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A raw material asfeedstock shouldbe renewablerather thandepletingwherevertechnically andeconomicallypracticableA raw material asfeedstock shouldbe renewablerather thandepletingwherevertechnically andeconomicallypracticableA raw material asfeedstock shouldbe renewablerather thandepletingwherevertechnically andeconomicallypracticable

Chapter Linear programming (LP) What is linear programming ? Linear programming refers to problems in which both the objective function and the constraints are linear Prob • Assign employees to schedules for adequation, satisfaction and productivity Prob • Products selection for the best advantage of existing resources and prices to maximum profit Prob • Products distribution from plants to warehouse for cost minimization within capacity limitation Prob • Bids offering to take into account profit for competition in operating constraints Canonical form for linear programs Minimize the objective function: 𝑛 𝑓 𝑋 = ෍ 𝑐𝑗 𝑥𝑗 𝑗=1 𝑛 ෍ 𝑎𝑖𝑗 𝑥𝑗 = 𝑏𝑖 Subject to: 𝑗=1 𝑥𝑗 ≥ 𝑏𝑖 ≥ 𝑖 = 1,2, ⋯ , 𝑚 𝑛>𝑚 𝑗 = 1,2, ⋯ , 𝑛 Canonical form for linear programs Matrix form Minimize: 𝑓 = 𝑐𝑇 𝑋 Subject to: 𝐴𝑋 = 𝑏 ቐ 𝑋≥0 𝑏≥0 𝑅𝑎𝑛𝑘 𝐴 = 𝑚 Note that all constraints are equalities 𝑋: variable vector 𝑐: objective function vector 𝐴: constraint matrix 𝑏: constraint vector Important theorems Theorem • The linear programs are convex problems • The local minimum of 𝑓 𝑋 is also the global minimum Theorem • Any point in the feasible region is linear combination of another two points “on” boundary • The optimal solution is only search “on” the boundaries (vertices) Theorem • The minimum solution is a vertex which is nearest the origin Example Maximize: 𝑓 𝑋 = 𝑥1 + 3𝑥2 Subject to: −𝑥1 + 𝑥2 ≤ ቐ 𝑥1 + 𝑥2 ≤ 𝑥1 , 𝑥2 ≥ Example 𝑥2 𝑥1 Definitions Dependent variables • That appears once in constraint matrix 𝐴 Number of dependent variables is 𝑚 Independent variables • That appears more than once in constraint matrix 𝐴 Number of independent variables is 𝑛 − 𝑚 Basic solution • That is a unique vector (also the vertex) determined by setting all of “independent variables” equal to zero, and solving the remaining “dependent variables” Equivalent systems 𝑛 System 𝐴 ෍ 𝑎𝑖𝑗 𝑥𝑗 = 𝑏𝑖 𝑗=1 𝑥𝑗 ≥ 0; 𝑏𝑖 ≥ 𝑖 = 1,2, ⋯ , 𝑚 𝑛>𝑚 𝑗 = 1,2, ⋯ , 𝑛 𝑛 System 𝐵 ෍ 𝛼𝑖𝑗 𝑥𝑗 = 𝛽𝑖 𝑗=1 𝑥𝑗 ≥ 0; 𝑏𝑖 ≥ 𝑖 = 1,2, ⋯ , 𝑚 𝑛>𝑚 𝑗 = 1,2, ⋯ , 𝑛 Systems 𝐴 and 𝐵 are equivalent if they have the “same” solution sets Elementary operations Elementary operations are the operations transform a given system 𝐴 into an equivalent system 𝐵 𝑛 System 𝐴 ෍ 𝑎𝑖𝑗 𝑥𝑗 = 𝑏𝑖 𝑗=1 There