Test by Bikram | Mock GATE | Test 4 | Question: 39

Question

A microprogrammed control unit supports $300$ instructions. Each instruction takes $15$ micro operations, $12$ flags are supported and $40$ vertical microprogrammed control signals are used.

Then, the size of $6$ control words is ______ bytes.

Comments

Correct answer is 17.25 B

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@Utkarsh Joshi

Please someone correct if wrong:
Hierarchy is

1 instruction contains many  microinstructions ,  1 Microinstruction contains microoperations , 1Microoperation performed by 1 control signal  

In control memory 1Word ==> 1 MIcroinstructions containing many Microoperations

In question shouldn't it be like

Each instruction take 12 micro instructions instead microoperations

Answer 1

Number of given instruction $= 300$

Each instruction have $15$  micro-operation

So total number of micro-operations  $=  300 * 15 = 4500$  

Number of $bit= \lceil \log_2 (4500) \rceil= \lceil 12.13 \rceil= 13 bits.$

Number of bits of flag $=\lceil \log_2 (12) \rceil=\lceil 3.58 \rceil= 4 bit$

Number of bit in vertical control signal is  $=\lceil \log_2 (40) \rceil=\lceil 5.321 \rceil= 6  bit$

Size of control word $= (13+4+6) = 23 bit$

It take 6 control words , hence total size is  $(23 *6 )/8 = 17.25 Bytes$

Comments

I wrote 18B in answer after rounding

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If we have 4 bits, it is $0.5$ bytes. So number of bytes can be a fraction unless at places where it is not possible like in byte addressing.

For numerical answers you should never enter any alphabet -- I'm not sure how GATE handles this.

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@Rishabh Gupta 2, I am having the same doubt. I too calculated considering flag bits as 12 .

Did you get any clue? Why flag bits are encoded too?

@VS , your explanation makes sense too but what if it is a case where we want to find Carry bit flag + Sign bit flag, then we should not encode it.Hence there could be several cases like that, we can't always gurantee that each time only one flag will be needed

@Arjun Sir, @Bikram Sir please clear this doubt.

Thanks in advance !