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Lecture-4-Mathematical Optimization_This.ppt
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- 2. Optimization Chemical Process Designand Simulation Dr. Imran Nazir Unar Department of Chemical Engineering MUET Jamshoro Lecture No. 6 – Mathematical Optimization
- 3. Lecture 6 –Objective Understand the different types of optimization problems and their formulation On completion of this part, you should:
- 4. Optimization Basics • Whatis Optimization? – The purpose of optimization is to maximize (or minimize) the value of a function (called objective function) subject to a number of restrictions (called constraints). • Examples 1.Maximize reactor conversion Subject to reactor modeling equations kinetic equations limitations on T, P and x
- 5. Optimization Basics • Examples(Continued) 2.Minimize cost of plant Subject to mass & energy balance equations equipment modeling equations environmental, technical and logical constraints
- 6. Optimization Basics • Examples(Continued) 3.Maximize your grade in this course Subject to extracurricular activities full-time-job requirements constant demand by other courses and/or your advisor/boss
- 7. Optimization Basics • Formulationof Optimization Problems min (or max) f(x1,x2,……,xN) subject to g1(x1,x2,……,xN) 0 ≤ g2(x1,x2,……,xN) 0 ≤ gm(x1,x2,……,xN) 0 ≤ h1(x1,x2,……,xN)=0 h2(x1,x2,……,xN)=0 hE(x1,x2,……,xN)=0 Inequality Constraints Equality Constraints Feasibility Any vector (or point) which satisfies all the constraints of the optimization program is called a feasible vector (or a feasible point) The set of all feasible points is called feasibility region or feasibility domain Any optimal solution must lie within the
- 8. Optimization Basics • Classificationof Optimization Problems – Linear Programs (LP’s) • A mathematical program is linear if f(x1,x2,……,xN) and gi(x1,x2,……,xN) 0 are ≤ linear in each of their arguments: f(x1,x2,……,xN) = c1x1 + c2x2 + …. cNxN gi(x1,x2,……,xN) = ai1x1 + ai2x2 + …. aiNxN where ci and aij are known constants. Linear Programs (LP’s) can be solved to yield a global optimum. Solver routines can guarantee a truly optimal solution.
- 9. Optimization Basics • Classificationof Optimization Problems – Non-Linear Programs (NLP’s) • A mathematical program is non-linear if any of the arguments are non-linear. For example: min 3x + 6y2 s.t. 5x + xy 0 ≥ – Integer Programming • Optimization programs in whichALL the variables must assume integer values. The most commonly used integer variables are the zero/one binary integer variables. • Integer variables are often used as decision variables, e.g. to choose between two reactor types. Non-Linear Programs (NLP’s) can be solved to yield a local optimum. Solver routines can not always guarantee a globally optimal solution.
- 10. Optimization Basics • Classificationof Optimization Problems – Mixed Integer Linear Programs (MILP’s) • Linear programs in which SOME of the variables are real and other variables are integers • Can be solved as individual LP’s by fixing the integer variables, thus a global optimum can be identified. – Mixed Integer Non-Linear Programs (MINLP’s) • Non-linear programs in which SOME of the variables are real and other variables are integers • Can be solved as individual NLP’s by fixing the integer variables, but depending on the nature of the NLP’s it may not be possible to find a global optimum.
- 11. Optimization Basics • Formulationof Optimization Problems – Step 1 • Determine the quantity to be optimized and express it as a mathematical function (this is your objective function) • Doing so also serves to define variables to be optimized (input variables or optimization variables) – Step 2 • Identify all stipulated requirements, restrictions, and limitations, and express them mathematically. These requirements constitute the constraints – Step 3 • Express any hidden conditions. Such conditions are not stipulated explicitly in the problem, but are apparent from the physical situation, e.g. non-negativity constraints
- 12. Optimization Example • HydrogenSulfide Scrubbing – Two variable grades of MEA. – First grade consists of 80 weight% MEA and 20% weight water. Its cost is 80 cent/kg. – Second grade consists of 68 weight% MEA and 32 weight% water. Its cost is 60 cent/kg. – It is desired to mix the two grades so as to obtain an MEA solution that contains no more than 25 weight% water. – What is the optimal mixing ratio of the two grades which will minimize the cost of MEA solution (per kg)?
