Before proceeding with its execution, each process must acquire all the resources it needs is called
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A:
Q: 2- Illustrate the types of scheduling.
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| 47. | Before proceeding with its execution, each process must acquire all the resources it needs is called |
| a. | hold and wait |
| b. | No pre-emption |
| c. | circular wait |
| d. | starvation |
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- 78. To ensure no preemption, if a process is holding some resources and requests another resource that cannot be immediately allocated to it : a. then the process waits for the resources be allocated to it b. the process keeps sending requests until the resource is allocated to it c. the process resumes execution without the resource being allocated to it d. then all resources currently being held are preempted76. To ensure that the hold and wait condition never occurs in the system, it must be ensured that : a. whenever a resource is requested by a process, it is not holding any other resources b. each process must request and be allocated all its resources before it begins its execution c. a process can request resources only when it has none d. all of the mentioned496. Before execution an entire process doesn't need to be in memory, in technique of a. Offline storage b. Online storage c. Virtual memory d. Procedural memory
- 59. A system is in the safe state if a. the system can allocate resources to each process in some order and still avoid a deadlock b. there exist a safe sequence c. all of the mentioned d. none of the mentioned81. Given a priori information about the ________ number of resources of each type that maybe requested for each process, it is possible to construct an algorithm that ensures that the system will never enter a deadlock state. a. minimum b. average c. maximum d. approximate107. To _______ to a safe state, the system needs to keep more information about the states of processes. a. abort the process b. roll back the process c. queue the process d. none of the mentioned
- Consider the following processes SAVINGSACCOUNT and USER, and composite process ||ACCOUNT_USER. SAVINGSACCOUNT = (create -> ACCOUNT), ACCOUNT = (withdraw -> ACCOUNT | deposit -> ACCOUNT). USER = (earn -> deposit -> USER | withdraw -> spend -> USER). ||ACCOUNT_USER = (SAVINGSACCOUNT || USER). Revise the composite process ||ACCOUNT_USER such that the same savings account is shared by two users a and b. Note that the account must be created at the same time for the two users.Assume that the following processes are the only processes in a computer system and that there are no input/output requests from all the given processes. Given the following arrival time and CPU burst time (also called service time) for each process, answer the following questions. (a) First Come First Service (FCFS) * (b) Preemptive Shortest Job first * (c ) Round-Robin with time quantum of 4 (Assume that new processes will be inserted at the end of the ready listProblem Statement: (The following problem is based on problem statements in a number of operating systems texts) For this lab, you'll be setting up 2 queues of PCBs. A process control block (PCB) is an internal structure that is used to hold information associated with a process. A process is an instance of a program that is brought into random-access memory (RAM) for execution. To simplify matters, you'll be setting up 2 queues: a ready queue and a wait queue and, PCBs each PCB will have 2 fields: a process ID or PID (a positive integer) a link field to point to the next PCB on a queue The ready queue will have the PCBs for the processes that will be run in a first-come, first-served fashion. The wait queue will have the PCBs for the processes that will be waiting for a resource in a first-come, first-served fashion. If the PCB at the head of the ready queue needs some resource (for example, I/O) then it will be removed from the ready queue and moved to the rear of…
- When a process is waiting for an I/O operation to finish, it is said to be in the "Blocked" state. The following states are entered by a service when it is terminated: a) Terminated; b) Suspended; c) Running; d) Ready to Go.83. A state is safe, if : a. the system does not crash due to deadlock occurrence b. the system can allocate resources to each process in some order and still avoid a deadlock c. the state keeps the system protected and safe d. all of the mentionedWhen a process fork a child process then it enters ___________state. a. waiting b. terminated c. ready d. new