6 votes 6 votes A byte addressable computer has a memory capacity of $2^{m}$ Kbytes and can perform $2^{n}$ operations. An instruction involving $3$ operands and one operator needs a maximum of $3m$ bits $m + n$ bits $3m + n$ bits $3m + n + 30$ bits CO and Architecture ugcnetjune2014iii co-and-architecture byte-addressable + – makhdoom ghaya asked Jul 10, 2016 • recategorized Nov 24, 2017 by Arjun makhdoom ghaya 7.0k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 6 votes 6 votes D. should be the ans... n(operator) m+10(operand1) m+10(operand2) m+10(operand3) So ans = n + 3m + 30 papesh answered Jul 11, 2016 • selected Sep 13, 2017 by Sanjay Sharma papesh comment Share Follow See all 2 Comments See all 2 2 Comments reply A. Dalal commented Jul 13, 2016 reply Follow Share why 10 bits for each operands ?????????? 0 votes 0 votes Sanjay Sharma commented Sep 13, 2017 reply Follow Share because of 2^m K bytes (1024 =2^10) 0 votes 0 votes Please log in or register to add a comment.
7 votes 7 votes Memory capacity is 2m Kbytes= 2m+10bytes. Whole memory can be represented by m+10 bits. 2n operations can be represented by n bits. Now instruction of form Instruction op1,op2,op3. Will take (3m+30+n) bits Option D Somnath119 answered May 26, 2017 • edited Jun 2, 2017 by Somnath119 Somnath119 comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes option is D To specify a particular operation, out of the 2n possible operations, one needs n bits. As the machine is byte addressable, to specify a particular byte we need (m+10) bits (2(n+10) bytes are there). So 3 addressable and 1 operations needs 3(m+10)+n=3m+n+30 bits Prasanjeet Ghosh answered May 18, 2018 Prasanjeet Ghosh comment Share Follow See 1 comment See all 1 1 comment reply `JEET commented Jan 3, 2020 i moved by `JEET Jan 3, 2020 reply Follow Share Good question asked in UGC first time. :D 0 votes 0 votes Please log in or register to add a comment.