Answer to Question #329316 in Assembler for artest 9

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. As 128 = 2⁷ < 198 <= 2⁸ = 256 8 bits are needed for representation of 198 opcodes.

b. For address part remains 24 - 8 = 16 bits.

c. There 256 - 198 = 58 additional codes for extra instructions are remained.

d. The largest unsigned binary number that may be represented by 16 bits is 2¹⁶ - 1 = 65535.