OSDI_Answers(操作系统设计与实现 答案)

OPERATING SYSTEMS: DESIGN AND IMPLEMENTATION THIRD EDITION PROBLEM SOLUTIONS ANDREW S. TANENBAUM Vrije Universiteit Amsterdam, The Netherlands ALBERT S. WOODHULL Amherst, Massachusetts PRENTICE HALL UPPER SADDLE RIVER, NJ 07458 SOLUTIONS TO CHAPTER 1 PROBLEMS 1. An operating system must provide the users with an extended (i.e., virtual) machine, and it must manage the I/O devices and other system resources. 2. In kernel mode, every machine instruction is allowed, as is access to all the I/O devices. In user mode, many sensitive instructions are prohibited. Operating systems use these two modes to encapsulate user programs. Run- ning user programs in user mode keeps them from doing I/O and prevents them from interfering with each other and with the kernel. 3. Multiprogramming is the rapid switching of the CPU between multiple processes in memory. It is commonly used to keep the CPU busy while one or more processes are doing I/O. 4. Input spooling is the technique of reading in jobs, for example, from cards, onto the disk, so that when the currently executing processes are finished, there will be work waiting for the CPU. Output spooling consists of first copying printable files to disk before printing them, rather than printing directly as the output is generated. Input spooling on a personal computer is not very likely, but output spooling is. 5. The prime reason for multiprogramming is to give the CPU something to do while waiting for I/O to complete. If there is no DMA, the CPU is fully occu- pied doing I/O, so there is nothing to be gained (at least in terms of CPU utili- zation) by multiprogramming. No matter how much I/O a program does, the CPU will be 100 percent busy. This of course assumes the major delay is the wait while data is copied. A CPU could do other work if the I/O were slow for other reasons (arriving on a serial line, for instance). 6. Second generation computers did not have the necessary hardware to protect the operating system from malicious user programs. 7. Choices (a), (c), and (d) should be restricted to kernel mode. 8. Personal computer systems are always interactive, often with only a single user. Mainframe systems nearly always emphasize batch or timesharing with many users. Protection is much more of an issue on mainframe systems, as is efficient use of all resources. 9. Arguments for closed source are that the company can vet the programmers, establish programming standards, and enforce a development and testing methodology. The main arguments for open source is that many more people look at the code, so there is a form of peer review and the odds of a bug slip- ping in are much smaller with so much more inspection. 2 PROBLEM SOLUTIONS FOR CHAPTER 1 10. The file will be executed. 11. It is often essential to have someone who can do things that are normally for- bidden. For example, a user starts up a job that generates an infinite amount of output. The user then logs out and goes on a three-week vacation to Lon- don. Sooner or later the disk will fill up, and the superuser will have to manu- ally kill the process and remove the output file. Many other such examples exist. 12. Any file can easily be named using its absolute path. Thus getting rid of working directories and relative paths would only be a minor inconvenience. The other way around is also possible, but trickier. In principle if the working directory is, say, /home/ast/projects/research/proj1 one could refer to the password file as ../../../../etc/passwd , but it is very clumsy. This would not be a practical way of working. 13. The process table is needed to store the state of a process that is currently suspended, either ready or blocked. It is not needed in a single process sys- tem because the single process is never suspended. 14. Block special files consist of numbered blocks, each of which can be read or written independently of all the other ones. It is possible to seek to any block and start reading or writing. This is not possible with character special files. 15. The read works normally. User 2’s directory entry contains a pointer to the i-node of the file, and the reference count in the i-node was incremented when user 2 linked to it. So the reference count will be nonzero and the file itself will not be removed when user 1 removes his directory entry for it. Only when all directory entries for a file have been removed will its i-node and data actually vanish. 16. No, they are not so essential. In the absence of pipes, program 1 could write its output to a file and program 2 could read the file. While this is less efficient than using a pipe between them, and uses unnecessary disk space, in most circumstances it would work adequately. 17. The display command and reponse for a stereo or camera is similar to the shell. It is a graphical command interface to the device. 