Question

Memory organization a) Suppose that a 32MB system memory is built from 32 1MB RAM chips. How many address lines are needed to select one of the memory chips? Suppose a system has a byte-addressable memory size of 4GB. How many bits are required for each address? Suppose that a system uses 16-bit memory words and its memory built from 32 1Mx 8 RAM chips. How large, in words, is the memory on this system? Suppose that a system uses 32-bit memory words and its memory is built from 16 1Mx 8RAM chips. How many address bits are required to uniquely identify each memory word? Suppose we have a 1024-word memory that is 16-way low-order interleaved. What is the size of the memory address offset field? b) c) d) e)

Answer 1

a)

32 = 2⁵

As we need to select between 32 different memory chips we need 5 bits to do so.

b)

As 4 GB = 2³² bytes, we need 32 bits for a byte addressable memory,

c)

Size of a memory word = 16 bits = 2 bytes

Size of a single ram chip = 1 MB

Note that 1M x 8 means an array of bits with 1M rows and 8 columns, which equals 1 MB of memory.

Total size of 32 memory chips = 32 x 1 MB = 32 MB

But each memory word is 2 bytes

So size of memory in words is

32 16 M

d)

Size of 1 memory chip = 1 MB

Size of 16 memory chips = 16 MB

Size of 1 memory word = 32 bits = 4 bytes

Number of memory words in 16 MB memory

16 x 1024 x 1024 bytes 4 bytes

4 MB = 2²²

To address 4 MB of memory, we need 22 bits .

e)

Low-order interleaving means that the low order bits of the address specify which memory bank contains the address of interest.

To address 1024 - word memory, we need 10 bits because 1024 = 2¹⁰

We are given a 16 way interleaved memory, it means that there are 16 memory banks, to select one memory bank, we need

16 = 2⁴ , so we need 4 bits to select a memory bank

So bits needed for offset within a memory bank = 10 - 4 = 6 bits