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The database system must control concurrent execution of transactions, to ensure that the database state remains consistent Before we examine how the database
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Silberschatz Korth Sudarshan: Database System Concepts, Fourth Edition
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read(A) temp := A * 01 A := A temp write(A) read(B) write(A) read(B) B := B + 50 write(B) B := B + temp write(B) Figure 156 Schedule 4 a concurrent schedule
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system can carry out this task, we must rst understand which schedules will ensure consistency, and which schedules will not Since transactions are programs, it is computationally dif cult to determine exactly what operations a transaction performs and how operations of various transactions interact For this reason, we shall not interpret the type of operations that a transaction can perform on a data item Instead, we consider only two operations: read and write We thus assume that, between a read(Q) instruction and a write(Q) instruction on a data item Q, a transaction may perform an arbitrary sequence of operations on the copy of Q that is residing in the local buffer of the transaction Thus, the only signi cant operations of a transaction, from a scheduling point of view, are its read and write instructions We shall therefore usually show only read and write instructions in schedules, as we do in schedule 3 in Figure 157 In this section, we discuss different forms of schedule equivalence; they lead to the notions of con ict serializability and view serializability
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T1 read(A) write(A)
read(A) write(A) read(B) write(B) read(B) write(B) Figure 157 Schedule 3 showing only the read and write instructions
Silberschatz Korth Sudarshan: Database System Concepts, Fourth Edition
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The McGraw Hill Companies, 2001
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Let us consider a schedule S in which there are two consecutive instructions Ii and Ij , of transactions Ti and Tj , respectively (i = j) If Ii and Ij refer to different data items, then we can swap Ii and Ij without affecting the results of any instruction in the schedule However, if Ii and Ij refer to the same data item Q, then the order of the two steps may matter Since we are dealing with only read and write instructions, there are four cases that we need to consider: 1 Ii = read(Q), Ij = read(Q) The order of Ii and Ij does not matter, since the same value of Q is read by Ti and Tj , regardless of the order 2 Ii = read(Q), Ij = write(Q) If Ii comes before Ij , then Ti does not read the value of Q that is written by Tj in instruction Ij If Ij comes before Ii , then Ti reads the value of Q that is written by Tj Thus, the order of Ii and Ij matters 3 Ii = write(Q), Ij = read(Q) The order of Ii and Ij matters for reasons similar to those of the previous case 4 Ii = write(Q), Ij = write(Q) Since both instructions are write operations, the order of these instructions does not affect either Ti or Tj However, the value obtained by the next read(Q) instruction of S is affected, since the result of only the latter of the two write instructions is preserved in the database If there is no other write(Q) instruction after Ii and Ij in S, then the order of Ii and Ij directly affects the nal value of Q in the database state that results from schedule S Thus, only in the case where both Ii and Ij are read instructions does the relative order of their execution not matter We say that Ii and Ij con ict if they are operations by different transactions on the same data item, and at least one of these instructions is a write operation To illustrate the concept of con icting instructions, we consider schedule 3, in Figure 157 The write(A) instruction of T1 con icts with the read(A) instruction of T2 However, the write(A) instruction of T2 does not con ict with the read(B) instruction of T1 , because the two instructions access different data items Let Ii and Ij be consecutive instructions of a schedule S If Ii and Ij are instructions of different transactions and Ii and Ij do not con ict, then we can swap the order of Ii and Ij to produce a new schedule S We expect S to be equivalent to S , since all instructions appear in the same order in both schedules except for Ii and Ij , whose order does not matter Since the write(A) instruction of T2 in schedule 3 of Figure 157 does not con ict with the read(B) instruction of T1 , we can swap these instructions to generate an equivalent schedule, schedule 5, in Figure 158 Regardless of the initial system state, schedules 3 and 5 both produce the same nal system state We continue to swap noncon icting instructions: Swap the read(B) instruction of T1 with the read(A) instruction of T2 Swap the write(B) instruction of T1 with the write(A) instruction of T2 Swap the write(B) instruction of T1 with the read(A) instruction of T2
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