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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 :
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Eduncle Best Answer
In microinstruction format, microoperation field is divided into two subfields F1, F2, each having 15 distinct micro-operations, condition field CD for 4 status bits, branch field BR having four options used in conjunction with address field AD. The address space is of 128 memory locations.i.e:
F1, F2 each having 15 distinct micro-operation. So, 4 bits are required for each.
Condition field have 4 status, it needs 2 bits for four different condition.
Branch field have 4 option so, 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.