Que1:

P No. AT BT
1 0 3
2 1 2
3 2 4
4 3 5
5 4 6

Solution:

LJF1

P No. AT BT CT TAT WT
1 0 3 3 3 0
2 1 2 20 19 17
3 2 4 18 16 12
4 3 5 8 5 0
5 4 6 14 10 4
        53/5=10.6 33/5=6.6

Que2: If the BT of  process is matching then schedule the process, which is having lowest AT?

P No. AT BT
1 3 3
2 1 1
3 2 3
4 4 2
5 6 6
6 2 4

Solution:

LJF2

P No. AT BT CT TAT WT
1 3 3 18 15 12
2 1 1 2 1 0
3 2 3 15 13 10
4 4 2 20 16 14
5 6 6 12 6 0
6 2 4 6 4 0
        55/6=9.16 36/6=6

 

     

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