1. Exercise Ms.Reema alOraini
CPU SCHEDULING
Exercise
Ms.Reema alOraini

2. Burst Time (millisecond)
Question 1
Consider the following set of processes:
Draw the Gantt Chart that illustrates the execution of these processes using preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5

3. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer P1

4. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 1 P1

5. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
P2 had higher priority so it will be execution
Timer 2 P1 P2 2

6. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
At time 3 p2 is finished with execution
Timer 3 P1 P2 2 3

7. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
So p1 will be continue execution
Timer 3 P1 P2 P1 2 3

8. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 4 P1 P2 P1 2 3

9. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 5 P1 P2 P1 2 3

10. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 6 P1 P2 P1 2 3

11. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
At time 7 p1 is finished with execution
And p3 does not arrive yet
Timer 7 P1 P2 P1 2 7 3

12. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
P3 is arrive so it will be execution
Timer 8 P1 P2 P1 P3 8 2 7 3

13. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 9 P1 P2 P1 P3 8 2 7 3

14. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
At time 10 p3 is finished with execution
Timer 10 P1 P2 P1 P3 8 2 7 10 3

15. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 P2 2 3 P3 8 5
TAT for p1 = 7-0=7 P1 P2 P1 P3 2 7 8 10 3

16. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 7 P2 2 3 P3 8 5
TAT for p2 = 3-2=1 P1 P2 P1 P3 2 7 8 10 3

17. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 7 P2 2 3 P3 8 5
TAT for p3 = 10-8=2 P1 P2 P1 P3 2 7 8 10 3

18. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 7 P2 2 3 P3 8 5
WT for p1=7-6=1 P1 P2 P1 P3 2 7 8 10 3

19. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 7 P2 2 3 P3 8 5
WT for p2=1-1=0 P1 P2 P1 P3 2 7 8 10 3

20. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 7 P2 2 3 P3 8 5
WT for p3=2-2=0 P1 P2 P1 P3 2 7 8 10 3

21. preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 7 P2 2 3 P3 8 5 P1 P2 P1 P3
Average TAT=(7+1+2)/3=3.3
Average WT=(1+0+0)/3=0.33 2 7 8 10 3

22. Burst Time (millisecond)
Question 1
Consider the following set of processes:
Draw the Gantt Chart that illustrates the execution of these processes using NON- preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5

23. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer P1

24. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 1 P1

25. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 2 P1

26. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 3 P1

27. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 4 P1

28. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 5 P1

29. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer
At time 6 p1 finished execution 6 P1 P2 6

30. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer
At time 7 p2 finished execution 7 P1 P2 6 7

31. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer
At time 8 p3 will be execution 8 P1 P2 6 7 8

32. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 9 P1 P2 p3 6 7 8

33. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority P1 6 1 P2 2 3 P3 8 5
Timer 10 P1 P2 p3 6 7 8 10

34. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 P2 2 3 P3 8 5
TAT for p1 = 6-0=6 P1 P2 p3 6 7 8 10

35. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 P2 2 3 P3 8 5
TAT for p2 = 7-2=5 P1 P2 p3 6 7 8 10

36. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 P2 2 3 5 P3 8
TAT for p3 = 10-8=2 P1 P2 p3 6 7 8 10

37. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 P2 2 3 5 P3 8
WT for p1 = 6-6=0 P1 P2 p3 6 7 8 10

38. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 P2 2 3 5 P3 8
WT for p2= 5-1=4 P1 P2 p3 6 7 8 10

39. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 P2 2 3 5 4 P3 8
WT for p3=2-2=0 P1 P2 p3 6 7 8 10

40. Non-preemptive Priority algorithm
Processes
Arrival time
Burst Time (millisecond)
Priority
Turnaround time
Waiting time P1 6 1 P2 2 3 5 4 P3 8
Average TAT=(6+5+2)/3=4.33
Average WT=(0+4+0)/3=1.33 P1 P2 p3 6 7 8 10

41. D. For the given set of processes , which algorithm is better? Why?
Non-preemptive Priority algorithm
preemptive Priority algorithm
Average TAT=(6+5+2)/3=4.33
Average WT=(0+4+0)/3=1.33
Average TAT=(7+1+2)/3=3.3
Average WT=(1+0+0)/3=0.33
By Comparing Waiting Time and Turnaround Time
So , the Preemptive algorithm is better

42. Burst Time (millisecond)
Question 2
Consider the following set of processes:
Draw the Gantt Chart that illustrates the execution of these processes using Round-Robin algorithm where time quantum=5 millisecond.
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3

43. Round-Robin algorithm where time quantum=5 millisecond.
Timer
Queue : p1
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 P1

44. Round-Robin algorithm where time quantum=5 millisecond.
Timer 1
Queue : p1
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 P1

45. Round-Robin algorithm where time quantum=5 millisecond.
Timer 2
Queue : p1
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 P1

46. Round-Robin algorithm where time quantum=5 millisecond.
Timer 3
Queue : P1 p2
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 P1

47. Round-Robin algorithm where time quantum=5 millisecond.
Timer 4
Queue : P1 p2
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 P1

48. Round-Robin algorithm where time quantum=5 millisecond.
Timer 5
Queue : P1 P2 p1
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 P1 5

49. Round-Robin algorithm where time quantum=5 millisecond.
Timer 6
Queue : P1 P2 p3
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
6-5=1 P1 P2 5 6

