Question

14. A digital computer has a memory unit with 32 bits per word. The instruction set consists of 110 different operations. All instructions have an operation code part (opcode) and two address fields: one for a memory address and one for a register address. This particular system includes eight general-purpose, user-addressable registers. Registers may be loaded directly from memory, and memory may be updated directly from the registers. Direct memory-to-memory data movement operations are not supported. Each instruction is stored in one word of memory.
How many bits are needed for the opcode?

Answer 1

Answer:

7 bits

Explanation:

Given

Instruction Set = 110 operation

Memory unit = 32 bits per word.

We get the required bits by using the following formula

2^n = 110

But 110 is not a factor of 2.

So, we pick the nearest decimal number greater than 110 that is a power of 2.

The number is 128

2^n = 110 becomes

2^n = 128

2^n = 2^7 ---- 2 cancels out

So,

n = 7

Hence, the required number of bits needed for the opcode is 7 bits