Question 4 Chapter 4 How many bits are required to address a

Question 4 (Chapter 4)

How many bits are required to address a 4M × 16 main memory if

a) Main memory is byte-addressable?

b) Main memory is word-addressable (Note each word is 16 bits in size)?

Solution

Please follow the data and description :

4M x 16 main memory :

Here 4M x 16 refers to the size 64Mb of memory that implies 67108864 bytes of data or 33554432 words where a word is usually of 2 bytes.

a) For Main memory is byte-addressable :

So calculating the data,

4M * 16 = 2^2 * 2^20 * (2^4 / 2^3) As considered 16 bits / 8 bits is a byte.

we have,

2^2 * 2^20 * 2^1 = 2^23

So from the power of 2 it implies that the number of bits that are required to address a 4M x 16 main memory if so a main memory is byte-addressable is => 23 bits.

b) For Main memory is word-addressable :

As given, assuming a word is of 16 bits or simply a 2 bytes long. We could calculte as,

4M * 16 = 2^2 * 2^20 * (2^4 / 2^4) As considered 16 bits / 16 bits is a word.

We have,

2^2 * 2^20 * 2^0 = 2^22

So from the power of 2 it implies that the number of bits that are required to address a 4M x 16 main memory if so a main memory is word-addressable is => 22 bits.

Hope this is helpful.