Derive version 6.00 DfW file saved on 09 Dec 2003 z(x, y):=x + 500b7y hCross:=APPROX(1) vCross:=APPROX(1) Precision:=Exact PrecisionDigits:=10 Notation:=Rational NotationDigits:=10 Branch:=Principal Exponential:=Auto Logarithm:=Auto Trigonometry:=Auto Trigpower:=Auto Angle:=Radian CaseMode:=Insensitive VariableOrder:=[x,y] OutputBase:=Decimal InputBase:=Decimal InputMode:=Character DisplayFormat:=Normal TimesOperator:=Dot DisplaySteps:=false hCTextObj"{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24\par } 2{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Risoluzione di un problema di Programmazione lineare intera mediante il metodo Branch and bound utilizzando l'ambiente Derive\par \par 1\'b0 fase-Risoluzione del rilassamento continuo di un problema di P.L.I.,ottenuto tralasciando il vincolo di interezza delle variabili.\par \par } CExpnObj8Utente z(x,y):=x+5*y8User 11*x+6*y<=668User 5*x+50*y<=2258`Usery>=08.`>Userx>=0N`{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 risoluzione del sistema dei vincoli\par } 8p8Solve(User,[x,y])3SOLVE([11*x+6*y<=66,5*x+50*y<=225,y>=0,x>=0],[x,y])HSimp(#6)Q?/[11*x+6*y<=66 AND x+10*y<=45 AND x>=0 AND y>=0]HSimp(#6)Q?/[11*x+6*y<=66 AND x+10*y<=45 AND x>=0 AND y>=0]{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 determinazione dei vertici della regione ammissibile\par \par } 8Risolvi(Utente,[x,y]) 'SOLVE([11*x+6*y=66,5*x+50*y=225],[x,y])$`T Simp(Utente){Gz? [x=15/4 AND y=33/8]d{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice B(15/4, 33/8)\par \par } 8User "z(15/4,33/8)"  Simp(Utente) 195/8 {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 valore della funzione z(B)=195/8\par } 8p*Risolvi(Utente,[x,y]) SOLVE([11*x+6*y=66,y=0],[x,y]):HJSempl(Risolvi(Utente,[x,y]))Q? [x=6 AND y=0]Zl{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice A(6,0)\par } 8|pUtentez(6,0) Sempl(Utente)6{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 valore della funzione z(A)=6\par } 8xRisolvi(Utente,[x,y])SOLVE([5*x+50*y=225,x=0],[x,y])P.Sempl(Risolvi(Utente,[x,y]))Q?[x=0 AND y=9/2]>P{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice C(0,9/2)\par } 8`Utentez(0,9/2) Sempl(Utente)45/2p{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 valore della funzione z(C)=45/2\par valore ottimo della funzione, detto CBUB O estremo superiore, \'e8 z(B)=195/8 ossia z(B)= 24,375.\par Arrotondando la soluzione ottima B(15/4;33/8) si ha:x= 3;y=4; sostituendo tale soluzione nella funzione si determina l'estremo inferiore o CBLB =23\par 2\'b0 fase Branching su x\par Ponendo x\'93\'bf e x\'92 \'bf +1; dove \'bf rappresenta il valore intero della soluzione, si suddivide la regione ammissibile in sottoregioni disgiunte\par Aggiungendo,infati, nel sistema dei vincoli la disequazione x\'933, si genera il sottoproblema P'\par } {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 --------------------------------------------------------------------------------------------------\par } {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24\par } 8pRisolvi(Utente,[x,y])8SOLVE([11*x+6*y<=66,5*x+50*y<=225,x<=3,x>=0,y>=0],[x,y])(Sempl(Risolvi(Utente,[x,y])){Gz?8[11*x+6*y<=66 AND x+10*y<=45 AND x<=3 AND x>=0 AND y>=0]{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Calcolo dei vertici della sottoregione ammissibile\par } 8& 6Risolvi(Utente,[x,y])SOLVE([x=3,y=0],[x,y])FHVSempl(Risolvi(Utente,[x,y])){Gz? [x=3 AND y=0]fx{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice D (3,0)\par } 8pUtentez(3,0) Sempl(#25)3{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Valore della funzione z=3\par } 8xRisolvi(Utente,[x,y])SOLVE([x=3,5*x+50*y=225],[x,y]) P:Sempl(Risolvi(Utente,[x,y])){Gz?[x=3 AND y=21/5]J\{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice E(3,21/5)\par } 8lUtente z(3,21/5) Sempl(#29){Gz?24{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Valore della funzione z=24\par \par } 8xRisolvi(Utente,[x,y])SOLVE([5*x+50*y=225,x=0],[x,y]) PPSempl(Risolvi(Utente,[x,y])){Gz? [x=0 AND y=9/2]`r{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice F (0,9/2)\par } 8Utente!z(0,9/2) Sempl(#33)"45/2 8 2{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Valore della funzione z=(45/2), ossia z=22,5.\par Soluzione ottima z=24=CBUB\par Aggiungendo nel sistema dei vincoli la disequazione x\'924, si genera il sottoproblema P''\par } 8H pX Risolvi(Utente,[x,y])#8SOLVE([11*x+6*y<=66,5*x+50*y<=225,x>=4,x>=0,y>=0],[x,y])Hh x Sempl(Risolvi(Utente,[x,y])){Gz?