 # Homework 3 Solutions.pdf - ISyE 6669 Homework 3 Solution...

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ISyE 6669Homework 3 Solution Set4.10.4 (2nd edition) or 4.12.4 (4th edition)Adding excess and artificial variables we obtain,minz=3x1+Ma1+Ma2s.t.2x1+x2-e1+a1= 63x1+2x2+a2= 4Eliminatinga1anda2fromz-3x1-Ma1-Ma2= 0 yieldsz+ (5M-3)x1+ 3Mx2-Me1= 10M.The simplex now yieldsBVzx1x2e1a1a2RHSz15M-33M-M0010Ma1021-1106a20320014(1)BVzx1x2e1a1a2RHSz10(6-M)3-M0(3-5M)310M3+ 4a100-13-11-23103x10123001343(2)This is an optimal tableau.Note, however, that the artificial variablea1is positive(a1=103). Thus, the original problem has no feasible solution.4.17 (Review Problem)(a)-c0 andb0.(b)b0 andc= 0. Also needa2>0 and/ora3>0 to ensure that whenx1is pivoted in afeasible solution results. If onlya3is strictly positive, then we also needb >0.(c)-c <0,a20,a30 ensures thatx1can be made artitrarily large.1
6.2.2BV={x2, s1}, B="1110#, B-1="011-1#, cBV= cBVB-1= "011-1#= Coefficient ofx1in row 0 =cBVB-1a1-c1= "21#+ 1 = 2Coefficient ofs2in row 0 =cBVB-1"01#= 1RHS of row 0 =cBVB-1b= "42#= 2Column forx1=B-1a1="011-1# "21#="11#Column fors2=B-1"01#="1-1#RHS of optimal tableau =B-1b="011-1# "42#="22#Thus the optimal tableau isz+2x1+s2=2x1+x2+s2=2x1+s1-s2=22
6.3.6(a)x1is non-basic so changing the coefficient ofx1in the objective function will only changethe coefficient ofx1in the optimal row 0. Let the new coefficient ofx1in the objectivefunction be 3 + Δ. The new coefficient ofx1in the optimal row 0 will becBV B-1a1-(3-Δ) = 3

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candy bar, optimal tableau
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