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what is the answer given??
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It's 22.
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4+log52+log(256*16)

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

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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

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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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