1.Consider a byte-addressable RAM for which memory addresses are 24 bits long. a

Question

1.Consider a byte-addressable RAM for which memory addresses are 24 bits long.

a.What is the capacity of RAM? In other words, what is the maximum number of bytes addressable? Express the result in Megabytes.

b.Suppose this RAM is divided into small blocks, each of size 16 bytes. How many memory blocks are possible?

Given a 24-bit address of a byte, we need to know which block the byte falls into and what the byte offset of the address is within the block. Draw a figure to show how the 24-bits address can be split into two parts that will help us identify the block number and the byte offset within the block. Indicate clearly the number of bits for each part.

Given a 24-bit address 40AA18 (written in hexadecimal), split the address and express which block the address falls into and what the byte offset is within that block. Express the block number and byte offset in decimal.   

Explanation / Answer

a).24bit machines, we can have 2^24 combination set of instructions.

so for a max capacity of a RAM in 24-bit machine, it should be able to accommodate 2^24 instructions set, which is

2^24 = 16777216  

so as 24 bits = 3 bytes, the RAM capacity should be 16777216  

16777216*4 bytes = 51539607552 bytes

which is same as -->

67108864/1024 = 50331648kB

16777216/1024 = 49152 MB

16384/1024 = 48GB

b).the number of blocks are 2^16/2^3=2^13

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