are 𝑚 equations 𝐸𝑖 𝑥𝑗 ≥ 0; 𝑏𝑖 ≥ • Multiplying any equation 𝐸𝑖 by a constant 𝑞 ≠ • Replacing any equation 𝐸𝑡 by the equation 𝐸𝑡 + 𝑞𝐸𝑖 , where 𝐸𝑖 is any other Example Minimize the objective function: 𝑓 𝑋 = 2𝑥1 + 𝑥2 Subject to: 𝑥1 + 𝑥2 = 3𝑥1 − 𝑥2 ≥ 2𝑥1 + 3𝑥2 ≤ 𝑥1 , 𝑥2 ≥ Example Transform to the canonical form Minimize Subject to: 𝑓 𝑋 = 2𝑥1 + 𝑥2 𝑥1 + 𝑥2 = 3𝑥1 − 𝑥2 − 𝑥3 = 2𝑥1 + 3𝑥2 + 𝑥4 = 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 ≥ 𝑥3 , 𝑥4 are “slack variables” Example Build the constraint matrix 𝐴 𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 11 0 33 −1 −1 22 𝒙𝟓 0 𝒙𝟔 Dependent variables: 𝑥4 adding “hidden variables”: 𝑥5 , 𝑥6 Independent variables: 𝑥1 , 𝑥2 , 𝑥3 Example Basic solution 𝑥1 𝑥2 𝑥3 𝑋𝑜 = 𝑥 = 𝑥6 𝑥4 Example There are “hidden variables”, simplex table of Phase Dependent 𝒄𝒋 variables 𝒙𝟓 𝟏 𝒙𝟔 𝟏 𝒙𝟒 𝟎 𝒇𝒊 − 𝒄𝒋 𝑿𝒐 𝟎 𝟎 𝟎 𝟎 𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 1 0 −1 −1 𝑓1 𝑋 = 𝑥5 + 𝑥6 𝟏 𝒙𝟓 0 𝟏 𝒙𝟔 Example There are “hidden variables”, simplex table of Phase Dependent 𝒄𝒋 variables 𝒙𝟓 𝟏 𝒙𝟔 𝟏 𝒙𝟒 𝟎 𝒇𝒊 − 𝒄𝒋 𝑿𝒐 𝟎 𝟎 𝟎 𝟎 𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 1 0 −1 −1 𝟏 𝒙𝟓 0 𝟏 𝒙𝟔 0 −1 𝑓1 𝑋 = 𝑥5 + 𝑥6 Example Change the basic solution Dependent 𝒄𝒋 variables 𝒙𝟓 𝟏 𝒙𝟔 𝟏 𝒙𝟒 𝟎 𝒇𝒊 − 𝒄𝒋 𝑿𝒐 𝟎 𝟎 𝟎 𝟎 𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 1 0 −1 −1 𝟏 𝒙𝟓 0 𝟏 𝒙𝟔 0 −1 𝑓1 𝑋 = 𝑥5 + 𝑥6 Example Change the basic solution Dependent 𝒄𝒋 variables 𝒙𝟓 𝟏 𝒙𝟏 𝟎 𝒙𝟒 𝟎 𝒇𝒊 − 𝒄𝒋 𝑿𝒐 𝟎 𝒙𝟏 𝟎 𝒙𝟐 𝟎 𝒙𝟑 𝑓1 𝑋 = 𝑥5 + 𝑥6 𝟎 𝒙𝟒 0 𝟏 𝒙𝟓 0 𝟏 𝒙𝟔 Example Elementary operations Dependent 𝒄𝒋 variables 𝒙𝟓 𝟏 𝒙𝟏 𝟎 𝒙𝟒 𝟎 𝒇𝒊 − 𝒄𝒋 𝑿𝒐 3 25 𝟎 𝒙𝟏 𝟎 𝒙𝟐 𝟎 𝒙𝟑 𝟎 𝒙𝟒 𝟏 𝒙𝟓 𝟏 𝒙𝟔 −13 11 3 −13 3 −13 0 0 −23 −43 0 𝑓1 𝑋 = 𝑥5 + 𝑥6 Example Change the basic solution Dependent 𝒄𝒋 variables 𝒙𝟓 𝟏 𝒙𝟏 𝟎 𝒙𝟒 𝟎 𝒇𝒊 − 𝒄𝒋 𝑿𝒐 3 25 𝟎 𝒙𝟏 𝟎 𝒙𝟐 𝟎 𝒙𝟑 𝟎 𝒙𝟒 𝟏 𝒙𝟓 𝟏 𝒙𝟔 −13 11 3 −13 3 −13 0 0 −23 −43 0 𝑓1 𝑋 = 𝑥5 + 𝑥6 Example Change the basic solution Dependent 𝒄𝒋 variables 𝒙𝟐 𝟎 𝒙𝟏 𝟎 𝒙𝟒 𝟎 𝒇𝒊 − 𝒄𝒋 𝑿𝒐 𝟎 𝒙𝟏 𝟎 𝟎 𝒙𝟐 𝒙𝟑 0 𝑓1 𝑋 = 𝑥5 + 𝑥6 𝟎 𝒙𝟒 0 𝟏 𝒙𝟓 𝟏 𝒙𝟔 Example Elementary operations Dependent 𝒄𝒋 variables 𝑿𝒐 𝟎 𝒙𝟏 𝟎 𝒙𝟐 𝟎 𝒙𝟑 𝟎 𝒙𝟒 𝟏 𝒙𝟓 𝟏 𝒙𝟔 −14 −14 4 −11 −14 𝒙𝟐 𝟎 𝒙𝟏 𝟎 1 𝒙𝟒 𝟎 0 𝒇𝒊 − 𝒄𝒋 𝑓1 𝑋 = 𝑥5 + 𝑥6 1 4 Example Elementary operations Dependent 𝒄𝒋 variables 𝑿𝒐 𝟎 𝒙𝟏 𝟎 𝒙𝟐 𝟎 𝒙𝟑 𝟎 𝒙𝟒 𝟏 𝒙𝟓 𝟏 𝒙𝟔 −14 4 −11 −1 −1 𝒙𝟐 𝟎 𝒙𝟏 𝟎 1 𝒙𝟒 𝟎 0 −14 −14 0 𝒇𝒊 − 𝒄𝒋 𝑓1 𝑋 = 𝑥5 + 𝑥6 4 Example 𝑓 𝑋 = 2𝑥1 + 𝑥2 Phase Dependent 𝒄𝒋 variables 𝑿𝒐 𝟐 𝒙𝟏 𝟏 𝒙𝟐 𝟎 𝒙𝟑 𝟎 𝒙𝟒 −14 −14 𝒙𝟐 𝟏 𝒙𝟏 𝟐 1 𝒙𝟒 𝟎 0 𝒇𝒊 − 𝒄𝒋 Example 𝑓 𝑋 = 2𝑥1 + 𝑥2 Phase Dependent 𝒄𝒋 variables 𝑿𝒐 𝟐 𝒙𝟏 𝟏 𝒙𝟐 𝟎 𝒙𝟑 𝟎 𝒙𝟒 −14 −14 −14 𝒙𝟐 𝟏 𝒙𝟏 𝟐 1 𝒙𝟒 𝟎 0 0 𝒇𝒊 − 𝒄𝒋 Satisfying optimality: all values 𝑓𝑖 − 𝑐𝑗 ≤ ... + 10

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