Auxiliary variable for the balance to pay the

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:= Auxiliary variable for the balance to pay the Discovery in i th month($). C i := Auxiliary variable for the balance to pay the Chase Bank in i th month($). T i := Auxiliary variable for the balance to pay the T.J. Maxx in i th month($). Minimize 36 i =1 ( d i + c i + t i ) subject to D 37 = 0 (Pay-off Discovery) D 1 = 30000 (Initial Discovery balance) C 37 = 0 (Pay-off Chase Bank) C 1 = 20000 (Initial Chase Bank balance) T 37 = 0 (Pay-off T.J Maxx) T 1 = 10000 (Initial T.J Maxx balance) D i +1 = (1 + 0 . 02)( D i - d i ) (Discovery equation) where i ∈ { 1 , . . . , 36 } C i +1 = (1 + 0 . 05)( C i - c i ) (Chase Bank equation) where i ∈ { 1 , . . . , 36 } T i +1 = (1 + 0 . 03)( T i - t i ) (T.J. Maxx equation) where i ∈ { 1 , . . . , 36 } d i + c i + t i 3000 for i ∈ { 1 , . . . , 36 } (Monthly budget constraints) d i , c i , t i 0 for i ∈ { 1 , . . . , 36 } (Nonnegativity) 2
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Problem 2: Intermediate Products (15 pt) Decision variables x p := The amount of product p that is produced ( p ∈ { A, B, C } ). Linear Program (with auxiliary variables) With the following auxiliary variable: s p := The amount of product p that is available for sale ( p ∈ { A, B, C } ). Maximize 25 s A + 40 s B + 80 s C subject to x A + 2 x B + 5 x C 100 (Labor hours limit) s A = x A - 2 x B - x C (Product A) s B = x B - 3 x C (Product B) s C = x C (Product C) x A , x B , x C , s A , s B , s C 0 (Nonnegativity) Linear Program (without auxiliary variables) Maximize 25 x A - 10 x B - 65 x C subject to x A + 2 x B + 5 x C 100 (Labor hours limit) x A , x B , x C 0 (Nonnegativity) - Note) In this case, you need to explain how you get the objective function in detail.
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  • Fall '19
  • Optimization, DI, AirTrain Newark, Chase Bank, T.J. Maxx

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