CO-350-1045-Midterm_exam

CO-350-1045-Midterm_exam - CO 350 NAME M— Section_l I.D...

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Unformatted text preview: CO 350 NAME: M— Section #: _l___ I.D. #: fl.— CO 350 Linear Optimization Midterm Examination - Spring 2004 June 17, 2001, 7—9 p.m. Instructors: . Tuncel (87 ion 1), C. B. Chua (Section 2) Instructions: 1. There are 14 pages including this cover page. Make sure that you have a complete copy. Answer all questions on these pages. Use backs of pages if necessary. 2. Calculators are not permitted. 3. Some questions require you to prove or disprove certain statements. In doing so, you may refer to any result from the lecture notes except the particular one you are trying to prove or disprove. NW. QM CO 350 NAME: 1. [15 points] [5] (a) Convert the following problem to standard inequality form: maximize 10:61 — 5362 subject to $1 “l" 3552 _>_ “—8 2381 ~ 7x2 = 12 (1)1 Z O. (”Vl‘ “Y! “ 3%? S3 Z’Y) -—' V”§<7 3:? lZ/ XI 7/“ <9 (/5 wwx 104w~ ([urvz) 1 €“L= -’><, - gun-F302 S g “'" 7N/ + 67M? ~7V? 4:42 RI 26 w 7/ e V, 7/ c) [5] (b) Convert the following problem to standard equality form: minimize 93:1 — 611:2 subject to $1 — 4x2 2 —20 3331 + 5.162 S 1 931 a $2 2 0- d?) “A” ”3X: 14qu 5+ —?<J +tm2< g. 20. 3% {MS’XZ é} QCI/IX2 7/0 €557) 1/”.ch a» 3% ~é Saw “V ’k +1 I'm +k3 =20 / ,, 5 3’3‘! *‘S’X’z We, 1::( / flfj m2), IKE/kg >0 ' CO 350 NAME: 3 [5] (c) Prove algebraically that the following linear programming problem is unbounded. maximize z = 5x1 + x2 — _:v3 — 7504 + :65 subject to 131 + 2323 — 2374 + 305 = 1 *8 332 + 1133 — 9.734 -v 1135 = 4 X‘ 351 , SE2 , $3 a 334 , $5 2 0- W 132; 1,75 .3 a sow aa— A 3( WC: $Ww rsi 2 i239 "’ {72% 4 3M6 2 :3 ”>0 «Hog —-Z:x¢1 +¢<¢g a “Lt Q‘fl “é W3 ‘ (MK) Paglfiélfivfl Ecé’owg / 3((‘63 :: i Z 3 j; +tF , <6 20‘ ° 5 ' CO 350 NAME: 4 2. [20 points] 4 -1 3 5 10 A:= 6 5’ 1 ,b:= 10 ,c:= 20 2 8 8 2 30 describe the data for an LP problem in standard inequality form. [5] (3.) Introduce slack variables $4, :35, 3:6 2 0 and transform the given problem to an LP problem in standard equality form. (of) : m4 mix. 4' 7cm z +30%; 64’ H’X/ " *2, +383 ’6 ’X"! :— Q 6’X/ +59%; ‘6 X3 +’Y~® :, ,0 4< ’X a ' 3c I / />(\b/ ‘1 / 4<§ I 1/ 7/ a 1%. w; WK ‘ CO 350 NAME: 5 [15] (b) Using the simplex method with the smallest subscript rule, solve the LP problem you obtained in part (a). { 6 £1 ,. I W, Z q. , M. ... ;_ ” (4 (mrzhal {94%; {gas wafi 1 i; {i . {a LNfi/l Mbu "' l ._ Z ' (O’X( ” ZO’XZ *gofl m C) H W ’1') Z + 333 Pkt‘ ' I? 6% +"Xz Mg +5“? .1, (s 73‘! fg’kc ’l’ 8%; *"k6 '5’ 2 1/1/10" iww IVA”: g}. 