An operating system comprises three user processes, and each of these processes consumes two units of the resource R. The minimal number of R units that must be employed in order to avoid deadlocks is
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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 isThree 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 isAn operating system uses a total of two units of the resource R for each of its three user processes. The operating system uses a total of 12 units of R. The absolute minimum number of R units that must be used to guarantee that there are no deadlocks is
- An operating system contains a total of three user processes, and each of these processes utilises a total of two units of the resource R. The minimum amount of R units that must be used to prevent any deadlocks from occurring isAn operating system contains 3 user processes each requiring 2 units of resource R .The minimum number of units of R such that no deadlocks will ever arise isAn operating system has three user processes, each of which uses two units of resource R. The smallest number of R units required to ensure that no deadlocks occur is
- An 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:Two 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 isProcess Scheduling Multiprocessor systems use multiple CPUs to perform various tasks. This increases throughput and reduces response time. In this problem, a multiprocessor system has a certain number of processors. Each processor has the ability to schedule a limited number of processes in one second. However, after this scheduling, the processor's ability is reduced to floor(ability/2). Given the processor's abilities and the number of processes, what is the minimum time required to schedule all the processes in the system? Example n = 5 (number of processors and size of ability[]) ability = 13, 1, 7, 2, 4] processes = 15 This optimal solution is: 1. First, the processor with ability = 7 schedules 7 processes in one second. Now, ability = (3, 1, 3, 2, 4] because 7 was reduced to floor(7/2). There are 15 - 7 = 8 remaining processes. 2. Second, the processor with ability = 4 is used. After that, ability = [3, 1, 3, 2, 2). Remaining processes = 8 - 4= 4.
- 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 isThere 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 isEach of the three user processes that make up an operating system utilizes a total of two units of the resource R. Twelve units of R are used overall by the operating system. To ensure that there are no deadlocks, the bare minimum number of R units that must be employed is