Answer to Question #328453 in Assembler for Nov

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

a. How many bits are needed for the opcode?

2^8 = 256,  2^7 = 128

198>128. Therefore, the opcode need 8 bits.

b. How many bits are left for the address part of the instruction?

24-8 = 16 bits

There are 16 bits available for the address part.

c. How many additional instructions could be added to this instruction set without exceeding the assigned number of bits?

256 -198 = 57

d. What is the largest unsigned binary number that the address can hold?

Memory size:  2^16 = 0xFFFF = 65535

The largest unsigned binary number 2^24 = 0xFFFFFF