Question

Consider a system that contains 32K bytes. Assume we are using byte addressing, that is assume that each byte will need to have its own address, and therefore we will need 32K different addresses. For convenience, all addresses will have the same number n , of bits, and n should be as small as possible. What is the value of n ?

Answer 1

Answer:

n = 15

Explanation:

Using

Given

Number of bytes = 32k

At the smallest scale in the computer, information is stored as bits and bytes.

In the cases when used to describe data storage bits/bytes are calculated as follows:

Number of bytes when n = 1 is 2¹ = 2 bytes

When n = 2, number of bytes = 2² = 4 bytes

n = 3, number of bytes = 2³ = 8 bytes

So, the general formula is

Number of bytes = 2^n

In this case

2^n = 32k

Note that 1k = 1024 bytes,

So, 32k = 32 * 1024 bytes

Thus, 2^n = 32 * 1024

2^n = 2^5 * 2^10

2^n = 2^15

n = 15.