Question

Assume we have a computer where the CPI is 1.0 when all memory accesses (including data and instruction accesses) hit in the cache. The cache is a unified (data + instruction) cache of size 256 KB, 4-way set associative, with a block size of 64 bytes. The data accesses (loads and stores) constitute 50% of the instructions. The unified cache has a miss penalty of 25 clock cycles and a miss rate of 2%. Assume 32-bit instruction and data addresses. Now, answer the following questions

a) What is the tag size for the cache?
b. How much faster would the computer be if all memory accesses were cache hits?

Answer 1

a. 16

b. 1.75

Step-by-step explanation:

Number of bits used for block offset = log 64 = 6.

Number of sets in the cache = 256K/(64 * 4) = 1K

Number of bits for index = log 1K = 10

Number of bits for tag = 32 - (10 + 6) = 16

Now,

CPI = CPIexecution + StallCyclesPerInstruction

For computer that always hits, CPI would be 1 i,e CPI = 1

Now let us compute StallCyclesPerInstruction for computer with non-zero miss rate

StallCyclesPerInstruction = (Memory accesses per instr) * miss rate * miss penalty

Memory accesses per instruction = 1 + 0.5 (1 instruction access + 0.5 data access)

StallCyclesPerInstruction = 1.5 * 0.02 * 25 = 0.75

Therefore, CPI = 1.75

Hence the computer with no cache misses is 1.75 times faster.