Final Exam Review. CPSC 457, Spring 2016 June 29-30, M. Reza Zakerinasab Department of Computer Science, University of Calgary

Transcription

1 Final Exam Review CPSC 457, Spring 2016 June 29-30, 2015 M. Reza Zakerinasab Department of Computer Science, University of Calgary

2 Final Exam Components Final Exam: Monday July 4, 8 am in ICT % of the final mark You can bring a non-programmable calculator and one letter size paper for notes. 25 multiple choice questions 1 pseudo code question 1 long answer question ~10 short answer questions 2

3 Review: Address Space What is an address space? Address space is an abstraction of physical memory. Physical addresses in memory are managed by the memory hardware and accessed by the CPU, creating a physical address space. However, processes use a set of virtual addresses to access memory words, called (logical) address space. What are the names of the two hardware registers used to translate a virtual address to a physical address? Base register (or relocation register) Limit register Address binding can be done at compile time, load time, or execution time. True or False? True 3

4 Review: Memory Allocation What are the main disadvantages of fixed memory partitioning? The number of partitions limits the number of active processes in the system. It also suffers from internal fragmentation and low memory utilization. What is the technique for removing external fragmentations? Briefly explain. Memory compaction. From time to time, the OS shifts the processes so that their physical memory partitions are contiguous and all free memory is together in one block. 4

5 Virtual Memory Virtual memory illusion for contiguous memory allocation Without the limits posed by the size of the physical memory Virtual (logical) memory is divided to memory pages A page is a fixed-length contiguous block of virtual memory Advantages No need to worry about the physical limit on process size More programs can be loaded at the same time better CPU utilization Less I/O time spent on swapping the entire program 5

6 Paging Exercise virtual address space =? physical address space =? 64KB 32KB page size =? # of pages =? frame size =? # of frames =? 4KB 16 4KB 8 Physical addresses of virtual addresses 0, 8192, 24576, and 36876? Not in memory K (36864) =

7 Resources A system consists of a finite number of resources to be shared among a number of competing processes. Resources: Physical resources: printers, drives, memory space, CPU cycles Logical resources: files, semaphore, monitors Preemptable resources (a.k.a. shareable resources) They can be taken away from the processing owning it with no ill effects. e.g., memory, a read-only file Non-preemptable resources (a.k.a. non-shareable resources) They cannot be taken away from its current owner without causing the computation to fail e.g., printer, DVD burner 7

8 Necessary Conditions for Deadlock A deadlock situation can arise if the following four conditions holds simultaneously: Mutual exclusion: Only one process at a time can use the resource. Others must wait. Hold and wait: A process holding a resource is waiting for another resource that is currently held by other processes. No preemption: A resource can be released only voluntarily by the process holding it. Circular wait: In a set {P 0, P 1,..., P n } of waiting processes, P 0 is waiting for resource held by P 1, P 1 is waiting for resource held by P 2, and so on. 8

9 Deadlock Handling, How? Prevent or avoid deadlocks so that they never occur Prevention: Ensure that at least one of the necessary conditions cannot hold Avoidance: requires, in advance, additional information concerning which resource a process will request and use during its lifetime Let it happen, detect and recover Detection examines the state of the system to determine whether a deadlock has occurred Recovery provides mechanism for getting out from a deadlock Ignore deadlock Used by most OSs since deadlocks occur infrequently in many systems 9

10 Deadlock Prevention If we can make sure at least one of the necessary conditions cannot hold, we can prevent the occurrence of a deadlock. 15

11 Review Define deadlock. A situation in which every process in the set is waiting for an event that can be caused only by another process in the set. If there is a circular wait among processes, we must have a deadlock situation. True or False False 16

12 Review What is the key for preventing deadlocks? Ensure that at least one of the necessary conditions of deadlock (mutual exclusion, hold and wait, no preemption, circular wait) cannot hold Which one of the following methods is used to prevent circular waiting among processes and resources? Spooling Request all resources initially Take resources away Order resources numerically 17

