Question

The memory unit of a computer has 512K words of has an instruction format with 4 fields: opcode, add and memory address. Given that there are 4 addressing mode, 32-bit instruction, answer the following questions: 32 bits each. The computer ress mode, register address 50 register and 1. How large is the opcode? 2. How large is the address mode field? 3. How large is the register field? 4. How large is the memory address field?

Answer 1

2) Given there are 4 addressing mode.

Address mode field = log₂4=2

4) Number of memory address=

Given there are 512 k words.

512k words= 2⁹X 2¹⁰=2⁽⁹⁺¹⁰⁾=2¹⁹

memory address=19

3) Given registers=50

Since 2⁵=32 which is less than 50 we will consider 2^6. Which is nearest value to 50

So 6 bits are required for register field. i.e register address bits=6.

1) Given Total instruction bits=32.

Total instruction bits=memory address+ register address bits + opcode + address mode field.

32=19+6+opcode+2;

Opcode=5

So the answers are:

1)5

2)2

3) 6

4)19