a. Two possible executions: Case 1: A B Initially 0 0 After T 1 0 1 After T 2 0 1 Consistency met: A=0 ∨ B=0 T Ȣ...

Question

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Answer 1

Expert Solution & Answer

Chapter 17, Problem 15E

Explanation of Solution

a.

Two possible executions:

Case 1:

A B
Initially 0 0
After T₁ 0 1
After T₂ 0 1

Consistency met: A=0 ∨ B=0

T Ȣ...

Explanation of Solution

b.

Any interleaving of T1 and T2 results in a non-serializable schedule.

Precedence graph

Explanation of Solution

c.

Explanation:

There is no parallel execution resulting in a serializable schedule.

From part a. it is known that a serializable schedule results in A = 0 ∨ B = 0.

Suppose if started with T1 read(A). Then when the schedule ends, no matter when run the steps of T2, B = 1...

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Answer 2

Expert Solution & Answer

Chapter 17, Problem 15E

Explanation of Solution

a.

Two possible executions:

Case 1:

A B
Initially 0 0
After T₁ 0 1
After T₂ 0 1

Consistency met: A=0 ∨ B=0

T Ȣ...

Explanation of Solution

b.

Any interleaving of T1 and T2 results in a non-serializable schedule.

Precedence graph

Explanation of Solution

c.

Explanation:

There is no parallel execution resulting in a serializable schedule.

From part a. it is known that a serializable schedule results in A = 0 ∨ B = 0.

Suppose if started with T1 read(A). Then when the schedule ends, no matter when run the steps of T2, B = 1...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

Chapter 17, Problem 15E

Explanation of Solution

a.

Two possible executions:

Case 1:

A B
Initially 0 0
After T₁ 0 1
After T₂ 0 1

Consistency met: A=0 ∨ B=0

T Ȣ...

Explanation of Solution

b.

Any interleaving of T1 and T2 results in a non-serializable schedule.

Precedence graph

Explanation of Solution

c.

Explanation:

There is no parallel execution resulting in a serializable schedule.

From part a. it is known that a serializable schedule results in A = 0 ∨ B = 0.

Suppose if started with T1 read(A). Then when the schedule ends, no matter when run the steps of T2, B = 1...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 17, Problem 15E

Explanation of Solution

a.

Two possible executions:

Case 1:

A B
Initially 0 0
After T₁ 0 1
After T₂ 0 1

Consistency met: A=0 ∨ B=0

T Ȣ...

Explanation of Solution

b.

Any interleaving of T1 and T2 results in a non-serializable schedule.

Precedence graph

Explanation of Solution

c.

Explanation:

There is no parallel execution resulting in a serializable schedule.

From part a. it is known that a serializable schedule results in A = 0 ∨ B = 0.

Suppose if started with T1 read(A). Then when the schedule ends, no matter when run the steps of T2, B = 1...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 17, Problem 15E

Explanation of Solution

a.

Two possible executions:

Case 1:

A B
Initially 0 0
After T₁ 0 1
After T₂ 0 1

Consistency met: A=0 ∨ B=0

T Ȣ...

Explanation of Solution

b.

Any interleaving of T1 and T2 results in a non-serializable schedule.

Precedence graph

Explanation of Solution

c.

Explanation:

There is no parallel execution resulting in a serializable schedule.

From part a. it is known that a serializable schedule results in A = 0 ∨ B = 0.

Suppose if started with T1 read(A). Then when the schedule ends, no matter when run the steps of T2, B = 1...

Answer 8

Chapter 17, Problem 15E

Explanation of Solution

a.

Two possible executions:

Case 1:

A B
Initially 0 0
After T₁ 0 1
After T₂ 0 1

Consistency met: A=0 ∨ B=0

T Ȣ...

Answer 9

Chapter 17, Problem 15E

Explanation of Solution

a.

Two possible executions:

Case 1:

A B
Initially 0 0
After T₁ 0 1
After T₂ 0 1

Consistency met: A=0 ∨ B=0

T Ȣ...

Answer 10

Explanation of Solution

b.

Any interleaving of T1 and T2 results in a non-serializable schedule.

Precedence graph

Answer 11

Explanation of Solution

b.

Any interleaving of T1 and T2 results in a non-serializable schedule.

Precedence graph

Answer 12

Explanation of Solution

c.

Explanation:

There is no parallel execution resulting in a serializable schedule.

From part a. it is known that a serializable schedule results in A = 0 ∨ B = 0.

Suppose if started with T1 read(A). Then when the schedule ends, no matter when run the steps of T2, B = 1...

Answer 13

Explanation of Solution

c.

Explanation:

There is no parallel execution resulting in a serializable schedule.

From part a. it is known that a serializable schedule results in A = 0 ∨ B = 0.

Suppose if started with T1 read(A). Then when the schedule ends, no matter when run the steps of T2, B = 1...

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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Answer 17

Ch. 17 - Prob. 1PE Ch. 17 - Prob. 2PE Ch. 17 - Prob. 3PE Ch. 17 - Prob. 4PE Ch. 17 - Prob. 5PE Ch. 17 - Prob. 6PE Ch. 17 - Prob. 7PE Ch. 17 - Prob. 8PE Ch. 17 - Prob. 9PE Ch. 17 - Prob. 10PE

a. Two possible executions: Case 1: A B Initially 0 0 After T 1 0 1 After T 2 0 1 Consistency met: A=0 ∨ B=0 T Ȣ...

Explanation of Solution

a. Two possible executions: Case 1: ABInitially00After T101After T201 Consistency met: A=0 ∨ B=0 T Ȣ...

Explanation of Solution

b. Any interleaving of T1 and T2 results in a non-serializable schedule. Precedence graph

Explanation of Solution

c. Explanation: There is no parallel execution resulting in a serializable schedule. From part a. it is known that a serializable schedule results in A = 0 ∨ B = 0. Suppose if started with T1 read(A). Then when the schedule ends, no matter when run the steps of T2, B = 1...

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Chapter 17 Solutions

a.

Two possible executions:

Case 1:

A B
Initially 0 0
After T₁ 0 1
After T₂ 0 1

Consistency met: A=0 ∨ B=0

T Ȣ...

b.

Any interleaving of T1 and T2 results in a non-serializable schedule.

Precedence graph

c.

Explanation:

There is no parallel execution resulting in a serializable schedule.

From part a. it is known that a serializable schedule results in A = 0 ∨ B = 0.

Suppose if started with T1 read(A). Then when the schedule ends, no matter when run the steps of T2, B = 1...