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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?

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MrRoyal

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

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

  • We have 110 instructions.

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.

  • We know that 2^(6) = 64

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.

  • In conclusion, 7 bits is enough to represent 110 opcodes.

Read more about opcodes at; https://brainly.com/question/10848659

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