1 The size of addressable memory is not limited by the size of the memory data bus, the size of instructions, the size of arithmetic operands (e.g., register size), or even the size of addresses …

(For a) & b), my take is $2^ {16}$ of whatever are addressable units - the bus size not being an indication (see e.g. Intel 8086↔8088). It does not read 16 address lines/signals, nor does it …

Dec 7, 2019 · I have an associative mapped cache with 10 tag bits and an offset of 7bits. What is the size of each main memory block in words (word addressable) and main memory size in …

It is word-addressable and direct-mapped with 32B lines. It is able to fetch one line at a time. Given the access pattern: 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, …, 32768, where each access …

I dont know if this is any important IRL, but I am trying to understand lines below. 4K words of main memory (this implies 12 bits per address). 4M X 16 means the memory is 4M long (4M = …

I am confused on byte addressable/word addressable architecture. I have studied MIPS implementation and I came to know that when the data is retrieved from main memory, it is …

Let's say I have a memory with 4,096 addressable location, doesn't matter how wide those memory addresses are (8 bit wide, or 16, or 64) I only care about their location.

Your logical address is made of 16 bits, that means you have a addressable space of $2^ {16}$ bits. The page size is typically a power of 2, $2^n$, in this case $2^n = 256 \Rightarrow n = 8$.

That means - even if the Memory (as a component - RAM for example) isn't byte-addressable by itself (which means that each address contains, for example, 4 bytes of data instead of 1), …

In MIPS processor, address bus is of 32 bits. So on addressing an instruction, a whole 32 bit instruction is fetched. How is it byte addressable then? I mean if on addressing a particular …