soln9-14.txt

operating system concepts sixth edition windows XP updat 操作系统课后答案

  Some systems allow different file operations based on the type of
  the file (for instance, an ascii file can be read as a stream while
  a database file can be read via an index to a block). Other systems
  leave such interpretation of a file's data to the process and provide
  no help in accessing the data. The method which is  better  depends
  on the needs of the processes on the system, and the demands the
  users place on the operating system. If a system runs mostly database
  applications, it may be more efficient for the operating system to
  implement a database-type file and provide operations rather than
  making each program implement the same thing (possibly in different
  ways). For general-purpose systems it may be better to only implement
  basic file types to keep the operating system size smaller and allow
  maximum freedom to the processes on the system.

Question 11.7:
  Explain the purpose of the open and close operations. 

Answer:   
  The open operation informs the system that the named file is about
  to become active.  The close operation informs the system that the
  named file is no longer in active use by the user who issued the
  close operation.

Question 11.14:
  Consider a file currently consisting of 100 blocks. Assume that the file
  control block (and the index block, in the case of indexed allocation)
  is already in memory. Calculate how many disk I/O operations are
  required for contiguous, linked, and indexed (single-level) allocation
  strategies, if, for one block, the following conditions hold. In the
  contiguous-allocation case, assume that there is no room to grow in
  the beginning, but there is room to grow in the end. Assume that the
  block information to be added is stored in memory.

    a. The block is added at the beginning. 
    b. The block is added in the middle. 
    c. The block is added at the end. 
    d. The block is removed from the beginning. 
    e. The block is removed from the middle. 
    f. The block is removed from the end. 

Answer:
           Contiguous Linked Indexed 
         a.    201       2      2 
         b.    101      52      2 
         c.      1       3      2 
         d.    198       2      1
         e.     98      51      1 
         f.      0     100      1

Question 11.17:
  Why must the bit map for file allocation be kept on mass storage rather
  than in main memory?

Answer: 
  In case of system crash (memory failure), the free-space list would not
  be lost as it would be if the bit map had been stored in main memory.

Question 11.22:
  How do caches help improve performance? Why do systems not use more
  or larger caches if they are so useful?

Answer: 
  Caches allow components of differing speeds to communicate more
  efficiently by storing data from the slower device, temporarily,
  in a faster device (the cache). Caches are, almost by definition,
  more expensive than the device they are caching for, so increasing
  the number or size of caches would increase system cost.

Question 11.24:
  Why is it advantageous for the user of an operating system to
  dynamically allocate its internal tables? What are the penalties to
  the operating system for doing so?

Answer: 
  Dynamic tables allow more flexibility in system use growth tables are
  never exceeded, avoiding artificial use limits. Unfortunately, kernel
  structures and code are more complicated, so there is more potential
  for bugs. The use of one resource can take away more system resources
  (by growing to accommodate the requests) than with static tables.

Question 12.1:
  State three advantages of placing functionality in a device controller
  rather than in the kernel. State three disadvantages.

Answer:

  Three advantages: Bugs are less likely to cause an operating system
  crash.  Performance can be improved by utilizing dedicated hardware and
  hard-coded algorithms. The kernel is simplified by moving algorithms
  out of it.

  Three disadvantages: Bugs are harder to fix a new firmware version or
  new hardware is needed Improving algorithms likewise requires a hardware
  update rather than just a kernel or device driver update. Embedded
  algorithms could conflict with the application s use of the device,
  causing decreased performance.

Question 12.4:
  Describe three circumstances under which blocking I/O should be used.
  Describe three circumstances under which nonblocking I/O should be used.
  Why not just implement nonblocking I/O and have processes busy-wait
  until their device is ready?

Answer: 

  Generally, blocking I/O is appropriate when the process will only
  be waiting for one spe-cific event. Examples include a disk, tape,
  or keyboard read by an application program.

  Non-blocking I/O is useful when I/O may come from more than one source
  and the order of the I/O arrival is not predetermined. Examples
  include network daemons listening to more than one network socket,
  window managers that accept mouse movement as well as keyboard input,
  and I/O-management programs, such as a copy command that copies data
  between I/O devices. In the last case, the program could optimize its
  performance by buffering the input and output and using non-blocking
  I/O to keep both devices fully occupied.  Non-blocking I/O is more
  complicated for programmers, because of the asynchronous rendezvous that
  is needed when an I/O occurs. Also, busy waiting is less efficient than
  interrupt-driven I/O so the overall system performance would decrease.

