Recall, the formula for average memory access time is AMAT = t_h + r_m middot t_
ID: 3815069 • Letter: R
Question
Recall, the formula for average memory access time is AMAT = t_h + r_m middot t_m, where t_h is the hit time, r_m is the miss rate, and t_m is the miss time (or miss penalty). Further recall that this formula can be applied recursively, e.g., the miss penalty for a given cache level is the AMAT for the next lower level of cache. Suppose a computer system has two levels of cache. L1 and L2, and a main memory. The L1 cache has a hit time of 0.5ns, the L2 cache has a hit time of 5.0ns, and main memory has a hit time of 50ns. Assume the miss rate for the L1 cache is 5 percent, the miss rate for the L2 cache is 25 percent, and that the data is always in main memory. What is the AMAT for this system? List and briefly describe the four primary cache design considerations. A computer has 64GB virtual memory and 512MB physical memory. Assume the page size is 16KB. How many entries would be needed for a traditional page table? How many entries would be needed for an inverted page table? A simple performance model of a hard drive involves queue time, controller time, seek time, rotational delay time, and transfer time. Briefly describe each of these performance terms. A small embedded processor has 128MB byte-addressable memory and a 16MB cache. The block size is 64 bytes, and the cache is 8-way set associative. How many bits are needed for the offset field? How many address bits does the memory require? How many cache blocks are there? How many cache sets are there? How many index bits are needed? How many tag bits are needed?Explanation / Answer
Solution:
1.)
Let's write down the things given in the question:
L1 cache hit time= 0.5 ns
L2 cache hit time= 5 ns
main memory access time= 50 ns
L1 miss rate= 5%,so hit rate will be= 95%
L2 miss rate= 25%,so hit rate will be= 75%
Now, AMAT= (0.95*0.5)+(0.05*((0.75*5)+(0.25*(0.5+5+50))))= 1.35625 ns
So, AMAT= 1.35625 ns
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