1 votes 1 votes A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have a operation code part (opcode) and an address part ( allowing for only one address). Each instruction is stored in one word of memory. a. How many bits are needed for the opcode? b. How many bits are left for the address part of the instruction? c. What is the maximum allowable size for memory? d. What is the largest unsigned binary number that can be accommodated in one word of memory? CO and Architecture co-and-architecture + – rahul sharma 5 asked May 26, 2017 retagged Nov 13, 2017 by Arjun rahul sharma 5 7.7k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply akash.dinkar12 commented May 26, 2017 reply Follow Share a) 8 bits b) 16 bits c) 16 MB d) 2^24 -1 0 votes 0 votes rahul sharma 5 commented May 27, 2017 reply Follow Share Yeah,I got the same. 1 votes 1 votes A_i_$_h commented Jun 26, 2017 reply Follow Share can u please explain the reason for each of these values the explanation please 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes opcode size = 8 bits address size = 24 - 8 =16 bits maximum allowable memory = (2 ^16)*24 bits since 2^16 addresses and each location can hold 24 bits. maximum unsigned binary = (2^24)-1 Niket Gangwar answered Sep 22, 2017 Niket Gangwar comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes i)For representing 150 operations we need 8 bits. So 8 bits for opcode. ii)leftover bits for address of instruction is 24-8=16bits iii)since memory unit is 24 bits, size of memory 2^24 i.e 16MB iv)largest unsigned int 2^24 - 1 shraddha priya answered Jul 10, 2017 shraddha priya comment Share Follow See 1 comment See all 1 1 comment reply rahul sharma 5 commented Nov 2, 2017 reply Follow Share For option c ,how is it 2^24B? 24 is the data width As address part in instruction is 16 bits,it means 2^16 memory locations can be addressed and hence 64 KB memory. I also got 16 MB earlier,now i am having some doubt.Can you re check? 0 votes 0 votes Please log in or register to add a comment.