Aplicaţie Practică Drum Critic
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Transcript of Aplicaţie Practică Drum Critic
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8/10/2019 Aplicaie Practic Drum Critic
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Aplicaie Practic
Folosind metoda drumului critic, s se determine durata lansrii unui nou produs i s
se identifice activitile critice, cunoscnd activitile i duratele acestora.
Activitai Predecesor necesar Durat(luni)
A Design produs - 5
B Cercetarea pieei - 1
C Analiza produciei A 2
D Model de produs A 3
E Vnzri brour A 2
F Cost producie C 3
G Testare produs D 4
H Testare vnzri B, E 2
I Pre H 1J Raport proiect F,G, I 1
S se determine:a) drumul critic;
b) timpii liberi t i
ai evenimentelor xi;
c) timpii limit t *
ievenimentelor x
i;
d) evenimentele critice i operaiile critice;e) rezerva liber de timp a fiecrei activitate;
f) rezerva sigur de timp a fiecrei activitate;
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g) rezerva total de timp a activitilor ( x1; x
2), (x
2; x
5), (x
5; x
7), (x
7; x
8)
a) Aplicnd alogritmul lui Bellman-Kalaba obinem:
x1 x2 x3 x4 x5 x6 x7 x8
x1 0 5 1 - - - - -x2 - 0 2 2 3 - - -
x3 - - 0 - - 2 - -x4 - - - 0 - - 3 -
x5 - - - - 0 - 4 -
x6 - - - - - 0 1 -x7 - - - - - - 0 1
x8 - - - - - - - 0
1 - - - - - - 1 0
2
- - - 4 5 2 1 0
3 - 8 4 4 5 2 1 0
4
13 8 4 4 5 2 1 0
5 13 8 4 4 5 2 1 0
lmax
= 13
x1 x2 x5 x7 x8
b) Considerm timpul de nceput al proiectului egal cu 0, deci vom avea t1=0
t2
= max (t1+ t
12) = max(0+5) = 5
t3= max (t
1+ t
13; t
2+ t
23) = max(0+1; 5+2) = 7
t4
= max (t2
+ t24
) = max (5+2) = 7
t5= max (t
2+ t
25) = max (3+5) = 8
t6= max (t
3+ t
36) = max (7+2) = 9
t7
= max (t5+ t
57; t
4+ t
47; t
6+ t
67) = max (8+4 ; 7+3 ; 9+1 ) = 12
t8= max (t
7+ t
78) = max (12+1) = 1
c) Timpii limit t*
i
t*
8 = 13
t*
7 = min (t*
8 - t 78 ) = min (13-1) = 12
t*
6 = min (t*
7 - t 67 ) = min (12-1) = 11
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t*
5 = min (t*
7 - t 57 ) = min (12-4) = 8
t*
4 = min (t*
7 - t 47 ) = min (12-3) = 9
t*
3 = min (t*
6 - t 36 ) = min (11-2) = 9
t*
2 = min (t*5 - t 25 ; t
*4 - t 24 ; t
*
3 - t 23 ) = min (8-3 ; 9-2 ; 9-2) = 5
t*
1 = min (t*
2 - t 12 ; t*
3 - t 13 ) = min(5-5 ; 9-1) = 0
d) Avnd n vedere rezultatele de lapunctul precedent observm c evenimentele critice sunt x1,
x2
, x5, x
7, x
8(pentru ele avem t
*
i = t i ) i activitile critice sunt (x1 , x 2 ) ; (x 2 , x 5 ) ;
( x5, x
7) ; (x
7, x
8). Obervm din nou cdrumul critic este { x
1, x
2, x
5, x
7, x
8}
de lungime 13.
e) Rezerva liber de timp a activitilor
R78
= (t8- t
7- t
78) = (13-12-1) = 0
R57
= (t7
- t5- t
57) = (12-8-4) = 0
R67
= (t7
- t6- t
67) = (12-9-1) = 2
R47
= (t7
- t4
- t47
) = (12-7-3) = 2
R36 = (t 6 - t 3 - t 36 ) = (9-7-2) = 0
R25
= (t5- t
2- t
25) = (8-5-3) = 0
R24
= (t4
- t2
- t24
) = (7-5-2) = 0
R23
= (t3- t
2- t
23) = (7-5-2) = 0
R13
= (t3- t
1- t
13) = (7-0-1) = 6
R12
= (t2
- t1- t
12) = (5-0-5) = 0
f) Rezerva sigur de timp a activitilor
R78
= (t8- t
*
7 - t 78 ) = (13-12-1) = 0
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R57
= (t7
- t*
5 - t 57 ) = (12-8-4) = 0
R67
= (t7
-t*
6 - t 67 ) = (12-11-1) = 0
R47
= ( t7
- t*
4 - t 47 )= (12-9-3) = 0
R36 = (t 6 - t *3 - t 36 ) = (9-9-2) = -2
R25
= (t5- t
*
2 - t 25 ) = (8-5-3) = 0
R24
= (t4
- t*
2 - t 24 )= (7-5-2) = 0
R23
= (t3
- t*
2 - t 23 ) = (7-5-2) = 0
R13
= (t3
- t*
1 - t 13 ) = (7-0-1) = 6
R12
=( t2
- t*
1 - t 12 ) = (5-0-5) = 0
g) Rezerva total de timp pentru activitile :
R*
12 = t*
2 - t*
1 - t 12 = 5-0-5 = 0
R*
25 = t*
5 - t*
2 - t 25 = 8-5-3 = 0
R*
57 = t*
7 - t*
5 - t 57 = 12-8-4 = 0
R*
78 = t*
8 - t*
7 - t 78 = 13-11-1 = 1
BIBLIOGRAFIE
Diaconia, Vlad, Optimizri liniare, Editura Sedcom Libris, Iai, 2001.
