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A computer has a cache, main memory, and a disk used for virtual memory. If a referenced word is in the cache, $10$ $ns$ are required to access it. If it is main memory but not in the cache, total $200$ $ns$ are needed to load it into cache, and then the reference is started again.

If the word is not in main memory, total $500$ $ms$ are required to fetch the word from the disk, followed by $200$ $ns$ to copy it to the cache and then the reference is started again.

The cache hit ratio is $99%$, and the main memory hit ratio is $90%$. What is the average time in ns to access a referenced word on this system?

Ignore the time it takes to determine whether or not a reference may be found in cache or main memory.The hit ratio is the percent of the time a reference is found in the given level of the memory hierarchy.

  1. $732000$ $ns$
  2. $485750$ $ns$
  3. $500012$ $ns$
  4. $23009$ $ns$
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This Qtn is telling about simultaneous access in memory hirarchy.

avg access time = .99 * 10 + 0.01 ( .90*200  +  .10 * (500*10^6  + 200)  )  ns  = 500011.9 ns

 

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10+1/100*90/100(200+10)+1/100*10/100(500000000+200+10)
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Tcache = 10ns
Tmainmemory = 200ns
Tdisk = 500ms
Tcachehit = 0.99
Tcachemiss = 0.01
Tmainmemhit = 0.90
Tmainmemmiss = 0.10
EAT = Tcachehit(Tcache)+(Tcachemiss*Tmainmemhit(Tcache+Tmainmem))+Tcachemiss*Tmainmemmiss(Tcache+Tmainmem+Tdisk)
         = 0.99(10)+(0.01*0.90(10+200))+0.01*0.10(10+200+500*10^6)ns
         = 500011.9 ns
Answer:

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