UGC NET Computer Science December 2019 | Question 61

Question 61

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A micro instruction format has microoperation field which is divided into 2 subfields Fl 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:

1. 1. 19
2. 2. 18
3. 3. 17
4. 4. 20

Explanation

Microprocessor instruction format, which is divided into three subfields F1, 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 ADF.

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.

micro-instruction field:

Instruction sub-field:	F1	F2	CD	BR	AD
Number of bits:	4 bits	4 bits	2 bits	2 bits	7 bits

So, option 1 is correct answer

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