3 votes 3 votes 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. GATE tbb-mockgate-4 numerical-answers co-and-architecture microprogramming + – Bikram asked May 14, 2017 • retagged Sep 11, 2020 by ajaysoni1924 Bikram 902 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Tuhin Dutta commented Dec 2, 2017 reply Follow Share Correct answer is 17.25 B 0 votes 0 votes jatin khachane 1 commented Jan 9, 2019 reply Follow Share @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 0 votes 0 votes Please log in or register to add a comment.
Best answer 3 votes 3 votes 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 $(23 *6 )/8 = 17.25 Bytes$ Bikram answered May 14, 2017 • selected Jan 17, 2019 by Rishi yadav Bikram comment Share Follow See all 16 Comments See all 16 16 Comments reply Show 13 previous comments Ashwani Kumar 2 commented Dec 15, 2019 reply Follow Share I wrote 18B in answer after rounding 0 votes 0 votes Arjun commented Dec 15, 2019 reply Follow Share 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. 2 votes 2 votes jiminpark commented Dec 26, 2021 reply Follow Share @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 ! 0 votes 0 votes Please log in or register to add a comment.