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Hw5 - ALEX BODEA ISE 330 Engineering Operations...

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Unformatted text preview: ALEX BODEA FEB 16, 2010 ISE 330: Engineering Operations ResearclinEVVORK‘b‘t5 Winter Semester, 2010 Homework #5 Due 2/ 16/ 10 HW4.2 (15 points) From Introduction to Operations Research, Ninth Edition, by F.S.Hillier and G.J.Lieberrnan, do 4.6-2 parts (a)-(f), p. 154. HW4.3 (15 points) From Introduction to Operations Research, Ninth Edition,by F.S.Hillier and G.J.Lieberman, do 4.6—8, p. 155. HW5.1 (points) From Introduction to Operations Research, Ninth Edition,by F.S.Hillier and G.J.Lieberman, do 4.7-6 p. 157. HW5.2 ( points) From Introduction to Operations Research, Ninth Edition,by F .S.Hillier and G.J.Lieberman, do 5.1-7, p. 189. HW5.3 ( points) From Introduction to Operations Research, Ninth Edition,by F .S.Hillier and G.J.Lieberman, do 5.2-1, p. 191. HW5.4 ( points) From Introduction to Operations Research, Ninth Edition,by F.S.Hillier and G.J.Lieberman, do 5.3-5, p. 193. G a) {53 H max/Woo 4-6—2M Max 2 = 4 x1 + 2 X2 + 3 X3 + 5 X4 subject to 1) 2 X1 + 3 X2 + 4 X3 + 2 X4 = 2) 8 X1 + 1 x2 + 1 X3 + 5 x4 = and X1 >= 0, X2 >= 0, X3 >= 0, X4 >= 0. $01ve Interactive1y by the simp1ex Method: BasIEqI Coefficient of I Right g) VariNoI 2: X1 X2 x3 x4 X5 X6 I S1de "—I—I—I ~10M —4M -5M -7M I ~600M Z I 0| 1I— 4 - 2 - 3 - 5 0 0 I 0 x5] 1I 0| 3 4 2 1 0 I 300 X6I 2| 0| /”6* 1 1 5 0 1 | 300 BasIEqI Coefficient of I Right VariNo: 2} x1 x2 x3 x4 x5 x6 I S1de ’—I—I—I -2.8M -3.8M —0.8M 1.25M I -225M 2 I 0| 1| 0 - 1.5 ; .5 - 2.5 0 + 0.5 I 150 X5I 1I OI 0 2.75 -3.75* 0.75 1 -0.25 I 225 x1I 2| 0| 1 0.125 0. 0.625 0 0.125 I 37.5 BasIqu Coefficient of | Right varINo! 2: x1 x2 x3 x4 X5 X6 I S1de —I_I_I 1M m I Z I 0| 1| 0 0.333 O —2 +0.67 +0.33 | 300 X3I 1| 0| 0 0.733 1 042 0.267 -0.07 I 60 x1I 2I 0| 1 0.033 0 Io.6?+o.o3 0.133 I 30 BasIEqI coefficient of I Right VariNo} Z} x1 X2 x3 x4 x5 x6 I S1de ‘T’I—I 1M m I Z I 0| 1I3.333 0.444 0 0 +0.56 +0.78 I 400 X3| 1| 0|-0.33 0.722 1 0 0.278 —0.11 I 50 X4| 2| 0I1.667 0.056 0 1 -0.06 0.222 I 50 I m .3? when (o’clolol‘éoo/Zao) [MI 500 MAW/Vila} whflok OF‘M‘NA gobsdom ‘. GIG/60160I0!0> / Z =LIoo nAm} a) Two , QMRSfi. Nam/‘03 4—6—2Ph Max Z = 4 x1 + 2 X2 + 3 x3 + 5 X4 subject to 1) 2 x1 + 3 X2 + 4 X3 + 2 x4 2) 8 X1 + 1 X2 + 1 X3 + 5 X4 and X1 >= 0, X2 >= 0, X3 >= 0, X4 >= 0. So1ve Interactiveiy by the simpiex Method: Phase 1: Bas|Eq| Coefficient of variNo: 2: X1 X2 X3 X4 X5 X6 —|—|_| Z I 0| 1| —10 -4 —5 -7 0 O X5| 1| 0| 3 4 2 1 0 x6| 2| 0| \8*/ 1 1 5 0 1 Bas|Eq| Coefficient of VariNo} 2: X1 x2 X3 X4 x5 X6 —I—|_| z | 0| 1| 0 -2. 