Computer Science: An Overview (13th Edition) (What's New in Computer Science)
13th Edition
ISBN: 9780134875460
Author: Glenn Brookshear, Dennis Brylow
Publisher: PEARSON
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Chapter 9, Problem 45CRP
Program Plan Intro
Wound wait protocol:
In this protocol, an older transaction is given priority.
- If older requires access to an item that is locked by a younger transaction, the younger transaction is forced to release all of its data items, and its activities are rolled back.
- The older transaction is given access to the item it required, and the younger transaction is forced to start again.
- If a younger transaction is repeatedly rolled back, it grows older in the process and ultimately become one of the older transactions with high priority.
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How is the "two-phase commit" protocol beneficial in a transactional context?
Let 2PL be a scheduler based on the two-phase locking protocol and let TS be a scheduler based on the timestamping protocol. Here is a schedule of three transactions:r1(X), w1(X), r2(X), w2(X), r3(Y), w3(Y), r1(Y), w1(Y)Determine the following:Is the schedule serial?Is it serializable?Can it be produced by 2PL?Then, identify the true statement from the list below.
a)
S is not serial and S cannot be produced by 2PL
b)
S is serial and S can be produced by 2PL
c)
S is not serializable and S cannot be produced by TS
d)
S is not serial and S cannot be produced by TS
which one ?
The first rule in the basic timestamping protocol for concurrency control is:
If transaction T asks to read the variable P, check for any younger transaction’s write.
If WriteTimestamp(P) > TimeStamp(T), then T is late to read (the value of P it needs is already overwritten, will be inconsistent read)
Explain why this rule states that the value of P that this transaction needs is already overwritten.
Chapter 9 Solutions
Computer Science: An Overview (13th Edition) (What's New in Computer Science)
Ch. 9.1 - Identify two departments in a manufacturing plant...Ch. 9.1 - Prob. 2QECh. 9.1 - Summarize the roles of the application software...Ch. 9.2 - Prob. 1QECh. 9.2 - Prob. 2QECh. 9.2 - Prob. 4QECh. 9.2 - Prob. 5QECh. 9.2 - Prob. 6QECh. 9.3 - Prob. 1QECh. 9.3 - What is a persistent object?
Ch. 9.3 - Identify some classes as well as some of their...Ch. 9.3 - Prob. 4QECh. 9.4 - Prob. 1QECh. 9.4 - Prob. 2QECh. 9.4 - Prob. 3QECh. 9.4 - Prob. 4QECh. 9.4 - Prob. 5QECh. 9.4 - Prob. 6QECh. 9.5 - Prob. 1QECh. 9.5 - Prob. 2QECh. 9.5 - Prob. 3QECh. 9.5 - Prob. 4QECh. 9.5 - Prob. 5QECh. 9.5 - Prob. 6QECh. 9.5 - Prob. 7QECh. 9.6 - Prob. 1QECh. 9.6 - Give an additional example of a pattern that might...Ch. 9.6 - Prob. 3QECh. 9.6 - How does data mining differ from traditional...Ch. 9.7 - Prob. 1QECh. 9.7 - Prob. 2QECh. 9.7 - Prob. 3QECh. 9.7 - Prob. 4QECh. 9 - Prob. 1CRPCh. 9 - Prob. 2CRPCh. 9 - Prob. 3CRPCh. 9 - Prob. 4CRPCh. 9 - Prob. 5CRPCh. 9 - Prob. 6CRPCh. 9 - Prob. 7CRPCh. 9 - Prob. 8CRPCh. 9 - Prob. 9CRPCh. 9 - Prob. 10CRPCh. 9 - Prob. 11CRPCh. 9 - Prob. 12CRPCh. 9 - Using the commands SELECT, PROJECT, and JOIN,...Ch. 9 - Answer Problem 13 using SQL. PROBLEM 13 13. Using...Ch. 9 - Prob. 15CRPCh. 9 - Prob. 16CRPCh. 9 - Prob. 17CRPCh. 9 - Prob. 18CRPCh. 9 - Prob. 19CRPCh. 9 - Empl Id Name Address SSN Job Id Job Title Skill...Ch. 9 - Empl Id Name Address SSN Job Id Job Title Skill...Ch. 9 - Prob. 22CRPCh. 9 - Prob. 23CRPCh. 9 - Prob. 24CRPCh. 9 - Prob. 25CRPCh. 9 - Write a sequence of instructions (using the...Ch. 9 - Prob. 27CRPCh. 9 - Prob. 28CRPCh. 9 - Prob. 29CRPCh. 9 - Prob. 30CRPCh. 9 - Prob. 31CRPCh. 9 - Prob. 32CRPCh. 9 - Prob. 33CRPCh. 9 - Prob. 34CRPCh. 9 - Prob. 35CRPCh. 9 - Prob. 36CRPCh. 9 - Prob. 37CRPCh. 9 - Prob. 38CRPCh. 9 - Prob. 39CRPCh. 9 - Prob. 40CRPCh. 9 - Prob. 41CRPCh. 9 - Prob. 42CRPCh. 9 - Prob. 43CRPCh. 9 - Prob. 44CRPCh. 9 - Prob. 45CRPCh. 9 - Prob. 46CRPCh. 9 - Prob. 47CRPCh. 9 - Prob. 48CRPCh. 9 - Prob. 49CRPCh. 9 - Prob. 50CRPCh. 9 - Prob. 51CRPCh. 9 - Prob. 52CRPCh. 9 - Prob. 53CRPCh. 9 - Prob. 54CRPCh. 9 - Prob. 55CRPCh. 9 - Prob. 56CRPCh. 9 - Prob. 57CRPCh. 9 - Prob. 58CRPCh. 9 - Prob. 59CRPCh. 9 - Prob. 60CRPCh. 9 - Prob. 61CRPCh. 9 - Prob. 62CRPCh. 9 - Prob. 1SICh. 9 - Prob. 2SICh. 9 - Prob. 3SICh. 9 - Prob. 4SICh. 9 - Prob. 5SICh. 9 - Prob. 6SICh. 9 - Prob. 7SICh. 9 - Prob. 8SICh. 9 - Prob. 9SICh. 9 - Prob. 10SI
