Recent questions tagged process-synchronization - GATE CSE Doubts

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Synchronization
Can anyone tell from where to perfect topics – Synchronization and Concurrency in OS? Or how to become self efficient to solve questions in exam

asked Jun 14 in Operating System by prajjwalsingh_11 (16 points) | 7 views

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Self-Doubt In OS, Process Synchronization
Consider the following implementation of a general (counting) semaphore. This implementation assumes the existence of binary semaphore operations upb and downb implemented with a test-and-set instruction. Procedure UP( S : semaphore): downb( ... preemptive version) c) Priority (you may assign whatever priorities to the processes you wish) d) Round-Robin

asked Jun 5 in Operating System by Chinmay Agnihotri (13 points) | 8 views

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Self-Doubt In OS, Process Synchronization
Consider the following implementation of a general (counting) semaphore. This implementation assumes the existence of binary semaphore operations upb and downb implemented with a test-and-set instruction. Procedure UP( S : semaphore): downb( ... preemptive version) c) Priority (you may assign whatever priorities to the processes you wish) d) Round-Robin

asked Jun 5 in Operating System by Chinmay Agnihotri (13 points) | 2 views

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Operating System Concepts (9th Ed) - Gagne, Silberschatz, and Galvin

asked May 22 in Operating System by kayes360 (6 points) | 29 views

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Operating System#GATE#2021
SJF scheduling algorithm is frequently used in Long Term scheduling Short term scheduling both A and B None of these.

asked Apr 19 in Operating System by sushmitagoswami (14 points) | 4 views

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GATE2017-1-27 Video Solution
A multithreaded program $P$ executes with $x$ number of threads and uses $y$ number of locks for ensuring mutual exclusion while operating on shared memory locations. All locks in the program are non-reentrant, i.e., if a thread holds a lock $l$, then it cannot re-acquire lock $l$ without releasing it. If ... : $x = 1, y = 2$ $x = 2, y = 1$ $x = 2, y = 2$ $x = 1, y = 1$

asked Apr 19 in Operating System by admin (3.6k points) | 2 views

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GATE2013-39 Video Solution
A certain computation generates two arrays a and b such that $a[i] = f(i)$ for $0 \leq i < n$ and $b[i] = g(a[i])$ for $0 \leq i < n$. Suppose this computation is decomposed into two concurrent processes $X$ and $Y$ such that $X$ computes the array $a$ and $Y$ computes the ... EntryY(R, S) { V(S); P(R); } ExitX(R, S) { V(R); P(S); } EntryY(R, S) { V(S); P(R); }

asked Apr 19 in Operating System by admin (3.6k points) | 3 views

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GATE2012-32 Video Solution
Fetch_And_Add(X,i) is an atomic Read-Modify-Write instruction that reads the value of memory location $X$, increments it by the value $i$, and returns the old value of $X$. It is used in the pseudocode shown below to implement ... take on a non-zero value when the lock is actually available works correctly but may starve some processes works correctly without starvation

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2006-61 Video Solution
The atomic fetch-and-set $x, y$ instruction unconditionally sets the memory location $x$ to $1$ and fetches the old value of $x$ in $y$ without allowing any intervening access to the memory location $x$ ... a pair of normal load/store can be used The implementation of $V$ is wrong The code does not implement a binary semaphore

asked Apr 19 in Operating System by admin (3.6k points) | 2 views

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GATE2009-33 Video Solution
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 ... at the same time Which of the above statements are TRUE? (I) only (I) and (II) (II) and (III) (IV) only

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2005-IT-41 Video Solution
Given below is a program which when executed spawns two concurrent processes : semaphore $X : = 0 ;$ /* Process now forks into concurrent processes $P1$ & $P2$ */ $\begin{array}{|l|l|}\hline \text{$P1$} & \text{$ ... (II) are true. (I) is true but (II) is false. (II) is true but (I) is false Both (I) and (II) are false

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2010-45 Video Solution
The following program consists of $3$ concurrent processes and $3$ binary semaphores. The semaphores are initialized as $S0=1, S1=0$ and $S2=0.$ ... $P0$ print '$0$'? At least twice Exactly twice Exactly thrice Exactly once

asked Apr 19 in Operating System by admin (3.6k points) | 2 views

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GATE1997-6.8 Video Solution
Each Process $P_i$, i = $1....9$ is coded as follows repeat P(mutex) {Critical section} V(mutex) forever The code for $P_{10}$ is identical except it uses V(mutex) in place of P(mutex). What is the largest number of processes that can be inside the critical section at any moment? $1$ $2$ $3$ None

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2006-78 Video Solution
Barrier is a synchronization construct where a set of processes synchronizes globally i.e., each process in the set arrives at the barrier and waits for all others to arrive and then all processes leave the barrier. Let the number of processes ... need not be inside a critical section The barrier implementation is correct if there are only two processes instead of three.