50. Round-Robin algorithm where time quantum=5 millisecond.
Timer 7
Queue : P1 P2 p3
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
6-5=1 P1 P2 P1 5 7 6

51. Round-Robin algorithm where time quantum=5 millisecond.
Timer 8
Queue : P1 P2 p3
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
6-5=1 P1 P2 P1 P3 5 6 7

52. Round-Robin algorithm where time quantum=5 millisecond.
Timer 9
Queue : P1 P2 p3
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
6-5=1 P1 P2 P1 P3 5 6 7 9

53. Round-Robin algorithm where time quantum=5 millisecond.
Processes
Arrival time
Burst Time (millisecond)
Waiting Time P1 6 P2 2 1 P3
WT for p1=(0-0)+(6-5)=1 P1 P2 P1 P3 5 6 7 9

54. Round-Robin algorithm where time quantum=5 millisecond.
Processes
Arrival time
Burst Time (millisecond)
Waiting Time P1 6 1 P2 2 P3
WT for p2=(5-2)=3 P1 P2 P1 P3 5 6 7 9

55. Round-Robin algorithm where time quantum=5 millisecond.
Processes
Arrival time
Burst Time (millisecond)
Waiting Time P1 6 1 P2 2 3 P3
WT for p3=(7-6)=1 P1 P2 P1 P3 5 6 7 9

56. Round-Robin algorithm where time quantum=5 millisecond.
Processes
Arrival time
Burst Time (millisecond)
Waiting Time P1 6 1 P2 2 3 P3
Average WT =(1+3+1)/3 =1.67 P1 P2 P1 P3 5 6 7 9

57. Burst Time (millisecond)
Draw the Gantt Charts that illustrates the execution of these processes using FCFS algorithm , preemptive SJF and Non- preemptive SJF
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3

58. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 P1

59. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 1 P1

60. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 2 P1

61. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 3 P1

62. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 4 P1

63. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 5 P1

64. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 6 P1 P2 6

65. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 7 P1 P2 6 7

66. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 8 P1 P2 P3 6 7

67. The FCFS algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 9 P1 P2 P3 6 7 9

68. The FCFS algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 P2 2 1 P3 P1 P2 P3
WT for p1 =0-0=0 6 7 9

69. The FCFS algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 P2 2 1 P3 P1 P2 P3
WT for p2 =6-2=4 6 7 9

70. The FCFS algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 P2 2 1 4 P3 P1 P2 P3
WT for p3 =7-6=1 6 7 9

71. The FCFS algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 P2 2 1 4 P3 P1 P2 P3
Average WT =(0+4+1)/3=1.67 6 7 9

72. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 P1

73. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 1 P1

74. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 2
P2 shorter than p1 P1 P2 2

75. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 3
6-2=4 P1 P2 P1 2 3

76. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 4
6-2=4 P1 P2 P1 2 3

77. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 5
6-2=4 P1 P2 P1 2 3

78. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 6
6-2=4 P1 P2 P1 2 3

79. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 7
6-2=4 P1 P2 P1 P3 2 3 7

80. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 8
6-2=4 P1 P2 P1 P3 2 3 7

81. The preemptive SJF (short job first) algorithm Gantt Chat is :
Timer
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3 9
6-2=4 P1 P2 P1 P3 2 3 7 9

82. The preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 P2 2 1 P3
WT for p1 =(0-0)+(3-2)=1 P1 P2 P1 P3 2 3 7 9

83. The preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 1 P2 2 P3
WT for p2= 2-2=0 P1 P2 P1 P3 2 3 7 9

84. The preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 1 P2 2 P3
WT for p3=7-6=1 P1 P2 P1 P3 2 3 7 9

85. The preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 1 P2 2 P3
Average WT=(1+0+1)/3=0.67 P1 P2 P1 P3 2 3 7 9

86. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer P1

87. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer 1 P1

88. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer 2 P1

89. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer 3 P1

90. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer 4 P1

91. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer 5 P1

92. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer 6
P2 is shorter than p3 P1 P2 6

93. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer 7 P1 P2 6 7

94. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer 8 P1 P2 P3 6 7

95. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond) P1 6 P2 2 1 P3
Timer 9 P1 P2 P3 6 7 9

96. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 P2 2 1 P3
WT for p1=0-0=0 P1 P2 P3 6 7 9

97. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 P2 2 1 P3
WT for p2=6-2=4 P1 P2 P3 6 7 9

98. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 P2 2 1 4 P3
WT for p3=7-6=1 P1 P2 P3 6 7 9

99. The Non-preemptive SJF (short job first) algorithm Gantt Chat is :
Processes
Arrival time
Burst Time (millisecond)
Waiting time P1 6 P2 2 1 4 P3
Average WT=(0+4+1)/3=1.67 P1 P2 P3 6 7 9

100. Question 2 cont.
D. For the given set of processes and algorithms which algorithm is better? Why?
preemptive SJF
(short job first) algorithm
Non-preemptive SJF
(short job first) algorithm
FCFS algorithm
Average WT =(0+4+1)/3=
1.67
Average WT=(1+0+1)/3=0.67
Average WT=(0+4+1)/3=1.67
By Comparing Waiting Time So , the Preemptive SJF algorithm is better