Question 12.5:
  Why might a system use interrupt-driven I/O to manage a single serial
  port but polling I/O to manage a front-end processor, such as a
  terminal concentrator?

Answer: 

  Polling can be more efficient than interrupt-driven I/O. This is the
  case when the I/O is frequent and of short duration. Even though a
  single serial port will perform I/O relatively infrequently and should
  thus use interrupts, a collection of serial ports such as those in a
  terminal concentrator can produce a lot of short I/O operations, and
  interrupting for each one could create a heavy load on the system. A
  well-timed polling loop could alleviate that load without wasting many
  resources through looping with no I/O needed.

Question 12.6:
  Polling for an I/O completion can waste a large number of CPU cycles
  if the processor iterates a busy-waiting loop many times before
  the I/O completes. But if the I/O device is ready for service,
  polling can be much more efficient than is catching and dispatching
  an interrupt. Describe a hybrid strategy that combines polling,
  sleeping, and interrupts for I/O device service. For each of these
  three strategies (pure polling, pure interrupts, hybrid), describe a
  computing environment in which that strategy is more efficient than
  is either of the others.

Answer:
  This is a good test question!

Question 12.8:
  How does DMA increase system concurrency? How does it complicate
  hardware design?

Answer: 
  DMA increases system concurrency by allowing the CPU to perform
  tasks while the DMA system transfers data via the system and memory
  busses. Hardware design is complicated because the DMA controller
  must be integrated into the system, and the system must allow the DMA
  controller to be a bus master. Cycle stealing may also be necessary
  to allow the CPU and DMA controller to share use of the memory bus.

Question 13.1:
  None of the disk-scheduling disciplines, except FCFS, is truly fair
  (starvation may occur).

    a. Explain why this assertion is true. 
    b. Describe a way to modify algorithms such as SCAN to ensure fairness.
    c. Explain why fairness is an important goal in a time-sharing system. 
    d. Give three or more examples of circumstances in which it is important
       that the op-erating system be unfair in serving I/O requests.

Answer: 

  a. New requests for the track over which the head currently resides can
     theoretically arrive as quickly as these requests are being serviced.

  b. All requests older than some predetermined age could be  forced
     to the top of the queue, and an associated bit for each could
     be set to indicate that no new request could be moved ahead of
     these requests. For SSTF, the rest of the queue would have to be
     reorganized with respect to the last of these  old  requests.

  c. To prevent unusually long response times.

  d. Paging and swapping should take priority over user requests. It may
     be desirable for other kernel-initiated I/O, such as the writing
     of file system metadata, to take precedence over user I/O. If the
     kernel supports real-time process priorities, the I/O requests of
     those processes should be favored.

Question 13.2:
  Suppose that a disk drive has 5000 cylinders, numbered 0 to 4999. The
  drive is currently serving a request at cylinder 143, and the previous
  request was at cylinder 125. The queue of pending requests, in FIFO
  order, is

        86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130

  Starting from the current head position, what is the total distance
  (in cylinders) that the disk arm moves to satisfy all the pending
  requests, for each of the following disk-scheduling algorithms?

    a. FCFS 
    b. SSTF
    c. SCAN 
    d. LOOK 
    e. C-SCAN

Answer: 
    a. The FCFS schedule is 
          143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130. 
       The total seek distance is 7081. 
    
    b. The SSTF schedule is 
          143, 130, 86, 913, 948, 1022, 1470, 1509, 1750, 1774. 
       The total seek distance is 1745. 
    
    c. The SCAN schedule is
          143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 130, 86.
       The total seek distance is 9769. 
    
    d. The LOOK schedule is 
           143, 913, 948, 1022, 1470, 1509, 1750, 1774, 130, 86.
       The total seek distance is 3319. 
    
    e. The C-SCAN schedule is 
          143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 86, 130.
       The total seek distance is 9813. 
    
    f.  (Bonus.) The C-LOOK schedule is 
           143, 913, 948, 1022, 1470, 1509, 1750, 1774, 86, 130.
        The total seek distance is 3363.
 
Question 13.4:
  Suppose that the disk in Exercise 13.3 rotates at 7200 RPM. 
    a. What is the average rotational latency of this disk drive? 
    b. What seek distance can be covered in the time that you found 
       for part a? 

Answer: 

    a. 7200 rpm gives 120 rotations per second. Thus, a full rotation 
       takes 8.33 ms, and the average rotational latency (a half rotation) 
       takes 4.167 ms. 
    
    b. Solving t =0.7561 + 0.2439 pL for t =4.167 gives L = 195.58,