With these characteristics for a processor:Base CPI: 1.2Processor speed: 2GHZMain Memory access time: 200nsL1 cache miss rate: 8%L2 cache direct-mapped miss rate: 2%L2 cache direct-mapped access speed: 15 cyclesL3 cache 8-way set associative miss rate: 0.5%L3 cache 8-way set associative speed: 40 cyclesa) What is the CPI for the system without the L2 cache (all misses access main memory)?Explain/show your work.b) What is the CPI for the system with the addition of the direct-mapped L2 cache?Explain/show your work.c) What is the CPI for the system with the further addition of the 8-way associative L3cache? Explain/show your work.

Below Here, base CPI refers to the CPI when data is found in cache. Hence, we have L1 cache access time, tL1 = 1.2 cycles L2 cache access time, tL2 = 15 cycles L3 cache access time, tL3 = 40 cycles L1 cache miss rate, mL1 = 0.08 L2 cache miss rate, mL1 = 0.02 L3 cache miss rate, mL1 = 0.005 Also, main memory access time in cycles = (access time in seconds) * (processor frequency in Hz) = (200 * 10-9) * (2 * 109) = 400 cycles Thus, main memory access time, tmemory = 400 cycles Thus, we use the cascaded formula to calculate the CPI.For Here, only L1 cache and main memory are in picture. Hence, CPI = tL1 + (mL1 * tmemory) = 1.2 + (0.08 * 400) = 33.2For Here, L1 cache, L2 cache and main memory are in picture. Hence, CPI = tL1 + (mL1 * [tL2 + (mL2 * tmemory)]) = 1.2 + (0.08 * [15 + (0.02 * 400)]) = 1.2 + (0.08 * [15 + 8]) = 1.2 + (0.08 * 23) = 3.04For Here, L1 cache, L2 cache, L3 cache and main memory are in picture. Hence, CPI = tL1 + (mL1 * [tL2 + (mL2 * [tL3 + (mL3 * tmemory)])]) = 1.2 + (0.08 * [15 + (0.02 * [40 + (0.005 * 400)])]) = 1.2 + (0.08 * [15 + (0.02 * [40 + 2])]) = 1.2 + (0.08 * [15 + 0.84]) = 1.2 + (0.08 * 15.84) = 2.4672