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# Hw3 - ALEX BODEA ISE 330 Engineering Operations Research...

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Unformatted text preview: ALEX BODEA January 26, 2010 ISE 330: Engineering Operations Research Homework #3 Winter Semester, 2010 Homework #3 Due 1/26/10 HW3.1 (15 points) From Introduction to Operations Research, Ninth Edition,by F.S.Hillier and G.J.Lieberman, do Problem 4.1-6, p 151. HW3.2 (15 points) From Introduction to Operations Research, Ninth Edition, by F.S.Hillier and G.J.Lieberman, do Problem 4.4-4, p 153. HW3.3 (15 points) From Introduction to Operations Research, Ninth Edition,by F.S.Hillier and G.J.Lieberman, do Problem 4.5-8., p 154. ALEX BODEA January 26, 2010 Homework #3 MaxZ=2 X1+ 3 X2 Subject to 1) -3 X1 + 1 X2 <= 1 2) 4X1+ 2X2<= 20 3) 4 X1 — 1.X2<= 10 4) -1X1+ 2X2<= 5 And X1 >= 0, X2' >= 0. The feasible region is computed as follows: Constraint #1:'-3 x1 + x2 <=-1 Constraint #2: 4 x1 + 2 x2 <= 2 ' Constraint #4: -x1+ 2 x2 <= By solving theline intersections we- get the-followingCPFs: (0, 0), (Q, 1;), (0.6, 2.8L (3, 4), (3.33, 3.33), (2.5, 0) After initializationlintroducing the slack variables. to. be the initial'basic variablesandselecting‘the decisionvariablesto. be theinitialnon-basicvariablessetequal to zero) theinitial B-F solution is‘(0,'0, 1, 20, 10, 5), The Simplex method then starts at the origin where .Z' = 2x1" + 3x2 = 20 + 3 0': 0‘ and travels along the feasible region edges. The negative coefficient having-the largest absolutevaluev corresponds with-x2, thereforeit will travel along x2- axls until it reaches a. boundary constraint. ‘X2'becomes-the entering basic variable and after- the ratio—test- x3~ becomes the leaving basic variable. After Gaussian elimination for x2 we have the following'table: ALEX BODEA January 26, 2010 ' Homework #37 BasIqu Coefficient of Var No 2 x1 X2 X3 X4 2 0| 1 -11 XZI ll 0 -3 X4 10 5 i 8| pxoi 4,! iii €53 Graphically thiscorresponds toarrivingat the next CPF solution (0, 1), where .th'eBF solution is“), 1,0, 18, 11, 13) and obviously Z = 3. HDOOD In the next iteration-the- negative coefﬁcient having-the "largest absolute value corresponds withxl, therefore it will travel along constraint #1 line until. it reaches a'boundary constraint- Xi'becomesthe entering basic variable. and after the-ratio test x6becomes~the leaving basic variable. After Gaussian elimination for x1 werhave the following table: Coeff'id em: of X3 __.____.—__———— Graphically this corresponds to arriving at the nextiCPF' solution- (05, 2.8), where the BF solution .is-.(0'.6, 2.8, 0, 12',- 10.4, 0) and obviousle =96. January 26, 2010 Homework #3 In the next iteration- the negative coefﬁcient having the largest absolute value corresponds‘with-x3, therefore it will travel. along; constraint #4 line until-it reaches a. boundary constraint. X3 becomes the entering basic variable and. after the ratio testx4 becomes the leavingzbasicvariabie. After Gaussian elimination for: x31we have the foilowing'table: Baleq coefficient of varl‘No 2 XL x2 x3 x4 _I__ _ Graphically this corresponds to arriving» at the next CPF solution-v ('3: 4), where-the BF solution is (3-, 4, 6, 0..