Process Optimization .


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Process Optimization. Mathematical Programming and Optimization of Multi-Plant Operations and Process Design Ralph W. Pike Director, Minerals Processing Research Institute Horton Professor of Chemical Engineering Louisiana State University.
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Prepare Optimization Mathematical Programming and Optimization of Multi-Plant Operations and Process Design Ralph W. Pike Director, Minerals Processing Research Institute Horton Professor of Chemical Engineering Louisiana State University Department of Chemical Engineering, Lamar University, April, 10, 2007

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Process Optimization Typical Industrial Problems Mathematical Programming Software Mathematical Basis for Optimization Lagrange Multipliers and the Simplex Algorithm Generalized Reduced Gradient Algorithm On-Line Optimization Mixed Integer Programming and the Branch and Bound Algorithm Chemical Production Complex Optimization

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New Results Using one script to compose and run a program in another dialect Cumulative likelihood dispersion rather than an ideal point utilizing Monte Carlo recreation for a multi-criteria, blended whole number nonlinear programming issue Global improvement

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Design versus Operations Optimal Design − Uses flowsheet test systems and SQP Heuristics for an outline, a superstructure, an ideal plan Optimal Operations On-line improvement Plant ideal booking Corporate production network advancement

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Plant Problem Size Contact Alkylation Ethylene 3,200 TPD 15,000 BPD 200 million lb/yr Units 14 76 ~200 Streams 35 110 ~4,000 Constraints Equality 761 1,579 ~400,000 Inequality 28 50 ~10,000 Variables Measured 43 125 ~300 Unmeasured 732 1,509 ~10,000 Parameters 11 64 ~100

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finish Optimization Programming Languages GAMS - G eneral A lgebraic M odeling S ystem LINDO - Widely utilized as a part of business applications AMPL - A M athematical P rogramming L anguage Others: MPL, ILOG enhancement program is composed as a streamlining issue optimize: y( x ) economic display subject to: f i ( x ) = 0 constraints

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Software with Optimization Capabilities Excel – Solver MATLAB MathCAD Mathematica Maple Others

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Mathematical Programming Using Excel – Solver Using GAMS Mathematical Basis for Optimization Important Algorithms Simplex Method and Lagrange Multipliers Generalized Reduced Gradient Algorithm Branch and Bound Algorithm

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Simple Chemical Process P – reactor weight R – reuse proportion

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Excel Solver Example Solver ideal arrangement Showing the conditions in the Excel cells with beginning qualities for P and R

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Excel Solver Example

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Excel Solver Example Not the base for C Not

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Use Solver with these estimations of P and R Excel Solver Example

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Excel Solver Example ideal Click to highlight to create reports

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Excel Solver Example Information from Solver Help is of restricted esteem

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Excel Solver Answer Report administration report organize values at the ideal imperative status slack variable

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Excel Sensitivity Report Solver utilizes the summed up lessened slope advancement calculation Lagrange multipliers utilized at affectability investigation Shadow costs ($ per unit)

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Excel Solver Limits Report Sensitivity Analysis gives restrains on factors to the ideal answer for stay ideal

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GAMS

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GAMS S O L V E S U M A R Y MODEL Recycle OBJECTIVE Z TYPE NLP DIRECTION MINIMIZE SOLVER CONOPT FROM LINE 18 **** SOLVER STATUS 1 NORMAL COMPLETION **** MODEL STATUS 2 LOCALLY OPTIMAL **** OBJECTIVE VALUE 3444444.4444 RESOURCE USAGE, LIMIT 0.016 1000.000 ITERATION COUNT, LIMIT 14 10000 EVALUATION ERRORS 0 C O N O P T 3 x86/MS Windows form 3.14P-016-057 Copyright (C) ARKI Consulting and Development A/S Bagsvaerdvej 246 A DK-2880 Bagsvaerd, Denmark Using default alternatives. The model has 3 factors and 2 limitations with 5 Jacobian components, 4 of which are nonlinear. The Hessian of the Lagrangian has 2 components on the corner to corner, 1 components beneath the slanting, and 2 nonlinear factors. ** Optimal arrangement. Diminished inclination not as much as resistance.

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GAMS Lagrange multiplier LOWER LEVEL UPPER MARGINAL - EQU CON1 9000.000 117.284 - EQU OBJ . . . 1.000 LOWER LEVEL UPPER MARGINAL - VAR P 1.000 1500.000 +INF . - VAR R 1.000 6.000 +INF EPS - VAR Z - INF 3.4444E+6 +INF . **** REPORT SUMMARY : 0 NONOPT 0 INFEASIBLE 0 UNBOUNDED 0 ERRORS values at the ideal 900 page Users Manual

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GAMS Solvers 13 sorts of improvement issues NLP – Nonlinear Programming nonlinear monetary model and nonlinear imperatives LP - Linear Programming direct financial model and straight requirements MIP - Mixed Integer Programming nonlinear financial model and nonlinear limitations with constant and whole number factors

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GAMS Solvers 32 Solvers new worldwide analyzer DICOPT One of a few MINLP streamlining agents MINOS a modern NLP enhancer created at Stanford OR Dept utilizes GRG and SLP

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Mathematical Basis for Optimization is the Kuhn Tucker Necessary Conditions General Statement of a Mathematical Programming Problem Minimize: y(x) Subject to: f i (x) <  0 for i = 1, 2, ..., h f i (x) = 0 for i = h+1, ..., m y(x) and f i (x) are twice ceaselessly differentiable genuine esteemed capacities.

