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prob.1 CSE 255 Query Optimization Problem CustomerName ( BranchName=‘Storrs’^ Balance > 1000 ( Branch x Account x Depositor)) Consider the Three Relations: Branch (BranchName, BranchCity, Assets) Account(AccountNumber, BranchName, Balance) Depositor(CustomerName, AccountNumber) Optimize the following Relational Express by Drawing the Initial Expression Tree Optimizing (Rewriting) the Tree The Query Finds all Customers who have Accounts at the Storrs Branch with Balance > 1000

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prob.2 CSE 255 Depositor BranchAccount CustomerName BranchName = Storrs ^ Balance > 1000 X X Create the Initial Query Tree

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prob.3 CSE 255 Algorithm 2 Algorithm 2 Problem Apply Algorithm 2 to the Schedule of Transactions on the next Slide In this Case, Wlock Implies Reading For this Problem: Draw the Precedence Graph Check to See if the Schedule is Serializable If So, Determine ALL Serial Schedules

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prob.4 CSE 255 Algorithm 2: Read/Write Lock Model Input: Schedule S for Transactions T 1, T 2, … T k Output: Is S Serializable? If so, Serial Schedule Method: Create a Directed Precedence Graph G: Suppose in S, T i :Rlock A. If T j : Wlock A is the Next Transaction to Wlock A (if it exists) then place an Arc from T i to T j. Repeat for all T i ’s. Note for all Rlocks before the Wlock on A! Suppose in S, T i :Wlock A. If T j : Wlock A is the Next Transaction to Wlock A (if it exists) then place an Arc from T i to T j. Further, if there exists T m :Rlock A after T i :Wlock A but before T j : Wlock A, then Draw an Arc from T i to T m. Review the Resulting Precedence Graph If G has Cycles - Non-Serializable If G is Acyclic - Topological Sort for Serial Schedule

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prob.5 CSE 255 Apply Algorithm 2 to Following Schedule T 1 T 2 T 3 T 4 (1) Rlock A (2)Rlock A (3)Wlock C (4) Unlock C (5)Rlock C (6) Wlock B (7)Unlock B (8)Rlock B (9)Unlock A (10)Unlock A (11)Wlock A (12)Rlock C (13)Wlock D (14)Unlock B (15)Unlock C (16)Rlock B (17)Unlock A (18)Wlock A (19)Unlock B (20)Wlock B (21)Unlock B (22)Unlock D (23)Unlock C (24)Unlock A

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prob.6 CSE 255 Algorithm 3 Input: Schedule S for Transactions T 1, T 2, … T k Output: Is S Serializable? If so, Serial Schedule Method: Create a Directed Polygraph Graph P: 1. Augment S with Dummy T o (Write Every Item) an Dummy T f (Read Every Item) 2. Create Initial Polygraph P by Adding Nodes for T o, T f, and Each T i Transaction, in S 3. Place an Arc from T i to T j Whenever T j Reads A in Augmented S (with Dummy States) that was Last Written by T i. Repeat this Step for all Arcs. Don’t Forget to Consider Dummy States! 4. Discover Useless Transactions - T is Useless if there is no Path from T to T f This is the “Initialization” Phase of Algorithm 3

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prob.7 CSE 255 Algorithm 3 Method: Reassess the Initial Polygraph P: 5. For Each Remaining Arc T i to T j (meaning that T j Reads Item A Written by T i ), Consider all other T T o and T T f that Also Writes A: If T i = T o and T j = T f then Add No Arcs If T i = T o and T j T f then Add Arc from T j to T If T i T o and T j = T f then Add Arc from T to T i If T i T o and T j T f then Add Arc Pair from T to T i and T j to T 6. Determine if P is Acyclic by “Choosing” One Transaction Arc for Each Pair - Make Choices Carefully 7. If Acyclic - Serializable - Perform Topological Sort without T o, T f for Equivalent Serial Schedule. Else - Not Serializable

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prob.8 CSE 255 Apply Algorithm 3 to Following Schedule T 1 T 2 T 3 T 4 (T 0 ) Write A, B (1) Wlock A (2)Rlock B (3)Unlock A (4) Rlock A (5)Unlock B (6) Wlock B (7)Rlock A (8)Unlock B (9)Wlock B (10)Unlock A (11)Unlock A (12)Wlock A (13)Unlock B (14)Rlock B (15)Unlock A (16)Unlock B (T F ) Read A, B

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prob.9 CSE 255 Algorithm 4: Optimistic CC Let T be a Transaction with Timestamp t Attempting to Perform Operation X on a Data Item I with Readtime t R and Writetime t W If (X = Read and t t W ) or (X = Write and t t R ) then Perform Operation If t > t R then set t R = t for Data Item I If t > t W then set t W = t for Data Item I If (X = Write and t R t < t W ) the Do Nothing since Later Write will Cancel out the Write of T If (X = Read and t < t W ) or (X = Write and t < t R ) then Abort the Operation 1st - T trying to Read Item Before it was Written 2nd - T trying to Write an Item Before it was Read

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prob.10 CSE 255 Algorithm 4 Algorithm 4 Problem Apply Algorithm 4 to the Schedule of Transactions on the next Slide Assume the Following Transaction Times T 1 = 175, T 2 = 150, T 3 = 200, T 4 = 225 Determine the RT and WT for A, B, C, and D at each Step Indicate when a Transaction Aborts or has No Effect Remember, RT=WT=0 for A, B, C, and D Prior to the 1st Step in the Schedule

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prob.11 CSE 255 T 1 =175T 2 =150 T 3 =200 T 4 =225 A B C D (1) Read A (2)Read A (3)Write C (4) Read C (5) Write B (6) Read B (7) Write A (8) Read C (9)Write D (10)Read B (11) Write A (12) Write B Algorithm 4 Problem Initially RT/WT of A, B, C, and D are all Zero (0)

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© 1997 UW CSE 11/13/97N-1 Concurrency Control Chapter 18.1, 18.2, 18.5, 18.7.

© 1997 UW CSE 11/13/97N-1 Concurrency Control Chapter 18.1, 18.2, 18.5, 18.7.

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