- Z. 0) and obviously Z = 18. We notice in the table above-there is nonegativecoefﬁcient in the Z rowto increase tmeaningthatthe last'solution is optimal. To- check our resuits we canuse the graphical'solution and we'see- we areigetting the same solution. (in! + ﬁes-l 4x! 92:2,“20 Ax! . x2¢=t¢ -x‘l * 2x2¢5 Z-ﬂ HI! 1183,12- 4-1-6 2 4 3X2 le+ - »ve Interactively by the-simplex Method: Number of FunetienaT Constraints: .1) Linear Programming Mode1: Number of Decision Variab1es: subject to MXZ= mm alums mu slung m“ j£.Aa mm R4523 9.9 gm 01\$ 9 2 m 0 gm... . i . . m R m m W 00001 m 00001 W 26422 W 84422 20 \$0 00 1% 5 00010 5 00010 S 00010 5 00010 X X X X f f f f o o o o, tm 00100 rm 00100 t“ 00100 t“ JlSJJ n w n D 00000 e . e e _ i i . . c3 01000 £3 31212 03 42P44 03 00100 i i i , i f , . Cl. _ . . . , .- - fX fx ﬁx 10 10 ﬁx 8 e _ _ _ m * m m m 2 31212 2 01000 V9 01000 2 01000 x . _ x , x . * . . . 1 23‘ 1 1 1 15 0 1 X ._44_ X 43m n 00 01 X 00001 Hglummmm Hgiummmm HAIMNNNN ﬂqlnmmmm WWAIMHﬂNM WWJINHHNM WWAIMHHNM WWJINHNMM mmJIHmmmm wmglummmm mm1lummmm umgnummlm BV . BV _ BV . an Page 1 — 0% kam‘tlt 1: ZN“ 37(7/ [mlo‘dcck A“ XH/LN/ 9&0 X‘ 4' XL 42/0 0wa ><,>/o7 x120 )(l’r'Lﬁ/‘77o ><rv0 >7 7w; ‘9 , M30 ’V ““7" "’7 Q”) 3“ Q70”) X4" X1 " 7/0 ALEX BODEA Januaty 26, 2010 Homework #3 % [LT] 4’77X1/ 77 %’ ZTl/:9ﬂ70 ' “”1 ’70 ’7 \WM» why/”w ”L X *M‘V’ *? NH'H 7‘5310 {emit H719 I Nanvbmswzxuh/ X'LCK) X} % " 7/7‘ [ '77X7/7p @) l J/L :[g Len , Q—EBKE) 7/ XI J’fl 'L X1, 2” U2») ~ 9t *lk1‘197‘l‘t’1-0 [) > ”70‘9”" X1,)“, 7 Not/\Lms'tt Kt“ ALEX BOD£A January 26, 2010' Homework #3 Q: \l yéxg :11; (a) (6.0 ’I’OI<)%7/) LXI ‘0 —§ ‘9) __67»@§X%L> Wm M o , XI\"1/1‘3" “35 (62) 4-4-4a1gebr Linear.Programming Model: Number of Decision variables: 2 Number of Functiona] Constraints:' 2 Max 2 = 2 x1 + 3 x2 subject to 1) 1 Xv + 2 X2 <= 30 2) 1 x1 + 1 x2 <= 20 and X1 >= 0, X2 >= 0; so1ve Interactive1y by the Simplex.Method: Maximize Z, subject to 0) Z — 2 X1 — 3 x2 + -0 X3 + ‘0 X4 =-0 1) 1 X1 + 2 x2'+ 1 X3 + 0 X4 = 30 2) 1 X1 + 1 x2 + 0 x3 + 1 x4 = 20 and X1 >= 0, X2 >= 0, X3 >= 0, X4 >= 0. Maximize 2, subject to- 0) Z — 0.5 X1 + O X2'+ 1 5 X3 + 0'X4 = 45 1) 0.5 X1 + 1 X2 + O 5 X3 + 0 X4 = 15 2) 0.5 x1 + 0 x2 - O 5 x3 + 1 x4 - 5 .and X1 >= 0, X2 >= 0, X3 >= 0, X4 >= 0. Maximize 2, subject to 0) Z + 0 x1 + 0 x2 + 1 X3 + 1 X4 = 50 1) 0 x1 + 1 x2 + 1 x3 - 1 x4 - 10 2) 1 x1 + 0 X2 — 1 x3 + X4 = 10 and X1 >= 0, X2 >= 0, X3 >= 0, X4->= 0; Page 1 424:4 Linear Programming Model: Number of Decision Variab1es; 2 Number of Functiona1 Constraints: 2 Max 2 = 2 X1 + 3 X2 subject to 1) 1 X1 + ZVXZ <= 30 2) 1 x1 + 1 x2 <= 20 and X1 >= 0, X2 >= 0. Soive Interactiveiy by the Simp1ex Method: BasIqu Coefficient of I Right VarINoI 2} x1 x2 x3 X4 I S1de Igle I I z I 0| 1| -2 -3 0 0 I 0 x3 I 1| 0| 1 2* 1 0 I 30 x4 I 2| 0| 1 1 0 1 I 20 BasIqu Coefficient of | Right varINo: 2: X1 X2 X3 X4 I Side —I’“I—I I z I OI 1I -O.5 O 1.5 0 I 45 x2| 1| OI 0.5 1 0.5 O I 15 X4I 2| OI 0.5* 0 —O.5 1 I 5 BasIEqI Coefficient of I Right VariNo} 2} X1 x2 X3 X4 1 S1de ___|__I__|_—_——__—___-—_______—————_| z I 0| 1| 0 O 1 1 I SO x2I 1| OI O 1 1 —1 | 10 x1I ZI OI 1 O -1 2 I 10 Page 1 4=4=4model (v) é;in2> (:LTDl.