Q7 ch-5 memory org.

Question

Let cache of 0.7 hit having average access time 9 times faster than that of memory.  if average access time increases 40% from 80ns. What would be new hit ratio.

A. 0.72     b. 0.36     c. 0.53     d. 0.80

 

My answer is 0.74

 

Am i correct?

Answer 1

Assuming hierarchical access:

$80 =  x + 0.3 \times 9x \implies x = \frac{800}{37}$

$1.4 \times 80 = \frac{800}{37} + (1-h) \frac{7200}{37} \implies  h = 0.53$

Answer 2

your formula be be different .

emat = cache time + miss rate ( memory access)

because every time there is a miss i have to go to main memory and even when there is a miss cache time has to be included.

let cache time to be x then main memory time will be 9x.

80= x+0.3(9x)

x=21.6  9x= 194.5

now new time will be 80 + 0.40*80= 80+32 = 112

112 = 21.6+ x( 194.5)

x=0.464.

which is actually miss rate. so hit rate = 1-miss rate . = 1-0.464 = 0.53

option c.

Comments

Just to mention, for write through cache, both cache and memory are usually updated simultaneously- so access time won't include cache time.

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Hey, can you plz tell why didnt you include cache + memory time in miss case and make it 10x instead of 9x in equation : 80= x+0.3(9x)

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lets take the concept according to miss rate . let p be the miss rate. ok  then hit rate =(1-p)

total time = cache *(1-p) + p ( service time)

service time include transmitting the page from main memory to cache . ( c+m)

cache time = c;

c(1-p) + p(c+m)

c-cp+cp+pm

c+pm. where c is cache time . p is miss rate and m is memory access. its simple in this way.

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plz explain why we have not multiply x with 0.7 as hit for cache is 0.7?