• Course: CS110
• Name: SooYoung Jung
• Homework Assignment #5
• Due Date: 11/12/00
• Date: 11/12/00
• Formatter: HTML

1. OSC, 6.15, pg 479, 4 Points
   Supermarkets are constantly faced with a problem similar to page reference in virtual memory systems. They have a fixed amount of shelf space to display an ever-increasing number of products. If an important new product comes along, say, 100% efficient dog food, some existing product must be dropped from the inventory to make room for it. The obvious replacement algorithms are LRU and FIFO. Which of these would you prefer?
   • In this case, LRU (Least Recently Used) will be appropriated.  All shelf spaces are assigned with certain products in supermarkets and it does not mean that oldest products (First in) are not selling well.  As a result, least recently unused products are occupied space which is wasting space.  Get rid of the product that is not used recently.

2. OSC, 6.16, pg 479, 3 Points
   Why are cache blocks always much smaller than virtual memory pages, often a hundred times smaller?
   • Cache memory is expensive.  Also, cache directly addresses memory but virtual memory is not.  To be used more effective on cache, it should be smaller.

3. T, 5.3, pg 237, 3 Points
   A disk is double interleaved, i.e., sectors are in the order 0, 3, 6, 1, 4, 7, 2, 5. It has eight sectors of 512 bytes per track, and a rotation rate of 300 rpm. How long does it take to read all the sectors of a track in order, assuming the arm is already correctly positioned, and 1/2 rotation is needed to get sector 0 under the head? What is the data rate? Now repeat the problem for a noninterleaved disk, i.e., sectors are in the order 0, 1, 2, 3, 4, 5, 6, 7, with the same characteristics. How much does the data rate degrade due to interleaving?
   • Interleaved disk:
     0.5 rotation + 2.625 (rotation to get all sectors from 0) = 3.125 rotation.
     300round : 60second = 3.125 : x second
     x = 60 * 3.125 / 300 = 0.625s,
     0.625 second needed to read all the sectors of a track in order.
     
     4096bytes : 0.625s = y byte : 1s
     y = 4096 / 0.625 = 6553.6 bytes/second.
     
     Noninterleaved disk:
     0.5 rotation + 0.875 (rotation to get all sectors from 0) = 1.375 rotation.
     300round : 60second = 1.375 : x second
     x = 60 * 1.375 / 300 = 0.275s,
     0.275 second needed to read all the sectors of a track in order.
     
     4096bytes : 0.275s = y byte : 1s
     y = 4096 / 0.275 = 14894.5 bytes/second.
     
     Degraded 44% due to interleaving.

4. T, 5.7, pg 237, 5 Points
   In which of the four i/o software layers (Interrupt drivers, Device drivers, Device-independent os software, and User level software) is each of the following done.
   1. Computing the track, sector, and head for a disk read.
   2. Maintaining a cache of recently used blocks.
   3. Writing commands to the device registers.
   4. Checking to see if the user is permitted to use the device.
   5. Converting binary integers to ASCII for printing.
       
   1. Device Drivers
   2. Device-independent os software
   3. Device Drivers
   4. Device-independent os software
   5. User level software

5. T, 5.9, pg 238, 3 Points
   Disk requests come to the disk driver for cylinders 10, 22, 20, 2, 40, 6, and 38, in that order. A seek takes 6msec per cylinder moved. How much seek time is needed for:
   1. First come, first served
   2. Closet cylinder next.
   3. SCAN
   In all cases the arm is initially at 20 and moving towards track 0.
    
