Aspire Faculty ID #12046 · Topic: NIMCET 2025 · Just now
NIMCET 2025

Consider a system with a CPU having 6 registers and 32-bit instructions. The maximum possible size of the main memory is 512 KB (1K = 210). Each instruction takes two registers and one memory address as operands. Which one of the following correctly gives the maximum possible distinct instructions that can be there in the instruction set of the CPU?

Solution

Memory size $=512\text{ KB}=2^{19}$ bytes $\Rightarrow$ address field needs $19$ bits.

Registers $=6 \Rightarrow$ bits per register $=\lceil \log_2 6\rceil=3$ bits. For two registers $\Rightarrow 2\times3=6$ bits.

Total operand bits $=19+6=25$.

Opcode bits $=32-25=7 \Rightarrow$ maximum distinct instructions $=2^{7}= {128}$.

Previous 10 Questions — NIMCET 2025

Nearest first

Next 10 Questions — NIMCET 2025

Ascending by ID
Ask Your Question or Put Your Review.

loading...