YORK UNIVERSITY Faculty of Science and Engineering Faculty of Liberal Arts and Professional Studies MATH 317 6. A Test #2 June 27, 212 Surname (print): Given Name: Student No: Signature: INSTRUCTIONS: 1. Please write your name, student number and final answers in ink. 2. This is a closed-book test, duration- 75 minutes. 3. No calculator is permitted. 4. There are seven questions on eight pages. Answer all the questions. 5. Show all work necessary to justify each answer you give. Clearly indicate each time you use the back of a page for your work. 6. Remain seated until we collect all the test papers. 7. Do the easiest questions first. GOOD LUCK! Question I Points I Scored I 1 4 2 1 3 7 4 14 5 12 6 7 7 6 Total: 6
Name: Student No: 1. (2 + 2 pts) Choose the one alternative that best completes the statement. (a) When two ratios in a simplex tableau tie for being smallest, i. there is a possibility for multiple optimum solutions. @a degenerate BFS will occur. iii. the problem has an unbounded solution. ANSWER: (t-'c.'), (b) In a simplex tableau that gives an optimum solution, a zero indicator for a nonbasic variable suggests i. the problem has an unbounded solution. ii. a degenerate BFS will occur. @there is a possibility for multiple optimum solutions. 2. (3 + 3 + 4 pts) Consider the following LP: ANSWER: (Jt-t_') max z = 2XI- X2 + X3, subject to 3XI + X2 + X3 < XI- X2 + 2X3 < XI+ X2- X3 < Xl, X2, X3 > 6 1 2 (a) Write the initial simplex tableau for the solution of the problem by the Simplex Algorithm. O.tftf Cc_ VtJ.A/,; " 1 2- CJM.J,& 3. fu rl_ ({ff-/j st- <- +, f:tc_e_ et/tj;fvcm.- > r-eckve-4j, fo (,- fl.. LP tla fo.,vcr IM -z. s.f. o2. X 1 - X z... -t- ")C 3. A::, -t- + )C3 "'\- " -=- " x.\- X2- -t 2 x '1 + 2- :X. + X - 'X -t- 1 r.l. So? cw:h.: iec;"i' M c z x, x2... JC3..4.!::>:L ":!. RttS' - { -1 1 1_ 1. () - v 1 1 - Continues... 1
Name: --------------- (b) After one iteration the simplex tableau for this problem is z XI X2 X3 S1 82 83 RHS 1-1 3 2 2 4-5 1-3 3 1-1 2 1 1 2-3 -1 1 1 Student No: Determine basic variables and write the basic feasible solution corresponding to this tableau. Ba._yt...'e.. \fakc: tv.. -\, A 3>., 1ke_ Cc-rre.'(1 OM..J Fe; i<; ANSWER:
3. (7 pts) The following LINGO program is given. Write the LP problem, which is solved using this program. MODEL: SETS: DAYS/1.. 5/:X,Y,DEMAND; ENDSETS MIN=@SUM(DAYS:3*X+15*Y); X(1)=4; @FOR( DAYS( I) I I # GT # 1:X(1)=.9*X(I-1)+ Y(l-1)- DEMAND( I); @FOR(DAYS(1):16*X- 5*Y 2 DEMAND; DATA: DEMAND= 4,6,7,75,5; ENDDATA END L-L "'z_ -=- (3 'l:l\ t- 5' ca- J '"{- (3aJ Xz_ + 15 ) t -\-... + ( 3oO 'X.s- + 15S') ==- 3 ( ')( + '):2-* ")( -\- "')(tot -t- :X: s-'yt- 5' ( 1 +- '--\- <cj 3 + '"( k 5" ') :J S.t. X" =- Lt, x2.-, "+-(fa- Goo, X -=-. 3 'X.l. + t;f-'-- 'foo J xt -=- o.ca 'X:> +- <i 3-5> -:x: =- o. g ')(_ '< -\- lf - s-oo-> G x,- 6''(1- )- 4 > G Xz.- s-'2-" G x3.- Su_ ::rw <f.) G x4-5u > r:rso <fl.f --- > 1\ G xs- - 5 s- s-oo/ '): x X )( X u u u "., '> 3.., 4> 'S)(J"''o,(f><a4>Cf?::o. Continues... 3
