Compute the overall (Cache) performance CPU: A processor runs at 3.0 GHz and has a CPI=1.5 for a perfect cache (i. e. without including the stall cycles due to cache misses). Assume that load and store instructions are 25% of the instructions. The processor has an I-cache (Instruction – cache) with a 5% miss rate and a Dcache (Data memory cache) with 4% miss rate. The hit time is 1 clock cycle for both caches.
Assume that the time required to transfer a block of data from the main memory to the cache, i. e. miss penalty, is 40 ns.
a. Compute the number of stall cycles per instruction
b. Compute the overall (cycle per instruction) CPI
c. Compute the average memory access time (AMAT) in ns.

See explanation.

Explanation:

Given:

Instruction miss rate = I-cache miss rate = 5% = 0.05

Data miss rate = D-cache miss rate = 4% = 0.04

CPI (without including the stall cycles) = 1.5

Miss Penalty = 40 ns × 3 GHz = 120 cycles

load and store instructions = 25% = 0.25

hit time = 1 clock cycle

a. Compute the number of stall cycles per instruction

 Instruction miss cycles = I-cache miss rate * miss penalty

                                        = 0.05 * 120

                                        = 6

Data miss cycles = D-cache miss rate * miss penalty * load and store  

                                                                                                instructions

                            = 0.04 * 120 * 0.25

                            = 1.2

Total memory stall cycles = Instruction miss cycles + Data miss cycles

                                          = 6 + 1.2

                                          = 7.2

number of stall cycles per instruction = 7.2

b. Compute the overall (cycle per instruction) CPI

CPI (stall)  =  1.5 + 7.2

                    = 8.7

Ideal CPU = CPU time with stalls  / CPU time with perfect cache

                    = I x CPI (stall) x Clock cycle/ I x CPI x Clock cycle

                      = 8.7 / 1.5

                     = 5.8

c. Compute the average memory access time (AMAT) in ns.

Combined misses per instruction = 0.05 + 0.25 * 0.04 = 0.06

Combined Miss Rate = Combined misses per  I/ 1.25 access per I

Combined Miss Rate = 0.06 / 1.25 = 0.048

Average Memory Access time (AMAT) = Hit time + Miss rate × Miss penalty

                                                                = 1 + 0.048 * 120

                                                               = 6.7

                                                                = 6.76 cycles

                                                                 = 6.76 / 3

                                                                 = 2.253 ns

AMAT = 2.253 ns

c. would be your answer.

mark brainliest if correct! hope i .

the answer is you.