13 Deadlock Avoidance: Safe State A system is in a safe state Only if there exists a safe sequence <P 1, P 2,..., P n > of all the processes in the systems such that For each P i, the resources that P i can still request can be satisfied by current available resources + resources held by all processes P j, with j < i. That is: If the resources that P i needs are not immediately available, then P i can wait until all the P j, with j < i, have finished. When P j is finished, P i can obtain needed resources, execute, return allocated resources, and terminate. When P i terminates, P i+1 can obtain the resources, and so on. If no such sequence exists, then the system state is unsafe. 18

14 Unsafe State & Deadlock Unsafe state!= Deadlock state From a safe state, the system can guarantee that all processes will finish. From an unsafe state, no such guarantee can be given. The idea of the avoidance algorithms is simply to ensure that the system will always remain in a safe state. For single instance per type case: Resource allocation graph algorithm For general cases Banker s algorithm 19

15 Banker s Algorithm: Data Structure For general cases (multiple instances per resource type) Let n = # of processes, m = # of resource types Array Size Description Available [] m the # of available resources of each type Max [][] n x m the maximum demand of each process Allocation [][] n x m the # of resources of each type currently allocated to each process Need [][] n x m the remaining resource needed by each process Request [][] n x m the current request of each process Need[i][j] = Max[i][j] - Allocation[i][j] 20

16 Banker s Algorithm To determine whether a request / set of requests can be safely granted. Resource-request algorithm Safety algorithm Resource-request algorithm If request > needed, return error. If request > current available, return wait. Pretend the allocation: update Available, Allocation i, Need i Run Safety algorithm If safe state, grant the request. Otherwise, wait and restore the values of Available, Allocation i, Need i 21

17 Banker s Algorithm Safety algorithm Update the current available resources Find a process that can be satisfied by the current available resource Mark the process finished and pretend to release its resources Repeat until no more process can be found If all processes finish, return safe. Otherwise, return unsafe. 22

18 A simple example Banker s example: 4 customers (processes) A, B, C, and D 10 credit units (resources) Since the banker knows that not all customers will need their maximum credit immediately Has Max A 0 6 B 0 5 Has Max A 1 6 B 1 5 B requests one more credit Has Max A 1 6 B 2 5 C 0 4 C 2 4 C 2 4 D 0 7 D 4 7 D 4 7 Free: 10 Free: 2 Free: 1 Safe Safe Unsafe We can use Banker s algorithm to avoid the unsafe state. 23

19 Banker s Example Processes P 0, P 1, P 2, P 3, P 4 Resources A (10 instances), B (5 instances), C (7 instances) At time T0: Allocation Max Need Available A B C A B C A B C A B C P P P P P (Request1 = (1, 0, 2)) <= (Need1 = (1, 2, 2)) (Request 1 = (1, 0, 2)) <= (Available = (3, 3, 2)) 24

20 Banker s Example Allocation Max Need Available A B C A B C A B C A B C P P P P P (Request1 = (1, 0, 2)) Pretend the allocation: Available = (3, 3, 2) - (1, 0, 2) = (2, 3, 0) Allocation1 = (2, 0, 0) + (1, 0, 2) = (3, 0, 2) Need1 = (1, 2, 2) - (1, 0, 2) = (0, 2, 0) 25

21 Banker s Example Safety Algorithm Finish = [false, false, false, false, false] Work = Available = (2,3,0) Find an index i such that both a.finish[i] == false b.need i <= Work i = 1 Work = Work + Allocation i = (2,3,0) + (3,0,2) Allocation Max Need Available A B C A B C A B C A B C P P P P P = (5,3,2) Finish = [false, true, false, false, false] 26

22 Banker s Example Safety Algorithm Finish = [false, true, false, false, false] Work = (5,3,2) Find an index i such that both a.finish[i] == false b.need i <= Work i = 3 Work = Work + Allocation i = (5,3,2) + (2,1,1) Allocation Max Need Available A B C A B C A B C A B C P P P P P = (7,4,3) Finish = [false, true, false, true, false] 27

