2015-Midterm 1-Solutions - Linear Programming and Integer Programming MATH 3205 MATH 2230 MATH 3205 MATH 2230 I Linear and Integer Programming

2015-Midterm 1-Solutions - Linear Programming and Integer...

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Unformatted text preview: Linear Programming and Integer Programming MATH 3205 / MATH 2230 MATH 3205 / MATH 2230 I Linear and Integer Programming / Mid-termInteger ProgrammingOROR I 1 Examination MATH Linear and3205 / MATH 2230 / Mid-term solution Linear and Integer Programming / OR I Mid-term solution Mid-term solution Solution October 24, 2014 October 24, 2014 October 24, 2014 1. 1. 1. 1. (a) (a) (a)marks) (5 (a) (b) (b) (b)marks) (5 (b) 900x1 + 1000x2 + 1200x3 , max x1 ,x2max ,x3 900x1 + 1000x2 + 1200x3 , x1 ,x2 ,x900x + 1000x + 1200x , maxs.t. 32x1 1 2x2 + x2 ≤ 180, 3 + 3 x1 ,x2 ,x3 s.t. 2x1 + 2x2 + x3 ≤ 180, x 1 + 2x2 + x3 ≤ 120, s.t. 2x1 + 2x2 + 3x3≤ 180, x1 + 2x2 + 3x3 ≤ 120, + x 2 + 3x ≤ 120, xx1+ 2x2 + 2x3 ≤ 160, 1 x1 + x2 + 3 3 ≤ 160, 2x x1+x22 x3 2x3 ≤ 160, , x , + ≥ 0. x1 x1 , x2 , x3 ≥ 0. x1 , x2 , x3 ≥ 0. Z 1 Z 0 1 0 0 0 0 0 (x1 ) (x2 ) (x3 ) x4 x5 x6 RHS Z -900 1 )-1000 2 ) -1200 3 ) 0 x4 0 x5 0 x6 0RHS (x (x (x (x ) ) x 1 1 )-900(x2-1000(x3-1200 4 0x5 0x6 0RHS 0 2 2 1 1 0 0 180 -900 0 180 0 1 2 -1000 2 -1200 1 00 1 01 0 00 0 120 2 3 2 1 2 2 1 3 1 0 0 1 0 0 180 120 0 1 1 2 0 0 1 160 0 1 1 2 1 3 2 0 0 1 0 0 1 120 160 1 1 2 0 0 1 160 Z 1 Z 0 1 0 0 0 0 0 R3 → R R3 , 3 3 R3 3 → R3 , → R3 − 2R3 +, R4 → R , 3 − 2R3 + R4 →4 R4 , − R 3 + R4 → R4 , − 2R3 + R2 → R2 , − R3 + R2 → R2 , 1200R3 + → R R − R3 + R2 R1 →2 , 1 , 1200R3 + R1 → R1 , 1200R3 + R1 → R1 , (x1 ) (x2 ) x3 x4 (x5 ) x6 RHS Z -500 1 )-200 2 ) 0 x3 0 x4 400 5 ) 0 x6 48000 (x (x (x RHS (x1 )-500 x x 1 5/3 (x2 )-200 3 0x4 0(x5 )400 6 0 RHS 48000 4/3 0 1 -1/3 0 140 -5005/3 0 1/3 -2004/301 0 00 1400-1/300 048000 2/3 1/3 40140 05/3 1/34/3 2/300 1 10 0-1/31/301 0 140 40 1/3 -1/3 -2/3 80 01/3 1/32/3-1/31 0 0 01/3-2/30 1 40 80 1/3 -1/3 0 0 -2/3 1 80 3R2 → 3R R2 , 5 2 3R2 5 → R2 , → R2 R2 , 5 + − R R3 → R3 , 3 2 − R2 3 + R3 → R3 , − R2+ R3 → R3 , − 3 R2 R4 → R4 , + 3 − R2 3 + R4 → R4 , − 500R1+ R4 1→ R4 , , + R → R1 3 2 500R2 + R1 → R1 , 500R2 + R1 → R1 , 1 1 1 Z 1 0 0 0 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 , Zx1 x1(x2(x2 )x3 x3(x4(x4 )(x5(x5 )x6 x6 RHS RHS ) ) ) 1 0 Z0 200 (x2 )0 