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The enter_CS() and leave_CS() functions to implement critical section of a process are realized using test-and-set instruction as follows:

void enter_CS(X)
{
    while(test-and-set(X));
}

void leave_CS(X)
{
    X = 0;
}

In the above solution, X is a memory location associated with the CS and is initialized to 0. Now consider the following statements:

 

  1. The above solution to CS problem is deadlock-free

  2. The solution is starvation free

  3. The processes enter CS in FIFO order

  4. More than one process can enter CS at the same time

Which of the above statements are TRUE?

  1. I only
  2. I and II
  3. II and III
  4. IV only

 

asked in Operating System by Veteran (66.2k points) 1154 2199 2523
edited by | 2.2k views

7 Answers

+14 votes
Best answer
The answer is A only.

The solution satisfies:

1) Mutual Exclusion as test-and-set is an indivisible (atomic) instruction (makes option IV wrong)

2) Progress as at initially X is 0 and at least one process can enter critical section at any time.

But no guarantee that a process eventually will enter CS and hence option IV is false. Also, no ordering of processes is maintained and hence III is also false.

So eliminating all the 3 choices remains A.
answered by Veteran (19.3k points) 37 146 207
selected by
1. Progress is not satisfied here. Suppose P1 is in leave_CS() and going to execute X=0 whereas P2 trying to enter critical section is iterating in the while loop. So this shows that P1 which is not interested in entering in to CS is blocking P2 which is interested to enter into CS. Hence Progress not satisfied.

2. Coming to option 2 of Starvation. Lets suppose P1,P2,P3 as three processes. Now P1 enters CS and P2,P3 are iterating the while loop. Now when P1 exexutes X=0,then immediately instead of P2, P3 gets into CS. This shows there is starvation.
how should we check for starvation in such questions?..Arjun sir can you please explain?
I guess we call option A deadlock free just because there is only one resource - a memory location X and there cannot be a deadlock with only one resource, we need at least two resources for circular wait to occur. So I feel there is no need to evaluate whether individual requirements of deadlock occur or not. Am I right?
This makes more sense to me.
+9 votes

I.It is a test and set(load the previous value of R0 and store 1 into it),instruction,atomic in nature.Only one process executing TSL can only enter CS as it makes the value of variable=1 and other process trying to enter the CS would find value of variable=1 and hence wont enter the CS.

TSL can be thought of as

1.Load Ro

2 Store 1

3.CMP R0,#0

4.JNZ step 1

1 and 2 as atomic.

ME is guaranteed.

II. TSl suffers unbounded waiting.This is because there is no strict alteration and number of processes are also not bounded.

Lets us suppose

1.TSL

2.while(TSl!=1);

3.CS

4. store variable=0

P1:12 CS| P2 wants to enter but has to wait due to variable being 1 |P1 :4 | P3 enters executes TSL 1 2  CS | P2 wants to enter but has to wait due to variable being 1 |P3 :4 |P4 enters

So poor P2 everytime has to starve unlucky guy.

 

III. From the above example only we see that P2 although came first was scheduled later so FCFS not maintained.

Iv. ME guaranteed,

A)I Only

answered by Loyal (3.6k points) 7 35 55
edited by
How it is deadlock free ? I mean here ME is satisfied , progress is not , because if P1 after coming out of CS got terminated before executing X=0; and all other process waiting for X=0 will starve . Hence a deadlock ??
p1 will not be terminated/preempted for ever, (no semaphores here) it will run at some later stage and will execute x=0, so some other process can enter cs , and only one process will enter the cs as TSL is atomic op.

 

but we dont know which one will enter cs, hence starvation (no bounded waiting, order)
+4 votes
A) First one is true as it is simple test and set.
B) seems to be true but it is not as when a process is in spin lock all the thread of a process is waiting for this thread to complete job , whenever it is executed by cpu so in every context switch you are waiting for spin lock to over. means you are starving because you can complete other thread in parallel. This can be remove by introducing the waiting que but there is no que mention so system is starving.
c) No possible as no order or Que order is mention in the question.
d) Not possible.
answered by Loyal (3.1k points) 6 17 37
+2 votes

The above solution is a simple Test and Set solution that makes sure that deadlock doesn’t occur, but it doesn’t use any queue to avoid starvation or to have FIFO order.
 

answered by Veteran (43.1k points) 20 212 553
But wouldnt there be a deadlock in tsl due to priority inversion?
deadlock means both processes are in blocked state, but in priority inveraion only one is blocked and the other higher priority process is in ready state
0 votes
  1. I think every one understood why option (a) is correct and else options are wrong.
  2. I would just like to add a point that would be beneficial. Option B and Option C are inter-related, as in if the processes are in FIFO then there will not be any starvation. So if FIFO then, the solution will be starvation free.
answered by (21 points)
0 votes
Let me explain why B is wrong

suppose p1 is in critical section (only one can enter into CS because of TSL) and many process(p2,p3,p4,p5....pn) is busy waiting in loop , let me pick one process from many process which are looping  and that picked process is p2, now

consider the case when p1 came out of CS then p3 enters and when p3 comes out then p4 enters because there is no mechanism maintained which ensures after how much time p2 will get turn similarly this goes on till pn and p2 will keep on starving hence it is not starvation free
answered by Loyal (3.5k points) 1 5 8
0 votes
Ans is (A)
answered ago by (121 points) 2
Answer:

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