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35 votes

A micro instruction is to be designed to specify:

  1. none or one of the three micro operations of one kind and
  2. none or upto six micro operations of another kind

The minimum number of bits in the micro-instruction is:

  1. $9$
  2. $5$
  3. $8$
  4. None of the above
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3 Answers

Best answer
70 votes
70 votes

Actually the given question incorporates the concept of horizontal μprogramming (also known as decoded form of control signals) and vertical μprogramming (also known as encoded form of control signals)

The $(a)$ part says :

none or one of the three micro operations of one kind 

This is referred to encoding form of vertical one since at most one signal can be active in vertical microprogramming since it involves use of external decoder to select one control signal out of the given control signals..

No of bits required for vertical  microprogramming given n number of control signals $=\lceil ( \log_{2} n )\rceil$

Here, $n = 3$

So, no of bits required for part $(a)$ $=\lceil( \log_{2} 3)\rceil=   2$

Now coming to $(b)$ part , it says :

none or upto six micro operations of another kind

at maximum we can have at most $6$ microoperations of another kind at a time. To accommodate that we need decoded form of control signals which is horizontal signals.

So, no of bits required for $(b)$ part

$=$ No of control signals of $(b)$ kind $=  6$

Therefore overall bits required to accommodate both $(a)$ and $(b),$
$ = 2 + 6=8-bits$

Besides this, address field, flags etc are also there in a control word. That is why it is asked in the question :

 minimum number of bits in the micro-instruction required

Hence, minimum no of bits required  $=  8-bits$

C is the correct answer.

edited by
50 votes
50 votes

In question it is given that there are two kinds of instruction, let me classify them in two groups G1, G2. In G1 there are 3 microinstructions and in other group G2 there are 6 microinstructions.

in G1: None or one microinstruction is active at a time,  we can implement Vertical programming here as atmost one signal active at a time.

in G2: none or more than one (upto 6) microinstructions active at a time ,  we need Horigontal programming as upto 6 are needed.

Therefore toal bits = 2+6 = 8.

(Note that, if we implement Horizontal in G1 also then G1 itself would need 3 bits, and total bits will be 9 (That is option A). But they ask for minimum, therefore we should implement Vertical programming whenever chances.)

Option C is right choice.

13 votes
13 votes
Answer: C

The first condition (None or any one out of 3 operations) gives 4 variations to consider. This uses 2 bits.
The second condition (Any combination of 6 other operations) gives 64 variations to consider. This uses 6 bits.
In total 8 bits.
Answer:

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