A lecture on optimization

1. Course: Project Engineering
Optimization
Dr. Mahendra Chinthala
Assistant Professor
Department of Chemical Engineering
NIT Rourkela
Lecture-30

2. Optimization in Chemical Engineering
Optimization: is the science of making best possible decision
 Optimization is the act of achieving the best possible result under given
circumstances.
Why optimization ?
 To improve the process to realize the maximize system potential.
 To attain new or improved designs; maximize profits and minimize cost of
production.
Advantages of optimization in chemical process industry
 Improved plant performance,
 minimizing waste generation,
 increasing product yield,
 Less equipment wear,
 reduce cost of production,
 reduce energy consumption,
 low maintenance costs and so on…

3. 3
optimization
Reduce
the cost
Safety &
reduce the
error
reproducibilit
y
Save the
time
Why Optimization is necessary?
Innovation
&
efficiency

4. Optimization problem
All optimization problems are stated in some standard form.
You have to identity the essential elements of a given problem
and translate them into a prescribed mathematical form.
The following are the requirements for the application of
optimization problems:
 Design or decision variables
 Objective function
 Constraints
 Process model

5. Design Variables
Design or decision variables are the variables that
influence the system being optimized. It is varied during
optimization in order to achieve optimization.
Ex. Reactor temperature, Feed rate, No. of plates in
distillation column, reflux ratio, batch time, reactor yield,
etc.
 If a problem involves many design variables, some of
these may be highly influence the process being
optimized.
 Choose these as design variables and others may be
constant.

6. Objective Function: An objective function expresses the main
aim of the model which is either to be minimized or maximized.
 It is defined in terms of design variables and other process
parameters.
 The objective function may be technical or economic, which needs
to be either maximized or minimized.
Examples of economic objectives: maximize profits, minimize cost of production.
Examples of technical objectives; maximize reactor yield, minimized size of an
equipment, minimize error during curve fitting, etc.
Technical objectives are ultimately related to economics.
 For example:in a manufacturing process, the aim may be to maximize the profit or
minimize the cost.
 The two exceptions are:
• No objective function
• Multiple objective functions.

7. Constraints:
 The constraints represent some additional functional relationships
among the decision or design variables and process parameters.
 The constraints originate as design variables must satisfy certain
physical phenomenon and certain resource limitations.
Examples:
Variable bounds: 0< x<1
Equality constraints : sum of mole fractions should be unity
x1 + x2 + x3 =1 ; y1+y2+y3=1
Inequality constraints:
 In a packed reactor, temperature should be less than catalyst
deactivation temperature.
 Acidic condition: pH <7

8. Process model
A process model is required that describes the manner
in which the decision variables are related.
The process model tells us how the objective function is
influenced by the design or decision variables.
A mode is a mathematical equation or a is a collection
of several equations that define how the decision
variables are related and the acceptable values these
variables can take.
Optimization studies are carried out using a simplified
model of a real system. Working with real system is
time consuming, expensive, risky.

9. Consider the problem as an optimization
task

10. Statement of an optimization problem
 An optimization problem can be stated as follows:
To find X =
which minimizes f(X)
Subject to the constraints
gi(X) ≤ 0 , i = 1, 2, …., m
lj(X) = 0, j = 1, 2, …., p
where X is an n-dimensional vector called the design vector, f(X) is
called the objective function, and gi(X) and lj(X) are known as
inequality and equality constraints, respectively.

11. Minimization or
maximization

12. Classification of optimization methods
 Based on Constraints
 Constrained optimization (Lagrangian method)
 Unconstrained optimization (Least Squares)
 Based on Nature of the design variables
 Static optimization
 Dynamic optimization
 Based on Physical structure
 Optimal control
 Sub-optimal control
 Based on the Permissible Values of the Design Variables
 Inter programming
 Real valued programming
 Based on the Number of Objective Functions
 Single objective
 Multi objective

13.  Based on Nature of variables
 Stochastic optimization
 Deterministic optimization
 Based On Separability Of The Functions
 Separable
 Non separable
 Based on the Nature of the Equations Involved
 Linear programming
 Quadratic programming
 Nonlinear programming

