Share
Explore BrainMass

Addressable Memory

32 bit computer
32 bit instruction: 2 fields:opcode field and immediate operand or operand address
opcode = 1 byte wide
express answer in powers of 2
1. give number of available unique instructions
2. give max indrectly addressable mem capacity in bytes.
3. give max directly addressable memory capacity in bytes.
4. give bits needed for the program counter and instruction register.
5. max value of an unsigned data operand.

Please show how you got the answers.

Solution Preview

32-bit computer
32-bit instruction: 2 fields:opcode field and immediate operand or operand address
opcode = 1-byte wide
express ans in powers of 2
1.give number of available unique instructions
2.give max indrectly addressable mem capacity in bytes.
3.give max directly addressable memory capacity in bytes.
4 give bits needed for the program counter and instruction register.
5 max value of an unsigned data operand.

Answers:

Here is the pictorial representation of first two lines of your question.

0 78 31

Opcode(bits) Operand (24 bits)

1. 2exp 32 unique instructions are possible.
How?
Ex: if the Machine is of 3 bits. Then the following combination if possible out of those 3 bits

000
001
010
011
100
101
111

Now each of these combinations can be an instruction to the machine. Like 000 can be used to instruct the machine to write. 001 to move; 010 to load ...

Solution Summary

The following posting addresses an addressable memory problem.

$2.19