1. Example : Optimal blending of gasoline
Stream 1
Stream 2
Stream 3
Stream 4
Product
1 kg/s
m2 = ?
m3 = ?
m4 = ?
Stream 1
99 octane
0 % benzene
p1 = 0.1 (1 + m1) $/kg
Stream 2
105 octane
p2 = $/kg
Stream 3
95 → 97 octane
p3 = $/kg
Stream 4
2 % benzene
p4 = $/kg
Product
> 98 octane
< 1 % benzene
Disturbance

2. Optimization problem Degrees of freedom uo = (m1 m2 m3 m4 )T
min J = i pi mi = 0.1(1 + m1) m m m m4
subject to
m1 + m2 + m3 + m4 = 1
m1 ¸ 0; m2 ¸ 0; m3 ¸ 0; m4 ¸ 0
m1 · 0.4
99 m m m m4 ¸ 98 (octane constraint)
2 m4 · (benzene constraint)

3. Exercise 1: Optimal operation
The Matlab function quadprog will be used to find optimal operation
Calculate the optimal operating point for the nominal values (octane in stream 3 equal 95)
Calculate the optimal operating point with disturbance (octane in stream 3 equal 97)
Report value of objective function and flowrates for both cases
How many degrees of freedom, active constraints and unconstrained degrees of freedom are there?

4. Optimal solution Nominal optimal solution (d* = 95):
u0,opt = ( )T ) Jopt= $
Optimal solution with d=octane stream 3=97:
u0,opt = ( )T ) Jopt= $
m1 + m2 + m3 + m4 = 1
m1 ¸ 0; m2 ¸ 0; m3 ¸ 0; m4 ¸ 0
m1 · 0.4
99 m m m m4 ¸ 98 (octane constraint)
2 m4 · (benzene constraint)
3 active constraints ) 1 unconstrained degree of freedom

5. Exercise 1 cont.: Optimal operation
There is one unconstrained degree of freedom, which can be used to optimize operation
What is the loss of using this unconstrained DOF to maintain the following constant at nominal set point: m1 m2 m3

6. Implementation of optimal solution
Available ”measurements”: y = (m1 m2 m3 m4)T
Control active constraints:
Keep m4 = 0
Adjust one (or more) flow such that m1+m2+m3+m4 = 1
Adjust one (or more) flow such that product octane = 98
Remaining unconstrained degree of freedom
c=m1 is constant at ) Loss = $
c=m2 is constant at ) Infeasible (cannot satisfy octane = 98)
c=m3 is constant at ) Loss = $
Easily implemented in control system

7. Exercise 2: Maximum gain rule
Assume that there is an implementation error of 0.01 kg/s in flowrates
Use the singular value rule to find the approximate losses for the following cases:
m1 constant
m2 constant
m3 constant
Optimal combination of m1 and m2 constant (use Null space method for finding optimal combination)

8. Maximum gain rule: Details
c = m2
Need “gain” and “span”
Can use any of m1, m2 and m3 as remaining d.o.f. (use m1)
m1 + m2 + m3 = 1
99m m m3 = 98
) 99m m (1 - m1 - m2) = 98
) 4m1 + 10m2 = 3
) Gain = - 0.4
Span is simply the difference of optimal values for d = 95 and d = 97

9. Basis: Want optimal value of c independent of disturbances )
Unconstrained degrees of freedom: C. Optimal measurement combination (Alstad, 2002)
Basis: Want optimal value of c independent of disturbances )
 copt = 0 ¢  d
Find optimal solution as a function of d: uopt(d), yopt(d)
Linearize this relationship: yopt = F d
F – sensitivity matrix
Want:
To achieve this for all values of  d:
Always possible if
Optimal when we disregard implementation error (n)

10. Optimal combination c = h1 m1 + h2 m2 Requirement: HF = 0
From optimization:  mopt = F  d
sensitivity matrix F = ( )T
) – 0.03 h1 – 0.06 h2 = 0
This has infinite number of solutions
Fix h1 = 1 to get h2 = 0.5
) Loss = 0

