Md(tc+1) = (Td-(Tc+1))-(Ri-1)
		 = Td - Tc - 1 - Ri + 1
		 = (Td-Tc) - Ri
? Inactive

Md(tc+1) = (Td-(Tc+1))-Ri
		 = Md(Tc) - 1
		 
A lasts	5 u.t, is ready to  0 deadline 9
B		3					3		   8
C		2					2		   11
D		2					7		   12

ABCD possible

			T1		T2		T3		T4		T5
Period		30		30		60		60		120
Duration	10		8		6		10		4

LCM (PPCM) ? = 120
Correct schedule 5 tasks on one processor?

T1 T2 T3 T4 T1 T2 T5 T5 P T1 T2 T3 T4 T1 T2 P

			T1		T2
Period		4		8 (réexécuter toutes les X secondes)
Duration	1		4

W? U(2) = 0.83

Schedule with Rate Monotonic
P(T1) = 1/4 priority
P(T2) = 1/8
T1 > T2

T1, T2

			T1		T2		T3
Period		2		3		6
Duration	1		1		1

W = 1/2 + 1/3 + 1/6 = 1

1>W>0.78

Correct schedule?

			T1		T2		T3
Period		20		5		10
Duration	1		2		4
Deadline	8		4		10

D <= T1

W = 1/20 + 2/5 + 4/10 = 0.85

__EDF: Earliest Deadline First__

	C1		D1		P1
T1	1		8		20
T2	2		4		5
T3	4		10		10

C = durée d'exécution
D = échéance
P = période

On s'arrête à chaque fin de période

At t=0, T2>T1>T3

LCM (périodes) = 20

W = 1/20 + 2/5 + 4/10 = 17/20

Process: time taken + time before = total
T2: 2 = 2
T1: 1 + 2 = 3
T3: 2 + 3 = 5
T2: 2 + 5 = 7
T3: 2 + 7 = 9
P(ause): 1 + 9 = 10 (tout a déjà été fait, on attend une nouvelle période)
T2: 2 + 10 = 12
T3: 3 + 12 = 15
T2: 2 + 15 = 17
T3: 1 + 17 = 18
P(ause): 2 + 18 = 20

__LLF: Least Laxity First__

	C1		D1		P1
T1	3		7		20
T2	2		4		5
T3	1		8		10

Ri <= Ci
At t=0, Ri=Ci

Md = (Td-Tc) - Ri
Td -> deadline
Tc -> current time
Ri -> Ci
Ri: durée restante tâche

1) Interval of analysis
LCM(P1) = 20

2) Workload
W = 3/20 + 2/5 + 1/10 + 13/10 < 1

3) Schedule
Md(T1)t=0 = (7-0)-3 = 4
Md(T2)t=0 = (4-0)-2 = 2
Md(T3)t=0 = (8-0)-1 = 7

On calcule la laxité a chaque periode

T2: 2 = 2
T1: 3 + 2 = 5 (on s'arrête, plus petite période)

T2: on l'a exécuté donc on change de fenêtre (deadline change)
Md(T2)t=5 = (9-5)-2 = 2
Md(T3)t=5 = (8-5)-1 = 2

On choisit arbitrairement entre les deux 2

T3: 1 + 5 = 6
T2: 2 + 6 = 8
Pause jusque 10

14 = 5 + 5 + 4
5 + 5 = 2 périodes, 4 = deadline
Md(T2)t=10 = (14-10)-2 = 2
Md(T3)t=10 = (18-10)-1 = 7
18 = 10 + 8
10 = 1 période, 8 = deadline

T2: 2 + 10 = 12
T3: 1 + 12 = 13
Pause jusque 15
T2: 15 + 2 = 17



A start time 4, deadline 8, CA = 3
A start time 3, deadline 8, CA = 3
A start time 1, deadline 8, CA = 4

A > B > C

C takes M1 at time 2
C releases M1 after 2 units of time
A takes M1 at time 2
A releases M1 after 1 unit of time