Name: Student No: 4. (1 + 4 pts) (a) Suppose that we have five different maximization LP's and assume that, in trying to solve them, after several iterations we have arrived to the following simplex tableaus: LP 1: LP 2: LP 3: LP 4: [ [ X! X2 X3 X4 rhs ] 7/4 1/8 55 z =55 1 1 1/2 28 X2 = 28 ' 1 1/4-1/8 9 X!= 9 X! X2 X3 X4 rhs ] -1 3/2 5 z =5 1 1/2 2 X!= 2 ' 1-1 1/2 1 X2 = 1 u X! X2 X3 X4 rhs 2 1 z = 1 1/2 1 1/2 5/2 X2 = 5/2 ' 1/2-1/2 1 3/2 X4 = 3/2 X! X2 X4 X3 a rhs [ LP 5: [ ] z=44m] 1+5M M 2+4M 6 2 1 1 2 X2 = 2 ' -5-1 -4 1 4 a=4 X! X2 X3 X4 rhs -3-9 17 z = bv 17 ] 1 4 1 8 X3 = 8 ' 1 2 1 X4 = where a is an artificial variable. Denote by AS = alternative optimal solutions, DEG = degeneracy, INFS = infeasible solution, UBS = unbounded solution, NONE = none of the above. Specify for each problem which one of the possible answers mentioned above is true? LP 1: NONE ; LP 2: f) B s ; LP 3: _._A--'-"S=<----_ LP 4: INJ=$ ; LP 5: Z> EG-- Note: No explanation is required. Evaluation: +2 for each correct answer, for each unanswered part, -1 for each incorrect answer - but zero is the lowest total score for this question. (b) For the unbounded LP from part (a), find a direction of unboundedness. L.P o1. is u"'e.of. LP. FrDiM 1tu_ ) x L e.,.kwi.a--t 1H.W'C:aifL, GM J X -1 =- JJJ C4t :t?.2.-.x 3 -=-1 'X" -=-.:lo $ - ): 2- =-{ + 3. ' Ust..e.-re... )C3 C 2A..<.._ {ku'" UMJ. /) O_ /J i..t.e-b-re.) a.. tft, re.d.c.. ef- 'j J -:::::: [ O 1 1 OJ Continues... 4
Name: Student No: 5. (1 + 2 pts) Consider the LP: max z = Xl +xz +2x3 subject to 2x 1 +xz +x3 < 2 3xl +4xz +2x3 > 8 x 1, Xz, x3 > (a) Solve the problem above by the Big-M Method using the largest coefficient rule. If there is a tie in choosing a leaving variable, priority should be given to an artificial variable. '"2.. "= XII ;- ')C 2-. t- c1?<:-.!:. - Mo-) s. f.. a '):-i -r -:x: 2- +- Jc 3. t- -==-..L X. A + 4 "l:2-2. X 3. - e 2. + C\._ -=:. _& (! ) S<J 'X-1 )')C2.) XJ..) h" ' '- 'a_ 'V.. ) # o+ f!:,\1 s. -::::. ( : Y Ut..r<. ft ::::- 2-) 1 # o-f. /VTJV G-2-= l(j UM.J t-c sv C\W J. CL. lo e.e...\u..c.: ki. f;.:cl'a.k: Va.-Yc.-'.fY 1<'2-' 'M <;jwctd v ';} Q-z_ M a cl- to "2- -==- XA + 'XL cl. ")( 3:. - M c.-_ + 3 t-j\'x" -+ L(f-.1\.x.L * 2.M X> - MeL M a-.8 M \IWVJC-.;_ -=- ( M,f) f'jc 11 +- ( 4 M -t- -i') -x.l- + ( \'-'1 -\- 2. 'yx: - M e.2.-.& M f{-u.-tu.) ch:a.r <;.:&x { t:s CD A.e.. 2- R+t S V RM-Lo M o -gm z. :::-&M -..t. ot &'\ = T =.t. -1 t1 <;:\A-u... Wwc.. va.v t' xl (JAA. tk.'vt.- & -=-& t-=cl (Q,) VM"f&_ i'<;..) ft. "lreji otaj- -& rl-- () ) u c;:{w<; ( )Ro, ) 1<-z + (4M-;-1) {:2-x.._ CJA<of Continues... A-() Px2.- to ocfcu f:gzcu..r 5