11. Singular value rule CV c span c G |Gs| Loss factor 1/|Gs|2 m1 0.0700
1.000
14.29
0.0049 m2
0.1310
-0.400
3.053
0.1073 m3
0.1910
-0.600
3.141
0.1013
m1-0.5m2
0.015
1.200
77.41
1.558e-4

12. Blending: Example of implementation of ”self-optimizing” constant setpoint policyFC OC
mtot.s = 1 kg/s
mtot m3
m4 = 0 kg/s
Octanes = 98
Octane m2
Stream 2
Stream 1
Stream 3
Stream 4
cs =
0.5
m1 = cs m2
Octane varies
Selected ”self-optimizing” variable: c = m1 – 0.5 m2
Changes in feed octane (stream 3) detected by octane controller (OC)
Implementation is optimal provided active constraints do not change
Price changes can be included as corrections on setpoint cs
Comment: Example is for illustration. Use MPC in practice (changing constraints)!

13. Dynamic simulations m3 m1 m2 m4
mtot

14. Details of dynamics P-control of tank level: m = Kc M with Kc=0.001
The tank has a variable mass M of 1000m (where m is the total flowrate)
Time constant in valves are 10s
Octane measurement has a delay of 60 s
Each flow may vary between 0 and two times nominal value
m4 is not used so inputs are [m1, m2, m3]' = Flowrates
Output:
y1 = Product flowrate
y2 = Octane number in product
y3 = c (a selected measurement, or a combination of two measurements)

15. Exercise 3: Linear model
Derive the linear model for the different control policies:
m1 constant
m2 constant
m3 constant
m m2 constant

16. Linear Model Recall ) 99m1 + 105 m2 + 95 m3 = 98 y1 + y2
m1 + m2 + m3 = m = y1
99m m m3 = X m = y2 y1
Linearize second equation
99m m m3 = y2* y1 + y2 y1*
) 99m m m3 = 98 y1 + y2
) = 98 (m1 + m2 + m3) + y2
) m1 + 7 m2 - 3 m3 = y2

17. Linear analysis
Linear model with controlled variable c = m m2 constant:
G = 1/(1000s + 1)(10s + 1) [ ;
1e-60s 7e-60s e-60s; ]
Steady state RGA:
c = m1
c = m2
0.3000
0.7000
1.0000
0.0000
0.7500
0.0000
0.2500
1.0000
c = m3
c = m m2
1.1667
0.0000
1.0000
0.1250
0.2500
0.6250
0.0417
0.5833
0.3750
0.8333
0.1667
0.0000

18. Controller tuning: SIMC tuning rules
Flow controller:
G(s) = 1/((1000s+1)*(10s+1))
k = 1
τ1 = /2 = 1005
θ = 10/2 = 5
K = τ1/(2 θ k) = 100.5
τI = min(τ1, 8θ) = 40
Octane controller
G(s) = k'*e-60s/((1000s+1)*(10s+1))
k = 7, 1, -3 depending on pairing
τ1 = /2 = 1005
θ = /2 = 65
K = τ1/(2 θ k) = 7.73/k
τI = min(τ1 , 8 θ) = 520
K = 1.10 if m1 controls octane
K = 7.73 if m2 controls octane
K = if m3 controls octane

19. Dynamic simulations Disturbance
Octane of stream 3 steps from 95 to 94 at t=0
Octane of stream 3 steps from 94 to 97 at t=3000
Profit/kg = (product price - raw material price - TV)/amount of product
TV is a function of controller usage
Product price = 0.2 when octane > 97.9
0.15 when octane < 97.9
Simulation time is s

20. c = m1: Profit/kg =

21. c = m2: Infeasible

22. c = m3: Profit/kg =

23. c = m m2: Profit/kg =

24. Selecting control structure
Loss calculations shows zero losses for linear combinations of two measurements (m m2 constant)
Singular value rule shows a small loss factor for linear combination of two measurements (m m2 constant)
Steady state RGA indicates interactions when using the linear combination m m2 constant. Are other combinations better?
Dynamic simulations shows that there is better to keep m3 constant than m m2 constant, but is it possible to overcome this by re-tuning the controllers?