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Kuhn Tucker Necessary Conditions Lagrange Function – changes over compelled issue to an unconstrained one λ i are the Lagrange multipliers x n+i are the slack factors used to change over the disparity requirements to correspondences.

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Kuhn Tucker Necessary Conditions Necessary conditions for a relative least at x *

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Lagrange Multipliers Treated as an: Undetermined multiplier – increase requirements by λ i and add to y( x ) Variable - L(x, λ ) Constant – numerical esteem figured at the ideal

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Lagrange Multipliers streamline: y(x 1 , x 2 ) subject to: f(x 1 , x 2 ) = 0

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Lagrange Multipliers Rearrange the incomplete subsidiaries in the second term

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Lagrange Multipliers ( ) = λ Call the proportion of halfway subordinates in the ( ) a Lagrange multiplier, λ Lagrange multipliers are a proportion of fractional subordinates at the ideal.

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Lagrange Multipliers Define L = y + λ f , an unconstrained capacity and by similar system Interpret L as an unconstrained capacity, and the incomplete subordinates set equivalent to zero are the vital conditions for this unconstrained capacity

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Lagrange Multipliers Optimize: y(x 1 ,x 2 ) Subject to: f(x 1 ,x 2 ) = b Manipulations give: ∂y = - λ ∂ b Extends to: ∂y = - λ i shadow cost ($ per unit of b i ) ∂ b i

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Geometric Representation of a LP Problem Maximum at vertex P = 110 A = 10, B = 20 max: 3A + 4B = P s.t. 4A + 2B < 80 2A + 5B < 120 target capacity is a plane no inside ideal

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LP Example Maximize:    x 1 + 2x 2             =   P Subject to:     2x 1 +  x 2 + x 3                        = 10 x 1 +  x 2          + x 4               = 6 - x 1 + x 2                 + x 5         =  2 -2x 1 + x 2                        + x 6   =  1 4 conditions and 6 questions, set 2 of the x i =0 and illuminate for 4 of the x i. Essential attainable arrangement: x 1 = 0, x 2 = 0, x 3 = 10, x 4 = 6, x 5 = 2, x 6 =1 Basic solution: x 1 = 0, x 2 = 6, x 3 = 4, x 4 = 0, x 5 = - 4, x 6 = - 5

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Final Step in Simplex Algorithm Maximize:             - 3/2 x 4 - 1/2 x 5        =  P - 10   P = 10 Subject to:                x 3 - 3/2 x 4 + 1/2 x 5       =  2      x 3 = 2                   1/2 x 4 - 3/2 x 5 + x 6   =  1     x 6 = 1 x 1            + 1/2 x 4 - 1/2 x 5        =  2      x 1 = 2      x 2     + 1/2 x 4 + 1/2 x 5        =  4      x 2 = 4           x 4 = 0            x 5 = 0 Simplex calculation trades factors that are zero with ones that are nonzero, each one in turn to land at the greatest

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Lagrange Multiplier Formulation Returning to the first issue Max:  (1+2 λ 1 + λ 2 - λ 3 - 2 λ 4 ) x 1 (2+ λ 1 + λ 2 + λ 3 + λ 4 )x 2 + λ 1 x 3 + λ 2 x 4 + λ 3 x 5 + λ 4 x 6 - (10 λ 1 + 6 λ 2 + 2 λ 3 + λ 4 ) = L = P Set halfway subordinates regarding x 1, x 2 , x 3 , and x 6 equivalent to zero (x 4 and x 5 are zero) and fathom coming about conditions for the Lagrange multipliers

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Lagrange Multiplier Interpretation (1+2 λ 1 + λ 2 - λ 3 - 2 λ 4 )=0 (2+ λ 1 + λ 2 + λ 3 + λ 4 )=0 λ 3 =-1/2 λ 4 =0 λ 2 =-3/2 Maximize: 0x 1 +0x 2  +0 x 3 - 3/2 x 4 - 1/2 x 5  +0x 6  =  P - 10    P = 10 Subject to:                x 3 - 3/2 x 4 + 1/2 x 5       =  2     x 3 = 2                   1/2 x 4 - 3/2 x 5 + x 6   =  1     x 6 = 1 x 1            + 1/2 x 4 -

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