Question

Suppose that you have a computer with a memory unit of 24 bits per word. In this computer, the assembly program’s instruction set consists of 198 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. A. How many bits are needed for the opcode? b. How many bits are left for the address part of the instruction? c. How many additional instructions could be added to this instruction set without exceeding the assigned number of bits? Discuss and show your calculations. D. What is the largest unsigned binary number that the address can hold?

Answer 1

The number of bits that are needed for this operation code (opcode) is equal to 8 bits.

Given the following data:

• Number of operations = 198.

• Instruction size, I = 24 bits.

How to calculate the number of bits.

The number of bits that are needed for this operation code (opcode) is given by this mathematical expression:

$2^n > 198\\\\2^8 > 198\\\\256 > 198$

n = 8 bits.

b. To determine the number of bits that are left for the address part of the instruction:

• Opcode, O = 8 bits.

• Instruction size, I = 24 bits.

Address = Instruction size - Opcode

Address = 24 - 8

Address = 16 bits.

c. To determine the maximum allowable size for memory:

Based on the memory unit, this is given by $2^{16}\times 24$

d. The largest unsigned binary number that can be accommodated in one word of memory is equal to $2^{24}-1$

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