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A process executes the code fork (); fork (); fork (); The total number of child processes created is
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We can also use direct formula to get the number of child processes. With n fork statements, there are always 2^n – 1 child processes.
מיין לפי
consider the 3 processes, P1, P2 and P3 shown in the table Process Arrival time Time unit required P1 0 5 P2 1 7 P3 3 4 The completion order of the 3 processes under the policies FCFS and RRS (round robin scheduling with CPU quantum of 2 time units) are
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מיין לפי
Consider the following table of arrival time and burst time for three processes P0, P1 and P2. Process Arrival time Burst Time P0 0 ms 9 ms P1 1 ms 4 ms P2 2 ms 9 ms The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of processes. What is the average waiting time for the three processes?
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Process P0 is allocated processor at 0 ms as there is no other process in ready queue. P0 is preempted after 1 ms as P1 arrives at 1 ms and burst time for P1 is less than remaining time of P0. P1 runs for 4ms. P2 arrived at 2 ms but P1 continued as burst time of P2 is longer than P1. After P1 completes, P0 is scheduled again as the remaining time for P0 is less than the burst time of P2.
P0 waits for 4 ms, P1 waits for 0 ms amd P2 waits for 11 ms. So average waiting time is (0+4+11)/3 = 5.
מיין לפי
Let the time taken to switch between user and kernel modes of execution be t1 while the time taken to switch between two processes be t2. Which of the following is TRUE?
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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 are randomly assigned. Method Used by P1 while (S1 == S2) ; Critica1 Section S1 = S2; Method Used by P2 while (S1 != S2) ; Critica1 Section S2 = not (S1); Which one of the following statements describes the properties achieved?
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It can be easily observed that the Mutual Exclusion requirement is satisfied by the above solution, P1 can enter critical section only if S1 is not equal to S2, and P2 can enter critical section only if S1 is equal to S2.
Progress Requirement is not satisfied. Let us first see definition of Progress Requirement.
Progress Requirement: If no process is executing in its critical section and there exist some processes that wishes to enter their critical section, then the selection of the processes that will enter the critical section next cannot be postponed indefinitely.
If P1 or P2 want to re-enter the critical section, then they cannot even if there is other process running in critical section.
מיין לפי
The essential content(s) in each entry of a page table is / are
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A page table entry must contain Page frame number. Virtual page number is typically used as index in page table to get the corresponding page frame number. See this for details.
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Consider a system with 4 types of resources R1 (3 units), R2 (2 units), R3 (3 units), R4 (2 units). A non-preemptive resource allocation policy is used. At any given instance, a request is not entertained if it cannot be completely satisfied. Three processes P1, P2, P3 request the sources as follows if executed independently. Process P1: t=0: requests 2 units of R2 t=1: requests 1 unit of R3 t=3: requests 2 units of R1 t=5: releases 1 unit of R2 and 1 unit of R1. t=7: releases 1 unit of R3 t=8: requests 2 units of R4 t=10: Finishes Process P2: t=0: requests 2 units of R3 t=2: requests 1 unit of R4 t=4: requests 1 unit of R1 t=6: releases 1 unit of R3 t=8: Finishes Process P3: t=0: requests 1 unit of R4 t=2: requests 2 units of R1 t=5: releases 2 units of R1 t=7: requests 1 unit of R2 t=8: requests 1 unit of R3 t=9: Finishes Which one of the following statements is TRUE if all three processes run concurrently starting at time t=0?
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We can apply the following Deadlock Detection algorithm and see that there is no process waiting indefinitely for a resource.
מיין לפי
The P and V operations on counting semaphores, where s is a counting semaphore, are defined as follows: P(s) : s = s - 1; if (s < 0) then wait; V(s) : s = s + 1; if (s <= 0) then wakeup a process waiting on s; Assume that Pb and Vb the wait and signal operations on binary semaphores are provided. Two binary semaphores Xb and Yb are used to implement the semaphore operations P(s) and V(s) as follows: P(s) : Pb(Xb); s = s - 1; if (s < 0) { Vb(Xb) ; Pb(Yb) ; } else Vb(Xb); V(s) : Pb(Xb) ; s = s + 1; if (s <= 0) Vb(Yb) ; Vb(Xb) ; The initial values of Xb and Yb are respectively
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Both P(s) and V(s) operations are perform Pb(xb) as first step. If Xb is 0, then all processes executing these operations will be blocked. Therefore, Xb must be 1.
If Yb is 1, it may become possible that two processes can execute P(s) one after other (implying 2 processes in critical section). Consider the case when s = 1, y = 1. So Yb must be 0.
מיין לפי
for (i = 0; i < n; i++) fork(); The total number of child processes created is
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If we sum all levels of above tree for i = 0 to n-1, we get 2^n - 1. So there will be 2^n – 1 child processes.
מיין לפי
What is the output of the following code? #include <stdio.h> #include <unistd.h> int main() { if (fork() || fork()) fork(); printf("1 "); return 0; }
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1. It will create two process one parent P (has process ID of child process)and other is child C1 (process ID = 0).
2. In if statement we used OR operator( || ) and in this case second condition is evaluated when first condition is false.
3. Parent process P will return positive integer so it directly execute statement and create two more processes (one parent P and other is child C2). Child process C1 will return 0 so it checks for second condition and second condition again create two more processes(one parent C1 and other is child C3).
4. C1 return positive integer so it will further create two more processes (one parent C1 and other is child C4). Child C3 return 0 so it will directly print 1.
מיין לפי