Conception et réalisation d'un système pour la gestion de patient

SECTION z De: termination du chemin critique et Calcul des dates

Etape1 :

1ere iteration

1. Etiquetons les nceuds :

M(n1)=M(n2)=M(n3)=M(n4)=M(n5)=M(n6)=M(n7)=M(ns)=M(n9)=M(n10)=M(n11) =M(n12)=M(n13)=M(n14)=M(n15)=M(n16)=0

2. Q= {a12, a23, a34, a45, a56, a67, a68, a79, a810, a911, a1012, a1113, a1213, a1314, a1415, a1516}

3. Q? 0

4. ast= a12; M(n2)= M(n1)+C12=0+12=12

2^eme iteration

2. Q= { a23, a34, a45, a56, a67, a6s, a79, as10, a911, a1012, a1113, a1213, a1314, a1415, a1516}

3. Q ? 0

4. ast= a23; M(n3)= M(n2) +C23=12+14=26

3^eme iteration

2. Q= {a34, a45, a56, a67, a6s, a79, aR10, a911, a1012, a1113, a1213, a1314, a1415, a1516}

3. Q ? 0

4. ast= a34; M(n4)= M(n3) +C34=26+3=29

4eme iteration

2. Q= {a45, a56, a67, a6s, a79, as10, a911, a1012, a1113, a1213, a1314, a1415, a1516}

3. Q? 0

4. ast= a45; M(n5)= M(n4)+C45=29+2=31

5^eme iteration

2. Q= {a56, a67, a6s, a79, as10, a911, a1012, a1113, a1213, a1314, a1415, a1516}

3. Q? 0

4. ast= a56 ; M(n6)= M(n5)+C56=31+3=34

6^eme iteration

2. Q= {a67, a68, a79, as10, a911, a1012, a1113, a1213, a1314, a1415, a1516}

3. Q ? 0

4. ast= a67 ;M(n7)= M(n6)+C67=34+5=39

47

7eme iteration

2. Q= {a68, a79, a810, a911, a1012, a1113, a1213, a1314, a1415, a1516}

3. Q# 0

4. ast= a79 ;M(n9)= M(n7)+C79=39+4=43

8eme iteration

2. Q= {a68, a810, a911, a1012, a1113, a1213, a1314, a1415, a1516}

3. Q# 0

4. ast= a911 ; M(n11)= M(n9)+C911=43+8=51

9eme iteration

2. Q= {a68, a810, a1012, a1113, a1213, a1314, a1415, a1516}

3. Q# 0

4. ast= a1113 ;M(n13)= M(n11)+C1113=51+3=54

10^eme iteration

2. Q= {a68, a810, a1012, a1213, a1314, a1415, a1516}

3. Q# 0

4. ast= a68 ; M(n8)= M(n6)+C68=34+4=38

11eme iteration

2. Q= {a810, a1012, a1213, a1314, a1415, a1516}

3. Q# 0

4. ast= a810 ; M(n10)= M(n8)+C810=38+3=41

12^eme iteration

2. Q= { 1012, a1213, a1314, a1415, a1516}

3. Q# 0

4. ast= a1012 ; M(n12)= M(n10)+C1012=41+7=48 13^eme iteration

2. Q= {a1213, a1314, a1415, a1516}

3. Q# 0

4. ast= a1213 ; M(n13)= M(n12)+C1213=48+7=55 14^eme iteration

². Q= {a1314, a1415, a1516}

3.Q#0

4. ast= a1314 ; M(n14)= M(n13)+C1314=55+60=115

15eme iteration

2. Q= {a1415, a1516}

3. Q# 0

4. ast= a1415 ; M(n15)= M(n14)+C1415=115+15=130

16^eme iteration

2. Q= {a1516}

3.Q#0

4. ast= a1516 ; M(n16)= M(n15)+C1516=130+5=135

17^eme iteration

2. Q= { }

3. Q=0

Donc M(n16) =135 est la longueur du plus long c hemin de n1 a n16. Etape2 :

ES(ni)=M(ni)

ES(n1)=0 ES(n2)=12 ES(n3)=26 ES(n4)=29 ES(n5)=31 ES(n6)=34

ES(n7)=39 ES(n8)=38 ES(n9)=43 ES(n10)=41 ES(n11)=51 ES(n12)=48

ES(n13)=55 ES(n14)=115 ES(n15)=130 ES(n1)=135

Etape 3:

LS(n16)=135 ; LS(n15)=min{(135-5)}=130 ; LS(n14)=min{(130-15)}=115 LS(n13)=min{(115-60)}=55; LS(n12)=min{(55-7)}=48; LS(n11)=min{(55-3)}=52 LS(n10)=min{(48-7)}=41; LS(n9)=min{(52-8)}=44; LS(n8)=min{(41-3)}=38 LS(n7)=min{(44-4)}=40; LS(n6)=min{(40-5), (38-4), (40-5)}=min{35,34}=34 LS(n5=min{(34-3}=31; LS(n4)=min{(31-2)}=29; LS(n3)=min{(29-3)}=26 LS(n2)=min{(26-14)}=12

Etape 4:

M={ n1, n2, n3, n4, n5, n6, n8, n10, n12, n13, n14, n15, n116 } {Cp}=( n1, _n2, n3, n4, n5, n6, n8, n1o, n12, n13, n14, n15, n116 )

La durée du projet sera de 135 jours.