14. Examples of optimization problems in chemical engineering
Optimal design of a can: Design a can which hold at least 500 ml of
liquid. Height = [ 7, 12] cm, Radius =[ 3, 7 ] cm. What dimensions for the
cylinder will use the least amount of material ?
Sol. We can minimize the material by minimizing the area, A
Objective Function: A = 2𝜋𝑟2
+ 2𝜋rh;
Constraint: V = 𝜋𝑟2ℎ ≥ 500 𝑚𝑙;
Bounds: 3≤ 𝑟 ≤ 7, 7≤ ℎ ≤ 10;

15. Example 2: Critical insulation
thickness
At critical thickness of insulation,
maximum heat dissipation from the tube
occurs, Resistance is minimum at critical
insulation.
Objective is to minimize the objective
function
Rc = k/h for cylindrical Cross
sections
Rc =2k/h for spherical Cross-
sections

16. 3. Chemical reactor design for series
reactions
How to maximize concentration of B,
CB(t)

17. Optimum design conditions
• An optimum design is based on the best or most
favorable conditions.
• In almost every case, these optimum conditions
can ultimately be reduced to a consideration of
costs or profits.
• Thus, an optimum economic design could be
based on conditions giving the least cost per unit
of time or the maximum profit per unit of
production.
• When one design variable is changed, it is often
found that some costs increase and others
decrease.
• Under these conditions, the total cost may go

18. Example 1:
To determine the optimum thickness of insulation for a given
steam-pipe installation .
 As the insulation thickness is increased, the annual fixed
costs increase, the cost of heat loss decreases, and all
other costs remain constant.
 Therefore, as shown in Fig, the sum of the costs must go
through a minimum at the optimum insulation thickness

19. Procedure for determining optimum condition.
1. Identify the parameter or design variable to be optimized.
ex ; Total cost per unit of production or unit of time,
Profit, Final product cost, etc.,
2. Identify the other variables affecting the design variable
3. Develop the objective function or relationship how the
design variable is related to other variables.
4. Identify whether the design variable has to be minimized
or maximized.
5. The objective function can be solved graphically or
analytically to give the desired optimum conditions.

20. Procedure with one variable
If the design variable which is to be optimized depends on
only one variable, the procedure then becomes simple
Consider the example where it is necessary to obtain the
optimum insulation thickness which gives the least total cost.
The primary variable involved is the thickness of the insulation,
and relationships can be developed showing how this variable
affects all costs.
where a, b, c, and d are constants and ‘x’ is the common
variable (insulation thickness)

21.  The graphical method for determining the optimum
insulation thickness is shown in Fig.
 The optimum thickness of insulation is found at the
minimum point on the curve obtained by plotting total
variable cost versus insulation thickness.
Graphical Method

22. Analytical method
 The slope of the total-variable-cost curve is zero at the
point of optimum insulation thickness.
 Therefore, if Eq. (3) applies, the optimum value can be
found analytically by merely setting the derivative of CT
with respect to ‘x’ equal to zero and solving for ‘x’
 (4)

 (5)
 This value of ‘x’ occurs at an optimum point or a point of
inflection.
 To determine whether the total variable cost (CT) is
minimized or maximized, the second derivative of eq 2
need to be evaluated.

23.  If the second derivative of Eq. 3 evaluated at the
given point, is greater than zero , then it indicates
the value occurs at a minimum .
 If the second derivative of Eq. 3 evaluated at the
given point, is less than zero, then it indicates the
value occurs at a maximum .
(6)
If ‘x’ represents a variable such as insulation thickness, its
value must be positive; therefore, if c is positive, the second
derivative at the optimum point must be greater than zero,
and represents the value of at the point where the total
variable cost is a minimum.
Minimization or maximization

24. Procedure with two or more variables
 When two or more independent variables affect the factor
being optimized, the procedure for determining the
optimum conditions may become rather tedious; however,
the general approach is the same as when only one
variable is involved.
 Consider the case in which the total cost for a given
operation is a function of the two independent variables
‘x’ and ‘y’, or
(7)
(8)
where a, b, c, and d are positive constants

25. GRAPHICAL PROCEDURE
 The relationship among CT, x and y could be shown as a
curved surface in a three-dimensional plot, with a minimum
value of occurring at the optimum values of x and y.
 However, the use of a three-dimensional plot is not practical
for most engineering determinations.
 The optimum values of ‘x ‘and ‘y’ in Eq. (8) can be found
graphically on a two-dimensional plot by using the method
indicated in Fig. 2.
(8)

