操作系统基础基本理论、基本方法复习

Chapter 1 Introduction1.Abstract View of System Components2.Multiprogrammed Batch SystemsSeveral jobs are kept in main memory at the same time, and theCPU is multiplexed among them.Chapter 2: Computer-System Structuresputer-System Architecture2. Interrupt Time Line For a Single Process Doing Output3.Moving-Head Disk Mechanism4.Direct Memory Access StructureSix Step Process to Perform DMA Transfer 5.Storage-Device Hierarchye of A Base and Limit Register7.Hardware Address ProtectionChapter 3: Operating-System Structures 1.MS-DOS Execution2.MS-DOS Layer StructureChapter 4: Processes1.Diagram of Process State2.Process Control Block (PCB)3.CPU Switch From Process to Process4.Representation of Process Scheduling5.Addition of Medium Term SchedulingChapter 5: Threads 1.Single and Multithreaded Processes2.Many-to-One Model3.One-to-one Model4.Many-to-Many ModelChapter 6: CPU Scheduling1.Alternating Sequence of CPU And I/O Bursts2.First-Come, First-Served (FCFS) SchedulingProcess Burst TimeP124P2 3P3 3●Suppose that the processes arrive in the order: P1 , P2 , P3The Gantt Chart for the schedule is:●Waiting time for P1= 0; P2= 24; P3 = 27●Average waiting time: (0 + 24 + 27)/3 = 173.Example of Non-Preemptive SJFProcess Arrival Time Burst TimeP10.0 7P2 2.0 4P3 4.0 1P4 5.0 4●SJF (non-preemptive)●Average waiting time = (0 + 6 + 3 + 7)/4 - 44.Example of Preemptive SJFProcess Arrival Time Burst TimeP10.0 7P 2 2.0 4P 3 4.0 1P 4 5.0 4● SJF (preemptive)● Average waiting time = (9 + 1 + 0 +2)/4 - 35.Example of RR with Time Quantum = 20 Process Burst TimeP 1 53P 2 17P 3 68P 4 24● The Gantt chart is:● Typically, higher average turnaround than SJF, but better response .6.Dispatch LatencyChapter 7: Process Synchronization 1.Implementation • Semaphore operations now defined aswait (S ):S.value--;if (S.value < 0) {add this process to S.L;block;} signal (S ):S.value++;if (S.value <= 0) {remove a process P from S.L;wakeup(P);}2.Deadlock•– two or more processes are waiting indefinitely for an event that can be caused by only one of the waiting processes.•Let S and Q be two semaphores initialized to 1P0P1wait(S); wait(Q);wait(Q); wait(S);signal(S); signal(Q);signal(Q) signal(S);3.Bounded-Buffer Problem•Shared datasemaphore full, empty, mutex;Initially:full = 0, empty = n, mutex = 1Problem Producer Processdo {…produce an item in nextp…wait(empty);wait(mutex);…add nextp to buffer…signal(mutex);signal(full);} while (1);Problem Consumer Processdo {wait(full)wait(mutex);…remove an item from buffer to nextc…signal(mutex);signal(empty);…consume the item in nextc…} while (1);4.Readers-Writers Problem•Shared datasemaphore mutex, wrt;Initiallymutex = 1, wrt = 1, readcount = 0 Writer Processwait(wrt);…writing is performed…signal(wrt);Reader Processwait(mutex);readcount++;if (readcount == 1)wait(rt);signal(mutex);…reading is performed…wait(mutex);readcount--;if (readcount == 0)signal(wrt);signal(mutex):5.Dining-Philosophers Problem•Shared datasemaphore chopstick[5];Initially all values are 1•Philosopher i:do {wait(chopstick[i])wait(chopstick[(i+1) % 5])…eat…signal(chopstick[i]);signal(chopstick[(i+1) % 5]);…think…} while (1);Chapter 8: Deadlocks1.Example of a Resource Allocation Graph2.Basic Facts•If graph contains no cycles ⇒ no deadlock.•If graph contains a cycle ⇒–if only one instance per resource type, then deadlock.–if several instances per resource type, possibility of deadlock.3.Safe State•When a process requests an available resource, system must decide if immediate allocation leaves the system in a safe state.•System is in safe state if there exists a safe sequence of all processes.•Sequence <P1, P2, …, P n> is safe if for each P i, the resources that Pi can still request can be satisfied by currently available resources + resources held by all the P j, with j<I.–If P i resource needs are not immediately available, then P i can wait until all P j 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 its needed resources, and so on.4.Banker’s Algorithm•Multiple instances.•Each process must a priori claim maximum use.•When a process requests a resource it may have to wait.•When a process gets all its resources it must return them in a finite amount of time. Data Structures for the Banker’s AlgorithmLet n = number of processes, and m = number of resources types.•Available:Vector of length m. If available [j] = k, there are k instances of resource type R j available.