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0m4r
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 [Esame di Marzo] Svolgimento

Ho risolto gli esercizi 2, 3 e 4 ma non so se sono corretti.
Se qualcuno ha fatto lo stesso potrebbe confrontare i risultati con i miei?

grazie

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31-03-2006 15:03
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GiObAT
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ciao allora ho svolto pure io:

2.1)uguale al tuo

2.2)facendo il sistema tra il II e il III vincolo: x1=-2/3 x2=5/3
sostituendo poi al sistema:
-2/3+5/3+x3=8 di base
1-x4=1 => x4=0 non di base
2/3+10/3+x5=4 =>x5=0 non di base
-2/3-5/3+x6=6 di base
quindi base: x1,x2,x3,x6

2.3)uguale al tuo

3)uguale al tuo

4)nel quarto hai fatto un errore xkè ti spunta un valore negativo in una
delle b. svolgi bene il problema ausiliare

spero di esserti stato d'aiuto

ps: si sa qualcosa sull'aula dell'esame del 5?
02-04-2006 16:54
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GiObAT
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percaso sai svolgere anke quello dello zaino?
02-04-2006 16:55
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0m4r
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Originally posted by GiObAT
2.2)facendo il sistema tra il II e il III vincolo: x1=-2/3 x2=5/3
sostituendo poi al sistema:
-2/3+5/3+x3=8 di base
1-x4=1 => x4=0 non di base
2/3+10/3+x5=4 =>x5=0 non di base
-2/3-5/3+x6=6 di base
quindi base: x1,x2,x3,x6

Questro errore l'ho corretto anche io, per sbaglio ho segnato la situazione del vertice ottimo anzichè di quello richiesto (almeno mi sembra)

Originally posted by GiObAT
4)nel quarto hai fatto un errore xkè ti spunta un valore negativo in una
delle b. svolgi bene il problema ausiliare

Vedrò di correggerlo, grazie

Originally posted by GiObAT
ps: si sa qualcosa sull'aula dell'esame del 5?

No, ancora niente... credo che si potrà trovare sul sifa, magari gia domani...

Originally posted by GiObAT
percaso sai svolgere anke quello dello zaino?

Purtroppo no.

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02-04-2006 17:06
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GiObAT
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non riesco a capire come fare con sto problema dello zaino :?

se ti serve il 6 te lo posto entro sera
02-04-2006 17:12
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0m4r
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Grazie, mi farebbe comodo

Il terzo esercizio l'hai svolto? E ti viene identico al mio?
Discutendone con un amico forse io ho commesso qualche errore sul verso delle disugualgianze...

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02-04-2006 17:16
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GiObAT
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ah si scusa non c'avevo fatto caso viene così:
min 8y1+y2-4y3+6y4
y1+y2-y3+y4<=3
y1+y2+2y3-y4<=1
y1,y3,y4>=0
y2<=0

y=(2,0,0,1,0,0) z=12
02-04-2006 18:19
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0m4r
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min 8y1+y2-4y3+6y4
perchè ti viene -4y3?

Per il resto mi pare ok.

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02-04-2006 18:41
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GiObAT
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sorry errore di battitura sono tutti positivi
02-04-2006 18:51
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GiObAT
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ecco il 4:
il valore mancante è 0 poichè dal nodo 3:
somma archi entranti=somma archi uscenti
x73=flusso da 7 a 3
x37=flusso da 3 a 7
1+5+x73+1=2+4+x37+5
allora: x73=x37+4
inoltre: x73+x37=4 (capacità)
mettendo a sistema x37=0 e x73=4

ti posto solo la prima parte xkè il flusso massimo ho provato a farlo diverse volte e mi torna sempre diverso :(
02-04-2006 20:18
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0m4r
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L'esercizio è il 6, non il 4 ;P

Cmq, non ho capito come hai ricavato questa equivalenza:
1+5+x73+1=2+4+x37+5
potresti spiegarmelo?

grazie

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02-04-2006 20:27
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GiObAT
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guarda l'allegato
in blu archi uscenti da 3
in rosso archi entranti in 3

blu=rossi =>
1+5+x37+1=2+4+x73+5
x37+7=11+x73
x37=4+x73

poi siccome x37+x73=4

mettendoli a sistema:
x37=4+x73
x37+x73=4

sostituendo nella seconda:
4+x73+x73=4 => x73=0
x37=4+0

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02-04-2006 20:54
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0m4r
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Scusami di nuovo, ma non capisco proprio come hai fatto a mettere quei valori sugli archi.
Si, lo so, sono messo malaccio per quanto riguarda questo argomento...

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02-04-2006 21:02
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GiObAT
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è una regoletta
Uij=capacità superiore arco i,j
Xij=flusso corrente inviato da i a j

[i]---(Uij,Xi)--->[j]

diventano 2 archi:

[i]----(Uij-Xij)--->[j]
[i]<---(Xij)----[j]

Uij-Xij=capacità residua
02-04-2006 21:11
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0m4r
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ho capito... grazie mille!

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