GATE CS 2016 Set 1 Q09 — Memory Hierarchy: Cache, Main Memory and Secondary Storage

A processor can support a maximum memory of

4\;\textsf{GB}

, where the memory is word-addressable (a word consists of two bytes). The size of address bus of the processor is at least _________bits.

View worked solution

Answer: 31

To find the minimum size of the address bus, we need to determine the number of unique addressable locations in the memory.

Given data:

Maximum memory capacity = $4\text{ GB} = 4 \times 2^{30}\text{ bytes} = 2^{32}\text{ bytes}$
Word size = $2\text{ bytes} = 2^1\text{ bytes}$

Since the memory is word-addressable, each unique address corresponds to one word. The total number of addressable locations (words) is:

\text{Number of addressable locations} = \frac{\text{Total memory size}}{\text{Word size}} = \frac{2^{32}\text{ bytes}}{2^1\text{ bytes}} = 2^{31}\text{ locations}

The number of address bits ( $n$ ) required to uniquely identify $2^{31}$ locations is given by:

2^n = 2^{31} \implies n = 31\text{ bits}

Thus, the size of the address bus of the processor is at least 31 bits.

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Published 10 Oct 2026

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