300 (x4 ) (x5 )0 90000 RHS 200 0 300 300 300 0 90000 x1 x3 x6 0 0 10 4/5 2000 3/5 1 -1/5 3000 84 8490000 -1/5 0 04/5 0 03/5 300 0 0 0 00 2/5 4/51 -1/5 3/52/5-1/5 12 12 84 -1/5 2/5 0 0 0 02/5 1 0 -3/5 0 -1/5-3/5 1 -3/5 0 0 00 -3/5 2/50 -1/5 -1/5 2/51 52 52 12 0 1 0 0 0 -3/5 0 -1/5 -3/5 1 52 x∗ = (84, 0, 12, 0, 0, 52), z ∗ = 900000. x∗ x∗ (84, 0, 12,12,0, 52), z ∗ = 900000. = = (84, 0, 0, 0, 0, 52), z ∗ = 900000. (c) (5 marks) (c) (c) (c) x∗ = (84, 0, 12, 0, 0, 52), z ∗ = 900000. −0.2 1 1 0.80.8 0 0 0.60.6 −0.2 0 0 1 0.8−0.2 0.4 0 . 0 0 B −1 A = B −1 A = 0 0 0.40.4 1 1 −0.20.6 0.4−0.2. 0 −1 −0.60.4−0.2 −0.2 1 1 0 . 1 −0.6 B A =−0.6 0 0 −0.2 −0.60.4 0 0 0 0 B −1 A = [B −1 B B −1 B −1 A = [B −1 A A −1 ], ], ∴ ∴ B −1 A = [B −1 A B −1 ], 0 −0.2 −0.6 −0.2 0.60.6 −0.2 0 0 0 B = −0.2 0.4 0 . = B −1 −1 −0.2 0.6 0.4−0.2. 0 −1 −0.2 1 1 0 B −0.2 −0.6 0.4 = −0.2 −0.6 ∴ (d)(d) (d) (5 marks) (d) −0.6 −0.2 −0.6 1 1 . 900 1000 1200 0 0 , C C = =900 1000 1200 0 0 0 0 , 1200 1000 , , 0 CBCB C900 9001200 0 1200 0 0 = = = 900 CB 2 2 900 1 0 0 , = 1 1 1200 2 2 1 0 0 0 A 2 2 2 1 1 , = 3 0 1 0 0 A = 1 1 2 3 0 1 , 0 1 2 0 0 1 0 , A =1 1 2 3 0 1 1 1 2 0 0 1 1 −0.2 0 00 1 1 −0.20 2 0.60.6 −1 −1 0 = −0.2 0.4 0 , B B = −0.2 0.6 0.4−0.2, 0 −1 −0.2 1 1 0 , B −0.2 −0.6 0.4 = −0.2 −0.6 0 ≥ CBCB BA − C ≥ −0.2 −0.6 B −1 −1 A − C 0. 0. , 1 −1 CB B A − C ≥ 0. LetLet 900ε+ ε be the new price for the low grade papa, remains optimal if if 900 + be the new price for the low grade papa, B B remains optimal Let 900 + ε be the new price for the low −1 grade papa, B remains optimal if [900 ε 1200 0]B A − [900 ε 1000 1200 0 0 [900 + ε+ 1200 0]B −1 A − [900 + ε+ 1000 1200 0 0 0] 0] =[0 2000.8ε 0 0]B 0.6ε − − 0.2ε =[0 200[900 + ε 012003000.6ε A −3000.2ε ε 0] 0] 1200 0 0 0] + + 0.8ε 300 + +−1 300[900 + 1000 ≥0, ≥0, =[0 200 + 0.8ε 0 300 + 0.6ε 300 − 0.2ε 0] ≥0, i.e., i.e., i.e., 200 + 0.8ε ≥ 0 200 + 0.8ε ≥ 0 ≥ −250 ε ε ≥ −250 2≥ 300 200 ≥ 0 0 ⇒ ε −500 −250 ≤ ≤ ε ≤ 1500. + 0.6ε ≥ 300 + 0.6ε + 0.8ε ≥ 0⇒ ε ≥ ε ≥ −250 −250 ε ≤ 1500. −500 ⇒ ⇒ 300 300 ≥ 0 − 0.2ε ≥ 0 1500 300 − 0.2ε + 0.6ε ≥ 0 ε⇒ε 1500≥ −500 ⇒ −250 ≤ ε ≤ 1500. ≤ ≤ε 300 − 0.2ε ≥ 0 ε ≤ 1500 B −1 1 1 2 0 0 1 0.6 −0.2 0 = −0.2 0.4 0 , −0.2 −0.6 1 CB B −1 A − C ≥ 0. Let 900 + ε be the new price for the low grade papa, B remains optimal if [900 + ε 1200 0]B −1 A − [900 + ε 1000 1200 0 0 0] =[0 200 + 0.8ε 0 300 + 0.6ε 300 − 0.2ε 0] ≥0, i.e., 200 + 0.8ε ≥ 0 ε ≥ −250 300 + 0.6ε ≥ 0 ε ≥ −500 ⇒ −250 ≤ ε ≤ 1500. ⇒ 300 − 0.2ε ≥ 0 ε ≤ 1500 2. (a) (8 marks) Optimal Solution: (x∗ , x∗ , x∗ ) = (0, 60, 0) and Z ∗ = 120 1 2 3 2 Bas Var Eq No Z Z ¯5 X ¯ X6 0 1 2 1 0 0 Bas Var Eq No Z Z X3 ¯ X6 0 1 2 Bas Var X1 -5M -3 2 3 X2 -4M -2 1 3 1 0 0 X1 M/3 - 1/3 2/3 -1/3 X2 - 4M/3 - 2/3 1/3 4/3 Eq No Z X1 X2 Z X3 X2 0 1 2 1 0 0 -1/2 3/4 -1/4 0 0 1 Bas Var Eq No Z X1 X2 Z X1 X2 0 1 2 1 0 0 0 1 0 0 0 1 Coefficient of X3 X4 -8M -4 M 3 0 5 -1 Coefficient of X3 X4 0 1 0 M 0 -1 Coefficient of X3 X4 0 1 0 -1/2 1/4 -3/4 Coefficient of X3 X4 2/3 4/3 1/3 -1/3 1/3 -2/3 3 ¯ X5 ¯ X6 Right Side 0 1 0 0 0 1 -180M 60 120 ¯ X5 8M/3 + 4/3 1/3 -5/3 ¯ X6 ¯ X5 M + 1/2 3/4 -5/4 ¯ X6 M + 1/2 -1/4 3/4 ¯ X5 M +1 1 -1 ¯ X6 M + 1/3 -1/3 2/3 0 0 1 Right Side - 20M + 80 20 20 Right Side 90 15 15 Right Side 100 20 20 Bas Var Eq No Z X1 X2 Z X4 X2 0 1 2 1 0 0 1 3 2 0 0 1 Coefficient of X3 X4 2 4 3 0 1 0 ¯ X5 M +2 3 1 (b) (8 marks) Optimal Solution: (x∗ , x∗ , x∗ ) = (0, 60, 0) and Z ∗ = 120 1 2 3 Phase 1: Bas Eq Coefficient of ¯ Var No Z X1 X2 X3 X4 X5 Z 0 1 -5 -4 -8 1 0 ¯5 X 1 0 2 1 3 0 1 ¯ X6 2 0 3 3 5 -1 0 Bas Var Z X3 ¯ X6 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 ¯ X6 Right Side M -1 0 120 60 60 ¯ X6 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 ¯ X5 8/3 1/3 -5/3 ¯ X6 0 0 1 Right Side -20 20 20 X2 0 0 1 Coefficient of X3 X4 0 0 1 1/4 0 -3/4 ¯ X5 1 3/4 -5/4 ¯ X6 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 ¯ X5 ¯ X6 ¯ X5 ¯ X6 ¯ X5 ¯ X6 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 ¯ X5 ¯ X6 Right Side 120 60 60 (c) (4 marks) The BF solutions of the two methods coincide. They are feasible only for the artificial ¯ ¯ problem until both artificial variables X5 and X6 are driven out of the basis, which in the two-phase method is the end of Phase 1. 3. 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 x7 = 5 ≥ 0, the ¯ problem is is the optimal tableau for Phase 1 and the artificial variable B( œ & !, the Since this infeasible. problem is infeasible. 5 Method 2 ( Use Big M method): Reformulation: max Z = 4x1 + 5x2 + 3x3 − M x7 ¯ s.t. x1 + x2 + 2x3 − x4 + x7 = 20 ¯ 15x1 + 6x2 − 5x3 + x5 = 50 x1 + 3x2 + 5x3 + x6 = 30 x1 , x2 , x3 , x4 , x5 , x6 , x7 ≥ 0. ¯ Bas Eq Var No Z Z X5 X6 ¯ X7 0 1 2 3 1 0 0 0 Bas Eq Var No Z Z X5 X6 ¯ X7 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 ¯ X7 0 0 1 0 0 0 0 1 M 1 0 0 0 20 50 30 X5 X6 ¯ X7 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 ¯ X7 Side 18 0 - 8M 1 8 0 80 0 6 5.1-20. (a) B& enters. 