Homework 2 Process Management and CPU Scheduling

 

COP 4610/CGS5765, Introduction to Operating Systems, Fall 2003, Florida State University

Points: 100 points

Due: Week 7, Tuesday, October 7, 2003

1.      (10 pnts) Problem 5 in the textbook on page 229.

a.       User name process

b.      Stack bottom thread

c.       Resources blocking me thread

d.      Primary memory allocated to me process

e.       Files allocated to me process

f.        Execution state thread

2.      (15 pnts) Problem 7 in the textbook on page 230.

3.      (15 pnts) Problem 4 in the textbook on page 281.

a.       Gantt chart

0 10 30 50 100 180

p2

p1

p3

p4

p0

 

b.      Turnaround time for p4 = 100

c.       Average wait time

 W(p_0) = 100
 W(p_1) = 10
 W(p_2) = 0
 W(p_3) = 30
 W(p_4) = 50
Average = (100+10+0+30+50)/5 = 38 

4.      (15 pnts) Problem 5 in the textbook on page 281.

a.       Gantt chart

0 20 70 150 160 180

p1

p4

p0

p2

p3

b.      Turnaround time for p2 = 160

c.       Average wait time

 W(p_0) = 70
 W(p_1) = 0
 W(p_2) = 150
 W(p_3) = 160
 W(p_4) = 20
 Average = (70+0+150+160+20)/5 = 80 

5.      (15 pnts) Problem 7 in the textbook on page 281.

SJN scheduling with variable arrival times

0 10 35 75 105 150 215

P0

p2

p1

p3

P4

p0

6.      (15 pnts) Problem 6 in the textbook on page 281.

Note that this is one solution. Other solutions are possible, depending on how you handle newly arrived processes.

RR scheduling with a time quantum of 15

a.       Gantt chart

0 15 30 45 60 75 85 100 115 130 140 145 160 175 190 205

p0

p1

p2

p0

p1

p2

p3

p4

p0

p1

p3

p4

p0

p4

p0

 

b.      Turnaround time for p3 = 65

c.       Average wait time

 W(p_0) = 0
 W(p_1) = 5
 W(p_2) = 20
 W(p_3) = 5
 W(p_4) = 15
 Average = (0+5+20+5+15)/5 = 9 

7.      (15 pnts) Problem 8 in the textbook on page 282 Here we assume the same system given in Problem 6 on Page 281.

Note that this is one solution. Other solutions are possible, depending on how you handle newly arrived processes.

RR scheduling with a time quantum of 15 & switch time of 5

a.       Gantt chart

0 15 20 35 40 55 60 75 80 95 100 110 115 130 135 150

p0

C

p1

C

p2

C

p0

C

p1

C

p2

C

p3

C

p4

150 155 170 175 185 190 195 200 215 220 235 240 255 260 275

C

p0

C

p1

C

p3

C

p4

C

p0

C

p4

C

p0

 

b.      Turnaround time for p3 = 115

c.       Average wait time

 W(p_0) = 0
 W(p_1) = 10
 W(p_2) = 30
 W(p_3) = 35
 W(p_4) = 50
Average = (0+10+30+35+50)/5 = 25