-3 (1R + 1F): Uses $4 + 6 = 10\text{ bits}$ for operands, leaving $6\text{ bits}$ for the opcode.
  • Available 6-bit opcodes = $4 \times 2^{6-4} = 16$
  • Used = $14$
  • Remaining 6-bit opcodes = $16 - 14 = 2$
• Type-4 (1F): Uses $6\text{ bits}$ for the operand, leaving $10\text{ bits}$ for the opcode.
  • Available 10-bit opcodes = $2 \times 2^{10-6} = 2 \times 16 = 32$

Thus, the maximum value of $N$ is 32.