Cn the Phse I problem be unfesible or unbounded? -No Phse I: min 1X AX + IX = b with b 0 X 1, X 0 By mnipulting constrints nd dding/subtrcting slck/surplus vribles, we cn get b 0 A fesible solution with X = 0 nd X = b exists, so the Phse I problem is fesible. The Phse I problem is bounded below by 0, becuse we re minimizing bounded. m i= 1 X i nd X I 0, so the LP is
4.7 mx 5X 1-2X 2 + 2X 1 + 4X 2 + 6 2X 1 + X 2 + 3 2 X 1, X 2 0 unrestricted Let s = + - with +, - 0 Add slck vrible to 1 st constrint ubtrct surplus vrible from 2 nd constrint Add rtificil vrible to 2 nd constrint mx 5X 1-2X 2 + + - 2X 1 + 4X 2 + X + 3 - X 3 + = 6 2X 1 + X 2 + 3 X + 3-3 - X 5 + X 6 = 2 X 1, X 2, X + 3, X 3, X 4,, 0 Phse I, min Eliminte -> W - = 0 nd 2X 1 + X 2 + 3 + - 3 - + = 2 W Z + - W 1 0 2 1 3-3 0-1 0 2 Z 0 1 5-2 1-1 0 0 0 0 0 0 2 4 1-1 1 0 0 6 0 0 2 1 3-3 0-1 1 2 6 / 1 2 / 3 Let + enter, nd leves Optiml to Phse I W Z X 1 X 2 + - W 1 0 0 0 0 0 0 0-1 0 Z 0 1 13/3-7/3 0 0 0 1/3-1/3-2/3 0 0 4/3 11/3 0 0 1 1/3-1/3 5 1/3 0 0 2 1 3-3 0-1 1 2/3 +
Phse I is complete, eliminte nd continue. Let X 1 enter, nd + leves Z X 1 X 2 + - Z 1 0-4.5-6.5 6.5 0 2.5-15/3 0 0 3-2 2 1 1 4 X 1 0 1 ½ 3/2-3/2 0-1/2 1 Let - enter, nd leves Z X 1 X 2 + - Z 1 0-14 ¼ 0 0-3 ¼ -3/4-18 - 0 0 3/2-1 1 ½ ½ 2 X 1 0 1 2 ¾ 0 0 ¾ ¼ 4 Phse II is optiml! X 1 = 4, X 2 = 0, = + - = 2, = 0, = 0 Z = 18
4.9 mx -2X 1-2X 2 + + +MX 8 +MX 9 X 1 + 2X 2 + + + 2X 1 + X 2 + 3 2 - X 1, X 2 0 + X 8 + X 9 = 6 = 3 = 2 X 1, X 2,,,,,, X 8, X 9 0 Z + X 8 X 9 Z 1 2-2 -1-4 -M -M X 1 * = 1 * = 1 All else = 0 Add 2 nd nd 3 rd constrints (*-M) to zero out (-1-M) Z Z 1 2+3M -2+2M -1+2M -1+2M 0 -M -M 0 0 5M 0 1 2 1 1 1 0 0 0 0 2 0 1-1 1 2 0-1 0 1 0 3 0 2-1 1 0 0 0-1 0 1 2 X 8 X 9 X 8 X 9 2 / 1 3 / 1 2 / 2 Let X 1 enter, nd X 9 leves Z 0-1-1/2M -2-1/2M -1+2M 0 -M 1+1/2M 0-1-3/2M -2+2M 0 5/2 ½ 1 1 0 ½ 0-1/2 1 X 8 0-1/2 ½ 2 0-1 ½ 1-1/2 2 X 1 1-1/2 ½ 0 0 0-1/2 0 ½ 1 X 8 X 9 TIE Let enter, nd X 8 leves X 8 X 9 Z 0-5/4-7/4 0 0-1/2 5/4 ½-M -5/4-M -1 0 1 ¼ ¼ 0 1 ½ ¼ -1/2-1/4 0 0-1/4 ¼ 1 0-1/2 ¼ ½ -1/4 1 X 1 1-1/2 ½ 0 0 0-1/2 0 ½ 1 Let X 7 enter, nd X 5 leves X 8 X 9 Z 0-15 -3 0-5 -3 0 3-M -M -1 0 11 1 0 4 2 1-2 -1 0 0-3 0 1-1 -1 0 1 0 1 X 1 1 5 1 0 0 1 0-1 0 1 This is optiml with X 1 * = 1, * = 0 nd ll the rest = 0
4.20 uppose tht either Phse I is completed, or the bounded optiml solution of the Big-M problem is found. Further, suppose tht there exists t lest one rtificil t positive level indicting tht the originl system Ax = b nd x 0 hs no solution. Cn you differentite between the following tow cses:. The system Ax = b is inconsistent b. The system Ax = B is consistent but Ax = b implies tht x 0. Apply the following procedure: At the end of phse I, for n rtificil vrible x s, in the bsis, find nonzero y sk nd replce s by k. One of the following will occur: 1) ll rtificil vribles could hve been eliminted from the bsis by this procedure; the degenerte bsic solution thus obtined would no longer contin rtificil vribles, thus there is no redundncy nd r(a) = m nd cse (b) holds. 