A micro instruction format has microoperation field which is divided into 2…
2019
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:
- A.
19
- B.
18
- C.
17
- D.
20
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Correct answer: A
Solution:
Microoperation fields: Each of F1 and F2 must encode 15 distinct microoperations. The number of bits required = ceil(log2(15)) = 4 bits per subfield → microoperation field = 4 + 4 = 8 bits.
Condition field: The field selects one of four status bits, so it needs to encode 4 possibilities → 2 bits.
Branch field: There are four branch options → 2 bits.
Address field: Address space of 128 memory words requires ceil(log2(128)) = 7 bits.
Total size = microoperation field (8) + condition (2) + branch (2) + address (7) = 19 bits.
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