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Similar questions
- The transactions below are supposed to have a write-read conflict (dirty reads). But I am having a hard time understanding how. T1: R(A), W(A), R(B), W(B), Abort T2: R(A), W(A), C The way I understand this is that transaction 1 (T1) reads A and then writes A. Then transaction 2 reads A and then writes A. Finally, transaction 1 reads B and then writes B. I don't see how this creates a write-read conflict. Can someone explain how this is the case?arrow_forwardW-timestamp(Q) is the biggest timestamp of any transaction that successfully completed write(Q). Instead, let's say we specified it as the timestamp of the most recent transaction that successfully executed write(Q). Is it possible that this adjustment in language will make a difference? Explain your response.arrow_forwardUnder what circumstances does the wait-die scheme perform better than the wound-wait scheme for granting resources to concurrently executing transactions?arrow_forward
- Draw a wait-for-graph for detecting a for deadlock cycle in the given transactions T1 and T2 considering (Read-Write) or (Write-Read) conflicting parameters.arrow_forwardConsider the following two transactions: T1 = R(A). W(A), R(B), R(C) T2 - W(C). R(B), R(A). W(B) Say that sharing lock, exclusive lock, and unlock actions are inserted by the scheduler, resulting in the following annotated transactions: T1 - X(A), R(A), W(A), S(B), R(B), S(C), U(A), R(C), U(B), S(C) T2 - X(C), W(C), X(B), S(A), U(C), R(B), R(A), U(A), W(B), U(B) The above schedule follows 2 PL so conflict serializability is guaranteed. O True O Falsearrow_forwardConsider what would happen if there was a problem with 2PC for a transaction. In Exercise 23.2a, you wrote down a list of things that could go wrong. For each one, explain how 2PC keeps transactions atomic even if one of them goes wrong.arrow_forward
- Given the following transactions (with basic operations r - read, w - write, a - abort, c - commit, data sets X, Y, Z) T1 = w1 (X), c1 Т2 %3 12 (Y), г2 (Z), r2 (X), с2 T3 = w3 (Y), w3 (Z), c3 T4 %3D г4 (X), г4 (Y), с4 Create a serializable history with the help of locks (2 PL protocol) or by skilfully arranging the basic operations, if possible.arrow_forward11. The deadlock in a set of transaction can be determined by a. Read-only graph b. Wait graph c. Wait-for graph d. All of the mentionedarrow_forwardDraw the overall diagram of Secure Hash Algorithm-1 (SHA-1) and explain the process of generating the message digest using SHA-1.arrow_forward
- Consider the following two transactions: T1-R(B), W(B), R(C), R(A) T2-R(B). W(B) Say that sharing lock, exclusive lock, and unlock actions are inserted by the scheduler, resulting in the following annotated transactions: T1= X(B), R(B), W(B), S(C), R(C), S(A), U(B), R(A), U(C), U(A) T2-X(B), R(B), W(B), U(B) Cascading rollback is not possible for the above schedule. True Falsearrow_forwardCheck whether the given schedule S is conflict serializable or not- S : R1(A) , R2(A) , R1(B) , R2(B) , R3(B) , W1(A) , W2(B), W3(B),W3(A). S: R2(X) , W3(X) commit, W1(X) Commit, W2(Y), R2(Z) Commit, R4(X), R4(Y), R5(Z), W1(Z). S: R4(A), R2(A), R3(A), W1(B), W2(A), R3(B), W2(B), W3(A), W4(B)arrow_forwardConsider the following schedule, where ri (v) means that transaction i reads object v and wi (v) means that transaction i writes to object v. r2(x) r3(y) r1(x) w3(y) w1(x) r4(y) r1(z) w4(y) r2(y) r3(z) w2(y) - List all pairs of conflicting operations in this schedule. - Draw the conflict graph for this schedule. - If the schedule is serializable, give an equivalent serial schedule.If it is not serializable, then explain why not.arrow_forward
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