2) p rtificil vribles remin in the bsis nd for ech of these i we hve y ik = 0 fir every j. Therefore, ech of the vectors of A my be expressed s liner combintion of the m-p vectors of A which remin in the bsis. In this cse, r(a) = m-p => cse () holds. X K X Z 0 0 X y sk 1 0
4.25 min 3X 1-3X 2 + X 1 + 2X 2 - - + -3X 1 - X 2 + + X 1, X 2,,, X 5, Phse I: min W = 6 or W - 6 = 0 = 5 = 4 0 Need to zero this out, W - 6 = 0 + X 1 + 2X 2 - - + = 5 W Z W 1 0 1 2-1 -1 0 0 6 Z 0 1-3 3-1 0 0 0 0 0 0 1 2-1 -1 0 1 5 0 0-3 -1 1 0 1 0 4 Let X 2 enter, nd X 6 leves W 0 0 0 0 0-1 0 Z -9/2 0 ½ 3/2 0-3/2-15/2 X 2 ½ 1-1/2-1/2 0 ½ 5/2-5/2 0 ½ -1/2 1 ½ 13/2 Phse I is optiml, is out of bsis continue without X B = (X 2, ) is fesible to originl LP Phse II: W 0 Z 3/2 X 2-1/2-1/2 If enters, there is no leving vrible, so the Phse II (& originl LP) is unbounded.
4.25 (Continued) min 3X 1-3X 2 + +M X 1 + 2X 2 - - + -3X 1 - X 2 + + X 1, X 2,,, X 5, = 5 = 4 0 Z - 3X 1 + 3X 2 - - M = 0 M(X 1 + 2X 2 - - + = 5) Add Z row nd (-M) + 1 st row Let X 2 enter, nd leves Z Z 1-3+M 3+2M -1-M -M 0 0 5M 0 1 2-1 -1 0 1 5 0-3 -1 1 0 1 0 4 If enters, there is no leving vrible, so the Big-M LP is unbounded with ll rtificil vribles equl to zero, so the (Cse B2) originl LP is unbounded Z Z 1-9/2 0 ½ 3/2 0-3/2-M -3 X 2 0 ½ 1-1/2-1/2 0 ½ 5/2 0-5/2 0 ½ -1/2 1 ½ 13/2
4.38 Let X i = inventory of item A t end of month i Y i = inventory of item B t end of month i i = 1,2,3,4 Feb Mrch April My June A: 100 400 X 1 500 X 2 600 400 B: 150 600 Y 1 600 Y 2 700 Y 3 600 Y 4 150 1$ holding costs/month 0.8 holding costs/month ) min inventory cost = 1(X 1 + X 2 + + ) + 0.8 (Y 1 + Y 2 + Y 3 + Y 4 ) torge X i + Y i 250 for i = 1,2,3,4 Production t = 0 X 1 100 + 400 500 for A, Mrch Inventory t inventory t-1 + Demnd t 0 X 2 X 1 + 500 500 for A, April 0 production t 500 for A, nd 650 for B 0 X 2 + 600 500 for A, My 0 + 400 500 for A, June 0 Y 1 100 + 600 650 for B, Mrch 0 Y 2 Y 1 + 600 650 for B, April 0 Y 3 Y 2 + 700 650 for B, My 0 Y 4 Y 3 + 600 650 for B, June Ending inventory for B 150 Y 4 150 X i, Y i 0 i = 1,2,3,4 (nother wy to write production constrints: Inventory t-1 + production t - Demnd t = inventory t+1 ) b) optiml solution: totl inventory cost is $600 Month Production A Inventory A (Xi) Production B Inventory B (Yi) Mrch 400 100 550 100 April 500 100 650 150 My 500 0 650 100 June 400 0 650 150
