Page 3: From Quarter 3 20 points: Analyze the performance of two scheduling mech
ID: 3915411 • Letter: P
Question
Page 3: From Quarter 3 20 points: Analyze the performance of two scheduling mechanisms: Exponential Averaging, and true ShortestBurstfirst, Calculate the Average Completion Time AND count the number of context switches. The workload consists of the following: P1: 1113,4,4 P2: 7(2,5); P3: 17/2,4,5,6). 1/0 between bursts for 4 time quantums. Exp Ave default 2, alpha-0.6 Processes are executed in FIFO order within each queue. Exec Ready 1,2,3 Pl P2 2 P3 2 Context switches Ave completion time True SJF: (shortest CPU-burst first) Time Executirn Read l/O:4 0 Completion time ave Context switches-Explanation / Answer
Processes are executed in FIFO order within each queue.
Executing (initial pred=2, alpha =0.6)
Time
0
2
6
8
12
14
18
20
24
26
30
32
36
38
42
44
48
50
54
56
60
62
Exec
P1
I/0
P2
I/0
P3
I/0
P1
I/0
P2
I/0
P3
I/0
P1
I/0
P2
I/0
P3
I/0
P1
I/0
P2
P3
Ready
1,2,3
P1
2
P2
2
P3
2
I/O:4
The next burst time is predicted with the above formula
P2 = 0.6*2+0.4*2 = 1.2+0.8 =2
P3 = 0.6*2+0.4*2 = 1.2+0.8 = 2 so like this we use the before value in the formula to calculate the future burst value , but in this example every time it is 2 , so in between every time I/o burst will take 4, which is depicted in the above table
P1=11,P2=7,P3=17
this will be the gantt chart
0 P1 2 I/O 6 P2 8 I/O 12 P3 14 I/O 18 P1 20 I/O 24 P2 26 I/O 30 P3 32 I/O 36 P1 38 I/O 42 P2 44 P3 46 I/O 50 P1 52 P2 53 I/O 57 P3 59 I/O 63 P1 65 I/O 69 P3 71 I/O 75 P1 77 I/O 81 P3 88
COMPLETION TIME = 53 + 77 + 88 /3 = 72.66
CONTEXT SWITCHES = 15
Shortest job First gantt cahrt will be like this
0 P2 7 I/O 11 P1 22 I/O 26 P3 43
completion time = 7+22+43/3 = 24 , context switches=3
Executing (initial pred=2, alpha =0.6)
Time
0
2
6
8
12
14
18
20
24
26
30
32
36
38
42
44
48
50
54
56
60
62
Exec
P1
I/0
P2
I/0
P3
I/0
P1
I/0
P2
I/0
P3
I/0
P1
I/0
P2
I/0
P3
I/0
P1
I/0
P2
P3
Ready
1,2,3
P1
2
P2
2
P3
2
I/O:4
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