4. Explain the code. Which process will execute first and why? semaphore s = 0; //Shared P1 { P2 { /* some code */ /* some code */ Wait(s); Signal (s); /* do some more code */ /* do some more code */
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- Ordinary Pipes allow communication in standard producer-consumer style. Producer writes to one end (the write-end of the pipe) and consumer reads from the other end (the read-end of the pipe). Write a simple program in C. Language, fork a process and let a pipe be between the parent and the child processes such that the the parent will write and the child read a message from keyboard. Also print the number of written bytes to the pipe.The following program consists of 3 concurrent processes and 3 binary semaphores. The semaphores are initialized as S0 =1, S1= 0, S2=0 Process P0 Process P1 Process P2 wait(S0); print '0' release(S1) release (S2);} wait(S1); release(S0); wait(S2); release(S0); How many times will process P0 print ‘0’?The following program consists of 3 concurrent processes and 3 binary semaphores. The semaphores are initialized as S0 = 0,S1 = 1 and S2 = 1. Process P0:while (true) { wait (S0); print '1';release(S1);release(S2);} Process P1:wait (S2);release (S0); Process P2:wait(S2);release(S0); How many times will process P0 print '1'?
- (a) FCFS with a single Ready Queue. A process gets scheduled once one of the CPUsis available. When both CPUs are available, we select CPU1. (b) FCFS with a Ready Queue per CPU. Processes A, B, and C go to CPU1 and processesD, E, and F go to CPU2. (c) HRRN with a single Ready Queue. A process gets scheduled once one of the CPUsis available. When both CPUs are available, we select CPU1.A single-threaded program cannot be deadlocked, and this is because one of the conditions of deadlock is being broken. Which one? a. Hold and wait b. Circular wait c. No preemption d. Mutual exclusionSuppose we have three processes P0, P1, and P2. Process P0 executes statement S0 and process P1 executes statement S1 and process P2 executes statement S2. Use wait and signal to ensure the following: S2 is always executed before S1 and S1 is always executed before S0S0.
- When a process is in the "Blocked" state, awaiting the completion of an I/O operation. When a service is terminated, it transitions to one of the following states: a) Terminated state, b) Suspended state, c) Running state, or d) Ready state.When a process enters an I/O queue then it enters ___________________ state. a. waiting b. terminated c. new d. readyDescribe the procedure through which a segment might be in the address space of two different processes at once.
- one of the following is considered an application of a stack a. execute printing jobs in a printer b. routing messages in a network system c. undo/redo operation in notepad d. binary searchWrite a segment of code which will initialize a pipe and then spawn off a child process. The parent process will read two integers in from the user and send them to the child process. It will then wait for the child process to exit. The child process will then add them together and print the result out to the console prior to exiting.Concurrent Server Programming TITLE: Quiz Game In this assignment you are asked to write a phyton system to support an online math contest. The contest consists of answering the maximum quantity of sums in 30 seconds. This is done in a concurrent way, which means, that there are three participants answering at the same time. The participant with most correct answers WINS. The jury is responsible in starting up the server, after that they wait for the clients. When the three clients are connected, they must send the message: READY to the server. When the server receives the three READY from the clients, the contest starts. At the end of the program, the server must show the results .Execution (SERVER) - Please provide me the complete code and solution to arrive to the results below.python3 MathinikServer 192.168.1.2 Connected192.168.1.2 is Anne192.168.1.4 Connected192.168.1.4 is Billy192.168.1.3 Connected192.168.1.3 is MarkMark is READYAnne is READYBilly is READYStarting Contest…Contest…