Question

Cache Layout: A processor has a separate D-cache and an I-cache.

D-cache: 64KB, 4-way set associative, block size of 1 word, write-back policy

I-cache: 32KB, direct mapped cache, block size of 1 word The processor uses the LRU algorithm for its replacement policy.

Answer the following questions. Make sure that you account for all the book -keeping bits.

A word is 4 bytes

(a) Calculate the number of tag, index and offset bits for the D-cache.

(b) Calculate the number of tag, index and offset bits for the I-cache.

(c) How many bits are needed to implement the D-cache?

(d) How many bits are needed to implement the I-cache?

Answer 1

Assuming the processor is 32 bit i.e. address size is 32 bit.

a. In D-cache, 1 word = 4 byte.

Number of blocks = D-cache size/ block size = 64KB / 4 byte = 16K blocks

Since it is 4-way set associative, therefore number of sets = Number of blocks/4 = 16K/4 = 4K sets

Since the block size is 1 word and since memory is word-addressable with 1 word = 4 byte

So, offset bits = log₂ ( Number of words in a block) = log₂ 1 = 0 bit

So no bit will be used for offset since every block has single word.

Index bits = log₂ ( Number of sets) = log₂ (4K) = log₂ (4*2¹⁰) = 12 bits

Number of tag bits = 32 - Index bits = 32 - 12 = 20 bits

b. Number of blocks = (size of cache)/block size = 32 KB/ 4 Byte = 8K

Since size of block = 1 word, hence there will be 0 bit for offset

Number of bits in index field in direct mapped cache = log₂ ( Number of blocks) = log₂ 8K = log₂ (8*2¹⁰) = 13 bits

Number of bits in Tag = 32 - 13 = 19 bits

c. Since D-cache used write-back policy, so it will use 1 bit as dirty bit for every cache block to indicate whether a particular block has been modified or flushed into main memory.

Number of bits needed to implement D-cache = (Number of blocks)*(Number of bits in Tag field + 1 bit due to dirty bit)

= 16K * (20 + 1) bits = 21*2¹⁴ bits

d. Since I-cache do not need dirty bit

So, Number of bits needed to implement I-cache = Number of blocks * Number of bits in tag field = 8K * 19 bits = 19*2¹³ bits

Please comment for any clarification.