Chapter 1 The Nature and Organization of Optimization Problems

Chapter 1

Chapter 1
The Nature and Organization of
Optimization Problems

8

WHY OPTIMIZE?
1. Improved yields, reduced pollutants

Chapter 1

2. Reduced energy consumption
3. Higher processing rates

4. Reduced maintenance, fewer shutdowns
5. Better understanding of process (simulation)
But there are always positive and negative factors to be
weighed

9

Chapter 1

10

Chapter 1

11

Chapter 1

OPTIMIZATION
• Interdisciplinary Field
Max Profit
Min Cost
Max Efficiency
• Requires
1. Critical analysis of process
2. Definition of performance objective

3. Prior experience (engr. judgment)

12

Chapter 1

13

Chapter 1

14

Chapter 1

15

Chapter 1
Min reflux to
achieve separation

Figure E1.4-3

Flooding
constraint

Optimal Reflux for Different Fuel Costs
16

Chapter 1

17

Chapter 1

18

Chapter 1

19

Chapter 1

20

Chapter 1

Material Balance Reconciliation

21

min  (mA  mCi  mBi ) 2

Least squares solution:
P

Chapter 1

i 1

opt. mA is the “average” value

any constraints on mA?

22

Chapter 1

23

THREE INGREDIENTS IN OPTIMIZATION PROBLEM
1. Objective function

economic model

Chapter 1

2. Equality Constraints 

3. Inequality Constraints

Process model

1. min f(x)

xnx1

3. g( x )  0

(m1 )

2. subject to h( x )  0

(feasible

(m2 )
2
region  :)
3

dependent variables 


independent variables 

relate to m1 and perhaps m2
24

Chapter 1

25

TABLE 1

Chapter 1

THE SIX STEPS USED TO SOLVE OPTIMIZATION PROBLEMS

1. Analyze the process itself so that the process variables
and specific characteristics of interest are defined, i.e.,
make a list of all of the variables.
2. Determine the criterion for optimization and specify
the objective function in terms of the above variables

together with coefficients. This step provides the
performance model (sometimes called the economic
model when appropriate).

26

Chapter 1

3. Develop via mathematical expressions a valid process
or equipment model that relates the input-output variables
of the process and associated coefficients. Include both
equality and inequality constraints. Use well-known
physical principles (mass balances, energy balances),
empirical relations, implicit concepts, and external
restrictions. Identify the independent and dependent
variables (number of degrees of freedom).

27

Chapter 1

4. If the problem formulation is too large in scope:
(A)Break it up into manageable parts and/or
(B)Simplify the objective function

5. Apply a suitable optimization technique to the
mathematical statement of the problem.
6. Check the answers and examine the sensitivity of the
result to changes in the coefficients in the problem and
the assumptions.

28

EXAMPLES – SIX STEPS OF OPTIMIZATION

Chapter 1

specialty chemical
100,000 bbl/yr.
2 costs inventory (carrying) or storage, production cost >

how many bbl produced per run?

Step 1
define variables
Q = total # bbl produced/yr (100,000)
D = # bbl produced per run
n = # runs/yr
29

Step 2
develop objective function

Chapter 1

inventory, storage cost = k1D
production cost

= k2

per run

(set up
cost)

+

k3

D

operating
cost per unit

(could be nonlinear)

C  k1 D  n (k2  k3 D )
Q
n
D

C  k1 D  k2

Q
 k3Q
D
30

Step 3
evaluate constraints

Chapter 1

integer
n
continuous
D>0
Step 4
simplification – none necessary

31

Step 5
computation of the optimum

Chapter 1

analytical vs. numerical solution
dC
k2Q
 k1  2  0
dD
D

D opt 

k1  1.0

Q  10 5

k2Q
k1

k2  10,000

D opt  31,622

k3  4.0

flat optimum
30,000  D  70,000
 good answer

check if minimum?
d 2C 2k2Q

0
2
3
dD
D

32

Chapter 1

33

supposecost per run  k2  k4 D1/ 2

Chapter 1

dC
k2Q
k4Q

 k1  2 
0
3/ 2
2D
dD
D
analytical solution?

