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jpascua
Posts: 197
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Posted 00:03 Nov 28, 2015 |
A couple of questions: 1. I'm going through the lecture notes again and I just want to confirm if this is correct: The question is to determine whether the following schedules are serializable. a. r1(x),w1(x),r2(y),r1(y),w2(y),w1(y),w2(x) => No. Conflict in r1(y) and w2(y) and w2(y) and w1(y). b. r1(x),w1(x),r1(y),w1(y),r2(y),w2(y),w2(x) = Yes. No conflicts. Thanks. 2. Also, just to refresh what was covered last Monday, how did we run through the schedule to check whether they're serializable? Did we check r1(x) against w1(x),r2(y),r1(y),w2(y),w1(y), and w2(x). Then w1(x) against r2(y),r1(y),w2(y),w1(y), and w2(x) ... and so forth (until the end) to check whether there's a conflict?
Last edited by jpascua at
00:59 Nov 28, 2015.
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cysun
Posts: 2935
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Posted 10:19 Nov 29, 2015 |
1. Your answers are correct, though the reasoning is a little off. Schedule (b) also has conflicting actions like w1(y) and r2(y). A schedule is not serializable only when conflicting actions form a cycle in the precedence graph. 2. Yes, you check each action against all the actions after it. |
AndersonChristoph
Posts: 34
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Posted 12:33 Dec 03, 2015 |
For number 1 part b, does r1(x) conflict with w2(x)? Likewise, does w1(x) conflict with w2(x)? I would go with yes, they're read-write & write-write conflicts (respectively). |
cysun
Posts: 2935
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Posted 19:56 Dec 03, 2015 |
Yes and yes. |