In the problem below, x and y are local application program values Read( ) and Writel ) fetch and stored thepersistent values in a database. The database values are read / written by two different transactions. The timesequence of these operations is shown in the table below. You cannot change the time sequence.Take the initial values of x and y to be x = 4, y = 2.a) In this nart the transactions are operating with isolation level READ UNCOMMITTED (no locks)TimeTransaction 1Transaction 2Read(y)y=y+52Read(y)y=y+ 1034Write(y)Read(x)7Read(x)8X=x+ y:Write(y)Commit9.1011x=x+yWrite(y)1213CommitFor the execution order shown above, what are the final values of x and y?Answer: x=b) There is a problem with the interleaved execution plan for these transactions.What is the problem, i.e., which concurrency problem does this demonstrate?Answer:c) Rewrite the sequence of operations using the standard technique for dealing with concurrencyproblems of this type for isolation level SERIALIZABLE. In the columns on the right specifythe type of requests made. (type your answer into the grid below) You cannot change the time sequence.TimeTrans # RequestsTransaction 1Transaction 2123456.7910111213d) Was the concurrency problem completely solved? If not, what happened and what should be done next?Answer:

NOTE: According to the rules, the first 3 questions are answerable. Given, T1 T2 R(y)…

In the problem below, x and y are local application program values Read( ) and Write( ) fetch and stored the persistent values in a database, The database values are read / written by two different transactions. The time sequence of these operations is shown in the table below. You cannot change the time sequence. Take the initial values of x and y to be x = 4, y = 2. a) In this part the transactions are operating with isolation level READ UNCOMMITTED (no locks) Time | Transaction 1 Transaction 2 Read(y) y=y+5 Read(y) yy+ 10 3 Write(y) Read(x) 6. Read(x) X=x+ y: Write(y) 8. 10 Commit 11 Xx+y Write(y) Commit 12 13 For the execution order shown above, what are the final values of x and y? Answer: x=

In the problem below, x and y are local application program values Read( ) and Write( ) fetch and stored the persistent values in a database, The database values are read / written by two different transactions. The time sequence of these operations is shown in the table below. You cannot change the time sequence. Take the initial values of x and y to be x = 4, y = 2. a) In this part the transactions are operating with isolation level READ UNCOMMITTED (no locks) Time | Transaction 1 Transaction 2 Read(y) y=y+5 Read(y) yy+ 10 3 Write(y) Read(x) 6. Read(x) X=x+ y: Write(y) 8. 10 Commit 11 Xx+y Write(y) Commit 12 13 For the execution order shown above, what are the final values of x and y? Answer: x=

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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