- 13. Optimization Example • HydrogenSulfide Scrubbing (Cont’d) – Objective function min z = 80x1 + 60x2 – Constraints • Water content limitation 0.20x1 + 0.32x2 0.25 ≤ • Overall material balance x1 + x2 =1 • Non-negativity x1 ≥ 0 x2 ≥ 0 MIXER Grade 1 x1 kg 0.80 MEA, 0.20 water 80 cents/kg Grade 2 x2 kg 0.68 MEA, 0.32 water 60 cents/kg 1 kg MEA solution water content 25 wt. % Variables (Basis 1 kg solution) x1 Amount of grade 1 (kg) x2 Amount of grade 2 (kg) z Cost of 1 kg solution (cents)
- 14. Optimization Example • HydrogenSulfide Scrubbing (Cont’d) – Feasibility region • The set of points (x1, x2) satisfying all the constraints, including the non-negativity conditions. • Constraint on water content 0.20x1 + 0.32x2 0.25 ≤ x2 x1
- 15. Optimization Example • HydrogenSulfide Scrubbing (Cont’d) – Feasibility region • Non-negativity constraints x1 0 , x2 0 x2 x1 x2 x1
- 16. Optimization Example • HydrogenSulfide Scrubbing (Cont’d) – Feasibility region • Mass balance constraint x1 + x2 = 1 x2 x1 x2 x1 x2 x1 0.20 x1 + 0.32 x2 = 0.25 x1 + x2 = 1
- 17. Optimization Example • HydrogenSulfide Scrubbing (Cont’d) – Feasibility region x2 x1 0.20 x1 + 0.32 x2 = 0.25 x1 + x2 = 1 Feasibility region is this heavy line Any optimal solution must lie within the feasibility region!
- 18. Optimization Example • HydrogenSulfide Scrubbing (Cont’d) – By plotting objective function curves for arbitrary values of z (here 70 and 75) we can evaluate the results: x2 x1 0.20 x1 + 0.32 x2 = 0.25 x1 + x2 = 1 z = 75 z = 70 Optimal Point Intersection between x1 + x2 = 1 and 0.20x1 + 0.32x2 = 0.25 In addition 70 < zmin < 75
- 19. Optimization Example • HydrogenSulfide Scrubbing (Cont’d) – Solving the two equations simultaneously yields the optimum amounts of the two MEA solutions along with the minimum cost of the mixture x2 x1 0.20 x1 + 0.32 x2 = 0.25 x1 + x2 = 1 z* = 71.6 Optimal point x2* = 0.42 x1* = 0.58
- 20. More Optimization Examples •Lab Experiment – Determine the kinetics of a certain reaction by mixing two species, A and B. The cost of raw materials A and B are 2 and 3 $/kg, respectively. – Let x1 and x2 be the weights of A and B (kg) to be employed in the experiment – The operating cost of the experiment is given by: OC = 4(x1)2 + 5(x2)2 – The total cost of raw materials for the experiment should be exactly $6. Minimize the operating cost!
- 21. More Optimization Examples •Coal Conversion Plant – What are the optimal production rates of gaseous and liquid fuels that maximize the net profit of the plant? Coal pre- treatment (maximum capacity 18 kg coal/s) Coal gasification (maximum capacity 4 kg coal/s) coal in 3x1 + 2x2 kg coal/s 2x1 kg coal/s for power generation of gasification plant (value of power breaks even with the cost of coal used in power generation) air x1 kg coal/s Coal liquefaction (maximum capacity 12 kg coal/s) Byproducts (negligible value) Gaseous Fuel x1 kg gas. fuel/s Net profit $3/kg of gaseous fuel 2x2 kg coal/s Byproducts (negligible value) Liquid Fuel x2 kg liquid fuel/s Net profit $5/kg of liquid fuel 3x1 + 2x2 18 ≤ x1 4 ≤ 2x2 12 ≤
- 22. More Optimization Examples •Coal Conversion Plant (Cont’d) – Objective function max z = 3x1 + 5x2 – Constraints • Pretreatment capacity 3x1 + 2x2 18 ≤ • Gasification capacity x1 ≤ 4 • Liquefaction capacity 2x2 ≤ 12 • Non-negativity x1 ≥ 0 x2 ≥ 0 0 2 4 6 8 10 0 2 4 6 8 x1 x2 x1 = 4 2x2 = 12 3x1 + 2x2 = 18
- 23. More Optimization Examples •Coal Conversion Plant (Cont’d) – Graphical solution Maximum profit Z = 36 for x1 = 2 and x2 = 6 0 2 4 6 8 10 0 2 4 6 8 x1 x2 x1 = 4 2x2 = 12 3x1 + 2x2 = 18 0 2 4 6 8 10 0 2 4 6 8 x1 x2 Z = 36 = 3x1 + 5x2 Z = 20 = 3x1 + 5x2 Z = 10 = 3x1 + 5x2
- 24. More Optimization Examples •Methanol Delivery – Supply methanol for three Methyl acetate plants located in towns A, B, and C – Daily methanol requirements for each plant: MeAc Plant location Tons/day A 6 B 1 C 10 – Methanol production plants MeOH plant 1 2 3 4 Capacity 7 5 3 2
