Question

Question 3. A computer has a memory space of 16 GB. a) How many address lines are required to span this address space, assuming it is byte- addressed? b) This computer has a block of 4 GB 32-bit-wide memory built using 512 MB static RAM chips that are each 8 bits wide. How many RAM chips are required to implement the memory?

Answer 1

a)

Since the memory is byte-addressable, so each byte needs to have a unique address.

Now,

16 GB

= 2⁴ GB

= (2⁴ * 2¹⁰) MB

= (2⁴ * 2¹⁰ * 2¹⁰) KB

= (2⁴ * 2¹⁰ * 2¹⁰ * 2¹⁰) B

= 2³⁴ bytes

Thus, number of address lines required to span this address space is

34

b)

Number of RAM chips required

= (Size of total memory) / (Size of 1 chip)

= (4 GB * 32) / (512 MB * 8)

Since 1 GB = 1024 MB, so continuing,

= (4 * 1024 MB * 32) / (512 MB * 8)

= 32