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?

The number of bits that are required for the given opcode is; 7 bits

We are given;

Number of bits per word = 32 bits per word

Number of different operations of instruction set = 110

Thus, to find how many bits will be required to represent 110 opcodes,we need to find the value of n from 2ⁿ that is big enough to represent 110 opcodes.

Thus,6 bits is not sufficient as 64 is less than 110

Let's try 2^(7) = 128

This is sufficient as 128 is more than the 110.

Read more about opcodes at; link

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

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