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A micro instruction format has microoperation field which is divided into $2$ subfields $F1$ and $F2$, each having $15$ distinct microoperations, condition field $CD$ for four status bits, branch field $BR$ having four options used in conjunction with address field $AD$. The address space is of $128$ memory words. The size of micro instruction is:






Solution

Each subfield $F1$ and $F2$ has $15$ microoperations.

To represent $15$ operations, we need:

$\lceil \log_2 15 \rceil=4$ bits

So,

$F1=4$ bits

$F2=4$ bits

Condition field $CD$ has $4$ options/status bits:

$\log_2 4=2$ bits

Branch field $BR$ has $4$ options:

$\log_2 4=2$ bits

Address space is $128$ memory words:

$\log_2 128=7$ bits

Total size of microinstruction:

$4+4+2+2+7=19$ bits



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