75 33.. —0. 75 O 1.25 x5| 1| 0| 0 2.75 3. 75* 0.75 1 -0.25 X1| 2| 0| 1 0.125 0712 0.625 0 0.125 Bas|Eq| Coefficient of Variuoi 2: x1 x2 x3 x4 x5 x6 —|_|_| z | 0| 1| 0 0 O 0 1 1 X3| 1| 0| 0 0.733 1 0.2 0 267 -0.07 Xll 2| 0| 1 0.033 0 0.6 ~0 03 0.133 Phase 2 Bas|Eq| Coefficient of | Right VarlNo: 2: x1 X2 X3 x4 } Side —|_|—| .1;— mm: 333;; 2 4| 323 X . . X1| 2| 0| 1 0.033 0 /8:%9 | 30 Baleq Coefficient of Right VarlNo Z X1 X2 X3 X4 | S1de Ri ht sgde -600 300 300 | Ri ht | sgde -225 225 37.5 Ri ht sgde 88o 300 300 (NBA (5? soluiflo“ (30,0) (.0 0 0 0) L/// L) Pkaqc l _/ 1... ml 07? gohbow (77")y0/C’OIO) L mm, L L///’ 4‘6-2 maximize 4 x1 + 2 x2 + 3 x3 + 5 x4 subject to 2 x1 + 3 x2 + 4 x3 + 2 x4 8 x1 + 1 x2 + 1 x3 + 5 x4 x1 >=0 x2 >=O x3 >=O x4 >=0 300 300 II II L) L? (“0391‘ . f. ‘7 ell gobjwm a0? L‘ M“: 4-6—2501 Processing... 4—6—2.txt Fi1e 4—6-2.txt up1oaded successfu11y. Disp1aying contents Line #0 : maximize Line #1 : 4 x1 + 2 x2 + 3 x3 + 5 x4 Line #2 : Line #3 : subject to Line #4 : 2 x1 + 3 x2 + Line #5 z 8 x1 + 1 x2 + Line #6 : Line #7 : x1 >=0 Line #8 : x2 >=O Line #9 : x3 >=0 Line #10 : x4 >= Line #11 : 300 300 XX #4?- || || Running g1pso1GLPSOL: GLPK LP/MIP So1ver 4.38 g1p_read_1 : reading prob1em data from /tmp/pnpf2brx9'... /tmp/phpf2 rx9:12: warnin : keyword end' m1551ng g1p_read_]p: 6 rows, 4 co umns, 12 non—zeros g1p_read_1p: 12 1ines were read Sca1in ... __ . A: min?aij| = 1.000e+00 maxlaijl = 8.000e+00 ratio = 8.000e+00 Prob1em data seem to be we11 sca1ed Crashing... Size of trian u1ar part = 5 0: obj = 7.508000000e+02 infeas = 4.500e+02 (1) * 1: obj = 4.000000000e+02 infeas = 0.000e+00 (0) OPTIMAL SOLUTION FOUND Time used: 0.0 secs Memory used: 0.0 Mb (30455 bytes) Writing basic so1ution to /tmp/F003022ee'... Prob1em: Rows: 6 Co1umns: 4 Non—zeros: Status: No. Row name St Activity Lower bound Upper bound Margina1 1 r.5 NS 300 300 = 0.555556 2 r.6 NS 300 300 = 0.777778 3 r.8 B 0 0 4 r.9 B 0 0 5 r.10 B 50 0 6 r.11 B 50 0 . CoTumn name Upper bound Margina1 —3.33333 —0.444444 COCO 1%) M {Mi-HA0) 4-6—8M Min 2 = 2 X1 + 1 x2 + 3 X3 subject to 1) 5 x1 + 2 x2 + 7 x3 = 420 2) 3 X1 + 2 X2 + 5 X3 >= 280 and X1 >= 0, x2 >= 0, X3 >= 0. So1ve