Question

The address of the last cell of a memory RAM is 3FFFFh.a total capacity of the main memory and 4M bits and the data bus can transfer the contents of two memory cells , respond
a)which register the size of the memory address(MAR) ?

b)where the maximum number of cells that can be stored in memory?

c)which the cell size of this machine ?

d)which the size of the memory buffer register (MBR ) ?

Answer 1

Hey there!:

Given the address of the last cell of a memory RAM is 3FFFFh then   :

There are 8 bits in a byte  

1024 bytes in a kilobyte  

 so 8*1024= 8192  

 so 8192 bits in a kb  

 32*8192 = 262144  

there are 262144 bits in 32KB .

a) 32 bit address registers must match 2^32 byte = 4 GB of physical memory.

However, with 32 Bit you can also address more than 4 GB, like Physical Address Extension (PAE) does.

The other way is that you can access 4 GB with less than 32 bit address register.

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b) Main Memory = 4M × 32Kbits, RAM chips = 32K × 4bit.

For this memory we require 4 × 2 = 8 RAM chips.

Each chip requires 18 address bits (ie. 218 = 256K).

And 1M × 8 bits requires 20 address bits (ie. 220 = 1M )

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c) The size of the storage cells is known as the word size for the computer.

In some computers, the word size is one byte while in other computers the word size is two, four, or even eight bytes. In our 4M main memory the size of cell is 4*2^20 bytes.

Each storage cell in main memory has a particular address which the computer can use for storing or retrieving data.  

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d) The size of MBR is the 4M bits of main memory.

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Hope this helps!