•Max: n x m matrix. If Max [i,j] = k, then process P i may request at most k instances of resource type R j.•Allocation: n x m matrix. If Allocation[i,j] = k then P i is currently allocated k instances of R j.•Need: n x m matrix. If Need[i,j] = k, then P i may need k more instances of R j to completeits task.Need [i,j] = Max[i,j] –Allocation [i,j].Safety Algorithm1) Let Work and Finish be vectors of length m and n, respectively. Initialize:Work = AvailableFinish [i] = false for i - 1,3, …, n.2) Find and i such that both:(a) Finish [i] = false(b) Need i≤WorkIf no such i exists, go to step 4.3) Work = Work + Allocation iFinish[i] = truego to step 2.4) If Finish [i] == true for all i, then the system is in a safe state.Resource-Request Algorithm for Process P iRequest = request vector for process P i. If Request i[j] = k then process P i wants k instances of resource type R j.1) If Request i≤Need i go to step 2. Otherwise, raise error condition, since process has exceeded its maximum claim.2) If Request i≤Available, go to step 3. Otherwise P i must wait, since resources are not available.3) Pretend to allocate requested resources to P i by modifying the state as follows:Available = Available = Request i;Allocation i= Allocation i + Request i;Need i= Need i–Request i;;•If safe ⇒ the resources are allocated to P i.•If unsafe ⇒P i must wait, and the old resource-allocation state isrestored5.Example of Banker’s Algorithm• 5 processes P0 through P4; 3 resource types A(10 instances),B (5instances, andC (7 instances).•Snapshot at time T0:•Allocation Max Available• A B C A B C A B C•P0 0 1 0 7 5 3 3 3 2•P1 2 0 0 3 2 2•P2 3 0 2 9 0 2•P3 2 1 1 2 2 2•P40 0 2 4 3 3•T he content of the matrix. Need is defined to be Max – Allocation.•Need• A B C•P0 7 4 3•P1 1 2 2•P2 6 0 0•P30 1 1•P4 4 3 1•T he system is in a safe state since the sequence < P1, P3, P4, P2, P0> satisfies safety criteria.Chapter 9: Memory Management1.Multistep Processing of a User2.Dynamic relocation using a relocation register3.Overlays for a Two-Pass4.Schematic View of Swapping5.Hardware Support for Relocation and Limit Registers6.Address Translation Architecture7.Paging Hardware With TLB8.Valid (v) or Invalid (i) Bit In A Page TableChapter 10: Virtual Memory 1.Transfer of a Paged Memory to Contiguous Disk Space2.Steps in Handling a Page Fault3.First-In-First-Out (FIFO) Algorithm•Reference string: 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5• 3 frames (3 pages can be in memory at a time per process)• 4 framesFIFO Replacement –Belady’s Anomaly–more frames ⇒ less page faultsOptimal Page ReplacementLeast Recently Used (LRU) AlgorithmChapter 11: File-System Interface1.Sequential-access File2.Access Lists and Groups●Mode of access: read, write, execute●Three classes of usersRWXa) owner access7 ⇒ 1 1 1RWXb) group access 6 ⇒1 1 0RWXc) public access 1 ⇒0 0 1●Ask manager to create a group (unique name), say G, and add some users to the group.For a particular file (say game) or subdirectory, define an appropriate access.Chapter 12: File System Implementationyered File System2. A Typical File Control Block3.In-Memory File System StructuresChapter 13: I/O Systems1.Interrupt-Driven I/O CycleChapter 14: Mass-Storage SystemsDisk Scheduling●Several algorithms exist to schedule the servicing of disk I/O requests.●We illustrate them with a request queue (0-199).98, 183, 37, 122, 14, 124, 65, 67Head pointer 531.FCFSIllustration shows total head movement of 640 cylinders.2.SSTFIllustration shows total head movement of 236 cylinders.3.SCANIllustration shows total head movement of 208 cylinders.4.C-SCAN5.C-LOOK6.MS-DOS Disk Layout§1．True-False Questions: (Please answer with ╳or√)( )1.Only one process can be in the state of running on any processor at any time.√( )2.Context-switch time is not overhead; the system still can do useful work while switching. ╳( )3.There is at least one thread belongs to a running process. √( )4.Many-to-one model provides more concurrency than the one-to-one threading model. √( )5.All processes, either I/O-bound or CPU-bound, must perform I/O formeaningful execution. √( )6.CPU scheduling deals with the problem of deciding which of the processes located in the hard disk is to be allocated the CPU. ╳( )7.FCFS is a kind of preemptive scheduling algorithm. ╳( )8.SJF is optimal – gives minimum average waiting time for a given set of processes √( )9.With multilevel queue scheduling, the RoundRobin should be used in the foreground for the interactive processes. √( )10.It is better to determine where the program will reside in memory before loading than making this decision until program execution.