merged_document.pdf - IE534 Homework2 Niechen Chen Solution Convert the minimize problem to a maximize problem max \u2212 = \u221271 \u2212 42 33 4 s.t 21 32 3

merged_document.pdf - IE534 Homework2 Niechen Chen Solution...

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IE534 Homework2 Niechen Chen Solution: Convert the minimize problem to a maximize problem: max −𝜁 = −7𝑥 1 − 4𝑥 2 + 3𝑥 3 + 𝑥 4 s.t 2𝑥 1 + 3𝑥 2 + 𝑥 3 − 𝑥 4 ≤ 20 −4𝑥 1 − 𝑥 2 + 0 ∙ 𝑥 3 − 3𝑥 4 ≤ 30 3𝑥 1 − 4𝑥 2 + 2𝑥 3 + 𝑥 4 ≤ 8 −3𝑥 1 + 4𝑥 2 − 2𝑥 3 − 𝑥 4 ≤ −8 Replace original variable as following: 𝑥 1 = 17 + 𝑥 1 ≥ 0 Add new constraints: 𝑥 1 ≤ 12 𝑥 2 = 7 + 𝑥 2 ≥ 0 𝑥 3 = 12 − 𝑥 3 ≥ 0 𝑥 5 − 𝑥 6 = 𝑥 4 𝑥 5 , 𝑥 6 ≥ 0 So the problem becomes: max – 𝜁 = −7(𝑥 1 − 17) − 4(𝑥 2 − 7) + 3(12 − 𝑥 3 ) + 𝑥 5 − 𝑥 6 which is equilibrium to : max – 𝜁 − 183 = −7𝑥 1 − 4𝑥 2 − 3𝑥 3 + 𝑥 5 − 𝑥 6 s.t 2𝑥 1 + 3𝑥 2 − 𝑥 3 − 𝑥 5 + 𝑥 6 ≤ 63 −4𝑥 1 − 𝑥 2 + 0 ∙ 𝑥 3 − 3𝑥 5 + 3𝑥 6 ≤ −45 3𝑥 1 − 4𝑥 2 − 2𝑥 3 + 𝑥 5 − 𝑥 6 ≤ 7 −3𝑥 1 + 4𝑥 2 + 2𝑥 3 − 𝑥 5 + 𝑥 6 ≤ −7 𝑥 1 + 0 ∙ 𝑥 2 + 0 ∙ 𝑥 3 + 0 ∙ 𝑥 5 + 0 ∙ 𝑥 6 ≤ 12
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IE534 Homework2 Niechen Chen 𝑥 1 , 𝑥 2 , 𝑥 13 , 𝑥 5 , 𝑥 6 ≥ 0 So we have: c = [ −7 −4 −3 1 −1 ] , A = [ 2 3 −1 −1 1 −4 −1 0 −3 3 3 −4 −2 1 −1 −3 4 2 −1 1 1 0 0 0 0 ] , b = [ 63 −45 7 −7 12 ] . Solution: Matlab code: >> c = [7,4,-3,-1]' >> A = [2,3,1,-1;-4,-1,0,-3] >> b = [20,30] >> Aeq = [3,-4,2,1] >> Beq = [8] >> lx = [-17;-7;-Inf;-Inf] >> ux = [-5;Inf;12;Inf] >> [x, zeta, flag] = linprog(c,A, b, Aeq, beq, lx, ux) x = -17.0000 5.4573 12.0000 56.8292 zeta = -190.0000 , flag = 1 So the optimal solution is : 𝑥 1 = −17, 𝑥 2 = 5.4573, 𝑥 3 = 12, 𝑥 4 = 56.8292 , the optimal value is -190. Solution: Decision variables 𝑥 1 : amount of Saudi Arabia crude refined per day (in 10 3 barrels) 𝑥 2 : amount of Venezuela crude refined per day (in 10 3 barrels)
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IE534 Homework2 Niechen Chen Objective function Minimize daily cost (in 10 3 dollars): 𝜁 = 20𝑥 1 + 15𝑥 2 − 10 × (0.3𝑥 1 + 0.4𝑥 2 − 2) − 14 × (0.4𝑥 1 + 0.2𝑥 2 1.5) − 8 × (0.2𝑥 1 + 0.3𝑥 2 − 0.5) = 9.8𝑥 1 + 5.8𝑥 2 + 45 Constraints Gasoline requirement: 0.3𝑥 1 + 0.4𝑥 2 ≥ 2 Jet fuel requirement: 0.4𝑥 1 + 0.2𝑥 2 ≥ 1.5 Lubricants requirement: 0.2𝑥 1 + 0.3𝑥 2 ≥ 0.5 Saudi availability: 𝑥 1 ≤ 9 Venezuela availability: 𝑥 2 ≤ 6 Non-negativity: 𝑥 1 , 𝑥 2 ≥ 0 Gusek Code: var x1 >= 0,<=9; var x2 >= 0,<=6; minimize zeta: 9.8*x1+5.8*x2+45; subject to c1: 0.3*x1+0.4*x2>=2; c2: 0.4*x1+0.2*x2>=1.5; c3: 0.2*x1+0.3*x2>=0.5; end; Optimal solution: 𝑥 1 = 2, 𝑥 2 = 3.5 Objective value: zeta = 84.9 (Minimum)
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Anmar Arif ------- IE 534 HW # 3 Q.1 Param: x ij : exchange rates Variables FD: Final value in Dollars S ixj i= 1..4, j= 1..4 , i ≠ j S 12 : Amount of Dollars exchanged to Euros S 13 : Amount of Dollars exchanged to Pounds S 14
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