1 votes 1 votes shefali1 asked Jul 10, 2017 shefali1 384 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments joshi_nitish commented Jul 10, 2017 reply Follow Share since 480 registers, 9 bits for register and there are two registers operand, therfore 18 bits are used for addresses, now for opcode we are left with 24-18=6 bits... total 2^6 for 2 addresses out of which only 48 are used, we are left with 2^6-48=16.. now, total 16*2^9 for one address(assume we used 'x' for one addresses), we are left with 16*2^9-x now (16*2^9-x)*2^9 for zero address which is already given as 2048, so finally (16*2^9-x)*2^9=2048......solving this you will get x=8188.. 3 votes 3 votes shefali1 commented Jul 10, 2017 reply Follow Share can u please suggest me any source from where i can get to how to solve these type of questions. 0 votes 0 votes rahul sharma 5 commented Jul 11, 2017 reply Follow Share William Stallings book exercise problems 0 votes 0 votes Please log in or register to add a comment.