26.  In this figure, the factor being
optimized is plotted against one of
the independent variables with the
second variable held at a constant
value. (varying ‘x’ for a constant ‘
y’)
 A series of such plots is made with
each dashed curve representing a
different constant value of the
second variable(y).
 As shown in Fig., each of the
curves (A, B, C, and gives one
value of the first variable at the
point where the total cost is a
minimum.
 The curve NM represents the
locus of all these minimum points,
and the optimum value of and y
occurs at the minimum point on
curve NM

27. ANALYTICAL PROCEDURE
 In Fig. 3, the optimum value of ‘x’ is found at the point
where
 Similarly, the same results would be obtained if y were
used as the abscissa instead of x.
 If this was done, the optimum value of y (i.e., y i ) would
be found at the point where is
Then from eq (8) as basis,
𝜕𝐶𝑇
𝜕𝑥 𝑦=𝑦𝑖
= 0
𝜕𝐶𝑇
𝜕𝑦 𝑥=𝑥𝑖
= 0

28.  At the optimum conditions, both of these partial
derivatives must be equal to zero;
 thus, Eqs. (9) and (10) can be set equal to zero
and the optimum values of
x = (cb/a2)1/3 and
y = (ab/c2) 1/3
can be obtained by solving the two
simultaneous equations.
 If more than two independent variables were
involved, the same procedure would be followed,
with the number of simultaneous equations being
equal to the number of independent variables.

29. Break-even chart
Break even chart is a chart that shows the sales volume or income at
which total costs equal sales. Losses will be incurred below this point,
and profits will be earned above this point. The chart plots revenue,
fixed costs, and variable costs on the vertical axis, and rate of
production on the horizontal axis.
Rate of production (units/day)

30.  There is a close relationship among operating time, rate of production,
and selling price.
 It is desirable to operate at a schedule which will permit maximum
utilization of fixed costs while simultaneously meeting market sales
demand and using the capacity of the plant production to give the best
economic results.
 The point where total product cost equals total income represents the
break-even point.

31. Ex1. Find the dimensions of a rectangle with perimeter 100 m whose
area is as large as possible
 The perimeter is 2x + 2y = 100.
 The function we want to maximize is the area, A = xy.
 Solving for y, we get y = 100 − 2x 2 = 50 − x.
 So the area can be written as a function of x, namely A(x) = xy = x(50
− x).
 The domain of this function is 0 < x < 50.
 We have A(x) = 50x − x 2
 A’(x) = 50 − 2x.
 Setting A’(x) = 0 we get x = 25 as the only critical point.
 Note that A’’(x) = −2 < 0 so by the second derivative test x = 25 is a
local maximum.
 It is also the global maximum because as you approach the endpoints
the area decreases.
 Thus, x = 25 and y = 50 − x = 50 − 25 = 25 are the dimensions that
maximize the area.
 So, among the rectangles with fixed perimeter, a square is the one that
maximizes the area
Examples of single independent variable

32. Problem 2. A cylindrical can is to be made to hold 1000 cm3 of oil. Find the
dimensions of the can that will minimize the cost of the metal when manufacturing
the can.
The volume is V = πr2h = 1000
and we want to minimize the total area A = 2πrh + 2πr2 .
We can solve h in terms of r by using πr2h = 1000.
We get h = 1000/ πr2 .
So A(r) = 2πr (1000/ πr2 ) + 2 πr2 = (2000/ r) + 2 πr2 is a function of r only.
The domain of this function is (0,∞).
So we get A’ (r) = − 2000 /r2 + 4πr.
Setting A’ (r) = 0 we get r = (500 /π)1/3 .
Next, A’’(r) = 4000/ r3 + 4π
so A’’((500 /π)1/3 ) > 0 so r = (500 /π)1/3 is a local minimum by the second
derivative test.
It is also global minimum, because as you approach the endpoints the surface
area increases. So the dimensions of the can that minimize the cost of the
metal is
r = (500 /π)1/3 and
h = 100/ (250π)1/3

35. Thank you