4.38 continued c) In the current system, production for B is constrined by 650 (binding constrints) for months 2, 3 nd 4. o the new system is cost effective becuse it would llow more production nd consequently less inventory, in ddition the new system would reduce mnufcturing cost (from 8 to 6.50$ per unit) so it is definitely worth it. New optiml solution (with new system), totl inventory cost is $400 Month Production A Inventory A (Xi) Production B Inventory B (Yi) Mrch 400 100 450 0 April 500 100 650 50 My 500 0 700 50 June 400 0 700 150 Note: Unit mnufcturing cost is reduced by $1.50, so totl mnufcturing cost is reduce by $3750. d) uppose we use new system, nd llow demnd to be bcklogged, t cost of $1.00/month for unstisfied demnd for product B. new formultion: Let X i = inventory of item A t end of month i Y i = inventory of item B t end of month i U i = Unstisfied demnd for item B t end of month i i = 1,2,3,4 min inventory cost = 1(X 1 + X 2 + + ) + 0.8 (Y 1 + Y 2 + Y 3 + Y 4 ) + 1(U 1 + U 2 + U 3 + U 4 ) torge X i + Y i 250 for i = 1,2,3,4 Production t Product A = 0 X 1 100 + 400 500 for A, Mrch 0 X 2 X 1 + 500 500 for A, April 0 X 2 + 600 500 for A, My 0 + 400 500 for A, June Production t Product B ` 0 Y 1 150 + 600 U 1 700 for B, Mrch with unstisfied Demnd = 0 Y 2 Y 1 + 700 U 2 700 for B, April 0 Y 3 Y 2 + 700 U 3 700 for B, My 0 Y 4 Y 3 + 600 U 4 700 for B, June Ending inventory for B 150 Y 4 150 X i, Y i, U i 0 i = 1,2,3,4 Optiml solution Month Production A Inventory A (Xi) Production B Inventory B (Yi) Mrch 400 100 450 0 April 500 100 650 50 My 500 0 650 0 June 400 0 700 150
4.40 Let X i = production of product i during qurter j = 1,2,3,4 Y i = inventory of product i t end of qurter j = 1,2,3,4 i = 1 (refrigertor) i = 2 (stove) i = 3 (dishwsher) inventory j-1 + production j - demnd j = inventory j min 4 3 j= 1 i= 1 5Y ij (holding cost) (holding cost) Y i,j-1 + X i,j d i,j = Y i,j for i = 1,2,3 nd j = 1,2,3,4 Demnd must be met: let d ij = demnd s given in text for production i durtin qurter j lso durtin qurter j lso Y i,0 = 0, no inventory to strt 2X 1,j +4X 2,j +3,j 18000 j = 1,2,3,4 Cpcity vilble (hrs) during ech qurter X 14 = 0 Y ij 150 for i = 1,2,3 nd j = 1,2,3,4 Y i,j-1, X i,j 0 for i = 1,2,3 nd j = 1,2,3,4 no refrigertors produced in 4 th qurter inventory must exceed 150 for ech product & ech qurter Optiml olution: Totl inventory cost is $15,000 Production schedule is: Qurter Refrigertor tove Dishwsher 1 1650 1650 1150 2 1000 1500 2000 3 3200 1200 1000 4 0 1500 2500
4.41 Let X i = number of cuts ccording to pttern i, i = 1,2,3,4 Minimizing totl length cut (mybe should be given costs for different sheets) min X 1 + X 2 + + 2X 1 + X 2 + 2 + 5 2000 sheets of 2 x 4 X 1 + X 2 + 1000 sheets of 4 4X 1 + 4 3000 length vilble of 11 sheet 4X 2 + 4 2000 length vilble of 10 sheet X 1, X 2,, 0 Optiml Tbleu: X 8 Z 0 0 0 0 0.8 3.2 0 0.2 4400 0 0 0 1-0.2 0.2 0-0.05 100 X 2 0 1 0 0-1 -1 0-0.25 500 0 0 0 0 0.8 3.2 1 1.2 600 1 0 1 0 0 0 0 0.25 500 Notice pttern 1 & 3 re linerly dependent, so multiple optim (switch) Also, 10 sheet uses ll tht is vilble
