Each of the periodic tasks is scheduled and executed according to a cyclic schedule. Choose appropriate frame
size for a set of tasks
Solutions:
Pi,Ei Pi,Ei Pi,Ei Pi= Perodicity of task
(6,1), (10,2), (18,2). Ei= Execution of task
Step 1: F(max)>ei
F(max)=2
Step 2: calculate probable frame
6=|1,2,3,6|
10=|1,2,5,10|
18=|1,2,3,6,9,18|
The probable frame is 2,3,5,6,9,10,18 (1 is not taking as a probable frame it is less vale of 2 )
Step 3: 2F-gratest common value of [pi,f]<Di if di is not given then (Pi=Di)
F 2 3 5 6 9 10 18
2F 4 6 10 12 18 20 36
p1=6 2 True 5 True 9False 6true 15false 18False 30 false Di<6
p2=10 2 True 5 True – 11False – – – Di<10
p3=18 2 True 3 True – – – – – Di<18
So above task only accepts the frame size 2 and 3 .