A computers memory is composed of 8k words of 32 bits each .how many bits are required for memory address if the smallest addressable memory unit is a word?
Answers
Answer:
For example, an 8-bit-byte-addressable machine with a 20-bit address bus (e.g. Intel 8086) can address 220 (1,048,576) memory locations, or one MiB of memory, while a 32-bit bus (e.g. Intel 80386) addresses 232 (4,294,967,296) locations, or a 4 GiB address
Answer:
The minimum number of bits necessary to address 8K words is 13.
Explanation:
You have the number of words to address that is 8000 words, a word is the smallest addressable memory unit.
8000 words can be addressed with units. Now you have to find the value of n that approximates to the number of words.
You have the number of words to address that is 8000 words, a word is the smallest addressable memory unit.
8000 words can be addressed with units. Now you have to find the value of n that approximates to the number of words.