soln9-14.txt

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

       so we can seek over 195 tracks (about 4% of the disk) during 
       an average rotational latency.
  
Question 13.13:
  Why is it important to balance file system I/O among the disks and
  controllers on a system in a multitasking environment?

Answer: 

  A system can only perform at the speed of its slowest bottleneck. Disks
  or disk controllers are frequently the bottleneck in modern systems as
  their individual performance cannot keep up with that of the CPU and
  system bus.  By balancing I/O among disks and controllers, neither an
  individual disk nor a controller is overwhelmed, so that bottleneck
  is avoided.

Question 13.17:
  The term  fast wide SCSI-II  denotes a SCSI bus that operates at a
  data rate of 20 megabytes per second when it moves a packet of bytes
  between the host and a device. Suppose that a fast wide SCSI-II disk
  drive spins at 7200 RPM, has a sector size of 512 bytes, and holds
  160 sectors per track.

  a. Estimate the sustained transfer rate of this drive in megabytes
     per second.

  b. Suppose that the drive has 7000 cylinders, 20 tracks per cylinder,
     a head switch time (from one platter to another) of 0.5 millisecond,
     and an adjacent cylinder seek time of 2 milliseconds. Use this
     additional information to give an accurate estimate of the sustained
     transfer rate for a huge transfer.

  c. Suppose that the average seek time for the drive is 8
     milliseconds. Estimate the I/Os per second and the effective
     transfer rate for a random-access workload that reads individual
     sectors that are scattered across the disk.

  d. Calculate the random-access I/Os per second and transfer rate for
     I/O sizes of 4 kilobytes, 8 kilobytes, and 64 kilobytes.

  e. If multiple requests are in the queue, a scheduling algorithm such
     as SCAN should be able to reduce the average seek distance. Suppose
     that a random-access workload is reading 8- kilobyte pages, the
     average queue length is 10, and the scheduling algorithm reduces
     the average seek time to 3 milliseconds. Now calculate the I/Os
     per second and the effective transfer rate of the drive.

Answer: 

  a. The disk spins 120 times per second, and each spin transfers a track
     of 80 KB. Thus, the sustained transfer rate can be approximated as
     9600 KB/s.

  b. Suppose that 100 cylinders is a huge transfer. The transfer rate
     is total bytes divided by total time. Bytes: 100 cyl * 20 trk/cyl
     * 80 KB/trk, i.e., 160,000 KB. Time: rotation time + track switch
     time + cylinder switch time. Rotation time is 2000 trks / 120 trks
     per sec, i.e., 16.667 s. Track switch time is 19 switch per cyl *
     100 cyl * 0.5 ms, i.e., 950 ms. Cylinder switch time is 99 * 2 ms,
     i.e., 198 ms. Thus, the total time is 16.667 + 0.950 + 0.198, i.e.,
     17.815 s. (We are ignoring any initial seek and rotational latency,
     which might add about 12 ms to the schedule, i.e.  0.1%.) Thus the
     transfer rate is 8981.2 KB/s. The overhead of track and cylinder
     switching is about 6.5%.

  c. The time per transfer is 8 ms to seek + 4.167 ms average rotational
     latency + 0.052 ms (calcu-lated from 1 / (120 trk per second *
     160 sector per trk)) to rotate one sector past the disk head during
     reading. We calculate the transfers per second as 1/(0.012219), i.e.,
     81.8. Since each transfer is 0.5 KB, the transfer rate is 40.9 KB/s.

  d. We ignore track and cylinder crossings for simplicity. For reads
     of size 4 KB, 8 KB, and 64 KB, the corresponding I/Os per second
     are calculated from the seek, rotational latency, and rotational
     transfer time as in the previous item, giving (respectively)
     1/(0.0126), 1/(0.013), and 1/(0.019).  Thus we get 79.4, 76.9,
     and 52.6 transfers per second, respectively.  Transfer rates are
     obtained from 4, 8, and 64 times these I/O rates, giving 318 KB/s,
     615 KB/s, and 3366 KB/s, respectively.

  e. From 1/(3+4.167+0.83) we obtain 125 I/Os per second. From 8 KB per
     I/O we obtain 1000 KB/s.

Question 13.18:
  More than one disk drive can be attached to a SCSI bus. In particular, a
  fast wide SCSI-II bus (see Exercise 13.17) can be connected to at most
  15 disk drives. Recall that this bus has a bandwidth of 20 megabytes
  per second. At any time, only one packet can be transferred on the bus
  between some disk s internal cache and the host. However, a disk can
  be moving its disk arm while some other disk is transferring a packet
  on the bus. Also, a disk can be transferring data between its magnetic
  platters and its internal cache while some other disk is transferring a
  packet on the bus. Considering the transfer rates that you calculated
  for the various workloads in Exercise 13.17, discuss how many disks
  can be used effectively by one fast wide SCSI-II bus.

Answer:

  For 8 KB random I/Os on a lightly loaded disk, where the random
  access time is calculated to be about 13 ms (see Exercise 13.17),
  the effective transfer rate is about 615 MB/s. In this case, 15 disks
  would have an aggregate transfer rate of less than 10 MB/s, which should
  not saturate the bus. For 64 KB random reads to a lightly loaded disk,
  the transfer rate is about 3.4 MB/s, so 5 or fewer disk drives would
  saturate the bus. For 8 KB reads with a large enough queue to reduce
  the average seek to 3 ms, the transfer rate is about 1 MB/s, so the
  bus bandwidth may be adequate to accommodate 15 disks.