$/[11*x+6*y<=66 AND x+10*y<=45 AND x>=4 AND y>=0]  {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Calcolo dei vertici della sottoregione ammissibile\par } 8  Risolvi(Utente,[x,y])%SOLVE([x=4,y=0],[x,y]) H Sempl(Risolvi(Utente,[x,y])){Gz?& [x=4 AND y=0]  {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice G(4,0)\par } 8 p Utente'z(4,0), <  Sempl(#39)(4L ^ {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Valore della funzione z=4\par } 8n p~ Solve(User,[x,y]))SOLVE([x=4,11*x+6*y=66],[x,y]) P Simp(Solve(User,[x,y]))?*[x=4 AND y=11/3]  {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice H(4,11/3)\par } 8 User+ z(4,11/3)0 `  Simp(#43),67/3p  {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 valore della funzione z=67/3; ossia z=22,33333333\par } 8 p Risolvi(Utente,[x,y])-SOLVE([11*x+6*y=66,y=0],[x,y]) H Sempl(Risolvi(Utente,[x,y])){Gz?. [x=6 AND y=0]  {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice I=(6,0)\par } 8 p Utente/z(6,0) $  Sempl(#47)064  {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Valore della funzione Z=6\par Soluzione ottima H(4,11/3) ; z=22,333333333=CBUB\par Il nodo termina perch\'e8 CBUB < CBLB\par 3\'b0 fase Branching su y\par Ponendo y\'93\'bf; y\'92\'bf+1 si generano due sottoproblemi derivanti dal nodo P'.\par Aggiungendo,infatti, nel sistemadei vincoli relativo al problema P?, il vincolo y\'934, si genera il sottoproblema P'''\par }   {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24\par }   {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24\par } 8  Risolvi(Utente,[x,y])1=SOLVE([11*x+6*y<=66,5*x+50*y<=225,x<=3,y<=4,x>=0,y>=0],[x,y])& 6  Sempl(#49) ףp= ?2A[11*x+6*y<=66 AND x+10*y<=45 AND x<=3 AND y<=4 AND x>=0 AND y>=0]8F V Risolvi(Utente,[x,y])3SOLVE([x=3,y=0],[x,y])f Hv Sempl(Risolvi(Utente,[x,y])){Gz?4 [x=3 AND y=0]  {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice L(3,0)\par } 8 p Utente5z(3,0)   Sempl(#53){Gz?63  {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Valore della funzione z=3\par } 8  Risolvi(Utente,[x,y])7SOLVE([x=3,y=4],[x,y])*H:Sempl(Risolvi(Utente,[x,y])){Gz?8 [x=3 AND y=4]J\{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice M(3,4)\par } 8lp|Utente9z(3,4) Sempl(#57):23{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Valore della funzione z=23\par } 8 Risolvi(Utente,[x,y]);SOLVE([y=4,x=0],[x,y])HSempl(Risolvi(Utente,[x,y])){Gz?< [x=0 AND y=4] {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Vertice P(0,4)\par } 80p@Utente=z(0,4)P` Sempl(#61)>20p{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Valore della funzione z=20;\par soluzione ottima M(3,4);z=23; CBUB=CBLB; l'algoritmo termina\par } 8Risolvi(Utente,[x,y])?=SOLVE([11*x+6*y<=66,5*x+50*y<=225,x<=3,y>=5,x>=0,y>=0],[x,y])(Sempl(Risolvi(Utente,[x,y])){Gz?@8[11*x+6*y<=66 AND x+10*y<=45 AND x<=3 AND x>=0 AND y>=5]{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq2\fcharset2 DfW5 Printer;}} \viewkind4\uc1\pard\f0\fs24 Il sistema non ha soluzioni\par } CDispOleObj CDispItem  ࡱ>    !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~Root EntryFhOle CONTENTSd CompObj_" FImmagine (Metafile)StaticMetafile9q      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~ƚ @JU <&  3x3&C 3x3(d HtHHHߜHߜtHHttHtHHHttttߜtߜtttHtHߜtߜHtttHHߜtߜttHtߜHHtttHHߜtߜtHHHttHtHߜtߜHttttߜtߜtHtHtߜHߜHtttHHHߜttHHHtttHtttHHttHHttHHtHttttHttߜttHtttߜHHHHttttttttHHtttHߜHHHttHߜHHttHHߜHttHHHHHtHHߜtŁHYHX}ttHttHHHHHHHHtHHHHttHtHHߜHߜHtHHHtHHHHHtHHHHtHHtߜttHHtHHtttߜHHHtttHHHtߜHHtHtߜHHߜHHHHHtttHtHtHHHtCONTENTS I{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq1\fcharset0 Derive Unicode;}} \viewkind4\uc1\pard\f0\fs24 La soluzione ottima del problema \'e8 z=23 , in corrispondenza del vertice M(3,4).\par \par Il grafico evidenzia le sottoregioni ammissibili, ottenute suddividendo la regione ammissibile OABC del problema rilassato di P.L. associato al problema di P.L.I.\par } Yy{\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq1\fcharset0 Derive Unicode;}} \viewkind4\uc1\pard\f0\fs24\par \par } {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq1\fcharset0 Derive Unicode;}} \viewkind4\uc1\pard\f0\fs24\par } {\rtf1\ansi\ansicpg1252\deff0\deflang1040{\fonttbl{\f0\fmodern\fprq1\fcharset0 Derive Unicode;}} \viewkind4\uc1\pard\f0\fs24\par } Ai