[운영체제] 기말고사 전 레포트[작성중] Study+more

"Busy Waiting"의 의미는 무엇인가? 이 방식과 다르게 기다리는 방식이 있다면 기술하시오.

>한 프로세스가 임계 구역에 있으면, 이 임계구역에 진입하려는 프로세스는 진입코드를 계속 반복하여 수행해야 한다. 이러한, 계속적인 반복은 하나의 CPU가 여러 프로세스들에 의해 공유되는 다중 프로그래밍 시스템에서 분명히 큰 문제가 된다.(194쪽)

>Busy Waiting의 필요성을 극복하기 위하여, wait와 signal 세마포 연산의 정의를 변경할 수 있다. 프로세스가 wait 연산을 실행하고 세마포 값이 양수가 아닌 것을 발견하면, 프로세스는 반드시 대기해야 한다. 그러나 Busy Waiting 대신에 프로세스는 자신을 봉쇄시킬 수 있다. 봉쇄 연산은 프로세스를 세마포에 연관된 대기 큐에 넣고, 프로세스의 상태를 대기 상태로 전환한다. 그 후에 제어가 CPU 스케줄러로 이전되고, 스케줄러는 다른 프로세스를 실행하기 위하여 선택한다. 세마포 S를 대기하며 봉쇄된 프로세스는 다른 프로세스가 signal연산을 실행하면 재시작되어야 한다. 프로세스는 wakeup 연산에 의하여 재시작 되는데 이것은 프로세스의 상태를 대기 상태에서 준비 완료 상태로 변경한다. 그리고 프로세스를 준비 완료 큐에 넣는다. (194쪽)


spinlock 가 왜 uniprocessor에서는 소용이 없고, multiprocessor에서는 필요한지 기술하시오.

> 단일 프로세서에서는 바쁜 대기(busy waiting) 상태는 다른 프로세스들이 생산적으로 사용할 수 있는 중앙처리장치 시간을 낭비하게 된다. 그러나 다중 프로세서 환경에서의 Spinlock은 유용하게 사용된다. Spinlock의 장점은 프로세스가 락에서 대기할 때 문맥교환(Context Switching)이 불필요하다는 점이다. 그러므로, 록이 매우 짧은 시간만 적용될 때는 Spinlock이 매우 유용하다.
현재 시스템의 상태가 아래와 같다.

ProcessAllocationMaxAvailable
자원A B C DA B C DA B C D
P00 0 1 20 0 1 21 5 2 0
P11 0 0 01 7 5 0 
P21 3 5 42 3 5 6 
P30 6 3 20 6 5 2 
P40 0 1 40 6 5 6 
은행원 알고리즘을 사용하여 다음 물음에 답하여라:

a. Need행렬의 내용은 무엇이 되는가?
ProcessAllocationMaxNeedAvailable
자원A B C DA B C DA B C DA B C D
P00 0 1 20 0 1 20 0 0 01 5 2 0
P11 0 0 01 7 5 00 7 5 0 
P21 3 5 42 3 5 61 0 0 2 
P30 6 3 20 6 5 20 0 2 0 
P40 0 1 40 6 5 60 6 4 2 

b. 시스템은 현재 safe상태에 있는가?
1 5 3 2 (P0)-> 2 3 5 6+0 5 3 0=2 8 8 6 (P2)-> 0 6 5 2+2 8 6 6=2 14 11 8 (P3)
->0 6 5 6+2 8 7 6=2 14 13 12 (P4)-> 1 7 5 0+2 7 8 12=3 14 13 12 (P1)

<P0, P2, P3, P4, P1>가 safety알고리즘을 만족한다. safe상태이다.
c. P1이 (0, 4, 2, 0)을 요청하게 되면, 그 용청을 즉시 들어줄 것인가?
<P0, P2, P3, P4, P1>에서 2 14 13 12-0 11 7 0=2 3 6 12  이기 때문에 safe상태이며,
리소스도 충분하다. 요청을 즉시 들어줄수 있다.

내부 단편화(internal fragmentation)와 외부 단편화(external fragmentation)사이의 차이점을 설명하고, 어떻게 다른지 설명하시오.