╳( )11.The logical and physical addresses are the same in compile-time andload-time address binding but different in execution-time address binding.√( )12.With memory management unit (MMU), the user program never sees the real physical addresses.√( )13.With overlay, the system is able to keep all the necessary instructions and data in memory at all time.╳( )14.Multiprogramming is impossible in fixed-sized partition because each process may require different sized memory space.√( )15.It is possible for paging scheme to generate external fragmentation, while the internal fragmentation is inevitable.√( )16.Both associative registers and page tables can be searched simultaneously.╳( )17. A bad page replacement choice increases the page-fault rate and slows process execution, but this will bot affect the correct execution of theprocesses.╳( )18.The working-set window should be fixed within the execution of a particular program.╳( )19.SSTF scheduling is a form of SJF scheduling; may cause starvation of some requests.√( )20.Since OS is only software, the changes in the design of the hardware has no influence on i╳t.( )21.When a process executes, it typically executes for a long time before it either finishes or needs to perform I/O.√( )22.Random access memory (RAM) can only be accessed by CPU, no other device can directly communicate with it.╳( )23.Being another kind of processor, device controllers cannot executeconcurrently with the CPU.╳( )24.The concept of cache memory can be applied only between CPU andmemory in the storage device hierarchy.√( )25. A program is a passive entity and a process is an active entity with a program counter specifying the next instruction to execute.√( )26.The execution of a process is not necessarily to be sequential. √( )27.With layered approach, an operating system can be debugged and verified without any concern for the lower layer. ╳( )28.During system design, although mechanisms are strongly affecting policies, it is better to separate the policy from the mechanism. √( )29.Two processes associated with same program are considered the same separate execution sequences. ╳( )30.There are five states a process can be in, and only one process can be in running state at any instance.√( )31.The long-term scheduler selects which processes should be brought in to the ready queue. √( )32.Context-switch time is an overhead because the system cannot do useful work while switching. √( )33.The parent processes and the children processes created by the parent can be executed concurrently. √( )34.There will be no race condition when the instructions involved can be executed atomically. ╳( )35.When one thread is executing inside its critical section, all other threads should stop running and wait. ╳( )36.Without proper control, it is possible for two processes to be in their critical sections at the same time. √( )37.When solving critical section problems, we could make possible assumption concerning the relative speed of the threads. ╳( )38.Semaphores can be implemented as synchronization tool that does not require busy waiting. √( )39. A system built with semaphores cannot guarantee mutual exclusion. ╳( )40. A semaphore is always as powerful as a monitor and it always has value 0 or 1. ( )41.Deadlock prevention differs from avoidance because deadlock avoidance make deadlock impossible. √( )42.If the processes in the system has no hold-and-wait condition, no deadlock can happen. √( )43.In order to prevent deadlock, there is no other ways except ensuring no cycle can happen in the resource allocation graph associated with this system. ╳( )44.It is impossible for a system with deadlock prevention scheme to become deadlocked. ╳( )45.If a resource-allocation graph contains cycle, that means this system isdeadlocked. ╳( )ually deadlock recovery method should be invoked before deadlock detection.√( )47.The system will not become deadlocked even the system has reached an unsafe state. √( )48.Overlay can be used when the entering process is larger than amount of memory allocated to it. √( )49. A process must be completely idle before swapping, particularly with pending I/O.