5: (0 ~ ; M L, ( 3"“; ~33 CL. 3: :9 6 {em-95 W W, 93; E {I (“DCS *L'fl'é'fig 2 + 201(2 4 '1 Q’XZ ‘ ; (O . ‘ '“I7VZ. “RX; 1‘4“, , ~— I ’43 /‘C2. _ ”23X3 14,36? : (k’ + ”9C7 4' ”’Y'b ‘ @L b 4‘ *e ‘ ) tsetse/0.x) 13 3,09%“! Cj<0 (71¢ GM ' CO 350 NAME: V 6 3. [20 points] Let describe the data for an LP problem in standard equality form. [10] (a) Introduce one artificial variable x5 2 0 and write down an auxiliary LP problem whose kiwi QWK ‘ CO 350 NAME: 7 [10] (b) Using the simplex method With the largest coefficient rule, solve the auxiliary LP problem you obtained in part (a). If the original LP problem is feasible then use the final tableau from Phase—I to make a feasible simplex tableau for the original LP problem. M» “all base: (5:: {M53 l (1 £0 3:2) Y: /€0v°5 Ma) @0395“ 8-" (, L13 422505.: W QM CO 350 NAME: 8 4. [15 points] [9] (a) Consider the following tableau. Assume that the initial basis was B = {5,6,7}. with AB = I. For each nonbasic variable which is eligible to enter the basis, find the corresponding leaving variable according to the Lexicographic Rule; do not pivot. Show er. 64km whats: 4Q (67:00) “(s (323876)). “k6(22:3>©)‘ 1:0" W}: MMIMJANMN EZYJ’WN‘ (O'lf‘7’0) CO; ll l,0\, it“ 7 z, / ””17“" """) \/' . 2 @H, «A / 1 no/ 3’ WAV(S’ "”6“" ’§*<z,13’“”4“19 raw 5am» (0.4mm: W3 ., :@ l752) (gN/‘te/ E<T Z61. W) “3) WZZ léavfis (lie £35595, \/ . N [M 09/ 0i CO 350 NAME: [6] (b) For each of the following statements, determine whether it is true or false (and mark T or F) Proof is not required. Each wrong answer will be assigned (-1) (instead of zero); however, negative total scores for part (b) will be treated as zero. g ’ (i) Every optimal solution of a linear programming problem is an extreme point of its feasible region. l: ‘ (ii) A linear programming problem is unbounded if and only if its feasible region is un— bounded. T- (iii) The feasible region of a linear programming problem is convex. I: (iv) Every convex set has at least one extreme point. (-2 (v) Every subset of a convex set is convex. (fivi) The intersection of any two convex sets in R" is convex. a) Fake, , 6f) [:45er (In) (ls/“Vt, ~ Q0) Faéx , (u) 9% ~ « 6/1) 77%., ‘ (O / M“ - co 350 NAME: 10 5. [10 points] [3] (a) Define the terms convex set, extreme point and basis. x 5043 E A $44 SL3 CWWX (F: \/ . H’XJI’YZ, 65/ AéEHZ/ 38“: >\ ’30 +0.8 7‘; 6 S- W [£67513 Léék A6 film“) FWK<A>z M A¥§e+ E f$ WM 6‘ ‘Bamki «JVLA’ / ""2 9 B C §(/ - / l7} 4 ‘8’ 1-: M ’ AC 5 CG 6 “g A 5/1/4417 ”We/(3% 2 j ‘l $04»— ([48 INS WC L33 ‘\ ”Kama" ”6L A). (KO/gxbnfwél/ ; > A3 I: flWfiMJul¢ L3. Q%((0V"5 <€W LW 4/3673”). WK WM 11 [7] (b) Let F be the set