Register

Peterson's solution

2,502 views
1 1 vote
Which of the following statement is/are True?
i)Bounded waiting is satisfied in Strict alternation mechanism.
ii)Bounded waiting satisfied in Peterson's solution.
iii)Progress in Strict alternation mechanism.
iv)Progress in Peterson's Solution

2 Answers

1 1 vote

statement 1,2,4 are TRUE...

1) IN STRICT alteration approch...both process has equal share to enter into CS ...so its true..

2)Perterson solution satisfy all..MUTEX,PROGRESS and bounded wait....so  statement 4 also true..

3)Progress is not gurantee in Strict alteration..as one process affects in decision of entering 2nd process into CS..

0 0 votes
Answer (ii) bound and waiting satisfied Peterson Solution
Voting & Accepting an answer helps improve the community
← Previous · ambiguityNext → · Combinatorics
← Previous in category · Ace test series ques...Next in category → · NIELIT2017 -SET-C-10...
Position:
Show:

Related questions

1 1 vote
1 1 answer
92
92 views
GO Classes asked Jul 29
92 views
GO Classes DPP | GATE CS | Operating System | Peterson's Solution
Two threads execute Peterson’s entry code. The following statements execute in this exact order:$T_0:$ $\texttt{flag[0] = true}$$T_0:$ $\texttt{turn = 1}$$T_1:$ $\texttt{...
6 6 votes
1 1 answer
155
155 views
GO Classes asked Jul 29
155 views
GO Classes DPP | GATE CS | Operating System | Peterson's Solution
Which of the following statements about Peterson’s solution are correct?Peterson’s solution uses atomic load and store operations. Peterson’s solution directly supports a...
4 4 votes
1 1 answer
186
186 views
GO Classes asked Jul 27
186 views
GO Classes DPP | GATE CS | Operating System | Peterson’s Solution
Two processes $P_i$ and $P_j$ use Peterson’s solution:$\texttt{flag[i] = TRUE;}$$\texttt{turn = j;}$$\texttt{while(flag[j] \&\& turn == j);}$$\texttt{Critical Section}$$\...
1 1 vote
1 1 answer
1.9k
1.9k views
Akriti sood asked Dec 5, 2016
1,874 views
Consider the following variation in Peterson’s solution for 2 processes.
Consider the following variation in Peterson’s solution for 2 processes.shared char want = {0,0}; shared int turn = 0; 1. P_i() 2. { while (1) { 3. turn = |1-i|; 4. want...