GATE CS 2016 Set 1 Q.19 (CS Q.9): Machine Instructions and Addressing Modes

A processor can support a maximum memory of 4 GB, where the memory is word-addressable (a word consists of two bytes). The size of the address bus of the processor is at least _____ bits.

Your answer

Official answer and solution

Official answer: 31
From: Official GATE 2016 answer key

Solution

Compute the number of unique word addresses in a 4 GB word-addressable memory.

Total memory capacity is:

\text{Capacity} = 4\text{ GB} = 4 \times 2^{30}\text{ bytes} = 2^2 \times 2^{30}\text{ bytes} = 2^{32}\text{ bytes}

The memory is word-addressable with a word size of 2 bytes ( $2^1$ bytes). Therefore, the total number of addressable words is:

\text{Number of words} = \frac{2^{32}\text{ bytes}}{2^1\text{ bytes/word}} = 2^{31}\text{ words}

To provide a distinct address for each of the $2^{31}$ words, the processor address bus must have at least 31 address lines:

\text{Address bus size} = \log_2(2^{31}) = 31\text{ bits}

Final answer

Common trap

Assuming byte-addressable memory and choosing 32 bits ( $4\text{ GB} = 2^{32}$ locations).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2016 Computer Science and Information Technology (CS) Session 1
Question: 19 of 65, printed as CS Q.9
Section: Computer Science and Information Technology

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