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It's 22.
4+log52+log(256*16)

4+6+12=>22??
seems right

Microinstructions present in control memory consists of three parts:

1. Conditional Flags – 16 flags – 2^4, requires 4 bits for its representation.
2. Control Signal – 52 control signal – 2^6=64, requires 6 bits for its representation.
3. Next Address – 256 instructions each having 16 micro operations = 256*16= 2^12 micro instructions to be stored in control memory. Therefore, bits required to address one microinstruction in control memory requires 12 bits.

Total bits required in each control word = 4+6+12 = 22 bits.

is it correct to say each instruction will need 52 control signals?

Total micro-operation = 256*16

These micro-operation have to be stored in Control Memory.

Number of bits are required to address one of the micro-operation (micro-instruction) = log(256*16)=12 bits

Control Word Format Includes :

Flags bits(4) + Control signals bits(6) + Next Micro operation address bits (12) = 22 bits

More Over, Control Memory Size = 2^12 * 22 bits
for vertical take log of the entries indivisually and then sum it up

log(256)+ log(16)+log(log16+log52 = 22
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