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homework_7 Version 0
👤 Author: by 353575688qqcom 2017-12-20 09:45:36


a.  1. The initial value of the counters is: 0


   2. Counters are add 1 when each page frame replace a new page.


   3. Counters are minus 1 when the replaced page was not appeared.


   4. Every time a new page need to replace page frame, first to determine all the page frames, select the current counter value is the smallest frame as a replacement pages, if there are multiple page frames of the same counter value, then select the page which first enter page frame as the replacement page.


b. reference string: 1, 2, 3, 4, 5, 3, 4, 1, 6, 7, 8, 7, 8, 9, 7, 8, 9, 5, 4, 5, 4, 2


  c represent counter


  M represent Miss, also page fault


  H represent Hit



































































































































































Page frame1



1(M, c=1)



1(c=1)



1(c=1)



1(c=1)



5(M, c=2)



5(c=2)



5(c=2)



5(c=1)



5(c=1)



5(c=1)



Page frame2



2(M, c=1)



2(c=1)



2(c=1)



2(c=1)



2(c=1)



2(c=1)



2(c=1)



2(c=1)



2(c=1)



Page frame3



3(M, c=1)



3(c=1)



3(c=1)



3(H, c=0)



3(c=0)



1(M, c=0)



6(M, c=0)



7(M, c=1)



Page frame4



4(M, c=1)



4(c=1)



4(c=1)



4(H, c=1)



4(c=1)



4(c=1)



4(c=1)



Page frame1



5(c=1)



5(c=1)



5(c=1)



5(c=1)



5(c=1)



5(c=1)



5(c=1)



5(H, c=1)



5(c=1)



5(H, c=0)



Page frame2



8(c=2)



8(c=2)



8(H, c=2)



8(c=2)



8(c=2)



8(H, c=1)



8(c=1)



8(c=1)



8(c=1)



8(c=1)



Page frame3



7(M, c=1)



7(H, c=1)



7(c=1)



7(c=1)



7(H, c=0)



7(c=0)



7(c=0)



7(c=0)



4(M, c=0)



4(c=0)



Page frame4



4(c=1)



4(c=1)



4(c=1)



9(M, c=2)



9(c=2)



9(c=2)



9(H, c=1)



9(c=1)



9(c=1)



9(c=1)



Page frame1



5(c=0)



2(M, c=0)



Page frame2



8(c=1)



8(c=1)



Page frame3



4(H, c=0)



4(c=0)



Page frame4



9(c=1)



9(c=1)



So there are 12 page faults


c.



































































































































































Page frame1



1(M)



1



1



1



1



1



1



1(H)



6(M)



6



Page frame2



2(M)



2



2



5(M)



5



5



5



5



5



Page frame3



3(M)



3



3



3(H)



3



3



3



7(M)



Page frame4



4(M)



4



4



4(H)



4



4



4



Page frame1



8(M)



8



8(H)



8



8



8(H)



8



8



4(M)



4



Page frame2



5



5



5



5



5



5



5



5(H)



5



5(H)



Page frame3



7



7(H)



7



7



7(H)



7



7



7



7



7



Page frame4



4



4



4



9(M)



9



9



9(H)



9



9



9



Page frame1



4(H)



4



Page frame2



5



2(M)



Page frame3



7



7



Page frame4



9



9



So there are 11 page faults.

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