Database 2PL protocol

Question

In 2PL protocol if all exclusive locks  are acquired by transaction in increasing order of their accesses then is the 2PL starvation free ?

Comments

I think 2PL does not prevent starvation..

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Infact 2PL leads to Deadlock.

T1                   T2

L_X(A)      

                    L_X(B)

L_X(B)         

                  L_X(A)

 

This leads to Deadlock hence infinite waiting. hence It is not starvation free.

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But the statement alludes to conservative 2PL which is deadlock free. According to the statement the transactions will be deadlock free. Will it be starvation free ?

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@ sumit chakraborty strict 2PL is not deadlock free. 

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@ sumit chakraborty  I have given example of basic 2PL. Still if you are considering conservative 2PL then it surely leads to starvation.

because if all locks are acquired by T1 initially then t2 has to starve until T1 unlocks. and if some operation of T1 is taking too much time then T2 will surely starve.

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@ Ashwin Kulkarni  i don't think your example comes under starvation problem..because  we assume that no process wants lo be in critical section or want to  use a resource infinitely. 

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@ Red_devil Deadlock is an infinite form of starvation. Hence I think in my example there is a deadlock means infinite waiting hence both processes will starve forever.

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@Ashwin Kulkarni your explanation is right for strict 2PL..let us talk about  conservative 2PL in which there is no deadlock.. now explain how there is starvation.

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i have already explained conservative 2PL above but let's take this in depth.

Let's say all locks are acquired by T1 initially, then T2 has to starve until T1 unlocks. and if some operation of T1 is taking too much time then T2 will surely starve.

T1                                    T2

L_X(A)

L_X(B)

 

Execution A  

                          

                               Now here B is taking too much time,

                               Hence T2 has to wait until execution of B

                              Then only T2 can aquire its locks and start its execution.

 

execution of B

                

Hence here Though lock on B in T1 is taken initially but B is taking too much time in T1 hence T2 has to  starve untill T1 unlocks B. Because In T2 also X(A) and X(B) should be acquire at start of its execution

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@Ashwin Kulkarni  so according to this Peterson's solution in OS should be not starvation free because we can't preempt any process while it is executing.

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You're talking about priority inversion problem in Peterson's solution.

Answer 1

It might help
https://gateoverflow.in/102854/2pl-protocol