>내 부 단편화: 기억 장치를 먼저, 아주 작은 공간(word)들로 분할하고 프로세스가 요청하면, 할당을 항상 이 분할된 크기의 정수 배로만 해주게 되는경우, 할당된 공간은 요구된 공간보다 약간 더 클수가 있으며, 이들 두 크기 사이의 남는 부분.
m워드 크기의 기억장소를 필요로 하는 어떤 작업이 n>m인 n워드
ex)메모리에 연속된 공간이 100KB가 남았을때, 99.9KB의 프로그램이 메모리 공간을 차지하면서 남아버린 100B의 공간
>외부 단편화:어떤 영역이 사용되지 않고 남아 있어서 이용할수 있지만, 대기중인 작업의 크기에 비해 기억장소가 작은 경우.
유휴 공간을 모두 합치면 충분한 공간이 되지만 그것들이 너무 작은 조각들로 여러곳에 분산되어 있을때 발생.
100KB, 500KB, 200KB, 300KB, 600KB 순으로 기억 장치 분할이 이루어져 있을 때 최초 적합, 최적 적합, 최악 적합의 알고리즘은 212KB, 417KB, 112KB, 426KB순의 프로세스를 어떻게 할당하는가? 어느 알고리즘이 기억 장치를 가장 효과적으로 사용하는가?
최초 적합: 100KB, 500KB, 200KB, 300KB, 600KB
1. 100KB, 500-212=282KB, 200KB, 300KB, 600KB,
2. 100KB, 500-212=282KB, 200KB, 300KB, 600-417=183KB
3. 100KB, (500-212=282)-112=170KB, 200KB, 300KB, 600-417=183KB
4. 100KB, (500-212=282)-112=170KB, 200KB, 300KB, 600-417=183KB ## 426KB를 올릴 공간이 없음(외부 단편화)
>Free Space: 170+200+300+183 = 853KB
최적 적합: 100KB, 200KB, 300KB, 500KB, 600KB
1. 100KB, 200KB, 300-212=88KB, 500KB, 600KB
2. 300-212=88KB, 100KB, 200KB, 500-417=83KB, 600KB ##메모리 리스트의 순서를 재 배치
3. 500-417=83KB, 300-212=88KB, 100KB, 200-112=88KB, 600KB ##메모리 리스트의 순서를 재 배치
4. 500-417=83KB, 300-212=88KB, 200-112=88KB, 100KB, 600-426=174KB ##메모리 리스트의 순서를 재 배치
>Free Space: 83+88+88+100+174 = 533KB  ##가장 효과적
최악 적합: 600KB, 500KB, 300KB, 200KB, 100KB
1. 600-212=388KB, 500KB, 300KB, 200KB, 100KB
2. 500-417=83KB, 600-212=388KB, 300KB, 200KB, 100KB ##메모리 리스트의 순서를 재 배치
3. (600-212=388)-112=176KB, 300KB, 200KB, 100KB, 500-417=83KB ##메모리 리스트의 순서를 재 배치
4. 300KB, 200KB, (600-212=388)-112=176KB, 100KB, 500-417=83KB ## 426KB를 올릴 공간이 없음(외부 단편화) && 메모리 리스트의 순서를 재 배치
>Free Space: 300+200+176+100+83 = 859KB
페이지의 크기는 항상 2의 제곱인 이유를 설명하시오.

기억장치에 페이지 테이블(page table)을 저장하는 페이징 시스템을 생각해 보자.
a. 만약 기억 장치 참조가 200 nsec 걸린다면 페이지로 이루어진 기억 장치의 참조는 얼마나 걸리는가?
페이지(메모리에 존재) 참조 시간+기억장치 참조시간
200+200 = 400nsec

b. 만약 연관 레지스터(associative register)를 가지고 있어 모든 페이지 테이블 참조의 75%를 연관 레지스터에서 찾는다면 기억 장치 접근 시간은 얼마나 줄어들게 되는가?(단, 연관 레지스터에 페이지 체이블 항목이 있다면 시간이 걸리지 않는다고 가정하시오)
<<메모리를 엑세스할 확률은 25%이며, 한번의 메모리 엑세스 시간은 총 400nsec가 걸린다.>>
(연관 메모리 참조의 경우를 제외한)메모리 엑세스 타임:400*0.25 = 100nsec
고로 원래 메모리 참조시간(400nsec)에서 100nsec로 줄어 들었음으로, 300nsec의 시간이 줄었음.

