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 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 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 is
- An 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 isAn 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 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 isFor 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
- 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 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.Here are the top five (5) factors that should be considered while formulating a process scheduling policy.