5.1-20. (b) B% leaves. (a) B& enters. (c) Ð%ß #ß %ß !ß #ß !ß !Ñ (b) B% Eq Bas leaves. 5.1-21. (c) #ß !ß VarÐ%ßNo %ß Z #ß !ß !Ñ 1 X 5.1-21. Z 0 X5 1 X1 2 X3 3 1 0 0 0 0 0 1 0 X2 43M/80 - 103/80 -43/80 9/16 39/80 Coefficient of X3 X4 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 ¯ X7 Since this is the optimal tableau for Big M method, Z = 35 − 5M = −∞, and the artificial variable x7 = 5 ≥ 0, the problem is infeasible. ¯ 4. (a) (12 marks) 5.2-1. Î B$ Ñ Î "" $ " B" œ F " , œ #( ' * ÏB Ò Ï # $ Î B$&Ñ Î "" $ " (a) Optimal Solution: B" œ F " , œ #( ' * Î $! Ñ Ï # $ Ï B& Ò Ð ! Ó Ð Ó ^ œ -B œ a ) % ' $ * bÐ $! Ñ œ **! Î &! Ó Ð Ó ! Ð ! Ó Ð &!Ó Ï Ò œ **! ^ œ -B œ a ) % ' $ * bÐ &! Ó Ð Ó ! Ï &! Ò Î "" $ (b) (8 marks) " (b) Shadow prices: -F F " œ #( a ' ) * b ' * Ï # $ Î "" $ " (b) Shadow prices: -F F " œ #( a ' ) * b ' * 5.2-2. Ï # $ # $ $ # # -œa 5.2-2. & ) ( % ' ! ! b, E œ Œ $ & % # % # $ $ # # - œ a & 0: F œ % " œ ! "! b,!E , B Œ œ B' œ " œ Iteration ) ( F ' Œ & F $ ŒB % #Œ% ! " ! 5.2-1. (a) Optimal Solution: ( " ÑÎ ")! Ñ Î &! Ñ $ #(! œ $! ÒÏ")! Ò Ï&! Ò "! ÑÎ ")!Ñ Î &!Ñ " $ #(! œ $! ÒÏ ")! Ò Ï &! Ò "! " Ñ Î "Þ$$ Ñ $ œ " Ò Ï"Þ$$ Ò "! Ñ Î #Þ'(Ñ " $ œ " "! Ò Ï #Þ'( Ò " ! #! , , œ Œ $! ! " " ! #! #! ! #! ,,œ ! Œ œŒ $! " $! Œ $! " " ! B " ! ! #! b so B #! Iteration 0:-F œ a !" ! b,Œ œ a &F œ Œ ( œ Œ ' Œ !, œ Œ # enters. F œ - , B ) ' % F ! " B( ! " $! $! " ! $ $ Revised ( % & ! ( leaves. -F œ a ! B#!coefficients: Œ !)" Œ & œ Œ' , so B! b, so B# enters. b, - œ a & " ! $ $ Revised B# coefficients: Œ œ , so B( leaves. ! " Œ & Œ & 5-10 5-10 7 (b) Defining equations: #B" #B# $B$ œ &ß B" B# B$ œ $ß B" œ ! 5.3-2. 5. (a)(14 marks) (b)(6 marks) " " (a) F " œ Œ " # Final constraint columns for ÐB" ß B# ß B$ ß B% Ñ: F " E œ Œ -F œ a $ " " % Œ $ " # #b # " " # " " œ Œ# " " ! $ " " ! Final objective coefficients for ÐB" ß B# ß B$ ß B% Ñ: -F F " E - œ a $ # bŒ " # " ! " ! a% $ " Right-hand side: F " , œ Œ " " & " Œ % œ Œ $ and ^ œ a $ " # $ " #b œ a$ ! # !b " # bŒ œ * $ Final tableau: (b) Defining equations: %B" #B# B$ B% œ &ß $B" B# #B$ B% œ %ß B" œ !ß B$ œ ! 5-14 8 ...
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