18. Windows has a call spawn that creates a new process and starts a specific program in it. It is effectively a combination of fork and exec. 19. If an ordinary user could set the root directory anywhere in the tree, he could create a file etc/passwd in his home directory, and then make that the root directory. He could then execute some command, such as su or login that reads the password file, and trick the system into using his password file, instead of the real one. PROBLEM SOLUTIONS FOR CHAPTER 1 3 20. The getpid, getuid, getgid, and getpgrp, calls just extract a word from the pro- cess table and return it. They will execute very quickly. They are all equally fast. 21. The system calls can collectively use 500 million instructions/sec. If each call takes 1000 instructions, up to 500,000 system calls/sec are possible while consuming only half the CPU. 22. No, unlink removes any file, whether it be for a regular file or a special file. 23. When a user program writes on a file, the data does not really go to the disk. It goes to the buffer cache. The update program issues SYNC calls every 30 seconds to force the dirty blocks in the cache onto the disk, in order to limit the potential damage that a system crash could cause. 24. No. What is the point of asking for a signal after a certain number of seconds if you are going to tell the system not to deliver it to you? 25. Yes it can, especially if the system is a message passing system. 26. When a user program executes a kernel-mode instruction or does something else that is not allowed in user mode, the machine must trap to report the attempt. The early Pentiums often ignored such instructions. This made them impossible to fully virtualize and run an arbitrary unmodified operating sys- tem in user mode. SOLUTIONS TO CHAPTER 2 PROBLEMS 1. It is central because there is so much parallel or pseudoparallel activity— multiple user processes and I/O devices running at once. The multiprogram- ming model allows this activity to be described and modeled better. 2. The states are running, blocked and ready. The running state means the pro- cess has the CPU and is executing. The blocked state means that the process cannot run because it is waiting for an external event to occur, such as a mes- sage or completion of I/O. The ready state means that the process wants to run and is just waiting until the CPU is available. 3. You could have a register containing a pointer to the current process table entry. When I/O completed, the CPU would store the current machine state in the current process table entry. Then it would go to the interrupt vector for the interrupting device and fetch a pointer to another process table entry (the service procedure). This process would then be started up. 4. Generally, high level languages do not allow one the kind of access to CPU hardware that is required. For instance, an interrupt handler may be required to enable and disable the interrupt servicing a particular device, or to 4 PROBLEM SOLUTIONS FOR CHAPTER 2 manipulate data within a process’ stack area. Also, interrupt service routines must execute as rapidly as possible. 5. The figure looks like this 1 23 4Blocked Running Ready 1. Process blocks for input 2. Scheduler picks another process 3. Scheduler picks this process 4. Input becomes available 5. Process is terminated New Terminated 5 0 0. New process made ready 6. It would be difficult, if not impossible, to keep the file system consistent using the model in part (a) of the figure. Suppose that a client process sends a request to server process 1 to update a file. This process updates the cache entry in its memory. Shortly thereafter, another client process sends a request to server 2 to read that file. Unfortunately, if the file is also cached there, server 2, in its innocence, will return obsolete data. If the first process writes the file through to the disk after caching it, and server 2 checks the disk on every read to see if its cached copy is up-to-date, the system can be made to work, but it is precisely all these disk accesses that the caching system is try- ing to avoid. 7. A process is a grouping of resources: an address space, open files, signal handlers, and one or more threads. A thread is just an execution unit. 8. Each thread calls procedures on its own, so it must have its own stack for the local variables, return addresses, and so on. 9. A race condition is a situation in which two (or more) process are about to perform some action. Depending on the exact timing, one or other goes first. If one of the processes goes first, everything works, but if another one goes first, a fatal error occurs. 