√§2 Multiple Choices, Single Answer Questions: (Please answer with capital letter( )1.Which of the following should not be contained in the stack of a process as temporary data?(A) method parameters (B) return addresses (C) global variables (D) localvariables( )2.Which of the following is not one of the information associated with a process in its process control block (PCB)? B(A) CPU registers (B) kernel (C) program counter (D) process state (E) I/O status ( )3.There are many scheduling queues associated with the processes running in a system, which are job queue, _______ queue, and device queues. A(A) ready queues (B) I/O queues (C) port queues (D) routine queues( )4.The selection of a process in the ready queue for entering the CPU for execution is called D(A) send (B) detach (C) transmit (D) dispatch( )5.The selection of a process to bring into the ready queue is happeninginfrequently, therefore, the execution of the ______-term scheduler may beslow. C(A) Short (B) medium (C) long (D) immediate( )6.When two processes wish to communicate, they need to establish a____________ between them before they can exchange messages. A(A) communication link (B) bridge (C) channel (D) password( )7.Which of the following should not be shared between threads? B(A) Code section (B) register set (C) Data section (D) other resources ( )8.Which of the following should not be one of the multithreading models? B(A) Many-to-one model (B) One-to-many model (C) One-to-one model (D)Many-to-many model( )9.In a system that do not support multiple kernel threads, we should use the _____________ model. B(A) Many-to-one (B) One-to-many (C) One-to-one (D) Many-to-many ( )10.For the criteria of CPU scheduling, which of the following is matching the goal of CPU utilization, throughput, turnaround time, and waiting time, respectively. C(A) max, min, max, min (B) min, max, max, min (C) max, max, min, min (D)min, min, max, max( )11.FCFS scheduling has convoy effect because many ______ processes waiting for one ______ process to get off the CPU. C(A) long, short (B) long, long (C) short, long (D) short, short( )12.SJF associates with each process the length of its next ______ to schedule the process. B(A) I/O interval (B) CPU burst (C) total process execution (D) program( )13.SJF cannot be implemented at the level of _______-term CPU scheduling. B(A) short (B) long (C) medium (D) temporary( )14.Dynamic loading D(A) is useful for executing small size code. (B) is better for frequentlyneeded program. (C) is loading the routine before it is needed. (D) obtaina better memory space utilization.( )15.In fixed-sized partition, the degree of multiprogramming is directly related to the number of ________ in the memory. E(A) CPUs (B) printers (C) registers (D) variables (E) partitions( )16.Suppose there are 40k, 50k, and 30k holes (in addressing sequence) left in the memory, which hole a 30k program will be allocated if first-fit, best-fit,and worst fit is used? (following the sequence) B(A) 50k, 30k, 40k (B) 40k, 30k, 50k (C) 40k, 50k, 30k (D) 30k, 40k, 50k ( )17.Alghough page table cannot be searched simultaneously, it uses the _________ property to locate the frame number in corresponding tableentry with one memory access. C(A) ordering (B) numbering (C) indexing (D) normalized( )18.Working set strategy prevents thrasing while keeping the degree ofmultiprocessing as _____ as possible. A(A) high (B) low (C) deep (D) short (E) long( )19.Two major determining factors of disk access time is the seek time and the __________. C(A) around time (B) feedback time (C) rotational latancy (D) executiontime (E) wait time( )20.There are four components comprise a computer system, including hardware, __________, application programs, and users. D(A) CPU (B) Windows (C) Hard disk (D) Operating Systems( )21.In multiprogrammed batch systems, several jobs are kept in main memory at the same time, and the CPU is _____________ among them. A(A) multiplexed (B) distributed (C) selecting (D) in-line (E) off-loaded ( )22.Which program loads the operating system into the memory? E(A) kernel (B) spooling (C) I/O (D) control (E) bootstrap( )23.Operating system is ___________ driven. D(A) data (B) signal (C) message (D) interrupt (E) software( )24.Device controller transfers blocks of data from buffer storage directly to main memory without CPU intervention is called D(A) Cycle