GATE CS 2020 Q44 — Machine Instructions and Addressing Modes

To find the maximum number of distinct R-type opcodes, we can use the concept of instruction space allocation (expanding opcodes).

1. Identify the parameters:

• Instruction size = $16$ bits, which gives a total of $2^{16} = 65,536$ possible instruction combinations.
• Number of registers = $64$, requiring $\log_2(64) = 6$ bits to specify each register.

2. Analyze I-type instructions:

• Each I-type instruction consists of: [Opcode] [Register (6 bits)] [Immediate (4 bits)].
• For each distinct I-type opcode, the number of combinations consumed is: $$2^6 \times 2^4 = 64 \times 16 = 1024 \text{ combinations}$$
• Since there are $8$ distinct I-type opcodes, the total combinations consumed by I-type instructions is: $$8 \times 1024 = 8192 \text{ combinations}$$

3. Analyze R-type instructions:

• Each R-type instruction consists of: [Opcode] [Register 1 (6 bits)] [Register 2 (6 bits)].
• Each distinct R-type opcode consumes: $$2^6 \times 2^6 = 64 \times 64 = 4096 \text{ combinations}$$

4. Calculate the maximum number of R-type opcodes ($N$):

The remaining instruction space is allocated to R-type instructions:

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