A digital computer has memory unit with $24$ bits word.The instruction set consists of $150$ different operations. All instructions have an operation code part and an address part. Each instruction is stored in one word of memory.
$Q1$ How many bits are needed for the OP-CODE and how many bit are left for the address of the instruction
$Q2$ What is the maximum available size for memory and the largest unsigned binary number that can be accommodated in one word of memory
24 bits word
word contains opcode bits + address bits.
Instructions set contain 150 different operations so to identify each uniquely would require at least 8 bits (2^8=256).
so 16 bits left for address. So for question no 1 Ans is B
16-bit for address so total memory that can be addressable is 2^16.
word size is 24 bit so largest unsigned binary number is 2^24 - 1.
So for question no 2 Ans is C
@Anmol 300 books will be dispatched this month...