UGC NET CSE | December 2019 | Part 2 | Question: 11

Question

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

• A. $19$
• B. $18$
• C. $17$
• D. $20$

Answer 1

Correct Answer : A) 19

We have two micro-operation subfields F1 and F2. Each of F1, F2 have 15 distinct micro-operations. Thus, 4 bits are required for each.

The condition field has four status conditions, Thus, 2 bits are required.

The branch field has four options. Thus, 2 bits are required.

There are 128 different memory words/locations. Thus, 7 bits are required for 128 different locations.

Thus, in total : 4 + 4 + 2 + 2 + 7 = 19 bits are required.

 

Answer 2

The correct answer is option A) i.e. 19 bits.

There are 3 fields in the microinstruction.

• Micro operation fields.
  F1,F2
• CD for status bits
• BR in conjunction with ADF  i.e we need to multiply the possible number of branch field with the Address field.

F1, F2 each are having 15 micro-operations.

# bits to represent it:

log15 + log15 = 4+4 = 8 bits

 

CD for status bits = log4 = 2 bits

 

BR in conjunction with ADF = log(4*128) = 9 bits

 

total bits = 8 + 2 + 9 = 19 bits