Question

2. Consider the following two transactions: (10 points)
T13: read(A);
read(B);
if A = 1 then B := B - 1;
write(B).
T14: read(B);
read(A);
if B = 1 then A := A - 1;
write(A).
Let the consistency requirement be A = 1 or B = 1, with A = 1 and B = 1 as the initial values.
a. Show that every serial execution involving these two transactions preserves the consistency of the database.
b. Show a concurrent execution of T13 and T14 that produces a non serializable schedule.
c. Is there a concurrent execution of T13 and T14 that produces a serializable schedule?

Answer 1

2). ANSWER:

GIVEN THAT:

T13 : read (*); read (B); if Azi then B:= 8-1; waite (B) T14: jead (6); Sead (A); it B-l then A:-A-1; Walite (A); V * Initial values ore A=I& Bul. 21 There are 2 possible executions : T13 T4 and T4T13 Case d: A B Initially i Allei T 0 After Ty i 0 => Toue Case a & A B Oui consistency mer : A = l O BE! True V False (B=0) We got Initially i After TH 0 1 After T30 1 Our consistency met SA - OY BEJ Toque True False v (A=0 ) We got'

Schedules TB Yead (A) Jead (6) read(6) slead() It is then B :: B-1 waite (6) A:-A-1 It B=1 then wolite (A) - Precedence graph for s * There is a so Schechule cliected Cycle in the gaaph, s is non- Seriali 20 ble. c) + There is no porallel executing gesulting in a sexuali qable Schedule * From cas, We knew that serializable Schedule Gesults in A -1 ) V (B=1). fa: Suppose, If we start with T3, No matter if we aun the steps of T14, A=l. suppase, Before completing TB, 9t we start Tiy pread (B) , A8 = 0. So, when TiB Completes, $31. Thus, Beln A=1 => 764 =O V B = 0 )