GATE CSE 2000 | Question: 2.22

Question

Suppose the time to service a page fault is on the average $10$ milliseconds, while a memory access takes $1$ microsecond. Then a $99.99\%$ hit ratio results in average memory access time of

• A. $1.9999$ milliseconds
• B. $1$ millisecond
• C. $9.999$ microseconds
• D. $1.9999$ microseconds 

Comments

#help

https://gateoverflow.in/77662/self_doubt             

---

effective memory access time = miss rate*(service time+memory acess time) +hit rate*memory access time .

---

From the above formula

$p=$miss rate$,1-p=$hit rate$,ma=$memory access time$,ps=$page fault time $(or)$ service time

$EMAT=p(ps+ma)+(1-p)\times ma$

$EMAT=p(ps)+p(ma)+ma-p(ma)$

$EMAT=p(ps)+ma$

Put the values $p=\frac{0.001}{100}=0.0001=1\times 10^{-4}=10^{-4},ps=10\times 10^{-3}sec,ma=10^{-6}sec$

we get $EMAT=10^{-4}(10\times10^{-3}sec)+10^{-6}sec$

$EMAT=10^{-4}\times(10^{-2}sec)+10^{-6}sec$

$EMAT=10^{-6}sec+10^{-6}sec$

$EMAT=1\mu sec+1\mu sec$

$EMAT=2\mu sec$

please correct me if I'm wrong$?$

---

lakshman bro, you are not wrong at all

Even the formula that you are deriving na

it is already there on internet

time = memory access time + page fault rate(page fault service time)

 

your answer is correct

1.99999 $\mu s$ is nearly equal to 2 $\mu s$

Answer 1

Average memory access time = (P*t1) + [(1-P)t2]
= (0.9999*1) + [(1-0.9999) *10000]
= (0.9999) + (0.0001 * 10000)
= 0.9999 + 1
= 1.9999 microseconds