Aspire Faculty ID #19198 · Topic: NIMCET 2026 · Just now
NIMCET 2026

A CPU uses a $16$-bit instruction format. If $4$ bits are used for the opcode and the remaining bits specify a single memory address, what is the maximum addressable memory space for this instruction format?

Solution

Total instruction size is:

$16$ bits

Opcode uses:

$4$ bits

So, remaining bits for memory address are:

$16-4=12$ bits

With $12$ address bits, maximum addressable memory locations are:

$2^{12}=4096$

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