10. One person calls up a travel agent to find about price and availability. Then he calls the other person for approval. When he calls back, the seats are gone. 11. A possible shell script might be: if [ ! –f numbers ]; echo 0 > numbers; fi count=0 while (test $count != 200 ) do count=‘expr $count + 1 ‘ PROBLEM SOLUTIONS FOR CHAPTER 2 5 n=‘tail –1 numbers‘ expr $n + 1 >>numbers done Run the script twice simultaneously, by starting it once in the background (using &) and again in the foreground. Then examine the file numbers. It will probably start out looking like an orderly list of numbers, but at some point it will lose its orderliness, due to the race condition created by running two copies of the script. The race can be avoided by having each copy of the script test for and set a lock on the file before entering the critical area, and unlocking it upon leaving the critical area. This can be done like this: if ln numbers numbers.lock then n=‘tail –1 numbers‘ expr $n + 1 >>numbers rm numbers.lock fi This version will just skip a turn when the file is inaccessible, variant solu- tions could put the process to sleep, do busy waiting, or count only loops in which the operation is successful. 12. Yes, at least in MINIX 3. Since LINK is a system call, it will activate server and task level processes, which, because of the multi-level scheduling of MINIX 3, will receive priority over user processes. So one would expect that from the point of view of a user process, linking would be equivalent to an atomic act, and another user process could not interfere. Also, even if another user process gets a chance to run before the LINK call is complete, perhaps because the disk task blocks looking for the inode and directory, the servers and tasks complete what they are doing before accepting more work. So, even if two processes try to make a LINK call at the same time, whichever one causes a software interrupt first should have its LINK call completed first. 13. Yes, it still works, but it still is busy waiting, of course. 14. Yes it can. The memory word is used as a flag, with 0 meaning that no one is using the critical variables and 1 meaning that someone is using them. Put a 1 in the register, and swap the memory word and the register. If the register contains a 0 after the swap, access has been granted. If it contains a 1, access has been denied. When a process is done, it stores a 0 in the flag in memory. 15. To do a semaphore operation, the operating system first disables interrupts. Then it reads the value of the semaphore. If it is doing a DOWN and the semaphore is equal to zero, it puts the calling process on a list of blocked processes associated with the semaphore. If it is doing an UP, it must check 6 PROBLEM SOLUTIONS FOR CHAPTER 2 to see if any processes are blocked on the semaphore. If one or more processes are blocked, one of then is removed from the list of blocked processes and made runnable. When all these operations have been com- pleted, interrupts can be enabled again. 16. Associated with each counting semaphore are two binary semaphores, M, used for mutual exclusion, and B, used for blocking. Also associated with each counting semaphore is a counter that holds the number of UPs minus the number of DOWNs, and a list of processes blocked on that semaphore. To implement DOWN, a process first gains exclusive access to the semaphores, counter, and list by doing a DOWN on M. It then decrements the counter. If it is zero or more, it just does an UP on M and exits. If M is negative, the pro- cess is put on the list of blocked processes. Then an UP is done on M and a DOWN is done on B to block the process. To implement UP, first M is DOWNed to get mutual exclusion, and then the counter is incremented. If it is more than zero, no one was blocked, so all that needs to be done is to UP M. If, however, the counter is now negative or zero, some process must be removed from the list. Finally, an UP is done on B and M in that order. 17. With round robin scheduling it works. Sooner or later L will run, and eventu- ally it will leave its critical region. The point is, with priority scheduling, L never gets to run at all; with round robin, it gets a normal time slice periodi- cally, so it has the chance to leave its critical region. 18. It is very expensive to implement. Each time any variable that appears in a predicate on which some process is waiting changes, the run-time system must re-evaluate the predicate to see if the process can be unblocked. With the Hoare and Brinch Hansen monitors, processes can only be awakened on a SIGNAL primitive. 19. The employees communicate by passing messages: orders, food, and bags in this case. In MINIX terms, the four processes are connected by pipes. 20. It does not lead to race conditions (nothing is ever lost), but it is effectively busy waiting. 21. If a philosopher blocks, neighbors can later see that he is hungry by checking his state, in test, so he can be awakened when the forks are available. 22. The change would mean that after a philosopher stopped eating, neither of his neighbors could be chosen next. With only two other philosophers, both of them neighbors, the system would deadlock. With 100 philosophers, all that would happen would be a slight loss of parallelism. 