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Consider the following four schedules due to three transactions (indicated by the subscript) using read and write on a data item x, denoted by $r(x)$ and $w(x)$ respectively. Which one of them is conflict serializable?

  1. $r_1(x)$; $r_2(x)$; $w_1(x)$; $r_3(x)$; $w_2(x)$;
  2. $r_2(x)$; $r_1(x)$; $w_2(x)$; $r_3(x)$; $w_1(x)$;
  3. $r_3(x)$; $r_2(x)$; $r_1(x)$; $w_2(x)$; $w_1(x)$; 
  4. $r_2(x)$; $w_2(x)$; $r_3(x)$; $r_1(x)$; $w_1(x)$;
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3 Answers

Best answer
26 votes
26 votes

(D)  make precedence graph for all the options, for option (D) only graph will be acyclic, hence (D) is CSS.

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In option D, there is no interleaving of operations. The option D has first all operations of transaction 2, then 3 and finally 1 There can not be any conflict as it is a serial schedule with sequence 2 --> 3 -- > 1
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