assign9.sol

# assign9.sol - Solutions 1 Exercise 11.4.1 from the course...

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Unformatted text preview: Solutions 1. Exercise 11.4.1 from the course notes. Solution: (a) The ﬁrst row is 5 1 5 l Z+<1+ﬁ)\$4+<0+1—2‘)\$5+(0+E)\$6=14+Z the cut is Emiri +3 >1 124 12“”5 12x5—'4' The second row is \$+o+11 + 1+11 + 1+59 *1 3 3 60 \$4 12 \$5 60 m6" +4 with cut 11 + 11 . + 59 > 3 609‘"4 12“5 60336 — 4 The third row is 1. 1 9 1 1122+(0+I6)€E4+(0+§)\$5+(—1+ﬁ)\$5—1+§ with cut The fourth row is l 5 l 1 m1+<0+ﬁ)m4+(0+ﬁ)m5+<O+ﬁ)16_1+1 with cut 1 + 5 + 1 >1 12CB4 12m5 12\$6 ‘ 4 (b) For the ﬁrst out we have 5 1 5 + ,_ 1 12“ 12"”5 12‘36 3— 4’ the leaving variable is 3. Compute mm 1712 112 512 _1 1 125’121’125 "1’ so 1:5 (or \$5) enters. For the second out we have ~—1—1—x———llm—§—9-os+s—-—§ 604 12° 606 ‘ 4 the leaving variable is 3. Compute ~ 11% i1? 3% i mm 1211’1211’1259 “11’ the entering variable is :35. For the third out we have 1_ 1‘ 9~+ _1 10954 215 10%6 8— 2 the leaving variable is 3. Compute .1710 12 510 _1 mm 121’121’129 _6’ the entering variable is \$5. For the last out we have 1 5 1 1 3954—5965 ‘ 123966”— “:1 1712112 512 _1 (a? E? a?) — 3’ again the entering variable is 935. (c) From the initail ILP we have m4 = 10—m1 ~5m3, m5 = 1—371 *332-1-(173 and m6 = —6m1+5m2, substituting into the cuts we get i.-3a:1— 2.732 + 2233 S 4, ii. 7301 — 4mg 3 2, iii. 6931 ~ 4932 S 1 and iv: \$1 5 1. 2. Exercise 11.4.2 from the course notes. Solution: The ﬁrst tableau for the relaxed LP is: z + x1 — 32:2 ' : 0 331 — (1:2 + \$3 = 2 2:31 + 4mg I + .734 = 15 Entering variable: x2 Leaving variable: m4 Pivoting gives the optimal tableau: Z + (5/2)\$1 -- (3/4)m4 = 45/4 (3/2):I:1 + :03 —— (1/4)a;4 = 23/4 (1/2):E1 + 1122 -- (ll/40374 = 15/4 Gomory cuts from either constraint are identical (1/2cc1 + 1/4234 2 3/4). Adding this cut to the tableau gives: » Z + (5/2)IL‘1 + (3/4)\$4 = 45/4 (3/2)x1 + \$3 + . (1/4)a:4 = 23/4 (1/2)x1 + x2 + (1/4)m4 = 15/4 —- (1/2)m1 — (1/4):l?4 + C135 = —3/4 Applying the dual simplex with leaving variable m5 and entering variable 5134 gives: Z ‘l' 1'1 —— 35135 = 9 £31 + {£3 -- 935 = 5 \$2 -- (135 = 3 2.1)] + (134 — 45E5 = 3 An optimal solution to the original problem is [0, 3]T. 3. Exercise 11.4.3 from the course notes. Solution: (80 (b) The second equation of the tableau is 1“ 1' _3 \$2+§L3~§l4-§. We split each of the coefﬁcients for \$3 and x4 and the right—hand-side value into their integer part and fractional part. 932 + \$013 + (—1 + \$134 = (1+ é). We further reorganize this equation to get: 1 1 1 §\$3+5\$4—§ =1"\$2+\$4. Since we want all the variables to be integers, 1 — 272 + x4 must be intege1. Thus, 2:33 + %m4 — — must be integer as well. We also need 1all the variables to be nonnegative, thus1 m—3—I—2 —ac4__ > 0 which implies2 :v—3—l—2 l:14 — — 1> ——.Since1—x3+%m4 ~ 1 1s an integer if it is larger than — ,it must be greater2 than 01 equal to 0 Thus the resulting Gomory cut is: 1 1 l 5:123 + 511:4 2 E. In the basic feasible solution for B = {1,2}, 333 and 1'4 are 0, thus the current solution violates this Gomory cut. The constraint %m3+%m4—% 2 —% is satisﬁed by all nonnegative solutions but we also know that the left— hand— side must be integer (from above). So integer solutions that satisfy 21r3+2 —a"4 — — 12 —— will satisfy 2m3+§1m4 — — > 0 although fractional solutions may not Add slack variable \$5 to the Gomory cut and make \$5 a basic variable, i.e., we add the following equation to our tableau: 1 1 1 "“5133 -‘ 5334 + \$5 : "E. The augmented tableau looks as follows: 2 + \$3 + —§ 3:4 2 12 m1 -- £133 + x4 = 2 \$2 —- \$123 — i334 = g - 59103 — 5564 + 335 = —§ This new tableau corresponds to a basic primal infeasible solution and a dual feasible basis. Dual simplex method will choose x5 as the leaving variable and :L4 as the entering variable since 3334:— — 111in{_—‘1——/2,—_-T/L§}. Pivoting 011 T4 and 235 gives us the following tableau: z + —%:C3 + 335 = 3 {1:1 1 + 2.15 = 1 £172 + \$3 — {135 = 2 {113 + \$4 - 21125 = 1 We have a primal feasible and dual feasible basis, so B = {1,2,4} is optimal for the new LP relaxation. Since the corresponding basic feasible solution, m* =' (1,2,0,1)T, 1s integral, (5* is also the optimal solution for the original ILP. ...
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