lec2 - IE521 Advanced Optimization Lecture 2 Dr Zeliha Akca...

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IE521 Advanced Optimization Lecture 2 Dr. Zeliha Akc ¸a October 2011 1 / 48
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Previous Lecture: Recall I System? Modeling? Model types? I Mathematical programming? min veya max f ( x 1 , x 2 , ..., x n ) s.t. g i ( x 1 , x 2 , ..., x n ) b i i S ( x 1 , x 2 , ..., x n ) X Decision Variables Objective Function Constraints I Solution? Feasible Solution? I Objective function value? Optimal solution? 1. Unique optimal 2. Multiple optimals: Limited in number, infinite number 3. None of the feasible solutions is optimal 4. Feasible solution set is empty 2 / 48
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Production Portfolio Problem: Recall I An oil company buys crude oil from 2 wells: Saudi Arabia, Venezuela I The company produces 3 different products using crude oil: - Jet fuel, - gasoline, - lubricant. I 1 barrel crude oil from each well yields the following daily amount of products. Also, minimum daily production amount of each product are given as follows gasoline Jet fuel lubricant Saudi Arabia 0.3 barrels 0.4 barrels 0.2 barrels Venezuela 0.4 barrels 0.2 barrels 0.3 barrels Max production 2000 barrels 1500 barrels 500 barrels 3 / 48
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Production Portfolio Problem: Recall I Maximum amount of crude oil that can be purchased from each well and unit cost: Maximum supply (daily) cost Saudi Arabia 9000 barrels $ 20 / barrels Venezuela 6000 barrels $15 / barrels I Objective: How can we produce the daily required amount of products with minimum cost using which amounts from each well? 4 / 48
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Production Portfolio Model I Decision variables x = barrels of crude oil purchased from Saudi Arabia well y = barrels of crude oil purchased from Venezuela well Min 20 x + 15 y s.t. gasoline demand 0 . 3 x + 0 . 4 y 2000 Jet fuel demand 0 . 4 x + 0 . 2 y 1500 Lubricant demand 0 . 2 x + 0 . 3 y 500 Max supply x 9000 Max supply y 6000 x 0 y 0 5 / 48
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Solution of Production Portfolio Model I Optimal solution x = 2000 , y = 3500 I Objective function value = 92500 I Constraints gasoline demand 0 . 3 x + 0 . 4 y = 2000 Jet fuel demand 0 . 4 x + 0 . 2 y = 1500 Lubricant demand 0 . 2 x + 0 . 3 y = 1450 Max supply x = 2000 9000 Max supply y = 3500 6000 6 / 48
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Graphical Solution minimize - x 1 - x 2 s.t. x 1 + 2 x 2 3 2 x 1 + x 2 3 x 1 0 x 2 0 1.5 3 1.5 3 x1+2x2 =3 x2 x1 2x1+x2 = 3 -x1-x2 = z =0 -x1-x2 = z = -1 -x1-x2 = z = -2 (1,1) 7 / 48
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Matrix and Vector Notation: Recall I An mxn matrix, includes n columns and m rows . It is an array consisting m*n numbers. A = a 11 a 12 ... a 1 n a 21 a 22 ... a 2 n ... ... ... ... a m 1 a m 2 ... a mn I A column vector is a matrix with n = 1 . I j th column vector: A j = a 1 j a 2 j ... a mj I A row vector is a matrix with m = 1 . I j th row vector: a > j = [ a j 1 a j 2 ... a jn ] a j = a j 1 a j 2 ... a jn 8 / 48
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Matrix Notation: Recall A R mxn A = [ A 1 A 2 ... A n ] = a > 1 a > 2 ... a > n 9 / 48
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Notation I Using the unit vector: Ax = A n X i = 1 e i x i = n X i = 1 Ae i x i = n X i = 1 A i x i I Therefore, Ax = a > 1 x a > 2 x ...
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