2015-Midterm 2-Solutions - Linear Programming and Integer Programming 6.2-1 Mid-term Examination 2(a Iteration 0 Since all coefficients are zero at the

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Unformatted text preview: Linear Programming and Integer Programming 6.2-1. Mid-term Examination 2 (a) Iteration 0: Since all coefficients are zero, at the current solution Ð!ß !Ñ, the three Solution resources (production time per week at plant 1, 2 and 3) are free goods. This means increasing them does not improve the objective value. Iteration 1: Ð!ß &Î#ß !Ñ. Now resource 2 has been entirely used up and contributes &Î# to profit per unit of resource. Since this is positive, it is worthwhile to continue fully 6.1-5. using this resource. 1. (a) Iteration 2: Ð!ß \$Î#ß "Ñ. Resources 2 and 3 are used up and contribute a positive minimize "&C" #&C# amount to profit.to ResourceC1 is a free good while resources 2 and 3 contribute \$Î# and " subject & " per unit of resource respectively. C % # (b) Iteration 0: Ð\$ß &Ñ. #Cactivities 1 and 2 (number of batches of product 1 and C" Both # \$ 2 produced per week) can ß be initiated ! give a more profitable allocation of the C" C# to resources. The current contribution of the resources required to produce one batch of ‡ ‡ (b) Optimal Solution: ÐC" ß C# Ñ œ Ð&ß %Ñ, so shadow prices for resources 1 and 2 are & and product 1 or 2 to the profit is smaller than the unit profit per batch of product 1 or 2 % respectively. respectively. Iteration 1: Ð\$ß !Ñ. Again activity 1 can be initiated to give a more profitable use of resources, but activity 2 is already being produced (or the resources are being used just as well in other activities). Iteration 2: Ð!ß !Ñ. Both activities are being produced (or the resources are being used just as profitably elsewhere). (c) Iteration 1: Since activities 1 and 2 can be initiated to increase the profit (give the same amount of resources), we choose to increase one of these. We choose activity 2 as the entering activity (basic variable), since it increases the profit by & for every unit of product 2 produced (as opposed to \$ for product 1). Iteration 2: Only activity 1 can be initiated for more profit, so we do so. Iteration 3: Both activity 1 and 2 are being used. Furthermore, since the coefficients for B\$ , B% and B& are nonnegative, it is not worthwhile to cut back on the use of(c) of the resources. Thus, we must be at the optimal solution. any 2. 6.3-1. (a) (a) minimize subject to [ œ #!C" "!C# &C" C# ' #C" #C# ) C" ß C# ! 1 (b) Primal: (b)(b) Primal: ÐB" ß B# Ñ œ Ð&Î#ß "&Î%Ñ is optimal with ^ œ %&. Infeasible corner point solutions are Ð!ß "!Ñ and Ð"!ß !Ñ. (x , x2 ) = (5/2, 15/4) is optimal with Z = 45. Infeasible corner point solutions are (0, 10)and(10, 0). ÐB" ß B# Ñ 1 Ð&Î#ß "&Î%Ñ is optimal with ^ œ %&. Infeasible corner point solutions are œ Dual: Ð!ß "!Ñ and Ð"!ß !