Web-Technologies
November 26, 2023Question 312 – Data-Structures
November 26, 2023Transactions
| Question 16 |
Two transactions T1 and T2 are given as
T1: r1(X)w1(X)r1(Y)w1(Y)
T2: r2(Y)w2(Y)r2(Z)w2(Z)
where ri(V) denotes a read operation by transaction Ti on a variable V and wi(V) denotes a write operation by transaction Ti on a variable V. The total number of conflict serializable schedules that can be formed by T1 and T2 is ____________.
T1: r1(X)w1(X)r1(Y)w1(Y)
T2: r2(Y)w2(Y)r2(Z)w2(Z)
where ri(V) denotes a read operation by transaction Ti on a variable V and wi(V) denotes a write operation by transaction Ti on a variable V. The total number of conflict serializable schedules that can be formed by T1 and T2 is ____________.
| 54 | |
| 55 | |
| 56 | |
| 57 |
Question 16 Explanation:
From the given transactions T1 and T2, total number of schedules possible = (4+4)!/4!4!
= 8!/(4×3×2×4×3×2)
= (8×7×6×5×4×3×2×1)/(4×3×2×4×3×2)
= 70
Following two conflict actions are possible:
∴# Permutations = 4 × 3 = 12
#permutations = 4 × 1 = 4
∴ Total no. of conflict serial schedules possible = 70 – 12 – 4 = 54
= 8!/(4×3×2×4×3×2)
= (8×7×6×5×4×3×2×1)/(4×3×2×4×3×2)
= 70
Following two conflict actions are possible:
∴# Permutations = 4 × 3 = 12
#permutations = 4 × 1 = 4
∴ Total no. of conflict serial schedules possible = 70 – 12 – 4 = 54
Correct Answer: A
Question 16 Explanation:
From the given transactions T1 and T2, total number of schedules possible = (4+4)!/4!4!
= 8!/(4×3×2×4×3×2)
= (8×7×6×5×4×3×2×1)/(4×3×2×4×3×2)
= 70
Following two conflict actions are possible:
∴# Permutations = 4 × 3 = 12
#permutations = 4 × 1 = 4
∴ Total no. of conflict serial schedules possible = 70 – 12 – 4 = 54
= 8!/(4×3×2×4×3×2)
= (8×7×6×5×4×3×2×1)/(4×3×2×4×3×2)
= 70
Following two conflict actions are possible:
∴# Permutations = 4 × 3 = 12
#permutations = 4 × 1 = 4
∴ Total no. of conflict serial schedules possible = 70 – 12 – 4 = 54