4.42 olve using lexicogrphic nd Blnd s rule mx X 1 + 2X 2 + X 1 + 4X 2 +6 + = 4 -X 1 + X 2 + 4 + = 1 X 1 + 3X 2 + + = 6 X 1, X 2,, X 4, X 5, X 6 0 Lexicogrphic (tck on I for lexico-min rtio test) rhs Z -1-2 -1 0 0 0 0 1 4 6 1 0 0 4 1 0 0-1 1 4 0 1 0 1 0 1 0 1 3 1 0 0 1 6 0 0 1 Tie in 1 st digit 4 / 4 ¼ 0 0 1 / 1 0 ¼ 0 6 / 3 Let X 2 enter, nd leves rhs Z -3 0 7 0 2 0 2 5 0-10 1-4 0 0 1-4 0 X 2-1 1 4 0 1 0 1 0 1 0 4 0-11 0-3 1 3 0-3 1 Let X 1 enter, nd leves 0 / 5 3 / 4 lowest Let enter, nd X 2 leves Optiml solution! rhs Z 0 0 1 0.6-0.4 0 2 X 1 1 0-2 0.2-0.8 0 0 0.2-0.8 0 X 2 0 1 2 0.2 0.2 0 1 0.2 0.2 0 0 0-3 -0.8 0.2 1 3-0.8 0.2 1 rhs Z 0 2 5 1 0 0 4 X 1 1 4 6 1 0 0 4 1 0 0 0 5 10 1 1 0 5 1 1 0 0-1 -5-1 0 1 2-1 0 1
4.42 (Continued) olve using Blnd s rule mx X 1 + 2X 2 + X 1 + 4X 2 +6 + = 4 -X 1 + X 2 + 4 + = 1 X 1 + 3X 2 + + = 6 X 1, X 2,, X 4, X 5, X 6 0 rhs Z -1-2 -1 0 0 0 0 1 4 6 1 0 0 4-1 1 4 0 1 0 1 1 3 1 0 0 1 6 4 / 4 1 / 1 6 / 3 Let X 1 enter, nd X 4 leves rhs Z 0 2 5 1 0 0 4 1 4 6 1 0 0 4 0 5 10 1 1 0 5 0-1 -5-1 0 1 2 Optiml olution!
4.45 how tht cycling cn never occur even in the presence of degenercy provided tht unique minimum is obtined in the computtion b i min 1 i m yik : y ik > 0 where b 1 1 = B b, yk = B k nd x k is the entering vrible. For the strting BF, consider the rows of the mtrix: 0 I M 1 c B B b 1 B b Bsic Vr RH The rows 2,, m (red from right to left) re lexicopositive ( 0). Also, since the min rtio test give unique leving vrible, the pivot now is the one tht is lexicogrphiclly smller thn the other rows scled by dividing by y ik > 0. Hence the rows 2,, m of the bove mtrix remin lexicopositive (by the lexicogrphic rule proof) nd the first row is lexicogrphiclly decresing )for minimiztion problem), which implies the bsis cnnot repet. Hence no cycling cn occur.
4.49 uppose we hve n optiml extreme point of minimiztion LP. In the presence of degenercy, is it possible tht this extreme point corresponds to BF such tht z j - c j > 0 for t lest one non-bsic vrible? Yes see exmple below If this were the cse, re we gurnteed of nther BF corresponding to the sme extreme point where z j - c j 0 for ll non-bsic vribles? Yes - this cn be gurnteed by the convergence rgument of the lexicogrphic rule. X 2 = 0 mx X 1 + X 2 X 1 2 (+ X 3 = 2) X 2 2 (- X 4 = 2) X 1 + X 2 0 (- X 5 = 0) = 0 = 0 X 1 The point (2,2) is the only fesible solution, nd so n optiml extreme point. If nd re non-bsic z j - c j > 0 for both nonbsic vribles (<0 for mx) X 1 X 2 rhs Z 0 0 0-2 -1 4 X 1 1 0 0-1 -1 2 X 2 0 1 0-1 0 2 0 0 1 1 1 0 Let enter, nd leves OPTIMAL TABLEAU! X 1 X 2 rhs Z 0 0 2 0 1 4 X 1 1 0 1 0 0 2 X 2 0 1 1 0 1 2 0 0 1 1 1 0