),, t<2 -+G_) Rx 3 : 2--l)l2-:tl; Student No: Name: (p ")(: )C..z...? :!:> -\ e..2... Q_ Rt-ts r3v Ra.b. o..\ _i._ -;z_ -- Lf MT1... ei.. Z-=-2 Lt '-{ s- Qi), t1.-1 ==- t:( y --l.f '1 -L( 'i " -., =- ' k... 2...L '):.2._'=c1- c1...:l -r::=4 s- --1 :; L _L Ci<- ) 2._ 2- - 3 L M- Z--=-j_, 2- s- J_...L _.L 'X?>=- ":f- J..., ol.l- _1._ t1 -i.l :C.z._-= -.L " L 2-.. S-t."'-c..L -s t's a,u r;p+- ( 1faie_ ') 1- / -xpf--=-1.) xpf-::::- o ) x ;rt--=-o 7_ wa;(-=- L. (b) How many optimal solutions does the above given LP have? Explain. 1L_ LP 6 9uL ot+ - 4-d. af- re.oj fo Jiff-e,r lo._ refah:4 to k'fterej- 'BFS. as o of: ) {(_. l.ni -<.:c. VCVV'. x ) 'X:z._ 1 ") -\ x2. 'x3.l.) -\. x2- ; ) V- 'X:2..' e 2..1 (k_ 'Cal-J. Re.caif tfu..l Je.se...e/e Oceu S eu=- 1L. /Vt 5: f.o CL B V h1j>peua fe f e_ e.-y"o' Continues... o+- c._ CLLJ- CPYY'e..C pow. "'9-6
Name: Student No: 6. (7 pts) Consider the LP: max z = 2xl + 3x2 subject to 2x1 + x2 < 1 (') Xl + X2 < 6 (..2.) -xl + X2 < 4 () X1,X2 > (Be very careful with your work because it will 2.. 1 6 Lj <:!. 1 (J l) 1\ h \ 1 PI / 1\ / \ la 1(.1,6 "" VJ rk 5'\ l/t o--,/),_ ;_\...------h' Vj c JJ. {L, ) """- lfo tt'\ \ f-';> 7 "\ 8 f IR..1" t i \ '""' 5 6 --- (b) Find the allowable range of change for the OFC of x2, for which the current optimal basis remains optimal. Justify your answer. J_ei z -:= ci..x., +X c...t..t r"ej-..st...'s ANSWER: [J. 1 J re.:.;:: -h.: t.<,-.e,ve,.r - 4: > C 'S.'\ 6:; C c2_ 7.v.tltv C. >O.. I Oo \. -- a vam-rc...- fp..r (.. s L..t, J (c) Find the allowable range of change for the RHS f the second constraint, for which the current optimal basis remains optimal. Justify your answer. 1U- -red- "-- {;-t.answer: [lf 1 g J t:.c (Le_ e. u_ :t-1 +- ')C:z._ =-6 H.<. a/ tv c:fr.e-e f. &} rc,k;f C(o 7 4) 9;( 7J()G). Sb/ ('2-) '('<;u:f ZJ R-11 -=-&;. (2-) fo C:. -='> fh1.s: =- L[. f-- Cont'"ues... e.>. /D fl. g1 7 t>..,t. t-s L..,) j.
Name: Student No: 7. (6 pts) Consider the LP: max z = 3xl + 4x2 subject to 2xl + x2 < 8 4xl + x2 < 1 X}, X2 2- In optimal tableau of the LP, the basic variables are x 1 and x 2. Use Important Formulas from Section 6.2 to find optimal simplex tableau and optimal solution to the problem. Note: No points will be given to any other answer._ - _ fl.+ (cn -1 N =-1 en'!_) )cno.v-::: e,v f:> ANSWER: l 'k 'I -\- JJ ')<:_N flv -==- B. 1 e... c v -== [ - ) )2.. " 3 Lt ],.g -=- [ " J T ) -=- \_ 4 1\ ) N-=- 1 2 c.nv [o o].., -" -==- r- 2- \ -.,.g -=- --- "z_ l c \ -:::. 1 \_ - J ">..t -A j J (; c r.,v - [-=- c.n.v ( - t;)-=- l L11 ( -=- :J..1-) c_ f!:n f!:,- ' N - c_ N (!, v = c I!> v - 1 -=- [ It J l } _\ J -t-i - J., u.z.. o- t-o...: A- trm ki"& {a._fku --z_ :x:" Jc. 2... x 3 x 4 R H s BY Ra.:h.' o t{ 3-5'.i +-- z -==J. +- - o -;, CD t "- - G ). )R R -+(s;_)r -xtt. o [). o -L 1. "'Z.'=:s' Jctr xl.- =- JS ( R xtr)