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斯坦福大学数据库系统实现第二版部分答案.pptx

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    Chapter 17
    17.1.1 0≤≤, initially A = 50, B = 25,with appropriate order of OUTPUT action to assure the consistency. a) B := A+B; A := A+B Action READ(A, t) READ(B, s) s:=t+s t:=t+s WRITE(A, t) WRITE(B, s) OUTPUT(B) OUTPUT(A) t 50 50 50 125 125 125 125 125 s Mem A Mem B Disk A Disk B 50 50 50 50 125 125 125 125 25 75 75 75 75 75 75 50 50 50 50 50 50 50 125 25 25 25 25 75 75 75 25 25 25 25 25 25 75 75 不能保持一致性 Crashes here
    17.1.1 b) A:=B+1; B:=A+1 Action READ(A, t) READ(B, s) t:=s+1 s:=t+1 WRITE(A, t) WRITE(B, s) OUTPUT(A) OUTPUT(B) t 50 50 26 26 26 26 26 26 s 25 25 27 27 27 27 27 Mem A Mem B Disk A Disk B 50 50 50 50 26 26 26 26 50 50 50 50 50 50 26 26 25 25 25 25 27 27 27 25 25 25 25 25 25 25 27 不能保持一致性
    17.1.1 • c) A:=A+B; B:=A+B; t Action 50 READ(A, t) 50 READ(B, s) 75 t := t + s s := t + s 75 WRITE(B, s) 75 OUTPUT(B) 75 WRITE(A, t) 75 OUTPUT(A) 75 s 25 25 100 100 100 100 100 Mem A Mem B 50 50 50 50 50 50 75 75 25 25 25 100 100 100 100 Disk A 50 50 50 50 50 50 50 75 Disk B 25 25 25 25 25 100 100 100 不能保持一致性
    17.2.2 • a) There are five events of interest, which we shall call: • A: database element A is written to disk. B: database element B is written to disk. • LA: the log record for A is written to disk. LB: the log record for B is written to disk. • $: the commit record is written to disk. • The undo rules imply the following constraints: • (1)$ must be last. (2)LA is before A. (3)LB is before B.(4)LA is before LB. • Note that (4) comes from the fact that we assume log records are written to disk in the order created. • We can conclude that LA is first; every other event must be preceded by something. Either A or LB can be next. In the first case, the only possible order of events is LA,A,LB,B, $. In the second case, A and B can then be written in either order, giving us the following two sequences: LA,LB,A,B,$ and LA,LB,B,A,$, or a total of 3 possible sequences.
    17.2.2 LA LA $ $ • n = 1 answer = 11=1 • n = 2 answer = 31=3(A,LB,B,选取一个盒子放A ,31, 剩余的盒子放(LB, B)) • n = 3 answer = 51∗31=15(选取一个盒子放A, 51,剩下LB,B,LC,C, 31) …∗51∗31=1∗3∗5∗…∗ 2−1 = (2n-1)!! ∗2 −1 −1 • answer = 2−1 1 1
    i. ii. At what points could the record written? For each possible crash point, how far back in the log we must look to find all possible incomplete transactions. Consider both the case that the record was or was not written prior to the crash. （a). i. ii. The can occur anywhere after / The only active transaction when the checkpoint began was S. If the crash occurs after the END CKPT, then we need only go back as far as the . However, if the crash occurs before the checkpoint ended, go back to (undo s) 17.4.4 ; ; ; ; ; ; ; ; ; ; ; ; ; ; ;
    17.4.4 ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; b) i. ii. The END CKPT record could appear anywhere after the record . The only active transaction when the checkpoint began was T. If the crash occurs after the END CKPT,  go back as far as the . if the crash occurs before the checkpoint ended, then any transaction that was active when the previous checkpoint ended may have written some but not all of its data to disk. In this case, the only other transaction that could qualify is S, so we must look back to the beginning of S, i.e., to the beginning of the log in this simple example (undo S)
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