4460-Lecture-8-20121

Optimization

CHEN 4460 – Process Synthesis, Simulation and Optimization

Dr. Mario Richard EdenDepartment of Chemical Engineering

Auburn University

Lecture No. 8 – Mathematical Optimization

October 16, 2012

Contains Material Developed by Dr. Daniel R. Lewin, Technion, Israel

Lecture 10 – Objectives

Understand the different types of optimization problems and their formulation

Be able to formulate and solve a variety of optimization problems in LINGO

On completion of this part, you should:

Optimization Basics

• What is 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 conversionSubject to reactor modeling

equationskinetic equationslimitations on T, P and x

Optimization Basics

• Examples (Continued)

2. Minimize cost of plantSubject to mass & energy balance

equationsequipment modeling

equationsenvironmental, technical

and logical constraints

Optimization Basics

• Examples (Continued)

3. Maximize your grade in this courseSubject to extracurricular activities

full-time-job requirements

constant demand by other courses and/or your advisor/boss

Optimization Basics

• Formulation of 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 feasibility region!

Optimization Basics

• Classification of 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.

Optimization Basics

• Classification of 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 which ALL 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.

Optimization Basics

• Classification of 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.

Optimization Basics

• Formulation of 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

Optimization Example

• Hydrogen Sulfide 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)?


• Hydrogen Sulfide 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 1x1 kg0.80 MEA, 0.20 water80 cents/kg

Grade 2x2 kg0.68 MEA, 0.32 water60 cents/kg

1 kg MEA solutionwater 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)


• Hydrogen Sulfide 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



• Non-negativity constraints x1 0 , x2 0

x2

x1

x2

x1



• Mass balance constraint x1 + x2 = 1

x2

x1

x2

x1

x2

x1

0.20 x1 + 0.32 x2 = 0.25

x1 + x2 = 1



x2

x1

0.20 x1 + 0.32 x2 = 0.25

x1 + x2 = 1

Feasibility region is this heavy line

Any optimal solutionmust lie within thefeasibility region!


• Hydrogen Sulfide 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


• Hydrogen Sulfide 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* = 71.6Optimal point

x2* = 0.42

x1* = 0.58

Optimization Software

• LINGO– Available on computers in Ross 306– To start entering a new optimization problem

type:

Model:

– Enter the objective function by typing:

min = ……; or max = ……;

– Then enter the constraints.– Each line must end by a semi-colon ;– The final statement in the problem should be

“end”

Optimization Software

• Resolving MEA Example in LINGO

LINGO Input

Model:

min = 80*x1 + 60*x2;

0.2*x1 + 0.32*x2 < 0.25;

x1 + x2 = 1;

x1 > 0;

x2 > 0;

end

LINGO Output

Rows= 5 Vars= 2 No. integer vars= 0 ( all are linear)Nonzeros= 10 Constraint nonz= 6( 4 are +- 1) Density=0.667Smallest and largest elements in absolute value= 0.200000 80.0000No. < : 1 No. =: 1 No. > : 2, Obj=MIN, GUBs <= 2Single cols= 0Optimal solution found at step: 0Objective value: 71.66667 Variable Value Reduced Cost X1 0.5833333 0.0000000E+00 X2 0.4166667 0.0000000E+00 Row Slack or Surplus Dual Price 1 71.66667 1.000000 2 0.0000000E+00 166.6667 3 0.0000000E+00 -113.3333 4 0.5833333 0.0000000E+00 5 0.4166667 0.0000000E+00

Value of objective function: 71.6667

Value of variable x1: 0.5833


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!


• Lab Experiment (Cont’d)

LINGO Input

Model:

min = 4*x1^2 + 5*x2^2;

2*x1 + 3*x2 = 6;

x1 > 0;

x2 > 0;

end

LINGO Output

Rows= 4 Vars= 2 No. integer vars= 0Nonlinear rows= 1 Nonlinear vars= 2 Nonlinear constraints= 0Nonzeros= 7 Constraint nonz= 4 Density=0.583

Optimal solution found at step: 4Objective value: 12.85714

Variable Value Reduced Cost X1 1.071429 0.0000000E+00 X2 1.285714 0.0000000E+00

Row Slack or Surplus Dual Price 1 12.85714 1.000000 2 0.0000000E+00 -4.285715 3 1.071429 0.1939524E-07 4 1.285714 0.0000000E+00

Value of objective function: 12.857




• 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

(maximumcapacity

18 kg coal/s)

Coalgasification(maximum

capacity4 kg

coal/s)

coal in

3x1 + 2x2kg coal/s

2x1 kg coal/s for power generation of gasificationplant (value of power breakseven with the cost of coal used in power generation)

air

x1 kg coal/s

Coalliquefaction(maximum

capacity12 kg coal/s)