Interactive1y by the simp1ex Method: BasIEqI Coefficient of I Right VarINo: 2: X1 X2 x3 x4 X5 X6 E S1de —_—I_—|_—| -8M -4M —12M IN I -700" Z I 0|—1|+ 2 + 1 + 3 0 + 0 0 I 0 x41 11 0| 5 2 fl? 1 o o | 420 . I _ X6I 2| 0| 3 2 \3i/, 0 1 I 280 BasIEqI Coefficient of | R1ght VarINoI 2: X1 X2 X3 X4 X5 X6 i S1de “I—I—I—OJM 0.8M —1.4M 24M | -28M 2 I 0|41I+ 0.2 — 0.2 0 0 +,Q. - O 6 | -168 X4I 1| 0| 0.8 —0.8 0 1 1.4* -1 4 I 28 X3I 2| 0| 0.6 0.4 1 0 - . 0 2 | 56 BasIEqI Coefficient of I Right VargNoi Z} X1 X2 X3 X4 X5 X6 I S1de _I_I_I———_—_—_1M—-_TI Z I OI-1I— 4}2.143 0 —0.43 0 + 0 I —180 X5I ll 0 0.571 ,0.57 0 0.714 1 -1 I 20 X3I.2I 0I07714’0.286 1 0.143 0 0 I 60 BasIEqI coefficient of I Right VarINoi 2: X1 X2 X3 x4 X5 X6 I S1de _l_l—| 1M 1M I z I OI—ll 0 0 0 -O.25 0.25 -O.25 | -175 X1IH1I 0| 1 21 0 1.25 1-75 -1.75 I 35 X3I 2I 0| 0 1 l —0.75 —1.25 1.25 | 35 O€H~®\ Qwhkfln‘. (?S’0'$€,0(0,€> L//// 1~~s176 hfl\f\ ‘fiko—QMGSC (ML¥VOA 4-6-8Ph Min 2 = 2 x1 + 1 X2 + 3 X3 subject to 1) 5 X1 + 2 X2 + 7 X3 = 420 2) 3 X1 + 2 x2 + 5 X3 >= 280 and X1 >= 0, X2 >= 0, X3 >= 0. So1ve Interactive1y by the Simp1ex Method: Phase 1: BasIEqI Coefficient of I Right VarINoI 2: x1 X2 x3 X4 X5 x6 I S1de —I"I_I I Z I 0|-1I -8 —4 —12 0 1 0 I —700 x4| 1| 0| 5 2 7 1 0 0 I 420 X6| 2| 0| 3 2 5* 0 —1 1 I 280 BasIqu Coefficient of | Right variNo: 2: X1 X2 X3 X4 X5 X6 } S1de —|_I—I I Z | 0|—1| —0.8 0.8 0 0 —1.4 2.4 | —28 x4| 1| 0| 0.8 —0.8 O 1 1.4 -1.4 | 28 x3| 2| OI 0.6 0.4 1 0 -0.2* 0.2 I 56 BasIEqI Coefficient of | Right Var:No{ 2: x1 x2 x3 x4 x5 x6 I S1de —l—I_l I z I 0I-1I —5 —2 -7 O 0 1 | -420 x4| 1| 0| 5 2 7 1 0 0 I 420 XSI 2| 0| —3 —2 -S* 0 1 —1 I —280 BasIEqI Coefficient of | Right VarlNog 2} X1 x2 x3 X4 X5 X6 : S1de —I“I_I I z | OI-1I -0.8 0.8 O 0 -1 4 2 4 I -28 x4] 1| 0| 0.8 -0.8 0 1 1 4* —1 4 I 28 x3| 2| 0| 0.6 0.4 1 0 —0 2 0 2 I 56 BasIEqI Coefficient of I Right VargNo: 2: x1 x2 x3 x4 X5 X6 I S1de _I—I_I I Z I 0I—1| 0 —0 0 1 0 1 | 0 XSI 1| 0I0.S71 —0.57 0 0.714 1 —1 I 20 X3I 2| 0|0.714 0.286 1 0.143 0 O I 60 Phase 2 BasIEqI Coefficient of | Right variNoi 2: X1 x2 x3 x4 : S1de —l_l_l I z | 0|—1I-0.14 0.143 0 0 I -180 x4| 1| 0|0.571*-0.57 0 1 | 20 X3I 2| 0|0.714 0.286 1 0 I 60 BasIEqI Coefficient of I Right varINo| ZI x1 x2 x3 X4 | S1de 4-6—8Ph 010 {m .. \16 a) L? {M 3d linilfize 2 x1 + 1 x2 + 3 x3 subject to 5 x1 + 2 x2 + 7 x3 = 420 3 x1 + 2 x2 + 5 x3 >= 280 x1 >=0 X2 >=0 x3 >=0 4-5.3 90b Jciow g0? L? 01038? Processing... 4-6—8.txt Fi1e 4-6—8.txt up1oaded successfu11y. Disp1aying contents Line Line Line Line Line Line Line Line Line Line Line Line #0 : #1 : #2 : #3 : #4 #5 #7 #9 #10 : #11 : minimize 2 x1 + 1 x2 + 3 x3 subject to : 5 x1 + 2 x2 + 7 x3 = 420 : 3 x1 + 2 x2 + 5 x3 >= 280 #6 : : x1 >=0 #8 : x2 >=0 x3 >=0 4-6-8501 Running g1pso1GLPSOL: GLPK LP/MIP So1ver 4.38 g1p_read_1p: reading probiem data from /tmp/pnpx7mubd'... /tmp/phpx7mubd:12: warnin : keyword ‘end' m1551ng g1p_read_1p: 5 rows, 3 co umns, 9 non—zeros g1p_read_1p: 12 1ines were read Scaii A: m1 n Prob1em data seem to be we11 sca1ed Crashing... Size of trian u1ar part = 5 * 0: obj = 1. 00000000e+02 infeas * 1: 0b] = 1.750000000e+02 infeas OPTIMAL SOLUTION FOUND Time used: 0.0 secs Memory used: 0.0 Mb (30006 bytes) . . . Writing basic so1ution to /tmp/FOO]K1Vk1'... Prob1 Rows: em: Co1umns: Non—zeros: 5 3 9 Status;,wHS:§§¥EEKE}eI»~»»_ obj = 175 (WEE (fibjééfive: St .___...__ _ _;——-» ActiVity IIH 0.000e+00 (0) 0.000e+00 (0) Lower bound Lower bound n?éij| = 1.000e+00 maxlaijl = 7.000e+00 ratio 7.000e+00 Upper bound Upper bound Marginai ——____ _.._____..____ .._ ___..__—___-_.— .——.——________- __-___...._____ -——-_--___——— / 3/ \ V,/ CHD<D < eps 4—7—6 So1ution av Max 2 = 56X1 + 4 x2 — 1 X3 + 3 X4 subject to 1) 3 x1 + 2 x2 — 3 x3 + 1 x4 <= 24 2) 3 x1 + 3 x2 + 1 x3 + 3 x4 <= 36 and x1 >= 0, x2 >= 0, x3 >= 0, x4 >= 0. So1ve Interactive1y by the simp1ex Method: BasIEqI coefficient of I Right VarINoI ZI x1 X2 x3 x4 XS x6 I Side —I—I——I I z I OI 1| —5 —4 1 —3 O 0 I 0 X5] 1I OI 3* 2 —3 1 1 0 I 24 X6I 2| 0| 3 3 1 3 0 1 I 36 BasIEqI Coefficient of I Right VarINoi 2: x1 x2 x3 X4 X5 X6 I S1de _I_I“I I Z I OI 1I 0 —0.67 —4 —1.33 1.667 0 I 40 X1I 1| 0| 1 0.667 -1 0.333 0.333 0 I 8 X6] 2| 0| 0 1 4 2* —1 1 I 12 BasIEqI Coefficient of | Right VarINoi 2: x1 X2 X3 X4 x5 x6 I Side —|_l_l I z I 0| 1| 0 —0 —1.33 0 1 0.667 I 48 x1| 1I OI 1 0 5 -1.67 0 0.5 —0.17 I 6 X4] 2I OI 0 0 5 2* 1 -0.5 0.5 I 6 BaleqI Coefficient of I Right VarlNo: 2: X1 x2 X3 x4 X5 X6 I S1de —I—-l—| . | Z I OI 1| 0 0.333 0 0.667(0.667) $_ 1) I 52 XlI 1I OI 1 0.917 0 0.833 . . I 11 x3I 2I OI 0 0.25 1 0.5 -0.25 0.25 I 3 b) §Ixot§ow @095 of MDKV<25'W\D\\ \m\vt$ Resource I 1 0.961 (@90va La I Einfj unfk IACYQAQL. 6;;’ Ivng vegoUTCCS vJA\ \anEQSfi. W OVJAQCX‘VK Swne‘kon Io‘ We cothsPonafi~$ gIAmIow {NW6 Page 1 Q> Constraint #1 Constraint #2 Variables 3x1+2x2-3x3+1x4<=24 3x1+3x2+1x3+3x4<=36 x1 . 