GATE CS 2016 Set 2 Q31 — Machine Instructions and Addressing Modes

To find the total memory consumed by the program, we first determine the size of a single instruction in bits, align it to the nearest byte, and then multiply by the total number of instructions.

1. Calculating the size of each field in an instruction:

• Opcode: Since there are $12$ instructions in the instruction set, the number of bits required is:$$\lceil \log_2(12) \rceil = 4 \text{ bits}$$
• Source Registers: There are $64$ registers, requiring $\lceil \log_2(64) \rceil = 6$ bits per register. For two source registers:$$2 \times 6 = 12 \text{ bits}$$
• Destination Register: One destination register requires:$$6 \text{ bits}$$
• Immediate Value: Given as:$$12 \text{ bits}$$

2. Total size per instruction:

Summing the bits of all five fields:

3. Byte-alignment:

Each instruction must be stored in a byte-aligned fashion (i.e., its size must be a multiple of $8$ bits). Therefore, we round up the size of each instruction to the nearest byte:

4. Total memory for 100 instructions: