3205 2016-Midterm-Solutions - Linear Programming and Integer Programming Mid-term Examination Solution 1(a True Hint Recall the definitions of polyhedra

3205 2016-Midterm-Solutions - Linear Programming and...

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Unformatted text preview: Linear Programming and Integer Programming Mid-term Examination Solution 1. (a) True. Hint: Recall the definitions of polyhedra set and convexity. (b) False. Hint: Let the obj equal 0. Then every BFS is optimal, and in general every BFS is clearly not adjacent. (c) False. Hint: Introduce two auxiliary nonnegative variables to reformulate the LP equivalently. (d) True. Hint: Let yi = |xi |, ∀i and a linear optimization problem in y follows. We can then form an optimal solution by taking x = y, but any x with xi = ±yi , ∀i will be a solution. (e) True. Hint: This means that improvements to this variable will increase the objective function at a nonzero constant rate and no upper bound is imposed on the variable. (f) True. Hint: There exists at least one feasible point (0, b). (g) False. Hint: Consider the case where the obj is a constant. 2. (a) True. According to certain Theorem. (b) False. Hint: The origin may not be a BFS solution, in which case Big M or two-phase methods will select the origin as an initial BFS, but this does not correspond to any BFS solution. (c) True. According to certain Theorem. (d) True. According to certain Theorem. (e) False. Hint: According to definition, λ should be in [0, 1] (f) False. 1 3. MATH 3205 / MATH 2230 MATH 3205 / MATH 2230 MATH 3205 /Programming MATH 2230 / OR Linear andand Integer I I Linear Integer Programming / OR Linear and Integer Programming / OR I Mid-term solution Mid-term solution Mid-term solution October 24,24, 2014 October 2014 October 24, 2014 1. 1. 1. (a)(a) 4. (a) max 900x + 1000x + 1200x max 900x 1 +1 1000x 2 +2 1200x 3, 3, xmax x21,x ,x32 ,x900x 1 ,x 3 1 + 1000x2 + 1200x3 , x1 ,x2 ,x3 s.t.s.t.2x12x +1 2x +2 x+3 x≤3 180, + 22x ≤ 180, s.t. 2x1 + 2x2 + x3 ≤ 180, x1 x +1 2x +2 3x + 22x + 33x≤3 120, ≤ 120, x1 + 2x2 + 3x3 ≤ 120, x1 x +1 x+2 x +2 2x + 32x≤3 160, ≤ 160, x1 + x2 + 2x3 ≤ 160, x1 ,xx12,,xx23, x≥3 0. ≥ 0. x1 , x2 , x3 ≥ 0. Z Z1 10 00 00 0 (b)(b) (a) (b) ) 1 ) (x2(x ) 2 ) (x3(x ) 3 ) x4 x4 x5 x5 x6 x6 RHS Z (x1(x RHS (x1 ) (x (x3 ) x40 x50 x60 RHS 2) 1 -900 -900-1000 -1000-1200 -1200 0 0 0 0 0 -900 -1000 0 2 2 -1200 1 1 01 1 00 0 00 0 180 0 2 2 180 21 22 13 10 01 00 180 0 1 2 3 0 1 0 120120 1 21 32 0 1 0 120 0 1 1 1 2 0 0 0 0 1 1 160160 1 1 2 0 0 1 160 R3 R3 R3 R , 3, R3 → → 3→ 3 R3 , 3− 2R + R → R , − 2R 3 3 +4 R4 →4 R4 , − 2R3 + R4 → R4 , −R +3 R R2 R , 2, −3 R +2 R→ 2 → − R3 + R2 → R2 , 1200R R1 R , 1, 1200R +1 R→ 3 +3 R 1 → (b) 1200R3 + R1 → R1 , Z Z1 10 00 00 0 ) 1 ) (x2(x ) 2 )x3 x3 x4 x4 (x5(x ) 5 )x6 x6 RHS Z (x1(x RHS (x1 ) (x x x (x5 ) x60 RHS 2) 1 -500 -500-200 -200 30 0 40 0 400 400 0 48000 48000 -500 -200 0 0 400 0 48000 0 5/35/3 4/34/3 0 0 1 1 -1/3 -1/3 0 0 140140 5/3 4/3 0 1 -1/3 0 140 0 1/31/3 2/32/3 1 1 0 0 1/31/3 0 0 40 40 1/3 2/3 1 0 1/3 0 40 0 1/31/3 -1/3 -1/3 0 0 0 0 -2/3 -2/3 1 1 80 80 1/3 -1/3 0 0 -2/3 1 80 3R3R 2 2 R2 R , 2, 3R2 → → 5 → 5 R2 , 5 R2 R 2 −R− R3 R , 3, +3 R→ 2 +R 3 → − 3 +3 R3 → R3 , 3R2 R 2 −R− R4 R , 4, +4 R→ 22 + R 4 → − 3 +3 R4 → R4 , 3 +R →R , 500R 500R 2 2 +1 R1 →1 R1 , 500R2 + R1 → R1 , 1 1 Z 1 0 0 0 (x1 ) -500 5/3 1/3 1/3 (x2 ) -200 4/3 2/3 -1/3 x3 0 0 1 0 x4 0 1 0 0 (x5 ) 400 -1/3 1/3 -2/3 x6 0 0 0 1 RHS 48000 140 40 80 3R2 → R2 , 5 R2 + R3 → R3 , − 3 R2 − + R4 → R4 , 3 500R2 + R1 → R1 , Z 1 0 0 0 x1 0 0 0 0 (x2 ) 200 4/5 2/5 -3/5 x3 (x4 ) 0 1 300 0 3/5 1 -1/5 0 -1/5 (x5 ) 300 -1/5 2/5 -3/5 x∗ = (84, 0, 12, 0, 0, 52), x∗ = (84, 0, 12, 0, 0, 52), 5. x6 0 0 0 1 RHS 90000 84 12 52 z ∗ = 900000. z ∗ = 900000. (c) (a) Optimal Solution: (x∗1 , x∗2 , x∗3 ) = (0, 60, 0)and Z ∗ = 120 1 0.8 0 0.6 −0.2 0 1 −0.2 0.4 0 B −1 A = . 0 0.4 Bas Eq Coefficient of Right ¯ ¯ 0 −0.6 0 −0.2 −0.6 1 Var No Z X1 X2 X3 X4 Side X5 X6 -5M -4M -8M ∵ B −1 A = [B −1 A B −1 ], Z 0 1 -3 -2 -4 M 0 0 -180M ∴ ¯ X5 1 0 2 1 3 0 60 0.6 0 −0.2 0 1 ¯6 X 2 0 3 3 5 -1 0 1 120 B −1 = −0.2 0.4 0 . Coefficient of Right Bas Eq −0.2 −0.6 1 ¯ ¯ Var No Z X1 X2 X3 X4 X5 X6 Side (d) M/3 - 4M/3 8M/3 - 20M  0  0 0 1 - 1/3 - 2/3 M + 4/3 + 80 Z C = 900 1000 1200 0 0 0 ,   X3 1 0 2/3 1/3 1 0 1/3 0 20 CB = 900 1200 0 , ¯6 X 2 0 -1/3 4/3 -1 -5/3 1 20 0 2 2 1 1 0 0 Bas Eq Coefficient of Right 1 2 3 0 1 0 A= , ¯ ¯ Var No Z X1 X2 X3 X4 X5 X6 Side 1 1 2 0 0 1 M M −0.2 +0 1/2 Z 0 1 -1/2 0 0 0.6 -1/2 + 1/2 90 X3 1 0 3/4 0 B −1 = 1 −0.2 1/4 3/4 -1/4 15 0.4 0 , X2 2 0 -1/4 1 0 -3/4 -5/4 3/4 15 −0.2 −0.6 1 −1 Bas Eq Right CCoefficient A − Cof≥ 0. BB ¯5 ¯6 Var No Z X1 X2 X3 X4 X X Side Let 900 + ε be the new price for the low grade papa, B remains optimal if M M Z X1 X2 i.e., 0 1 2 1 0 0 0 1 0 0 ε 12002/30]B −1 A-1/3 1 + 1/3 0 0 100 [900 + − [900 + ε+ 1000 1200 0] 0 4/3 1/3 1 -1/3 20 =[0 200 + 0.8ε 0 300 + 0.6ε 300 − 0.2ε 0] 1/3 -2/3 -1 2/3 20 ≥0, 1 3 200 + 0.8ε ≥ 0 ε ≥ −250 300 + 0.6ε ≥ 0 ε ≥ −500 ⇒ −250 ≤ ε ≤ 1500. ⇒ 300 − 0.2ε ≥ 0 ε ≤ 1500 Bas Var Eq No Z X1 X2 Z X4 X2 0 1 2 1 0 0 1 3 2 0 0 1 ¯6 X Right Side 0 1 0 ¯5 X M +2 3 1 M -1 0 120 60 60 (b) Optimal Solution: (x∗1 , x∗2 , x∗3 ) = (0, 60, 0) and Z ∗ = 120 Phase 1: Bas Eq Coefficient of Var No Z X1 X2 X3 X4 Z 0 1 -5 -4 -8 1 ¯ X5 1 0 2 1 3 0 ¯6 X 2 0 3 3 5 -1 ¯5 X 0 1 0 ¯6 X 0 0 1 Right Side -180 60 120 X2 -4/3 1/3 4/3 Coefficient of X3 X4 0 1 1 0 0 -1 ¯5 X 8/3 1/3 -5/3 ¯6 X 0 0 1 Right Side -20 20 20 X2 0 0 1 Coefficient of X3 X4 0 0 1 1/4 0 -3/4 ¯5 X 1 3/4 -5/4 ¯6 X 1 -1/4 3/4 Right Side 0 15 15 X2 -2 0 1 Coefficient of X3 X4 -4 0 1 1/4 0 -3/4 ¯5 X ¯6 X ¯5 X ¯6 X ¯5 X ¯6 X Bas Var Z X3 ¯6 X Eq No 0 1 2 Bas Var Z X3 X2 Eq No 0 1 2 Phase 2: Bas Eq Var No Z 0 X4 1 X2 2 Z 1 0 0 Z 1 0 0 Z 1 0 0 X1 1/3 2/3 -1/3 X1 0 3/4 -1/4 X1 -3 3/4 -1/4 Coefficient of X3 X4 2 4 3 Bas Var Z X3 X2 Eq No 0 1 2 Z 1 0 0 X1 -1/2 3/4 -1/4 X2 0 0 1 Coefficient of X3 X4 0 -1/2 1 1/4 0 -3/4 Bas Var Z X1 X2 Eq No 0 1 2 Z 1 0 0 X1 0 1 0 X2 0 0 1 Coefficient of X3 X4 2/3 -1/3 4/3 1/3 1/3 -2/3 4 Right Side 0 15 15 Right Side 90 15 15 Right Side 100 20 20 Bas Var Z X4 X2 Eq No 0 1 2 Z 1 0 0 X1 1 3 2 X2 0 0 1 Coefficient of X3 X4 2 0 4 1 3 0 ¯5 X ¯6 X Right Side 120 60 60 6. 4.6-17. Method 1: Reformulation: maximize ^ œ %B" &B# $B$ subject to B" B# #B$ B% B( œ #! "&B" 'B# &B$ B& œ &! B" $B# &B$ B' œ $! B" ß B# ß B$ ß B% ß B& ß B' ß B( ! Since this is the optimal tableau for Phase 1 and the artificial variable x¯7 = 5 ≥ 0, the problem Since thisis isinfeasible. the optimal tableau for Phase 1 and the artificial variable B( œ & !, the problem is infeasible. Method 2 ( Use Big M method): 5 Reformulation: max Z = 4x1 + 5x2 + 3x3 − M x¯7 s.t. x1 + x2 + 2x3 − x4 + x¯7 = 20 15x1 + 6x2 − 5x3 + x5 = 50 x1 + 3x2 + 5x3 + x6 = 30 x1 , x2 , x3 , x4 , x5 , x6 , x¯7 ≥ 0. Bas Eq Var No Z Z X5 X6 ¯7 X 0 1 2 3 1 0 0 0 Bas Eq Var No Z Z X5 X6 ¯7 X 0 1 2 3 1 0 0 0 Bas Eq Var No Z Z X5 X6 X3 0 1 2 3 1 0 0 0 X1 X2 -4 1 15 1 -5 1 6 3 X1 -M -4 1 15 1 X1 - 3M/5 - 17/5 3/5 16 1/5 X2 -M -5 1 6 3 X2 M/5 - 16/5 -1/5 9 3/5 Coefficient of X3 X4 -3 2 -5 5 0 -1 0 0 Coefficient of X3 X4 - 2M -3 M 2 -1 -5 0 5 0 Coefficient of X3 X4 0 0 0 1 M -1 0 0 6 Right Side X5 X6 ¯7 X 0 0 1 0 0 0 0 1 M 1 0 0 0 20 50 30 X5 X6 ¯7 X Right Side 0 0 1 0 0 0 0 1 0 1 0 0 -20M 20 50 30 X5 0 0 1 0 X6 2M/5 + 3/5 -2/5 1 1/5 Right ¯7 X Side 18 0 - 8M 1 8 0 80 0 6 (a) B& enters. (b) B% leaves. (c) Ð%ß #ß %ß !ß #ß !ß !Ñ 5.1-21. Bas Eq Var No Z Z X5 X1 X3 0 1 2 3 X1 1 0 0 0 0 0 1 0 Coefficient of X3 X4 X2 43M/80 - 103/80 -43/80 9/16 39/80 X5 X6 3M/80 7M/16 M + 17/80 + 13/16 -1 -3/80 -7/16 0 1/16 1/16 0 -1/80 3/16 0 0 0 1 Right Side 35 0 - 5M 1 5 0 5 0 5 ¯7 X Since this is the optimal tableau for Big M method, Z = 35 − 5M = −∞, and the artificial variable x¯7 = 5 ≥ 0, the problem is infeasible. 5.2-1. 7. Î B$ Ñ (a) Optimal Solution: B" œ F " , œ Ï B& Ò ^ œ -B œ a ) % ' (b) Shadow prices: -F F " œ - œ a& ) ( % ' ! " Iteration 0: F œ F " œ Œ ! -F œ a ! " ÑÎ ")! Ñ Î &! Ñ $ #(! œ $! "! ÒÏ ")! Ò Ï &! Ò Î "" $ ' * Ï # $ " Ñ Î "Þ$$ Ñ $ œ " Ò Ï "! #Þ'( Ò Î $! Ñ Ð ! Ó Ð Ó * bÐ &! Ó œ **! Ð Ó ! Ï &! Ò $ 5.2-2. Î "" $ ' * Ï # $ " #( " #( a ' *b ) ! b, E œ Œ # $ $ & $ % # # # % ! B " , BF œ Œ ' œ Œ " B( ! ! b, - œ a & " Revised B# coefficients: Œ ! ) ( % " ! ! #! ,,œŒ " $! ! #! #! œŒ Œ " $! $! ' ! ! b, so B# enters. ! $ $ œ Œ , so B( leaves. Œ " & & 5-10 7 ...
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