11 3 3 )Q 0 2 3 x3 3 -3 1 x4 0 Maximize:5x1+4x2-x3+3x4 MicrosofiExcel12.0SenslfivityReport b<\\OW0L\3\’L (twat gof CL WorksMet: [4-7-6.xlsx]4.7-6 Report Created: 2I16I2010 9:07:57 AM Adjustable Cells L L Final Reduced Objective Allowable Allowable Cell Name Value Com Coefficient Increase Decrease $863 x1 Variables 11 0 5 1E+30 11363636364 C\ $B$4 x2 Variables 0 6333333333 4 0333333333 1E+3O C 1. $855 x3Variables 3 O -1 2566666667 1333333333 C, .5 $B$6 x4 Variables 0 41666666667 3 0566666667 1E+30 C L, Constraints Final Shadow Constraint Allowable Allowable Cell Name Value Pn'oe R.H. Side Increase Decrease $057 3x1+2x2-3x3+1x4 <=24 24 0666666667 24' 12 132 ‘3‘ $F$7 3x1+3fl+1x3+3x4<=36 36 1 36 1E+30 12' b7, 3kmcxow Wk“? " Qegovfce/ 1'. 0.987 Resmwce 2. '. \ Allowable «an a: Max 5;“ + M, w“ a ><q (O‘oéecyflVb SN‘WSV‘WQ C‘ CL 0"; CL; .- (Dmflmolwa—K Maximize: 2 x1 + 1 x2 Subjectto: 0x1+1x2<=10 2x1+5x2<=60 1x1 +1x2<=18 3x1 +1x2<=44 x1 >=O x2 >=0 Defining Equation #1&#2 #1&#3 #1&#4 #1&#5 #1&#6 #2&#3 #2&#4 #2&#5 #2&#6 #3&#4 #3&#5 #3&#6 #4&#5 #4&#6 #5&#6 CP (5,10) (8,10) (1133,10) (0,10) DNE (10,8) (12.31,7.08) (0,12) (30,0) (13,5) (0,18) (13,0) (0.44) (1467.0) mm Addingslacks 0x1+1x2+1x3=10 2x1+5x2+1x4=60 1x1+1x2+1x5=18 3x1+1x2 +1x6=44 x1 =0 x2 =0 9 Feasibie Basic Solution Y (5.10,0.0,3,19) N (010,060.10) N (11.33,1o,o.—12.67,-3.33,0) y (0,10,0,10,8,34) u DNE y (10.8,2,0.0,6) M (12.31,7.08,2.92,0-136,0) .0 (0,12,-2,0,6,32) N (30,0,10.0,-12,—46) 7 (13,55,000) N (0,18,-8,-30,0,26) N (18,0,10.24.0,-10) N (0,44,-34,—160,—26,0) y (14.67,o,103067,3.33,0) )1 (0,0.10.60,18,44) Equation# #1 0x1+1x2=10 #2 2x1+5x2=60 #3 1x1+1x2=18 #4 3x1+1x2=44 #5 x1=0 #6 x2=0 Non-basic variables x3 x4 x3 x5 x3 x1 nla x4 x4 fi 66665161’356566 if) c 5151 II II H II CD 5: o 5.7.4 5—2—1 Mode1 maximize 8 x1 + 4 x2 + 6 x3 + 3 x4 + 9 x5 I subject to (Mew/r” / 1 1 2 2 3 3 3 4 0 5 = 180 4§113§212§311§411§5 2:270’Q280V’WZ' 1 x1 + 3 x2 + 0 x3 + 1 x4 + 3 x5 <= 180 \\ x1 >=o fl-(Lsouvce } xg >:8 , X >- wm wgffffwwufm/a // 4 \‘ $3 4 ’5 \ o u 45 \ \ l ‘ X| \' 2 q 1 : L -g, <3 J) 3 . 27 \, 0 K X5 L ' 3 1 ‘3 ‘0 ,/ IR ‘7} ¥ ‘- «M- \/ ““r/ X9) {5 B 180 A (X X0: = B) \b vv‘AUIL ‘12— t 2770 \g«: *3 \ “ '3 ‘ \Bw '60 2 7— Xl : 'Q ’73 O ’ > lg X6 21 c3 '2"! ,_ $0 7, ,3 \° HBO go 9V L6 8 51 5O 2— : (1% X g : &_C> 8 31 ’30 -; (5%0 36 fl; 9 9 <5 = D35 \ 2N] ’T ’v T ...
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