23. Variation 1: readers have priority. No writer may start when a reader is active. When a new reader appears, it may start immediately unless a writer is currently active. When a writer finishes, if readers are waiting, they are all PROBLEM SOLUTIONS FOR CHAPTER 2 7 started, regardless of the presence of waiting writers. Variation 2: Writers have priority. No reader may start when a writer is waiting. When the last active process finishes, a writer is started, if there is one, otherwise, all the readers (if any) are started. Variation 3: symmetric version. When a reader is active, new readers may start immediately. When a writer finishes, a new writer has priority, if one is waiting. In other words, once we have started reading, we keep reading until there are no readers left. Similarly, once we have started writing, all pending writers are allowed to run. 24. It will need nT sec. 25. If a process occurs multiple times in the list, it will get multiple quanta per cycle. This approach could be used to give more important processes a larger share of the CPU. 26. The CPU efficiency is the useful CPU time divided by the total CPU time. When Q ≥T, the basic cycle is for the process to run for T and undergo a pro- cess switch for S. Thus (a) and (b) have an efficiency of T /(S + T). When the quantum is shorter than T, each run of T will require T /Q process switches, wasting a time ST /Q. The efficiency here is then T + ST /Q T ��������� which reduces to Q /(Q + S), which is the answer to (c). For (d), we just sub- stitute Q for S and find that the efficiency is 50 percent. Finally, for (e), as Q → 0 the efficiency goes to 0. 27. Shortest job first is the way to minimize average response time. 0 < X ≤ 3: X, 3, 5, 6, 9. 3 < X ≤ 5: 3, X, 5, 6, 9. 5 < X ≤ 6: 3, 5, X, 6, 9. 6 < X ≤ 9: 3, 5, 6, X, 9. X > 9: 3, 5, 6, 9, X. 28. For round robin, during the first 10 minutes each job gets 1/5 of the CPU. At the end of 10 minutes, C finishes. During the next 8 minutes, each job gets 1/4 of the CPU, after which time D finishes. Then each of the three remain- ing jobs gets 1/3 of the CPU for 6 minutes, until B finishes, and so on. The finishing times for the five jobs are 10, 18, 24, 28, and 30, for an average of 22 minutes. For priority scheduling, B is run first. After 6 minutes it is finished. The other jobs finish at 14, 24, 26, and 30, for an average of 18.8 minutes. If the jobs run in the order A through E, they finish at 10, 16, 18, 22, and 30, for an average of 19.2 minutes. Finally, shortest job first yields finishing times of 2, 6, 12, 20, and 30, for an average of 14 minutes. 8 PROBLEM SOLUTIONS FOR CHAPTER 2 29. The first time it gets 1 quantum. On succeeding runs it gets 2, 4, 8, and 15, so it must be swapped in 5 times. 30. The sequence of predictions is 40, 30, 35, and now 25. 31. Yes. Two-level scheduling could be used if memory is too small to hold all the ready processes. Some set of them is put into memory, and a choice is made from that set. From time to time, the set of in-core processes is adjusted. This algorithm is easy to implement and reasonably efficient, cer- tainly a lot better than say, round robin without regard to whether a process was in memory or not. 32. There are three ways to pick the first one, four ways to pick the second, three ways to pick the third and four ways to pick the fourth, for a total of 3 × 4 × 3 × 4 = 144. Note that a thread can be chosen a second time. 33. The fraction of the CPU used is 35/50 + 20/100 + 10/200 + x/250. To be schedulable, this must be less than 1. Thus x must be less than 12.5 msec. 34. This pointer makes it easy to find the place to save the registers when a pro- cess switch is needed, either due to a system call or an interrupt. 35. When a clock or keyboard interrupt occurs, and the task that should get the message is not blocked, the system has to do something strange to avoid los- ing the interrupt. With buffered messages this problem would not occur. Notification bitmaps provide provide a simple alternative to buffering. 36. While the system is adjusting the scheduling queues, they can be in an incon- sistent state for a few instructions. It is essential that no interrupts occur dur- ing this short interval, to avoid having the queues accessed by the interrupt handler while they are inconsistent. Disabling interrupts prevents this prob- lem by preventing recursive entries into the scheduler. 37. When a RECEIVE is done, a source process is specified, telling who the receiving process is interested in hearing from. The loop checks to see if that process is among the process that are currently blocked trying to send to the receiving process. Each iteration of the loop examines another blocked pro- cess to see who it is. 38. Tasks, drivers and servers get large quanta, but even they can be preempted if they run too long. Also if a driver or server is not allowing other processes to run it can be demoted to a lower-priority queue. Even though they are given large quanta, all system