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2015-3-10 Video Solution
Two processes $X$ and $Y$ ... The proposed solution guarantees mutual exclusion and prevents deadlock The proposed solution fails to prevent deadlock and fails to guarantee mutual exclusion

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2007-58 Video Solution
Two processes, $P1$ and $P2$, need to access a critical section of code. Consider the following synchronization construct used by the processes: /* P1 */ while (true) { wants1 = true; while (wants2 == true); /* Critical ... waiting. It requires that processes enter the critical section in strict alteration. It does not prevent deadlocks, but ensures mutual exclusion.

asked Apr 19 in Operating System by admin (3.6k points) | 2 views

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GATE2013-34 Video Solution
A shared variable $x$, initialized to zero, is operated on by four concurrent processes $W, X, Y, Z$ as follows. Each of the processes $W$ and $X$ reads $x$ from memory, increments by one, stores it to memory, and then terminates. Each of the ... initialized to two. What is the maximum possible value of $x$ after all processes complete execution? $-2$ $-1$ $1$ $2$

asked Apr 19 in Operating System by admin (3.6k points) | 2 views

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GATE2010-23 Video Solution
Consider the methods used by processes $P1$ and $P2$ for accessing their critical sections whenever needed, as given below. The initial values of shared boolean variables $S1$ and $S2$ ... ? Mutual exclusion but not progress Progress but not mutual exclusion Neither mutual exclusion nor progress Both mutual exclusion and progress

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2001-2.22 Video Solution
Consider Peterson's algorithm for mutual exclusion between two concurrent processes i and j. The program executed by process is shown below. repeat flag[i] = true; turn = j; while (P) do no-op; Enter critical section, perform actions, then exit critical section Flag[i] = false; ... = i flag[j] = true and turn = j flag[i] = true and turn = j flag[i] = true and turn = i

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2016-2-48 Video Solution
Consider the following two-process synchronization solution. ... synchronization solution. This solution violates mutual exclusion requirement. This solution violates progress requirement. This solution violates bounded wait requirement.

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2000-1.21 Video Solution
Let $m[0]\ldots m[4]$ be mutexes (binary semaphores) and $P[0]\ldots P[4]$ be processes. Suppose each process $P[i]$ executes the following: wait (m[i]; wait (m(i+1) mod 4]); ........... release (m[i]); release (m(i+1) mod 4]); This could cause Thrashing Deadlock Starvation, but not deadlock None of the above

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2019-23 Video Solution
Consider three concurrent processes $P1$, $P2$ and $P3$ as shown below, which access a shared variable $D$ that has been initialized to $100$ ... possible values of $D$ after the three processes have completed execution are $X$ and $Y$ respectively, then the value of $Y-X$ is ____

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE1998-1.31 Video Solution
A counting semaphore was initialized to $10$. Then $6 P$ (wait) operations and $4V$ (signal) operations were completed on this semaphore. The resulting value of the semaphore is $0$ $8$ $10$ $12$

asked Apr 19 in Operating System by admin (3.6k points) | 2 views

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GATE2014-2-31 Video Solution
Consider the procedure below for the Producer-Consumer problem which uses semaphores: semaphore n = 0; semaphore s = 1; void producer() { while(true) { produce(); semWait(s); addToBuffer(); semSignal(s); semSignal(n); } } void consumer() { while(true ... the buffer is empty. The starting value for the semaphore $n$ must be $1$ and not $0$ for deadlock-free operation.

asked Apr 19 in Operating System by admin (3.6k points) | 3 views

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GATE2018-40 Video Solution
Consider the following solution to the producer-consumer synchronization problem. The shared buffer size is $N$. Three semaphores $empty$, $full$ and $mutex$ are defined with respective initial values of $0, N$ and $1$. Semaphore $empty$ denotes the number of available slots in the buffer, for ... $P: empty, \ \ \ Q:full, \ \ \ R:full, \ \ \ S:empty$

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2003-81 Video Solution
Suppose we want to synchronize two concurrent processes $P$ and $Q$ using binary semaphores $S$ and $T$. The code for the processes $P$ and $Q$ ... $1$ $V(S)$ at $W, V(T)$ at $X, P(S)$ at $Y, P(T)$ at $Z, S$ and $T$ initially $1$

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2005-IT-42 Video Solution
Two concurrent processes $P1$ and $P2$ use four shared resources $R1, R2, R3$ and $R4$, as shown below. $\begin{array}{|l|l|}\hline \textbf{P1} & \textbf{P2} \\ \text{Compute: } & \text{Compute;} \\ \text{Use $ ... binary semaphores are used to enforce the above scheduling constraints, what is the minimum number of binary semaphores needed? $1$ $2$ $3$ $4$

