Asked Nov 17, 2019

b. How many bits are required to address a 4M × 16 main memory if main memory is byte-addressable?
c. How many bits are required to address a 1M × 8 main memory if main memory is byte-addressable?


Expert Answer

Step 1

A memory is used to store some data on the computer. The data is stored in memory has a unique address so that it can be easily accessed.

The memory chip is divided into equal parts called cells, every cell has a unique address.

Byte addressable means, A byte is a memory unit for storage, when the size of a cell is 8 bits then the address is called byte address. A memory chip is full of such bytes and a binary address always points to a single byte only.

Computer counts by base 2:

 21 = 2                         

 22 = 2*2 = 4              

 23 = 2*2*2 = 8                       = 1 byte

 210 = 1,024                 = 1 kilobyte (KB) = 1K

 220 = 1,048,576                      = 1 megabyte (MB)    =1M


For example, the memory chip configuration is represented as 32K × 16

Here 32K is indicated the number of cells in the memory chip

16 indicates the size of the cell in bits that can be stored in the cell.

Hence the address is required


Here K is kilobytes. The 1M is represented in the power of 2 as 210.

32 can be represented as 25.

16 bits are equal to 2 bytes and it can be represented as 21.


32K × 16 = 25 * 210 * 21 = 216

Hence in byte-addressable memory, 16 bits are required to address a  32K × 16

Step 2


How many bits are required to address a 4M × 16 main memory when the main memory is byte-addressable.

Here M is Megabytes. The 1M is represented in the power of 2 as 220.

4 can be represented as 22.

16 bits are equal to 2 bytes and it can be represented as 21.


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