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4 votes
4 votes
Suppose that a cache is 20 times faster than main memory and cache memory can be used 80% of the time. The speed up factor that can be achieved by using the cache is

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Best answer
8 votes
8 votes

Let Cache access time is X, then main memory access time will be 20X .

Speed up factor = (Time taken to get data W/O Cache)/(Time taken to get data with Cache)

Time taken to get data W/O Cache: 20X

Time taken to get data with Cache: .8X + .2(20X) =4.8X

Then speed up will be 20X/(4.8X) = 4.166 (Answer)

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4 votes
4 votes

Soverall = ((1-f) + ∑$\frac{f}{s}$)-1 
 
           = ((1-.8) + ( $\frac{.8}{20}$) -1

           = 4.166

3 votes
3 votes
Hierarchical access to main memory..
T​c : Cache Access Time, T​m : Memory Access time, Tm = 20T​c, Hit Ratio = 0.80

Effective access Time = Tc + 0.20 *Tm = Tc + 0.20*20Tc = 5*Tc
Speed Up = Memory Access Time Without Cache / Memory Access Time With Cache
= Tm/Effective Access time = 20*Tc/5*Tc

Speed up = 4
Answer:

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