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2006-IT-55 Video Solution
Consider the solution to the bounded buffer producer/consumer problem by using general semaphores $S, F,$ and $E$. The semaphore $S$ is the mutual exclusion semaphore initialized to $1$. The semaphore $F$ corresponds to the number of free slots in the buffer and is ... Signal $(F)$ in the Consumer process (I) only (II) only Neither (I) nor (II) Both (I) and (II)

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2007-IT-10 Video Solution
Processes $P1$ and $P2$ use critical_flag in the following routine to achieve mutual exclusion. Assume that critical_flag is initialized to FALSE in the main program. get_exclusive_access ( ) { if (critical _flag == FALSE) { critical_flag = TRUE ; critical_region () ; critical_flag = ... Both (i) and (ii) are false (i) is true (ii) is false Both (i) and (ii) are true

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2004-48 Video Solution
Consider two processes $P_1$ and $P_2$ accessing the shared variables $X$ and $Y$ protected by two binary semaphores $S_X$ and $S_Y$ respectively, both initialized to 1. $P$ and $V$ denote the usual semaphore operators, where $P$ decrements the semaphore value, and $V$ increments the semaphore ... $P(S_X), P(S_X); P(S_Y), P(S_Y)$ $P(S_X), P(S_Y); P(S_X), P(S_Y)$

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2015-1-9 Video Solution
The following two functions $P1$ and $P2$ that share a variable $B$ with an initial value of $2$ ... $B$ can possibly take after the execution is______________________.

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE1996-2.19 Video Solution
A solution to the Dining Philosophers Problem which avoids deadlock is to ensure that all philosophers pick up the left fork before the right fork ensure that all philosophers pick up the right fork before the left fork ensure that one particular philosopher ... the right fork, and that all other philosophers pick up the right fork before the left fork None of the above

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2008-IT-53 Video Solution
The following is a code with two threads, producer and consumer, that can run in parallel. Further, $S$ and $Q$ are binary semaphores quipped with the standard $P$ and $V$ operations. semaphore S = 1, Q = 0; integer x; producer: ... be lost Values generated and stored in '$x$' by the producer will always be consumed before the producer can generate a new value

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2019-39 Video Solution
Consider the following snapshot of a system running $n$ concurrent processes. Process $i$ is holding $X_i$ instances of a resource $R$, $1 \leq i \leq n$. Assume that all instances of $R$ are currently in use. Further, for all $i$, process $i$ ... $\text{Min}(X_p,X_q) \leq \text{Max} \{Y_k \mid 1 \leq k \leq n, k \neq p, k \neq q\}$

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE1998-1.30 Video Solution
When the result of a computation depends on the speed of the processes involved, there is said to be cycle stealing race condition a time lock a deadlock

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2006-79 Video Solution
Barrier is a synchronization construct where a set of processes synchronizes globally i.e., each process in the set arrives at the barrier and waits for all others to arrive and then all processes leave the barrier. Let the number of processes ... at the beginning of the barrier and re-enabled at the end. The variable process_left is made private instead of shared

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE1996-1.19, ISRO2008-61 Video Solution
A critical section is a program segment which should run in a certain amount of time which avoids deadlocks where shared resources are accessed which must be enclosed by a pair of semaphore operations, $P$ and $V$

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2000-20 Video Solution
a. Fill in the boxes below to get a solution for the reader-writer problem, using a single binary semaphore, mutex (initialized to $1$) and busy waiting. Write the box numbers ($1$, $2$ and $3$), and their contents in your answer book. L1: int R = 0, W ... ...../*do the write*/ wait( mutex); W=0; signal (mutex); } b. Can the above solution lead to starvation of writers?

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2007-IT-56 Video Solution
Synchronization in the classical readers and writers problem can be achieved through use of semaphores. In the following incomplete code for readers-writers problem, two binary semaphores mutex and wrt are used to obtain synchronization wait (wrt) writing is performed signal ... ), wait (mutex), signal (wrt) signal (mutex), wait (mutex), signal (mutex), wait (mutex)

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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GATE2004-IT-65 Video Solution
The semaphore variables full, empty and mutex are initialized to $0$, $n$ and $1$, respectively. Process P1 repeatedly adds one item at a time to a buffer of size $n$, and process P2 repeatedly removes one item at a time from the same buffer using the programs given below. ... P(empty), V(full) P(empty), V(full), P(empty), V(full) P(empty), V(full), P(full), V(empty)

asked Apr 19 in Operating System by admin (3.6k points) | 1 view

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