- 25. More Optimization Examples •Methanol Delivery (Cont’d) – Shipping cost (100 $/ton) – Schedule the methanol delivery system to minimize the transportation cost MeOH Plant MeAc Plant A MeAc Plant B MeAc Plant C 1 2 1 5 2 3 0 8 3 11 6 15 4 7 1 9
- 26. More Optimization Examples •Methanol Delivery (Cont’d) – We define the transportation loads (tons/day) define the transportation loads (tons/day) going from each MeOH plant to each MeAc plant going from each MeOH plant to each MeAc plant as follows: as follows: – Total transportation cost (Z) MeOH Plant MeAc Plant A MeAc Plant B MeAc Plant C 1 X1A X1B X1C 2 X2A X2B X2C 3 X3A X3B X3C 4 X4A X4B X4C Z = Z = 2X1A + X1B + 5X1C + 3X2A + 0X2B + 8X2C + 11X3A 2X1A + X1B + 5X1C + 3X2A + 0X2B + 8X2C + 11X3A + 6X3B + 15X3C + 7X4A + X4B + 9X4C + 6X3B + 15X3C + 7X4A + X4B + 9X4C
- 27. More Optimization Examples •Methanol Delivery (Cont’d) – Objective function min Z = 2X1A + X1B + 5X1C + 3X2A + 0X2B + 8X2C + 11X3A + 6X3B + 15X3C + 7X4A + X4B + 9X4C – Constraints • Availability/supply X1A + X1B + X1C = 7 X2A + X2B + X2C = 5 X3A + X3B + X3C = 3 X4A + X4B + X4C = 2 • Requirements/demand X1A + X2A + X3A + X4A = 6 X1B + X2B + X3B + X4B = 1 X1C + X2C + X3C + X4C = 10
- 28. More Optimization Examples •Methanol Delivery (Cont’d) – Constraints • Non-negativity X1A ≥ 0 X1B ≥ 0 X1C ≥ 0 X2A ≥ 0 X2B ≥ 0 X2C ≥ 0 X3A ≥ 0 X3B ≥ 0 X3C ≥ 0 X4A ≥ 0 X4B ≥ 0 X4C ≥ 0
- 29. Mixed Integer Programs •Use of 0-1 Binary Integer Variables – Commonly used to represent binary choices – Dichotomy modeling 0 if the event does NOT occur 1 if the event does occur x
- 30. Mixed Integer Programs •The Assignment Problem – Assignment of n people to do m jobs – Each job must be done by exactly one person – Each person can at most do one job – The cost of person j doing job i is Cij – The problem is to assign the people to the jobs so as to minimize the total cost of completing all the jobs. – We can assign integer variables to describe whether a certain person does a certain job or not
- 31. Mixed Integer Programs •The Assignment Problem (Cont’d) – The event of person j doing job i is designated Xij – The objective function can be written as: – Since exactly one person will do job i, and each person at most can do one job, we get: 1 1 min m n ij ij i j C X 1 1 , 1,...., n ij j X i m 1 1 , 1,...., m ij i X j n
- 32. Mixed Integer Programs •Plant Layout – An Assignment Problem – Four new reactors R1, R2, R3 and R4 are to be installed in a chemical plant – Four vacant spaces 1, 2, 3 and 4 are available – Cost of assigning reactor i to space j (in $1000) is – Assign reactors to spaces to minimize the total cost Reactor Space 1 Space 2 Space 3 Space 4 R1 15 11 13 15 R2 13 12 12 17 R3 14 15 10 14 R4 17 13 11 16
- 33. Mixed Integer Programs •Plant Layout (Cont’d) – Let Xij denote the existence (or absence) of reactor i in space j, i.e. if Xij =1 then reactor i exists in space j – Objective function min Z = 15X11 + 11X12 + 13X13 + 15X14 + 13X21 + 12X22 + 12X23 + 17X24 + 14X31 + 15X32 + 10X33 + 14X34 + 17X41 + 13X42 + 11X43 + 16X44
- 34. Mixed Integer Programs •Plant Layout (Cont’d) – Constraints • Each space must be assigned to one and only one reactor X11 + X12 + X13 + X14 = 1 X21 + X22 + X23 + X24 = 1 X31 + X32 + X33 + X34 = 1 X41 + X42 + X43 + X44 = 1 • Each reactor must be assigned to one and only one space X11 + X21 + X31 + X41 = 1 X12 + X22 + X32 + X42 = 1 X13 + X23 + X33 + X43 = 1 X14 + X24 + X34 + X44 = 1
- 35. Mixed Integer Programs •Plant Layout (Cont’d) – Optimal assignment policy Reactor R1 in space 2 Reactor R2 in space 1 Reactor R3 in space 4 Reactor R4 in space 3 – Minimum cost Cost = 11 + 13 + 14 + 11 = $49,000
- 36.