—?EL)< 2 x1 + 3 X2 sub'ect to 1 H- + 2 X2 <= 30 1 x1 + 1 x2 <; 20 X1 >= 0 X2 >= 0 Processin ... 4:4-4mode .txt F11e 4-4-4m9delatxt upIQadgd successfuIIya DispIaying cantents Line #0 : Maximize Line #1 : 2 x1 + 3vX2 Line #2 : Line #3‘: subject to Ling #4 2 1 x1 + 2 x2 <= 39 Line #5 : 1 x1 + 1 x2 {a 20 Line #6 : x1.>= 0 Line #7 : x2 >= 0 Running ngsoTGLP\$OLz GLPK LP/MIP Soivef 4.38 91p=reéd;1?: reading pyoblem data from '/tmp/ghpof1YDv*... _tmp/9hprzYQv;§z warning; Kgngrd gnd' missing glp_read_1p: 4 rows, 2 columns, 6 non-zeros glp_read_1p:.8 lines were read Scalin ... _ A: minfaijl = 1.000e+00 maxlaijl = 2.000e+00‘rat10-= 2.000e+06» Probiem data seem to be we11 scaTed Crashing... Sig: Qf'triangu1ar part = 4 * 0: obj = 0.000000000e+00 infeas * 2: 0b] = 5.000000000e+01 infeas OPTIMAL SOLUTION FOUND Time-used: OLG-secs Memory used: 0.0 Mb (29916 bytes) _ Writing basic soTution to /tmp/F00K1Kuf6'... 0.000e+00 (0) 0-000e+00 (0) Margina] ProbTem: Rows: 4 »C01umns: 2 Non—zeros: 6 Status: OPTIMAL _ Objective: obj = 50 (MAXimum) No. Row name St Activity Lower bound Upper bound 1 r.5 NU 3O 30 2 r.6 NU 20- 20 3 r.7 B 10 0 4 r.8' B 10 O No. CoTumn name St Activity Lower bound Upper bound 1 x1 3 -10 0 2 x2 3 10 0 'Page 1 ALEXEMJDEA January 26, 2010 4-5-3 Homework#3 1near Programming Mode]: Number of Decision var‘iab‘I es: 4 Number of Functiona1 Constraints: 2 Max 2 = 50 X1.+ 25 X2 + 20 X3 + 40 X4 subject to 1) 2 X1 + 1 X2 + 0 X3 + -0 X4 <= 30 2) 0 X1 + 0 X2'+ 1 X3 + 2 X4 <= 20 and X1 >= 0, X2 >= 0, X3 >= 0, X4 >= 0. Solve Interactively by the Simplex Method: BasIEqI Coefficient of | Right VarINoI ZI x1 X2 x3 x4 X5 x6 I -s1de -l-I-I I Z | 0| ll -50 -25 ~20 —40 0 0 I 0 XSI 1I 0| 2* 1 0 0 1. 0 I 30 XGI 2I OI O -0 1 2 70 1 I 20 BasIEqI Coefficient of I R1" ght Vanilla: ZI' X1 X2 X3 X4 X5 X6 i s1de “1‘1“: I Z I 0| 1I O 0 —20. —40 25 0 I 750 X1I 1| OI 1 v0.5 70 0 -O.S -0 I 15 x6I ZI OI 0 0- 1 2* 0' 1 I 20 BasIEqI Coefficient of I Right var'INoI 2: x1 x2 X3 X4 x5 x6 I s1de -—-1-_1—_I-_——_—___—____-__—_—__-—“_—_—-__——I z I OI 1| 0 A0 0 -0 25 20 I 1150 X1I lI OI 1 0;5 0 0- 0 5 O I 15 x4I 2I OI 0 '0 0.5 1 0 0.5 I 10 I/givw) IOIQ TAT“ “I 4b Sank «(I TAP 9o lakong '. 73’40 (010(6r01010) £21400 (old/\$0,0(30m) -Z':7§0 (t5(0,o'0,0’7/o) f,“ \kfo (l§(0,%,010r0) %: [KO (Klo,ol (0,9,9 2’: 7gb (0"3'0,9(0)0)l0) 2; us» (0,\$o,0,10,0,0) 241'“ (0130/ u,0,0,0) Page 1 ...
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Hw3 - ALEX BODEA ISE 330 Engineering Operations Research...

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