If there are m-input lines and n-output lines for a decoder that is used to…

Concept: A decoder takes an m-bit input code and asserts exactly one output line for each distinct code, so a fully decoded unit has n = 2m outputs. Independently, uniquely addressing a byte-addressable memory of capacity C bytes means every single byte must have its own select line, so the decoder must supply C distinct outputs. Minimising m + n therefore reduces to finding the smallest m whose decoder still covers the whole capacity.

Application:

1. Capacity in bytes: 1 KB = 210 = 1024 bytes, and byte-addressable means each of those 1024 bytes needs its own unique address.

2. Required outputs: one select line per byte gives n = 1024.

3. Required inputs: substitute into n = 2m to get 2m = 1024 = 210, so m = log21024 = 10.

4. Add the two line counts: m + n = 10 + 1024 = 1034.

Cross-check: Take m = 9 instead: 29 = 512 outputs, which reaches only 512 of the 1024 bytes, so half the RAM stays unaddressable — 10 is genuinely the smallest usable input count. Once m is fixed at 10 the output count is forced to 210, so the pair (10, 1024) is the only minimal choice and no smaller total than 1034 exists. Note also that the two counts are on very different scales: the inputs grow logarithmically while the outputs grow exponentially, which is why the total is dominated by n.