Question

A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. How many bits are needed for the opcode?

Answer 1

Answer:

8 bit

Explanation:

we have given that the instruction consist 150 different operation

we know that $2^{8}$ = 256

but we have only 150 operation

so now we check $2^{7}$=128

which is less than 150 so we need 8 bit for the opcode

so the 8 bit are needed for the opcode because when we take 7 bit then $2^{7}$=128 which is less than 150 and not useful so we need 8 bit