For each of its three user processes, an operating system uses a total of two units of the resource R. A total of 12 units of R are used by the operating system. To ensure that there are no deadlocks, the bare minimum number of R units that must be used is
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Q: An operating system contains 3 user processes, each process requiring 2 units of the resource R. The…
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For each of its three user processes, an
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- An operating system has a total of three user processes, and each of these processes uses a total of two units of the resource R. The total number of units that R is used up by the operating system is twelve. The absolute minimal number of R units that must be used to eliminate the possibility of any deadlocks happening isAn operating system has 3 user processes running in a set, each requiring 2 units of resource R. The minimum number of units of R such that no deadlocks will ever arise is:Three user processes make up an operating system, and each of these processes uses a total of two units of the resource R. The bare minimum number of R units that must be employed to avoid any deadlocks is
- There are a total of three user processes in an operating system, and each of these processes utilizes a total of two units of the resource R. The operating system consumes a total of twelve units of R, making this the total amount of units that are used up. The very bare minimum amount of R units that must be used in order to do away with the potential of any deadlocks occurring isAn operating system contains 3 user processes, each process requiring 2 units of the resource R. The minimum number of units ofR required such that there is no possibility of deadlock on R is A 6 B 3 С 5 D 4Two units of resource R are used by both of the multiple user processes that comprise an operating system. The operating system consumes twelve R instances altogether. The absolute required amount of R subunits that must be used to prevent deadlocks is
- A single-CPU system has four processes, P1, P2, P3 and P4 in the ready queue. The execution times and I/O needs for these processes are given below. All times are in ms. (hint: when a process starts an IO operation, it is removed from the ready queue and put back at the end of the queue only when it completes its IO). Process P1: Arrives at 0ms, needs 23ms of CPU time. Performs I/O for 5ms after 10ms of its execution time, then another 5ms after 20ms of its execution time. Process P2: Arrives at 3ms, needs 12ms of CPU time. Performs I/O for 10ms after 5ms of its execution time. Process P3: Arrives at 1ms, needs 15ms of CPU time. Performs I/O for 1ms after 7ms of its execution time. Process P4: Arrives at 2ms, needs 8ms of CPU time. Performs I/O for 5ms after 4ms of its execution time. We assume that the CPU is idle if no one of these four processes is using it. Using Round Robin scheduling algorithm with a Time Quantum of 4ms and ignoring the context switch time, determine: a) The…An operating system uses the banker’s algorithm for deadlock avoidance when managing the allocation of three resource types X, Y and Z to three processes P0, P1 and P2. The table given below presents the current system state. Here, the Allocation matrix shows the current number of resources of each type allocated to each process and the Max matrix shows the maximum number of resources of each type required by each process during its execution Allocation Max X Y Z X Y Z P0 0 0 1 8 4 3 P1 3 2 0 6 2 0 P2 2 1 1 3 3 3 There are 3 units of type X, 2 units of type Y and 2 units of type Z still available. The system is currently in safe state. Consider the following independent requests for additional resources in the current state- REQ1: P0 requests 0 units of X, 0 units of Y and 2 units of Z REQ2: P1 requests 2 units of X, 0 units of Y and 0 units of Z Write a program to check whether: Only REQ1 can be permitted Only REQ2 can be permitted…4 Processes (P1,P2,P3,P4) arrived at time 0. The Duration of the processes are (6,8,4,2). All processes have a common X% wait time.The first process finishes at around 10.00. What is the common wait percentage for the processes (X). IVO Wait Time 80% 30% 60% Processes CPU Utilization 1 0.2 0.70 0.4 2 0.35 0.90 0.55 0.5 0.95 0.70 4 0.7 0.98 0.80 Select one: O a. Less than 30% O b. 30% O c. 60% v O d. More than 80% O e. 80% O f. None of the mentioned
- An operating system uses the Banker's algorithm for deadlock avoidance when managing the allocation of three resource types X, Ý, and Z to three processes PO, P1, and P2. The table given below presents the current system state. There are 3 instances of type X, 2 instances of type Y and 2 instances of type Z still available. The system is currently in a safe state. Answer the following questions! Allocation Max X Y z X Y PO 0 1 8 4 3 (a) If PO requests 2 instances of Z during the P1 3 2 current state, can it be granted? Explain! (b) If P1 requests 2 instances of X during the P2 2 1 1 3 3 3 current state, can it be granted? Explain! COAn operating system uses the banker's algorithm for deadlock avoidance when managing the allocation of three resource types X, Y and Z to three processes P0O, P1 and P2. The table given below presents the current system state. Here, the Allocation matrix shows the current number of resources of each type allocated to each process and the Max matrix shows the maximum number of resources of each type required by each process during its execution Allocation Max Y X Y PO 8 4 3 P1 3 6 P2 2 1 1 3 3 3 There are 3 units of type X, 2 units of type Y and 2 units of type Z still available. The system is currently in safe state. Consider the following independent requests for additional resources in the current state- REQ1: PO requests 0 units of X, 0 units of Y and 2 units of Z REQ2: P1 requests 2 units of X, 0 units of Y and 0 units of Z Write a program to check whether: A. Only REQ1 can be permitted B. Only REQ2 can be permitted C. Both REQ1 and REQ2 can be permitted D. Neither REQ1 nor REQ2…Real Time Scheduling: Select all statements below that are true EDF scheduling assigns the highest priority to a process with the smallest remaining time until its deadline. RM scheduling assigns a higher priority to processes with shorter periods. Rate Monotonic (RM) scheduling works by dividing the total amount of time available into an equal number of shares, and then each process must request a certain share of the total when it tries to start. If a process has period p, it is activated every p units of time. A real-time system is schedulable under Earliest Deadline First (EDF) when its overall CPU utilization is less than or equal to 1. A real-time system is schedulable under RM when its overall CPU utilization is less than 1.
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