of x satisfying 11:1 — 484984543 + $4 = 3 9/ 2x2 + 1048576x3 = 2 $1 7 $2 7 {E3 3 134 Z 0 (i) Find two extreme points of F. Justify your answer. “0 ~ OL x3 «m, J 3 (ML/g 0(: ‘1 fwsw Ft") ’Xz + Pkg :2 _ p: must—~35 \‘IJ'WZJ kg) 40.1 2/0 .; V (”09¢ E¢$Jfi; I3, I I: 3 (lab 3 V5 ‘ [44,92 «We» GWWWA 5:) E) .3 g j WWW. ' O J CLmJOK basis B< 2 r: E p, (/3 - r0 @ng ~§0U&‘7% defer/u M04 5) 52 13: ‘3 £3 Needle”, D) kg )3 an adrm, [wt/pt cm g/ 53 q diam» . V , \- k / a» we; CCNrés A (h) éérflw‘“ SQKJCS’ M 5 s C éa‘xiék S0 ems fc’rJ'NP/Tcy F‘v’v‘r’i“‘5 ,/ (ii) Find a point of F that is not an extreme point. Justify your answer. m 564* F: B Yfl‘eéozc/«lolfi Nafm ofi~ovi LP 4: g a WK 4+» (44W; LP 4444 W CM ‘qlC-wkafi,‘ WGM)’ 0 M33: Quzfigj é [ff/ ergl 6 7: o 7" 3/3 L I vhvk D [M GL0 (LC at€m$9fl fifl?’ afl @XM’V‘L. fbM/E, t . co 350 NAME: &1 §(6?( 0 8f 6L 12 6. [10 points] Prove that the set {3: : Ax = b, :1: 2 0} is unbounded if and only if there exists 0 such that the linear programming problem T maximize c cc subject to A7: = b a: _>_ 0 is unbounded. 2:25" gm gm. W, m W / (L‘ JU/Ménzl ’EZ/Ké g 3% :7/ng2/flfl. (099% (Ht, @‘oycA-Nt, {3mm Z ya, ' WZ (a _! _ Céé”"’”3‘ -; ”‘V‘: We" . - ”l “4621.3. IVY/~60 )9 WWW Ad“ ((03% m 655% 0d X, gx/VCQ'. ' “MM? "2‘ £3 6431/5 6 WM MQM‘CJ6CN”IYVF,t//6 (xx/INA (71M W lg \ M QW’l @ 7. [10 points] Suppose that in an LP problem we replace the unrestricted (free) variable :31 by ul — 121, where U1 and 111 are nonnegative variables, to create a new LP problem. Prove that if the point (ul,vl,$2, x3, . . . ,mn) is an extreme point of the feasible region of the new LP problem, then we cannot have both ul > 0 and 211 > 0. 69w MW‘X ’ 061“ Cowblravts A~~ wows} an “an al?"‘ 0* 35 A128 :2 b j ’X“ 20. @dea ZéwM/jjj? 383:0 Vjfifi. EA QEGAOQCIJFODW. all Na 3%)“; (”cw Fow+“\M 99:? ( CflW~3 I 2'2;- {niece A; 1'1? ”A2/ [fl/“FMS §QPV_0L9¢4-;’ I y , 653% ’x’. 2:»- u; ::~~ o or“ 192 3. V: a o ,3 we 5mm? Awe (56374 l (1,70 E U, >©A (gym UM) ”We" MD]: a (We die. bag '3 ( O/Uérmrlvg Eafi (“CE/WA$ M (MGM w’ 96M) - CO 350 NAME: 6% Qf’ék 14 8. (BONUS) [10 points] Suppose that the matrix [A l D] is m by n with rank m. Define [ cc y* ] to be a basic feasible solution of the system (Ax + Dy = b, y 2 0) if there exist u* and v* such that u* 56* = u* — v" and [ 21* ] is a basic feasible solution of the system y* [A l~—A][:]+Dy=b, [:]20,y20. Z Prove that if [ a: 11* ] is an extreme point of the set {[3}: Ax+Dy= b, we} then it is a basic feasible solution of the system( (Am + Dy: b, y >0 Slum {w 8 (w each—m Fcinir N 8': ET] Aégjg ...
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