다음 프로그래밍의 기법과 구조 가운데 어느 것이 요구 페이지(demand page)환경에 좋고, 어느 것이 좋지 않은가? 이유를 설명하시오.
a. 스택(stack):
b. 해시 기호 테이블(hashed symbol table):
c. 순차 검색(sequential search):
d. 이진 탐색(binary search):
e. 순수 코드(pure code):
f. 벡터 연산(vector operation):
g. 간접 접근(indirection):

다음의 페이지 참조열을 생각해 보시오.
1, 2, 3, 4, 2, 1, 5, 6, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6

1, 2, 3, 4, 5, 6또는 7개의 페이지 프레임이 있다고 가정할때, 다음의 교체 알고리즘(replacement algorithm)에 대해 각각 얼마나 많은 페이지 부재가 일어나는가? 단, 모든 프레임이 처음에는 비어 있다고 가정한다. 즉, 처음 참조되는 페이지들은 각기 한번의 페이지 부재를 일으킨다. (4개의 프레임을 사용한다고 가정한다.)
LRU 교체 (Page Fault: 10개)
Frame↓1234215612376321236
F1nullnullnull4215612376321236
F2nullnull33421561237632123
F3null222342156123763312
F41111134225612276661
PFooooxxooxxoooxxoxxx

FIFO 교체 (Page Fault: 14개)
Frame↓1234215612376321236
F11111

2345612

376
F2null222

3456123

761
F3nullnull33

4561237

612
F4nullnullnull4

5612376

123
PFooooxxoooooooxxooox

최적(oprimal) 교체 (Page Fault: 8개)
Frame↓1234215612376321236
F11111

11


7


1


F2null222

22


2


2


F3nullnull33

33


3


3


F4nullnullnull4

56


6


6


PFooooxxooxxxoxxxoxxx


번호가 0부터 4999인 5000개의 트랙을 가진 디스크 드라이브가 있다. 헤드는 현재 실린더 143을 처리하고 있고 방금 전에 실린더 125의 요청을 끝냈다. 만일 요청 큐가 FIFO순으로 다음과 같이 유지될 때
86, 1470, 913, 1774, 948, 1509, 1022, 1850, 130
다음과 같은 스케줄링 알고리즘에서 이 요청들을 만족시키기 위한 헤드의 총 이동 거리는 얼마인가?
a. 선입 선처리(FCFS)
143->86->1470->913->1774->948->1509->1022->1850->130
    57 +1384  +557  +861   +826   +561   + 487   +828   +1720  = 7281
b. 최소 탐색 시간 우선(SSTF)
143->130->86->913->948->1022->1470->1509->1774->1850
   13    +44  +827  +35   +74     +448    +39    +265    +76       = 1821
c. SCAN
143->913->948->1022->1470->1509->1774->1850->5000->130->86
   770  +35   +74     +448    +39     +265   +76     +3150  +4870 +44   = 9771
d. LOOK
143->913->948->1022->1470->1509->1774->1850->130->86
   770  +35   +74     +448    +39     +265   +76     +1720  +44   = 3471
e. C-SCAN (헤드가 첫 자리로 복귀시 걸리는 거리는 0라고 가정함)
143->913->948->1022->1470->1509->1774->1850->5000=>0->130->86
   770  +35   +74     +448    +39     +265   +76     +3150        +130  +44   = 5031
f. C-LOOK (헤드가 첫 자리로 복귀시 걸리는 거리는 0라고 가정함)
143->913->948->1022->1470->1509->1774->1850=>0->130->86
   770  +35   +74     +448    +39     +265   +76           +130  +44   = 1881

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