homework 2 sarah diesburg operating systems cop 4610

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Homework 2 Sarah Diesburg Operating Systems COP 4610

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Page 1: Homework 2 Sarah Diesburg Operating Systems COP 4610

Homework 2

Sarah Diesburg

Operating Systems

COP 4610

Page 2: Homework 2 Sarah Diesburg Operating Systems COP 4610

Why does OS loader set kernel mode? OS needs to run with full privileges

Sets more restrictive user mode bit for user programs

Protects OS from errant user programs

Page 3: Homework 2 Sarah Diesburg Operating Systems COP 4610

Three Processes

Process A

Process B

Process C

Time

0 100 200 300 400 500 600 700 800

Page 4: Homework 2 Sarah Diesburg Operating Systems COP 4610

First In, First Out

Process A

Process B

Process C

FIFOTime

0 100 200 300 400 500 600 700 800

BC A

Page 5: Homework 2 Sarah Diesburg Operating Systems COP 4610

First In, First Out

Process A

Process B

Process C

FIFO

Average response time

= [(300-50) + (500-150) + 0]/3 = 200

Time

0 100 200 300 400 500 600 700 800

BC A

Page 6: Homework 2 Sarah Diesburg Operating Systems COP 4610

First In, First Out

Process A

Process B

Process C

FIFO

Average wait time

= [(300-50) + (500-150) + 0]/3 = 200

Time

0 100 200 300 400 500 600 700 800

BC A

Page 7: Homework 2 Sarah Diesburg Operating Systems COP 4610

First In, First Out

Process A

Process B

Process C

FIFO

Average turnaround time

= [(500-50) + (900-150) + (300-0)]/3 = 500

Time

0 100 200 300 400 500 600 700 800

BA BC A

Page 8: Homework 2 Sarah Diesburg Operating Systems COP 4610

Round Robin (Time slice = 100)

Process A

Process B

Process C

RRTime

0 100 200 300 400 500 600 700 800

C

Page 9: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C

Page 10: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A

Page 11: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A

Page 12: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C

Page 13: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C

Page 14: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B

Page 15: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B

Page 16: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B A

Page 17: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B A

Page 18: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B A C

Page 19: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B A C

Page 20: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B A C B

Page 21: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B A C B

Page 22: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Average response time

= [(100-50) + (300 – 150) + 0]/3 = 66.7

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B A C B

Page 23: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process C

RR

Average wait time

= {[(100–50) + (400–200)] + [(300–150) + (600–400)]

+ [0 + (200–100) + (500–300)]}/3 = [250 + 350 + 300]/3 = 300

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B A C B

Page 24: Homework 2 Sarah Diesburg Operating Systems COP 4610

Process A

Process B

Process CRR

Average turnaround time= [(500 – 50) + (900 – 150) + (600 – 0)]/3 = 600

Round Robin (Time slice = 100)

Time

0 100 200 300 400 500 600 700 800

C A C B A C B

Page 25: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJNTime

0 100 200 300 400 500 600 700 800

C

Page 26: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJNTime

0 100 200 300 400 500 600 700 800

C

Page 27: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJNTime

0 100 200 300 400 500 600 700 800

C A

Page 28: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJNTime

0 100 200 300 400 500 600 700 800

C A

Page 29: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJNTime

0 100 200 300 400 500 600 700 800

C A

Page 30: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJNTime

0 100 200 300 400 500 600 700 800

C A B

Page 31: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJNTime

0 100 200 300 400 500 600 700 800

C A B

Page 32: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJN

Average response time

= [(300-50) + (500-150) + 0]/3 = 200

Time

0 100 200 300 400 500 600 700 800

C A B

Page 33: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJN

Average wait time

= [(300-50) + (500-150) + 0]/3 = 200

Time

0 100 200 300 400 500 600 700 800

C A B

Page 34: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Job Next

Process A

Process B

Process C

SJN

Average turnaround time

= [(500-50) + (900-150) + (300-0)]/3 = 500

Time

0 100 200 300 400 500 600 700 800

C A B

Page 35: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTFTime

0 100 200 300 400 500 600 700 800

C

Page 36: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTFTime

0 100 200 300 400 500 600 700 800

C

Page 37: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTFTime

0 100 200 300 400 500 600 700 800

C A

Page 38: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTFTime

0 100 200 300 400 500 600 700 800

C A

Page 39: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTFTime

0 100 200 300 400 500 600 700 800

C A C

Page 40: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTFTime

0 100 200 300 400 500 600 700 800

C A C

Page 41: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTFTime

0 100 200 300 400 500 600 700 800

C A C B

Page 42: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTFTime

0 100 200 300 400 500 600 700 800

C A C B

Page 43: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTF

Average response time

= [0 + (500-150) + 0]/3 = 117

Time

0 100 200 300 400 500 600 700 800

C A C B

Page 44: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process C

SRTF

Average wait time

= [0 + (500 – 150) + (250 – 50)]/3 = 183

Time

0 100 200 300 400 500 600 700 800

C A C B

Page 45: Homework 2 Sarah Diesburg Operating Systems COP 4610

Shortest Remaining Time First

Process A

Process B

Process CSRTF

Average turnaround time= [(250 – 50) + (900 – 150) + (500 – 0)]/3 = 483

Time

0 100 200 300 400 500 600 700 800

C A C B