At a minimum, how many bits are needed in the MAR with each of the following memory sizes? a. 1 million bytes
b. 10 million bytes
c. 100 million bytes
d. 1 billion bytes

Using the bit-byte relationship, the minimum bit memory allocation for the memory sizes stated are 20, 24, 27 and 30 bits respectively.

To obtain the minimum number of bits required for each of the given memory sizes :

We use the relation :

Where, n = number of bits

For 1 million byte :

n = 20

2.) For 10 million bytes :

All integers before 24 equates to a value less than 10,000,000

n = 24

3.) For 100 million bytes :

n = 27

All integers before 27 equates to a value less than 100,000,000

4. For 1,000,000,000 bytes :

n = 30

All integer values before 30 equated to a value less than 1,000,000,000

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a. 20 bits

b. 24 bits

c. 27 bits

d. 30 bits

Explanation:

Using base 2 numbers where a location in RAM is represented using an address.

Hence, given that with 2 bits one can store 4 things maximum, with 3 bits, one can store between 5 and 8 things.

A. Therefore to represent 1 million bytes -1 addresses, we need 20 bits. This depicts 1,048,576 to be exact. This is a megabyte or 1 MB.

B For 10 million bytes we have 1 million different addresses using 20 bits.

This equates to 10 million is 10 x bigger than 2^20.

With each addition of one bit to the address, the addresses to represents doubled e double the number of addresses that we can represent (and reach).

For instance : 1 million -> 2 million -> 4 million -> 8 million -> 16 million = 2^20 -> 2^21 -> 2^22 -> 2^24

Thus, 24 bits equates to 10 million addresses. 2^23 < 10 million < 2^24 where 2^24 = 16,777,216.

C. In 100 million bytes, 27 bits are required.

Thereby 2^26 = 67,108,864 bytes, not enough for 100 million. 2^27 = 134,217,728 and this is enough.

D. 1 billion bytes is equal to 1000 times 1 million or 1000 * 2^20.

Therefore in base 2 notation, it is (2^20 * 2^10) = 2^30 = 1,073,741,824.

Hence 30 bits are required to represent 1 billion bytes. This equates to 1GB.

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