table - A machine has a 32 bit address and an 8KB machine -

table - A machine has a 32 bit address and an 8KB machine -

a machine has 32-bit address space , page size of 8kb. page table exclusively in hardware, 1 32-bit word per entry. when process scheduled, page table copied hardware memory, @ rate of 1 word / 100 ns. if each process runs 100ms (including time load page table), fraction of cpu time devoted loading page tables?

yes homework problem. looked , far i've found that

size of pagetable(bytes) = number of entries *size of entry. hence x= 32*8192(bits) = 262144

i know wrong reply , it's 52; 4288ms using in latter equation of

time load page table / total time = cpu utilize

a solution can't quite understand posted here.

http://www.cs.inf.ethz.ch/ssw/exercises/solution_04.pdf

the tricky part here calculating size of page table. 1 time have that, multiply times time takes load each table entry. not relevant how big table entry is--only how long takes load.

so think 32 bit address space. how many 8k chunks there in it. that's part of document referenced subracting bits. takes 13 bits describe 8k. notice 2 ^ 13 = 8k (quick calculate noting 1024 takes 10 bits. sort of mnemonic , easy remember. 8 takes 3 bits , 10 + 3 = 13. or utilize calculator see powerfulness of 2 equals 8k.)

32 bits whole address space less 13 gives 19, there 2 ^ 19 pages.

now multiply 2 ^ 19 times 100 ns , bingo. have it.

table size 32-bit memory-address