Question

3. (12 points) Consider a cache has lines of 16 bytes and a total size of 16 kB. The main memory is 16MB and a word takes 4 bytes. For the hexadecimal main memory addresses FFF666, show the following information in hexadecimal format a. Tag and word values for associative cache b. Tag, set and word values for a two-way set-associative cache

Answer 1

Line size = 16 bytes = 2⁴ bytes

Cache size = 16 KB = 2⁴*2¹⁰ bytes = 2¹⁴ bytes

Main memory size = 16 MB = 2⁴*2²⁰ bytes = 2²⁴ bytes

So, the address size is of 24 bits.

I'm assuming that the memory is byte addressable.

a) ASSOCIATIVE:

As line size is 2⁴ bytes so, line offset is of 4 bits.

Remaining bits = 24-4 = 20 bits.

Therefore, TAG is of 20 bits.

TAG(20 bits)

OFFSET(4 bits)

Address = FFF666 = 11111111111101100110 0110

Therefore TAG = 11111111111101100110 (in binary) = FFF66(in hex)

WORD = 0110 ( in binary) = 6(in hex)

b) SET ASSOCIATIVE:

As line size is 2⁴ bytes so, line offset is of 4 bits.

Number of cache lines = Cache size/ line size = 2¹⁴ bytes/2⁴ bytes = 2¹⁰

Number of sets = number of cache lines / associativity = 2¹⁰ / 2 = 2⁹ [ as 2 way set associative ]

Therefore Set Index is of 9 bits.

Remaining bits = 24-4-9 = 11 bits.

Therefore, TAG is of 11 bits.

TAG(11 bit)

SET(9 bit)

OFFSET(4 bit)

Address = FFF666 = 11111111111 101100110 0110

Therefore TAG = 11111111111 (in binary) = 7FF(in hex)

SET = 101100110 = 166(in hex)

WORD = 0110 ( in binary) = 6(in hex)