GATE 2020 CS Q.20: solution and answer

GATE 2020 · CSQ.20 · 1 mark · NAT

If there are m input lines and n output lines for a decoder that is used to uniquely address a byte addressable 1 KB RAM, then the minimum value of m + n is ____.

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Answer and solution

ANSWER

m + n = 1034

Official answer: 1034Source: Official answer key, GATE 2020 CS (IIT Delhi)

A byte-addressable 1 KB RAM has $1 KB = 2^{10} = 1024$ bytes, each requiring a unique address.
The decoder must provide exactly one output line per addressable byte, so $n = 1024$ .
For a decoder, the number of input (address) lines m must satisfy $2^{m} \geq n$ so that every output line can be uniquely selected.
The minimum m occurs when $2^{m} = 1024$ , i.e., $m = lo g_{2} (1024) = 10$ .
Minimum value of $m + n = 10 + 1024 = 1034$ .

KEY IDEA

A decoder that uniquely addresses every byte of a 1 KB RAM needs one output line per byte and enough input (address) lines to select among all outputs of a full binary decoder ( $n = 2^{m}$ ).

WATCH OUT FOR

Confusing "byte addressable 1 KB" with the number of address bits alone (giving m + m = 20) or forgetting that n must equal the full 1024 output lines (one per addressable byte), not some smaller count.

From the GATE 2020 CS paper. Every question of this paper · Download the full paper · All 22 Memory & cache questions, with notes

GATE 2020 CS Q.20: solution and answer

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