edited by
732 views
1 votes
1 votes
A demand paging system,with page table held in integer,takes $5ms$ to service a page fault if an empty page is available,or if the page to be replaced  is not dirty.It takes $15ms$ if the replaced page is dirty. Memory access time is $1\mu s.$ Assume we want an effective access time of $2\mu s$ and that the page to be  replaced is dirty $60\%$ of the time. The approximate maximum acceptable  page fault rate to meet this time requirement will be______$\%?$
$\text{(Correct to two decimal places).}$
edited by

2 Answers

1 votes
1 votes
2 = (1 - p)*1 + p * [ (0.4 * 5000 ) + ( 0.6 * 15000 )]

2 = 1 - p + p [ 2000 + 9000 ]

2 -1 = -p + p * 11000

1 = p [11000 - 1 ]

1 = p * 10999

p = ( 1 / 10999)  * 100 %

p = 0.00909

Therefore p  ≈ 0.01
0 votes
0 votes
2*10^-6=.6((1-p)*10^-6+p*15*10^-3)+.4((1-p)*10^-6 +p*5*10^-3) this is equation for the given problem.here i converts all the units into second.for understanding purpose first we for solve 60% dirty then for remaining pages then combinly it will give effective memory access time. after solving equation p=1/10999 = 0.0000909 = 0.00909 %

Related questions

0 votes
0 votes
1 answer
2