Step 6
Sensitivity of the optimum
subst Dopt into C
C opt  2 k1k2Q  k3Q

C opt
kQ
 2
k1
k1

C opt
k1Q

k2
k2

C opt
Q
k3

C opt
kk
 1 2  k3
Q
Q

31,620
3.162
100,000
4.316

34

D

opt



k2Q
k1

Chapter 1

k2Q  1
D opt

k1 2k1
k1

k2Q 1
D opt

k1 2k2
k2

D opt
0
k3

1.581
0

k2Q 1
D opt

k1 2Q
Q

k1  1.0

 15,810

k2  10,000

0.158
k3  4.0

Q  100,000
35

RELATIVE SENSITIVITY (Percentage change)
opt
opt
opt
C
C
C


/
ln
SkC1 

k1 / k1
 ln k1

Chapter 1

C opt  463, 240

k1  1.0

kQ
C opt
 2  31,620
k1
k1

SkC1  0.0683

SkD1  0.5

SkC3  0.863

SQD  0.5

SkC2  0.0683

SkD2  0.5

SQC  0.932

SkD3  0

abs. sens. on D

k1  k2  Q  k3

opt
C
k1

31620 (1.0)
C
Sk1 

 0.0683
opt
k1 C
463,240

abs. sens. on C

k3  k1  Q  k2

36

PIPELINE PROBLEM

Chapter 1

variables
V
p
f
Re
D

parameters



L
m
pipe cost
electricity cost
#operating days/yr
pump efficiency

37

Equality Constraints

Chapter 1



 D2

v m

4
Re  Dv  / 
2 L
p  2  v
f
D
f  .046 Re 0.2

38

Chapter 1

39

min (Coper + Cinv.)
subject to equality constraints

Chapter 1

L 2
p  2 f v
D

need analytical formula for f
f  .046 Re0.2

pump power cost  Co m

smooth tubes

p



m  mass flow rate 

 D2
4

v

substituting for ∆p,
 4.8 0.2  2.8 2.0
Coper  Co D  m 
Cinv  C1 D

1.5

(a nnua lized)

40



 2.8 D 4.8  C1D1.5
Total cost  TC  Co  0.2  2 m

Chapter 1

(constraint eliminated by substitution)
d (TC )
0
dD
solving,
(D opt )6.3

necessary condition for a minimum

Co 0.2 2 2.8
  m

C1



C
(D opt )   o 
 C1 



0.16

 .32m.45  .03

opt velocity V opt 



4
(sensitivity analysis)

m

2

Dopt

41

optimum velocity

Chapter 1

non-viscous liquids
gases (effect of ρ)

3 to 6 ft/sec.
30 to 60 ft/sec.

at higher pressure, need to use different
constraint (isothermal)




p1  


 fL ln  p2  
2 p1 
2
p 
 .323  
 S1V1 
p1  p2 
D
24










S1   gV1  upstream velocity

or use Weymouth equation
for large L, ln ( ) can be neglected
exceptions: elevation changes, slurries (settling),
extremely viscous oils (laminar flow, 42
f different)

Chapter 1

Heat Exchanger Variables
(given flow rate of one
1. heat transfer area
fluid, inlet
2. heat duty
temperatures, one
3. flow rates (shell, tube)
outlet temp., phys.
4. no. passes (shell, tube)
props.)
5. baffle spacing
6. length
7. diam. of shell, tubes
8. approach temperature
9. fluid A (shell or tube, co-current or countercurrent)
10.tube pitch, no. tubes
11.velocity (shell, tube)
12.∆p (shell, tube)
13.heat transfer coeffs (shell, tube)
14.exchanger type (fins?)
15.material of construction
43