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Addressable Memory

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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.

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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
See Also This Related BrainMass Solution

Numerical: Direct Mapped Cache and Byte Addressable Memory

Given, a byte addressable main memory of 2^16 bytes and block size of 8 bytes, using direct mapped cache consisting of 32 lines.

a. How many total memory bytes can be stored in the cache?
b. How is a 16 bit memory address divided into tag, line number and byte number?
c. Into what line would byte with the following address be stored?
0001 0001 0001 1011.

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