大连交大os第二章作业.doc

大连交大os第二章作业.doc

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第二章作业

14.What’sthebiggestadvantageofimplementingthreadinuserspace?

What’sthebiggestdisadvantage?

Solution:

Thefirstadvantageisthatauser-levelpackagecanbeimplementedonanoperatingsystemthatdoesnotsupportthreads.Thesecondadvantageisthattheyalloweachprocesstohaveitsowncustomizedschedulingalgorithm.

Thedisadvantageistheproblemofhowblockingsystemcallsareimplemented.Anotherproblemisthatifathreadstartsrunning,nootherthreadinthatprocesswilleverrununlessthefirstthreadvoluntarilygivesuptheCPU.

35.MeasurementsofcertainsystemhaveshownthattheaverageprocessrunsforatimeTbeforeblockingonI/O.AprocessswitchrequiresatimeS,whichiseffectivelywasted（overhead）.Forround-robinschedulingwithquantumQ,giveaformulafortheCPUefficiencyforeachofthefollowing:

（a）Q=∞CPUefficiency=T/（T+S）

（b）Q>TCPUefficiency=T/（T+S）

（c）S

（d）Q=SCPUefficiency=50%

（e）Qnearly0CPUefficiencyisnearly0

36.Fivejobsarewaitingtoberun.Theirexpectedruntimesare9,6,3,5,andX.Inwhatordershouldtheyberuntominimizeaverageresponsetime？

（YouranswerwilldependonX.）

Solution:

IF（X<=3）:

X,3,5,6,9

IF（X>=3ANDX<=5）:

3,X,5,6,9

IF（X>=5ANDX<=6）:

3,5,X,6,9

IF（X>=6ANDX<=9）:

3,5,6,X,9

IF（X>=9）:

3,5,6,9,X

37.FivebatchjobsAthroughE,arriveatacomputercenteratalmostthesametime.Theyhaveestimatedrunningtimesof10,6,2,4,and8minutes.Their（externallydetermined）prioritiesare3,5,2,1,and4,respectively,with5beingthehighestpriority.Foreachofthefollowingschedulingalgorithms,determinethemeanprocessturnaroundtime.Ignoreprocessswitchingoverhead.

（a）Roundrobin

（b）Priorityscheduling

（c）First-come,first-served（runinorder10,6,2,4,8）

（d）Shortestjobfirst

For（a）,assumethatthesystemismultiprogrammed,andthateachjobgetsitsfairshareoftheCPU.For（b）through（d）assumethatonlyonejobatatimeruns,untilitfinishes.AlljobsarecompletelyCPUbound.

Solution:

processes

Arrivaltime

Processingtime

priority

A

0

（1）

10

3

B

0

（2）

6

5（HIGH）

C

0（3）

2

2

D

0（4）

4

1

E

0（5）

8

4

（a）Roundrobin（quantum=2）

A

2

A

A

E

A

B

C

D

E

B

D

E

B

E

A

Turnaroundtime:

A:

10+6+2+4+8=30

B:

30-4*2=22

C:

2+2+2=6

D:

2*8=16

E:

30-2=28

Meanturnaroundtime=（30+22+6+16+28）/5=80/5=20.4

（b）priorityscheduling

6

A

C

B

E

D

8

10

2

4

Turnaroundtime:

A:

6+8+10=24

B:

6

C:

24+2=26

D:

26+4=30

E:

6+8=14

Meanturnaroundtime=（24+6+26+30+14）/5=100/5=20

（c）FCFS

10

C

D

A

B

E

6

2

4

8

Turnaroundtime:

A:

10

B:

10+6=16

C:

16+2=18

D:

18+4=22

E:

22+8=30

Meanturnaroundtime=（10+16+18+22+30）/5=96/5=19.2

（d）Shortedjobfirst

10

C

D

A

B

E

6

2

4

8

Turnaroundtime:

A:

2+4+6+8+10=30

B:

2+4+6=12

C:

2

D:

2+4=6

E:

2+4+6+8=20

Meanturnaroundtime=（30+12+2+6+20）/5=70/5=14

41.Asoftreal-timesystemhasfourperiodiceventswithperiodsof50,100,200,and250mseceach.Supposethatthefoureventsrequire35,20,10,andxmsecofCPUtime,respectively.Whatisthelargestvalueofxforwhichthesystemisschedulable?

Solution:

35/50+20/100+10/200+x/250<=1max（x）=12.5