A micro instruction format has micro operation field which is divided into 2 subfields F1 and F2, each having 15 distinct micro operations, 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
Answer: (A)
Explanation:
=> In given Microprocessor instruction format, which is divided into two sub-fields F1, F2 each having 15 distinct micro-operations, So, 4 bits are required for each.
=> Condition field CD for four status bits it needs 2 bits for four different condition.
=> Branch field BR having four options used in conjunction with address field AD it needs 2 bits for four option..
=> Now there are 128 different memory location, So, there 7 bits are required for 128 different location.
So, in Total : 4 + 4 + 2 + 2 + 7 = 19 Bits required.
micro-instruction field:
| Instruction sub-field: | F1 | F2 | CD | BR | AD |
| Number of bits: | 4 bits | 4 bits | 2 bits | 2 bits | 7 bits |
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