Question

Assume a 2²² byte memory:

a. What are the lowest and highest addresses if memory is byte-addressable?

b. What are the lowest and highest addresses if memory is word-addressable, assuming a 16-bit word?

c. What are the lowest and highest addresses if memory is word-addressable, assuming a 32-bit word?

Explain with Steps please

Answer 1

For a memory system, which stores data in bytes, if adress is allocated to every byte then it is known as byte-addressable memory. But if address is allocated to every word (a word can be defined a group of bytes), then it is known as word-addressable memory.

Given that memory size is 2²² Bytes. In order to access n bytes, log₂n bits are needed. Similarly to access 2²² Bytes, log₂2²² = 22 bits are needed. With 22 bits, the possible decimal number range is from 0 to 2²² - 1.

(a) So when memory is byte-addressable, a maximum of 2²² bytes will be present in memory. To distinguish these 2²² Bytes, 22 bits are needed. So

• Lowest Address (in decimal) = 0
• Highest Address (in decimal) = 2²² - 1 .

(b) Given that the word size is 16-bits i.e., 2 Bytes (1 Byte = 8 bits). It means all the bytes of memory will be divided into a group of 2 bytes each, where each group is a word and address will be allocated to every word. So the memory is now containing (2²² B / 1 word ) = ( 2²² B / 2 B ) = 2^22-1 words = 2²¹ words.

So to distinguish 2²¹ words, log₂2²¹ = 21 bits are needed. With 21 bits, the possible decimal number range is from 0 to 2²¹ - 1 and each of these will be allocated to a word.So

• Lowest Address (in decimal) = 0
• Highest Address (in decimal) = 2²¹ - 1

(c) Given that the word size is 32-bits i.e., 4 Bytes (1 Byte = 8 bits). It means all the bytes of memory will be divided into a group of 4 bytes each, where each group is a word and address will be allocated to every word. So the memory is now containing (2²² B / 1 word ) = ( 2²² B / 4 B ) = ( 2²² B / 2² B ) = 2^22-2 words = 2²⁰ words.

So to distinguish 2²⁰ words, log₂2²⁰ = 20 bits are needed. With 20 bits, the possible decimal number range is from 0 to 2²⁰ - 1 and each of these will be allocated to a word.So

• Lowest Address (in decimal) = 0
• Highest Address (in decimal) = 2²⁰ - 1