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homework8 Version 0
👤 Author: by 1146715683qqcom 2018-12-23 15:48:05
Answer:

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

  2. Counters are add 1 when each page frame replace a new page.(new page是指之前没有出现过的页面)

  3. Counters are minus 1 when the replaced page was not appeared.(也就是在之后的页面中都不再出现)

  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.

  5. 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































































































































































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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