Memory Access Time [15 points] Consider a MIPS processor that includes a cache,
ID: 3826978 • Letter: M
Question
Memory Access Time [15 points] Consider a MIPS processor that includes a cache, a main memory, and a hard drive. Access times of cache memory, main memory, and hard drive are 5 ns, 200 ns, and 1000 ns, respectively. Assume that cache memory is divided into instruction cache and data cache. Assume that data cache has a 90% hit rate. Assume that main memory has a 98% hit rate and hard drive is perfect (it has a 100% hit rate). Use the following formula for average access time calculations: AMST = tcache + MRcache(tMM + MRMM tV M ) (b) How many clock cycle per instruction (CPI) on average are needed for load and store word instructions. Memory Access Time [15 points] Consider a MIPS processor that includes a cache, a main memory, and a hard drive. Access times of cache memory, main memory, and hard drive are 5 ns, 200 ns, and 1000 ns, respectively. Assume that cache memory is divided into instruction cache and data cache. Assume that data cache has a 90% hit rate. Assume that main memory has a 98% hit rate and hard drive is perfect (it has a 100% hit rate). Use the following formula for average access time calculations: AMST = tcache + MRcache(tMM + MRMM tV M ) (b) How many clock cycle per instruction (CPI) on average are needed for load and store word instructions. Memory Access Time [15 points] Consider a MIPS processor that includes a cache, a main memory, and a hard drive. Access times of cache memory, main memory, and hard drive are 5 ns, 200 ns, and 1000 ns, respectively. Assume that cache memory is divided into instruction cache and data cache. Assume that data cache has a 90% hit rate. Assume that main memory has a 98% hit rate and hard drive is perfect (it has a 100% hit rate). Use the following formula for average access time calculations: AMST = tcache + MRcache(tMM + MRMM tV M ) (b) How many clock cycle per instruction (CPI) on average are needed for load and store word instructions.Explanation / Answer
Solution:
AMST= (0.9*5)+(0.1*((0.98*200)+(0.02*1000))= 26.1 ns
So, CPI= Instruction cout*AMST= 10^6*26.1*10^-9= 0.0261
Please note that 10^6 is instruction count because it's MIPS processor which means million instructions per second.
I hope this helps. Don't forget to give a thumbs up if you like this.