stealing (B) Memory protection (C) Buffering (D) Directmemory access( )25.The only large storage media that the CPU can access directly is A(A) main memory (B) control storage (C) flash memory (D) hard disk(E) associative memory( )26.The greatest advantage by dividing operating system into layers is D(A) flexibility (B) convenience (C) effectiveness (D) modularity (E) easeof control( )27. A process includes its text section, ___________, program counter and stacks.D(A) data section (B) queue (C) memory (D) registers (E) variables( )28.The context of a process is represented in the _______of a process. B(A) kernel (B) PCB (C) control storage (D) queue( )29. A semaphore is an __________ that can only be accessed via two indivisible operations: P and V. D(A) condition variable (B) monitor variable (C) concurrent variable (D)integer variable( )30.The value of a _________ semaphore can range between 0 and 1. B(A) constant (B) binary (C) counting (D) fixed( )31.Deadlock avoidance scheme requires that the system has some additional_________ information available. C(A) detailed (B) structure (C) a priori (D) database( )32.Banker’s algorithm is a kind of B(A) Deadlock Recovery (B) Deadlock Avoidance (C) Deadlock Detection(D) Deadlock Prevention( )33.When obtaining the wait-for graph form the resource-allocation graph, thenodes of type _________ is removed. A(A) resource (B) process (C) lock (D) request( )34.Which of the following strategies does not need to search the entire list of freeholes before applying it? A(A) first fit (B) best fit (C) worst fit (D) better fit( )35.The allocated memory may be slightly larger than requested memory; this sizedifference causes B(A) medium fragmentation (B) internal fragmentation (C) externalfragmentation (D) patched fragmentation§3. Fill-in-the-blank Questions:( )1. A process can be temporarily brought out of memory to a backing store, andthen brought back into memory for continued execution. This is called____________( )2.________-fit and ________-fit better than worst-fit in terms of speed andstorage utilization.( )3.It requires _______ memory accesses for every instruction access if the pagetable is placed in main memory.( )4.When referencing the memory, the ________________should be searchedfirst and the page table next.( )5.It is important to keep the page-fault rate ________ in a demand-pagingsystem.( )6.OPT is difficult to implement since it requires ______ knowledge of thereference string( )7. A page is busy swapping pages in and out is called ____________.( )8.________ algorithm is also called “elevator algorithm”?§4. Short Answer Question:( )1.Suppose there are three processes entering the queue in the order specifiedbelow. The arrival time and the next CPU burst time are provided below. Pleasecalculate the average waiting time if using FCFS , SJF (preemptive version), andRoundRobin scheduling (with time quantum = 10). You need to show the detailof your calculation and complete the Gantt Chart below to demonstrate theexecution sequence.( )2.Consider following sequence of request queue for disk I/O0 20 40 60 80 0 20 40 6035, 53, 92, 43, 15 With disk head originally at cylinder 40, draw the paths of head movement by SSTF, SCAN and C-LOOK scheduling and find out their total head movement distances. SSTF: SCAN:(current head direction: right) C-LOOK: (reading direction:right) ( )3.Complete the logical address layout below showing a 32-bit machine with 16K page size. Show the layout If two-level paging is used, with 8 bits for the outer page. Calculate the size of the page table for each case. ( ) ( ) (8) ( ) ( ) ( )4.Consider the paging scheme with associative registers on the right. Please fill-in the page table and the associative registers with proper numbers. ( )5.Evaluate FIFO and LRU page replacement algorithm by running it on the memory reference string below and completing the tables provided. Compare the number of page faults resulted. Reference string: 5, 2, 0, 4, 2, 3, 4, 5, 2 ( )6. Please complete the state transition diagram on the right.( )7.A solution to the critical-section problem must satisfy three requirements: ____________, ____________, ____________. ( )8.The classical definition of P and V operation of semaphore is given below. eliminated the problem of busy waiting. P(S) { V(S) { while S ≤ 0 S++; ; // no-op; } S--; } ( )9. Fill-in the blanks provided in the diagram on。