Test: Process Synchronization- 1 - Computer Science Engineering (CSE) MCQ

fork() returns 0 in child process and process ID of child process in parent process. In Child (x), a = a + 5 In Parent (u), a = a – 5; Therefore x = u + 10. The physical addresses of ‘a’ in parent and child must be different. But our program accesses virtual addresses (assuming we are running on an OS that uses virtual memory). The child process gets an exact copy of parent process and virtual address of ‘a’ doesn’t change in child process. Therefore, we get same addresses in both parent and child. See this run for example.

Option A can cause deadlock. Imagine a situation process X has acquired a, process Y has acquired b and process Z has acquired c and d. There is circular wait now. Option C can also cause deadlock. Imagine a situation process X has acquired b, process Y has acquired c and process Z has acquired a. There is circular wait now. Option D can also cause deadlock. Imagine a situation process X has acquired a and b, process Y has acquired c. X and Y circularly waiting for each other. 

Consider option A) for example here all 3 processes are concurrent so X will get semaphore a, Y will get b and Z will get c, now X is blocked for b, Y is blocked for c, Z gets d and blocked for a. Thus it will lead to deadlock. Similarly one can figure out that for B) completion order is Z,X then Y. 

Let us talk about the operation P(). It stores the value of s in x, then it fetches the old value of x, stores it in y and sets x as 1. The while loop of a process will continue forever if some other process doesn't execute V() and sets the value of s as 0. If context switching is disabled in P, the while loop will run forever as no other process will be able to execute V().

Processes can run in many ways, below is one of the cases in which x attains max value

Semaphore S is initialized to 2

Process W executes S=1, x=1 but it doesn't update the x variable.

Then process Y executes S=0, it decrements x, now x= -2 and signal semaphore S=1

Now process Z executes s=0, x=-4, signal semaphore S=1
Now process W updates x=1, S=2 Then process X executes X=2 
So correct option is D

Concepts:

V(S): signal operation will increment the semaphore variable, that is, S++. 

P(S): wait operation will decrement the semaphore variable., that is, S--. 

Data:

Initial counting semaphore (I) = 7

Signal operation = x V

Wait operation = 20 P

Final counting semaphore (F) = 5

Calculation:

5 = 7 + x V + 20 P

5 = 7 + x (+1) + 20 (-1)

x = 5 - 7 + 20

∴ x = 18

Use SRTF, for the minimum achievable average waiting time : Gantt chart is

Since, TAT = CT - AT and WT = TAT - BT, so WT = CT - AT - BT = CT - (AT+BT) Therefore, Avg, WT = {(26-0-16) + (46-0-20) + (10-0-10)} / 3 = {10 + 26 + 0} / 3 = 36 / 3 = 12

In a multitasking system, a computer executes several programs simultaneously by switching them back and forth to increase the user interactivity. Processes share the CPU and execute in an interleaving manner. This allows the user to run more than one program at a time. Option (B) is correct.

The processes should execute in SJF manner to get the lowest waiting time. So, the order should be 5, 9, 12, 18. Option (B) is correct.

Waiting or Blocked state is when a process has requested some input/output and is waiting for the resource. So, option (D) is correct.

In Critical section problem:

• No assumptions may be made about speeds or the number of CPUs.
• No two processes may be simultaneously inside their critical sections.
• Processes running outside its critical section can't block other processes getting enter into critical section.
• Processes do not wait forever to enter its critical section.

Option (C) is correct.

Multilevel Feedback Queue Scheduling (MLFQ) keep analyzing the behavior (time of execution) of processes and according to which it changes its priority of execution of processes. Option (C) is correct. 

Total no of characters = 200 (each character having one space) Time taken to print one character = 6 ms; Time taken to print one space = 2 ms. character printing = 200 * 6 = 1200 ms space printing = 200 * 2 = 400 ms total printing time = 1200 + 400 = 1600 ms = 1.6 s. The printing speed of the dot matrix printer in characters per second = 200 / 1.6 = 125 / s. So, option (E) is correct.

For switching from a CPU user mode to the supervisor mode Software interrupts occurs. Software interrupts is internal interrupt triggered by some some software instruction. And external interrupt is caused by some hardware module. Option (C) is correct.

Number of times YES printed is equal to number of process created. Total Number of Processes = 2n where n is number of fork system calls. So here n = 2, 24 = 4
fork ();   // Line 1
fork ();   // Line 2

So, there are total 4 processes (3 new child processes and one original process). Option (C) is correct.

The values stored in registers, stack pointers and program counters are saved on context switch between the processes so as to resume the execution of the process. There's no need of saving the contents of TLB as it is invalidated after each context switch. So, option (B) is correct