Ñ. Dual: ÐC" ß C#(y1œy2 ) = (1/2, 7/2) optimal with [ = 45. .Infeasible corner point solutions are (0, 4), (0, 0)and(6/5, 0). Ñ , Ð"Î#ß (Î#Ñ is is optimal with W œ %& Infeasible corner point solutions are Ð!ß %Ñ, Ð!ß !Ñ and Ð'Î&ß !Ñ. ÐC" ß C# Ñ œ Ð"Î#ß (Î#Ñ is optimal with [ œ %&. Infeasible corner point solutions are Ð!ß %Ñ, Ð!ß !Ñ and Ð'Î&ß !Ñ. 6-9 6-9 2 (c)(c) Primal BS Ð!ß &ß "!ß !Ñ Ð!ß !ß #!ß "!Ñ Ð%ß !ß !ß 'Ñ Ð&Î#ß "&Î%ß !ß !Ñ Ð!ß "!ß !ß "!Ñ Ð"!ß !ß \$!ß !Ñ Feasible? Yes Yes Yes Yes No No ^ %! ! #% %& )! '! Dual BS Ð!ß %ß #ß !Ñ Ð!ß !ß 'ß )Ñ Ð'Î&ß !ß !ß #)Î&Ñ Ð"Î#ß (Î#ß !ß !Ñ Ð%ß !ß "%ß !Ñ Ð!ß 'ß !ß %Ñ (d) (d) Primal: Ð!ß !ß #!ß "!Ñ 10) Primal:(0, 0, 20, Dual: Dual:(0,'ß )Ñ Ð!ß !ß 0, −6, −8) Primal: Ð!ß &ß "!ß !Ñ 0) Primal:(0, 5, 10, Dual: Dual:(0,#ß !Ñ0) Ð!ß %ß 4, −2, Primal:(5/2, 15/4, 0, Primal: Ð&Î#ß "&Î%ß !ß !Ñ 0) Dual:(1/2, 7/2, 0, 0) Dual: Ð"Î#ß (Î#ß !ß !Ñ 6.3-2. (a) minimize [ œ )C" %C# subject to C" C# " \$C" C# # C" ß C# ! 3 6-10 Feasible? No No No Yes Yes Yes 3. We denote by x1 : the units of Product 1 produced. x2 : the units of Product 2 produced. yi = 1 if any Product i is produced and yi = 0 otherwise, for i = 1, 2. Then, the integer programming model is: max z = 2x1 + 5x2 − 10y1 − 20y2 s.t. 3x1 + 6x2 ≤ 120, x1 ≤ 40y1 , x2 ≤ 20y2 , x1 ≥ 0, x2 ≥ 0; x1 , x2 be integer numbers; y1 , y2 = 0 or 1. 4. We deﬁne: x1 : x : 2 x3 : x4 : y1 = the the the the number number number number of of of of soldiers; tanks; planes; trains; 1, If any soldiers are manufactured, 0, otherwise. y2 = 1, If any tanks are manufactured, 0, otherwise. y3 = 1, If any planes are manufactured, 0, otherwise. y4 = 1, If any trains are manufactured, 0, otherwise. Then, the proﬁt is: = (12x1 + 10x2 + 8x3 + 15x4 ) − (6x1 + 5x2 + 4x3 + 8x4 ) − (200y1 + 150y2 + 120y3 + 100y4 ) = z 6x1 + 5x2 + 4x3 + 7x4 − 200y1 − 150y2 − 120y3 − 100y4 . The resource constraints are: 3x1 + 5x2 + 2x3 + 6x4 ≤ 150; 4x1 + x2 + 3x3 + 4x4 ≤ 160; The machine tool rentage constraints are: xi ≤ M i yi , , i = 1, 2, 3, 4, 4 where Mi are suﬃciently large constants. The either-or constraints for economical productions are: xi ≤ Ni zi ; i = 1, 2, 3, 4, 5 − xi ≤ Ni (1 − zi ); where Ni are suﬃciently large constants. In addition, the requirements 1-3 can be fulﬁlled by: y1 + y2 + y3 + y4 ≤ 3; y = y3 ; 1 y3 + y4 ≤ 1. Hence, the linear IP model is: max z= 6x1 + 5x2 + 4x3 + 7x4 − 200y1 − 150y2 − 120y3 − 100y4 3x1 + 5x2 + 2x3 + 6x4 ≤ 150; 4x1 + x2 + 3x3 + 4x4 ≤ 160; xi ≤ Ni zi ; i = 1, 2, 3, 4, 5 − xi ≤ Ni (1 − zi ); y1 + y2 + y3 + y4 ≤ 3; y = y3 ; 1 y3 + y4 ≤ 1. xi ≤ Mi yi , , i = 1, 2, 3, 4; xi ≥ 0, integers; yi ∈ {0, 1}, , i = 1, 2, 3, 4; zi ∈ {0, 1}, , i = 1, 2, 3, 4. 5 The Clearwater Bay Golf & Countη Club 5. :千 P戶:這uf? is iÌJ Ý e.g raJ 了 fr;二 I ,,J/ ~.魚 前kdUi 此以 主也~ rw:;，Ji到~立一←一一一一一一一 一 一 一←一且在胎盤-1=-~~及三乏主土~~t~Å~ …一一一一 一 … j一 ~~尸品;而n.e1，- 01'&- K::-w\斗必 LA 呼叫住抖守 rri::e; s' 川志可知'" b(..e吶 IJ A- ~I I ζ 二主 I I "3 …~ -.:; :;)舟，卻之 0.; ~斗。/' ..x.,..:: I X\ 二。 再三之 4立5 之抖。 7r r X ,::: I ;(~玄 _Th1.~~~~玲也斗-.-2;;1.但恰在主..:> 6 X l- - ~也且三一 ...
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