Question 13.25:
  You can use simple estimates to compare the cost and performance
  of a terabyte storage system made entirely from disks with one that
  incorporates tertiary storage. Suppose that magnetic disks each hold
  10 gigabytes, cost $1000, transfer 5 megabytes per second, and have
  an average access la-tency of 15 milliseconds. Suppose that a tape
  library costs $10 per gigabyte, transfers 10 megabytes per second,
  and has an average access latency of 20 seconds. Compute the total
  cost, the maximum total data rate, and the average waiting time for
  a pure disk system. If you make any assumptions about the workload,
  describe and justify them. Now, suppose that 5 percent of the data are
  fre-quently used, so they must reside on disk, but the other 95 percent
  are archived in the tape library. Further suppose that 95 percent of
  the requests are handled by the disk system and the other 5 percent are
  handled by the library. What are the total cost, the maximum total data
  rate, and the average waiting time for this hierarchical storage system?

Answer:

  First let s consider the pure disk system. One terabyte is 1024 GB. To
  be correct, we need 103 disks at 10 GB each. But since this question
  is about approximations, we will simplify the arithmetic by rounding
  off the numbers. The pure disk system will have 100 drives. The cost
  of the disk drives would be $100,000, plus about 20% for cables,
  power supplies, and enclosures, i.e., around $120,000. The aggregate
  data rate would be 100   5 MB/s, or 500 MB/s. The average waiting time
  depends on the workload.  Suppose that the requests are for transfers
  of size 8 KB, and suppose that the requests are randomly distributed
  over the disk drives. If the system is lightly loaded, a typical request
  will arrive at an idle disk, so the response time will be 15 ms access
  time plus about 2 ms transfer time. If the system is heavily loaded,
  the delay will increase, roughly in proportion to the queue length.

  Now let s consider the hierarchical storage system. The total disk
  space required is 5% of 1 TB, which is 50 GB. Consequently, we need
  5 disks, so the cost of the disk storage is $5,000 (plus 20%, i.e.,
  $6,000). The cost of the 950 GB tape library is $9500. Thus the
  total storage cost is $15,500. The maximum total data rate depends on
  the number of drives in the tape library. We suppose there is only 1
  drive. Then the aggregate data rate is 6   10 MB/s, i.e., 60 MB/s. For
  a lightly loaded system, 95% of the requests will be satisfied by
  the disks with a delay of about 17 ms. The other 5% of the requests
  will be satisfied by the tape library, with a delay of slightly more
  than 20 seconds. Thus the average delay will be (95 0.017 +5   20)/
  100, or about 1 second. Even with an empty request queue at the tape
  library, the latency of the tape drive is responsible for almost all
  of the system s response latency, because 1/ 20 th of the workload is
  sent to a device that has a 20 second latency. If the system is more
  heavily loaded, the average delay will increase in proportion to the
  length of the queue of requests waiting for service from the tape drive.

  The hierarchical system is much cheaper. For the 95% of the requests
  that are served by the disks, the performance is as good as a pure-disk
  system.  But the maximum data rate of the hierarchical system is much
  worse than for the pure-disk system, as is the average response time.

Question 14.3:
  Explain why a doubling of the speed of the systems on an Ethernet
  segment may result in decreased network performance. What changes
  could be made to ameliorate the problem?

Answer:

  Faster systems may be able to send more packets in a shorter amount
  of time. The network would then have more packets traveling on it,
  resulting in more collisions, and therefore less throughput relative
  to the number of packets being sent. More net-works can be used,
  with fewer systems per network, to reduce the number of collisions.

Question 14.4:
  Under what circumstances is a token-ring network more effective than
  an Ethernet network?

Answer:
  A token ring is very effective under high sustained load, as no
  collisions can occur and each slot may be used to carry a message,
  providing high throughput. A token ring is less effective when the
  load is light (token processing takes longer than bus access, so any
  one packet can take longer to reach its destination), or sporadic.

Question 14.8:
  The original HTTP protocol used TCP/IP as the underlying network
  protocol. For each page, graphic, or applet, a separate TCP session was
  contructed, used, and torn down. Because of the overhead of building
  and destroying TCP/IP connections, there were performance problems
  with this implementation method. Would using UDP rather than TCP have
  been a good alternative? What other changes could be made to improve
  HTTP per-formance?

Answer:
No one answer.

Question 14.9:
  Of what use is an address resolution protocol? Why is the use of such
  a protocol better than making each host read each packet to determine
  to whom it is destined? Does a token-ring network need such a protocol?

Answer:
  An ARP translates general-purpose addresses into hardware interface
  numbers so the interface can know which packets are for it. Software
  need not get involved. It is more efficient than passing each packet
  to the higher layers. Yes, for the same reason.

Question 14.10:
  What are the advantages and disadvantages of making the computer
  network transparent to the user?

Answer: No answer.

Question 14.11:
  What are two formidable problems that designers must solve to implement
  a network-transparent system?

Answer:

  No answer

Question 14.14:
  Is it always crucial to know that the message you have sent has arrived
  at its destination safely? If your answer is  yes,  explain why. If
  your answer is  no,  give appropriate examples.

Answer: 

  No answer.

Question 14.16:
  Consider a distributed system with two sites, A and B. Consider whether
  site A can distinguish among the following:

    a. B goes down. 
    b. The link between A and B goes down. 
    c. B is extremely overloaded and its response time is 100 times
       longer than normal. What implications does your answer have for
       recovery in distributed systems?

Answer:

  No answer