Average access time, Made Easy- Psus Book

Question

Two level memory contains cache and main memory. Cache access time is 20ns and main memory access time is 120ns/word.The size of block is 4 words. Main memory is referred  20% of times.What is the average access time:

a) 120ns. b) 20ns c) 116ns d) 150ns

Answer 1

Here memory hierarchy is followed , so for each main memory reference block need to be copied back to cache

So 1 word memory access time =120 ns

so 4 word memory access time=Tm=4*120=480 ns

Hc=cache hit ratio= 80%=0.8

(1-Hc)=memory hit ratio=cacge miss ratio=20%=0.2

Tc=20ns

Tm=480ns

Average Access time=H_c*T_C + (1-H_c)*(T_m+T_c)=(0.8*20)+0.2*(480+20)=116ns

Comments

as there is no mention of hierarchy ..how are u so sure to take it as hierarchy memory system?

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Though There is no mention of memory hierarchy but memory hierarchy meaning is inferenced from the question ...it clearly says it consist of cache and memory ....and memory is referred only 20% of the time that means cache is referred 80% of time... This is implicitly giving memory hierarchy meaning

Answer 2

I think its b. 20ns