Memory-Control Unit

Question

If size of MAR &MBR are $32$ bit and $16$ bit respectively then what is the main memory size.
if memory is byte addressable .

Comments

my approach -----------
 If an MAR is of size 32 bits, that means it can hold upto 2^32 numbers and it hence can refer up to 2^32 bytes of memory = 4GB of memory .If memory is byte addressable.
and if memory is word addressable then 2^32 words and word size here is 2 B so memory capacity =2^32 * 2B=8 GB.
MBR/MDR is used to send the value to be written to/read off the memory. If I have a MBRs of size 16 bits, it means a maximum of 2 bytes can be written to/read off the memory at a time.
m i right ??

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yes , right

Answer 1

MAR=Memory Adress register

it holds the number of adress bit and here it is 32

so the all possible adresss=2^32

so memory capacity 2^32=4GB

and MBR is the data register ,it indicates the amount of data can be written at a time

it can write 16 bit of data....

so

2 Byte(as it is Byte adressable)