   1. 10 + 12 + 2 + 18 + 38 + 34 + 32 = 146 cylinders movement.
      146 * 6 = 876 msec
   2. 0 + 2 + 12 + 4 + 4 + 36 + 2 = 60 cylinders movement.
      60 * 6 = 360 msec
   3. 20 -> 22 -> 38 -> 40 -> 10 -> 6 -> 2
      0 + 2 + 16 + 2 + 30 + 4 + 4 = 58 cylinders movement.
      58 * 6 = 348 msec

6. T, 5.12, pg 238, 3 Points
   The clock interrupt handler on a certain computer requires 2 msec (including process switching overhead) per clock tick. The clock runs at 60Hz. What fraction of the CPU is devoted to the clock?
•  2 * 60 = 120 msec/sec

7. T, 5.14, pg 238, 3 Points
   Consider how a terminal works. The driver outputs one character and then blocks. When the character has been printed, an interrupt occurs and a message is sent to the blocked drive, which outputs the next character and then blocks again. If the time to pass a message, output a character, and block is 4msec, does this method work well on 110 baud lines? How about 4800 baud lines?
   •  110 baud lines : 11 interrupts/second so 4 * 11 = 44msec.
   •  4800 baud lines : 480 interrupts/second so 4 * 480 = 1920msec.  The method will not work on 4800 baud lines because the system can only take care of 250 interrupts/second.

8. t.7.2, pg 313.
   2 Points
   Write a UNIX pipeline that prints the eighth line of file Z on standard output.
   •  head -8 Z | tail -1

9. t.7.5, pg 313.
   2 Points
   When the UNIX shell starts up a process, it puts copies of its environment variables, such as HOME, on the process' stack, so the process can find out what its home directory is. If this process should later fork, will the child automatically get these variables too?
   • Yes. fork will make exact copy of the parent process.  As a result, the child will have same variables on the exact copy of the stack.

10. t.7.7, pg 313.
    2 Points
    To what hardware concept is a signal closely related? Give two examples of how signals are used.
    • Signal is similar to hardware interruption.
      keyboard interruption produced by the ^C character.  Job control uses signals to start and stop subprocesses on demand (exceptional events).

11. t.7.13, pg 313.
    3 Points
    If a UNIX process runs for 1 sec without competition from other processes, how long does it take for its CPU usage counter to get back down to 0?
    •  CPU usage is divided in half every second so first second it is 30, 2:15, 3:7.5, 4:3.5 5:1.5 6:0.75 7:0
      So it is 7 seconds.

12. t.7.14, pg 314. 2 Points
    About how long does it take to fork off a child process under the following conditions: text size = 100K bytes, data size = 20K bytes, stack size = 10K bytes, process table size = 1K, user structure = 5K. The kernel trap and return takes 1 msec, and the machine can copy one 32-bit word every 500nsec. Text segments are shared.
    • 20 + 10 + 1 + 5 = 36K needed to be copied since text segments (100K) are shared.
      Machine can copy one 32-bit word every 500 nsec = two 32-bit words(8bytes) every 1microsec.
      8 bytes : 1 microsec = 36K = 36864bytes : x microsec
      x = 36864 / 8 = 4608 microsec.
      And also, the kernel trap and return takes 1 msec which is 1000microsec.
      Therefore 4608 + 1000 = 5608 microsec.

13. t.7.19, pg 314. 4 Points
    When the file /usr/ast/work/f is opened, several disk accesses are needed to read i-node and directory blocks. Calculate the number of disk accesses required under the assumption that the i-node for the root directory is always in memory, and all directories are one block long.
    • 8 disk accesses are required under the assumption that the i-node for the root directory is always in memory.
      4 accesses are needed to read directory block (/, /usr, /usr/ast, /usr/ast/work) and
      4 accesses are needed to read i-node (/usr, /usr/ast, /usr/ast/work, /usr/ast/work/f).

14. s.21.6, pg 694 4 Points
    Does 4.2BSD UNIX give scheduling priority to I/O or CPU-bound processes? For what reason does it differentiate between these categories, and why is one given priority over the other? How does it know which of these categories fits a given process?
•  4.2BSD UNIX gives scheduling priority to I/O-bound processes because I/O-bound processes have more work with users.  When system can schedule more time on I/O-bound processes, system will respond more often to users so users will feel the system is fast.