1 votes 1 votes A $\text{ROM}$ is used to store the table for Addition of $2$ number of $16$ bit unsigned integers. The size of the $\text{ROM}$ required for the above application in Giga Bytes is _______________. atul_21 asked Jan 1, 2018 edited Aug 3, 2021 by soujanyareddy13 atul_21 2.5k views answer comment Share Follow See all 16 Comments See all 16 16 Comments reply Ashwin Kulkarni commented Jan 1, 2018 reply Follow Share 68 GB ?? 1 votes 1 votes SHUBHAM SHASTRI commented Jan 1, 2018 i edited by SHUBHAM SHASTRI Jan 1, 2018 reply Follow Share i am getting as 2^16*2^16*(16)=(2^32)*16=4Gb*17=68 Gb that is 8.5 GB @Ashwin ..your answer is in Gb ..not in GB... 1 votes 1 votes Bad_Doctor commented Jan 1, 2018 reply Follow Share @shubham shastri Can you please explain? 0 votes 0 votes atul_21 commented Jan 1, 2018 reply Follow Share Answer is 8.5 GB . They are asking for addition. 0 votes 0 votes SHUBHAM SHASTRI commented Jan 1, 2018 reply Follow Share the formula is 2^n * 2^m * Max(m,n)...when addition is required and formula is 2^n*2^m*(m+n)...when multiplication is required... 0 votes 0 votes SHUBHAM SHASTRI commented Jan 1, 2018 reply Follow Share @atul_21 how answer is 8.5GB?? 0 votes 0 votes Bad_Doctor commented Jan 1, 2018 reply Follow Share @shubham Please give me a link or tell me how you got the formula? Thanks :-) 0 votes 0 votes SHUBHAM SHASTRI commented Jan 1, 2018 reply Follow Share say you want addition of 2 numbers which are 2 bits each ... so with 2 bits we can have 2^2=4 words .... like 0,1,2,3 these are my 1st operands similarly for 2nd operand 2^2 =4 words ...exactly same as 0,1,2,3... now total additions possible are (0,0)(0,1)(0,2)(0,3)(1,0)(1,1)(1,2)(1,3).....like that SO TO STORE ALL WE HAVE A TABLE WHICH WILL HAVE 2^n*2^n entries ... now for addition ,substraction and division ..output will be of same bits as that of operands ... so we get formula as SIZE OF RAM=ENRIES*EACH WORD SIZE=2^n*2^m*Max(n,m) 3 votes 3 votes Bad_Doctor commented Jan 1, 2018 reply Follow Share Everything is great. But suppose we add 2 4-bit number. Then Maximum bit in output can be 5-bit(considering carry). 1 votes 1 votes Bad_Doctor commented Jan 1, 2018 reply Follow Share So it will be 68Gb. 0 votes 0 votes atul_21 commented Jan 1, 2018 reply Follow Share @ SHUBHAM SHASTRI please check this . I am confused now. 0 votes 0 votes SHUBHAM SHASTRI commented Jan 1, 2018 reply Follow Share bro every thing is correct ony look at this ... 2^16*2^16*(16+1)= 2^32*17=4G(LOCATIONS)*17 bits ....but thing is that me and Ashwin have calculated it in bits ....ashwin wrote it as 68GB ..it should bw 68 Giga-bits..... given answer is in BYTES... so answer is 68 giga-bits/8= 8.5 Giga-Bytes.... its simply matter of bits and bytes.. 1 votes 1 votes atul_21 commented Jan 2, 2018 reply Follow Share Oh thanku so much. . :) Shubham Ashwin 0 votes 0 votes atul_21 commented Jan 2, 2018 reply Follow Share https://gateoverflow.in/188500/block-replacement-in-cache Bro,plz have a look on this one. 0 votes 0 votes Ashwin Kulkarni commented Jan 2, 2018 reply Follow Share Ohh thanks @shubham I have calculated in bits :( 68Gb = 8.5GB 0 votes 0 votes PiratedVirus commented Dec 10, 2018 reply Follow Share Just small additional info: For all such kinds of questions, Size of ROM = Size of 1 result * Number of all possible results Here, same question but with multiplication https://gateoverflow.in/2725/gate1996-1-21 0 votes 0 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes total entry=216*216=232 size of each entry is....Max(n,m) bits therefor ...total size =232*16/8 =8Gbytes considering carry:- 232 *17/8 =8.5 Gbytes hs_yadav answered Jan 1, 2018 selected Jan 2, 2018 by atul_21 hs_yadav comment Share Follow See all 0 reply Please log in or register to add a comment.