Byproducts(negligible value)

Gaseous Fuelx1 kg gas. fuel/sNet profit $3/kg of gaseous fuel

2x2 kg coal/s

Byproducts(negligible value)

Liquid Fuelx2 kg liquid fuel/sNet profit $5/kg of liquid fuel

3x1 + 2x2 ≤ 18

x1 ≤ 4

2x2 ≤ 12


• Coal Conversion Plant (Cont’d)– Objective function max z = 3x1 + 5x2– Constraints

• Pretreatment capacity 3x1 + 2x2 ≤ 18• Gasification capacity x1 ≤ 4• Liquefaction capacity2x2 ≤ 12• Non-negativity x1 ≥ 0

x2 ≥ 0

0

2

4

6

8

10

0 2 4 6 8x1

x2

x1 = 4

2x2 = 12

3x1 + 2x2 = 18


• 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 8x1

x2

x1 = 4

2x2 = 12

3x1 + 2x2 = 18

0

2

4

6

8

10

0 2 4 6 8x1

x2

Z = 36 = 3x1 + 5x2

Z = 20 = 3x1 + 5x2

Z = 10 = 3x1 + 5x2


• Coal Conversion Plant (Cont’d)

LINGO Input

Model:

max = 3*x1 + 5*x2;

3*x1 + 2*x2 <= 18;

x1 <= 4;

2*x2 <= 12;

x1 > 0;

x2 > 0;

end

LINGO OutputRows= 6 Vars= 2 No. integer vars= 0 ( all are linear)Nonzeros= 11 Constraint nonz= 6( 3 are +- 1) Density=0.611Smallest and largest elements in absolute value= 1.00000 18.0000No. < : 3 No. =: 0 No. > : 2, Obj=MAX, GUBs <= 2Single cols= 0

Optimal solution found at step: 1Objective value: 36.00000

Variable Value Reduced Cost X1 2.000000 0.0000000E+00 X2 6.000000 0.0000000E+00

Row Slack or Surplus Dual Price 1 36.00000 1.000000 2 2.000000 0.0000000E+00 3 0.0000000E+00 1.500000 4 0.0000000E+00 1.000000 5 2.000000 0.0000000E+00 6 6.000000 0.0000000E+00

Value of objective function: 36

Value of variable x1: 2

Value of variable x2: 6


• 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


• 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


• 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 going from each MeOH plant to each MeAc plant as follows:plant 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 + 2X1A + X1B + 5X1C + 3X2A + 0X2B + 8X2C + 11X3A 11X3A

+ 6X3B + 15X3C + 7X4A + X4B + 9X4C+ 6X3B + 15X3C + 7X4A + X4B + 9X4C


• 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


• Methanol Delivery (Cont’d)– Constraints

• Non-negativity X1A ≥ 0X1B ≥ 0 X1C ≥ 0

X2A ≥ 0

X2B ≥ 0

X2C ≥ 0

X3A ≥ 0

X3B ≥ 0

X3C ≥ 0

X4A ≥ 0

X4B ≥ 0

X4C ≥ 0

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 occurx


• 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


• 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

minm n

ij iji j

C X

1

1 , 1, ....,n

ijj

X i m

1

1 , 1, ....,m

iji

X j n


• 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


• 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


• Plant Layout (Cont’d)– Constraints

• Each space must be assigned to one and only one reactor

X11 + X12 + X13 + X14 = 1X21 + X22 + X23 + X24 = 1X31 + X32 + X33 + X34 = 1X41 + X42 + X43 + X44 = 1

• Each reactor must be assigned to one and only one space

X11 + X21 + X31 + X41 = 1X12 + X22 + X32 + X42 = 1X13 + X23 + X33 + X43 = 1X14 + X24 + X34 + X44 = 1


• Plant Layout (Cont’d)– Solve using LINGO– Optimal assignment policy

Reactor R1 in space 2Reactor R2 in space 1Reactor R3 in space 4Reactor R4 in space 3

– Minimum cost

Cost = 11 + 13 + 14 + 11 = $49,000

Lecture 10 – Summary

Understand the different types of optimization problems and their formulation

Be able to formulate and solve a variety of optimization problems in LINGO

On completion of this part, you should:

Other Business

• Homework– SSLW: 24.1 plus problems posted on class webpage– Due Tuesday October 23– LINGO software is available on class webpage as zip-file

• Next Lecture – October 23– Heat and Power Integration (SSLW p. 252-261)

• Review of Midterm Exam– Thursday October 18 during lab sessions– You will get your tests back to look at during solution

review

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4460-Lecture-8-20121