23 Banker s Example Safety Algorithm finish = [false, true, false, true, false] Work = (7,4,3) Find an index i such that both a.finish[i] == false b.need i <= Work i = 0 Work = Work + Allocation i = (7,4,3) + (0,1,0) Allocation Max Need Available A B C A B C A B C A B C P P P P P = (7,5,3) finish = [true, true, false, true, false] If we continue the algorithm, we see that all the processes can finish. Therefore, the state after granting the request is a safe state. 28

24 Deadlock Detection Single instance per resource type: Resource-allocation graph: a cycle implies a deadlock Wait-for graph: Remove the resource nodes and collapse the corresponding edges from the resource-allocation graph. A deadlock exists if and only if the wait-for graph contains a cycle 29

25 Deadlock Detection Multiple instance per resource type: Use the Safety Algorithm from the Banker s Algorithm If we cannot finish all the processes using current available resources and Safety Algorithm, we have a deadlock. Why deadlock detection instead of avoidance? Deadlock avoidance incurs considerable overhead Running safety algorithm per request Deadlock detection can run periodically E.g., once per hour Or whenever CPU utilization drops below a threshold 30

26 Deadlock Detection: Example Processes P 0, P 1, P 2, P 3, P 4 Resources A (7 instances), B (2 instances), C (6 instances) At time T 0 : Allocation Need Available A B C A B C A B C P P P P P

27 Deadlock Detection: Example Safety Algorithm Finish = [false, false, false, false, false] Work = Available = (0,0,0) Find an index i such that both a.finish[i] == false b.need i <= Work i = 0 Work = Work + Allocation i = (0,0,0) + (0,1,0) Allocation Need Available A B C A B C A B C P P P P P = (0,1,0) Finish = [true, false, false, false, false] 32

28 Deadlock Detection: Example Safety Algorithm Finish = [false, false, false, false, false] Work = Available = (0,1,0) Find an index i such that both a.finish[i] == false b.need i <= Work No Process Found Allocation Need Available A B C A B C A B C P P P P P DEADLOCK 33

29 Deadlock Recovery Process termination The system reclaims all resources allocated to the terminated processes. Abort all deadlocked processes Definitely breaks the deadlock cycle, but at a great cost --- all results of partial computations are lost. Abort one process at a time until the deadlock cycle is eliminated Pick the process whose termination will incur the minimum cost, and then invoke a deadlock detection algorithm. Repeat this process if the deadlock persists. How to select the victim? The one with least priority The one with least resource usage 34

30 Deadlock Recovery Process rollback Rollback processes to checkpoints right before acquiring the resource(s) involved in the deadlock. Problem? Additional rollback mechanism is needed. Resource preemption Preempt some resources Give the preempted resources to other processes until the deadlock cycle is broken. Problem? Select a victim Rollback the victim Prevent starvation 35

31 Ignoring Deadlocks No prevention, avoidance, detection or recovery Ignore the deadlocks If there is a deadlock, let it be Worst case scenario Gradually more and more resources are blocked due to deadlock Manual system restart is needed Doesn t seem good But much cheaper than prevention, avoidance or detection As deadlocks are infrequent in many systems 36

32 Deadlock - Review How to detect a deadlock? If we have single instance per resource type, we can use resource allocation graph or wait-for graph. If there is a cycle in one of these graphs, then we have a deadlock. If we have multiple instance per resource type, we use the detection (safety) algorithm to detect if we are in a deadlock. Name and briefly explain the three main ways for recovering from deadlocks. Process termination: terminate all the deadlocked processes or terminate them one by one until deadlock is broken. Process rollback: rollback the processes to a previous checkpoint and release the resources taken afterward. Resource preemption: preempt some resources that are involved in deadlock and give them to other processes so they can proceed. 37

33 Good luck in your exam 38