examples

Software and examples

Optimisation Software

LINGO

Used to solve optimisation problems

To start entering a new optimisation 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

Optimisation 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=0.667

Smallest and largest elements in absolute value= 0.200000 80.0000

No. < : 1 No. =: 1 No. > : 2, Obj=MIN, GUBs

More Optimisation 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. Minimise the operating cost!

Lab Experiment (Contd)

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= 0

Nonlinear rows= 1 Nonlinear vars= 2 Nonlinear constraints= 0

Nonzeros= 7 Constraint nonz= 4 Density=0.583

Optimal solution found at step: 4

Objective 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

Value of variable x1: 1.071

Value of variable x2: 1.286


Coal Conversion Plant

What are the optimal production rates of gaseous and liquid fuels that maximise 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


Coal Conversion Plant (Contd)

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



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



LINGO Input

Model:

max = 3*x1 + 5*x2;

3*x1 + 2*x2 : 2, Obj=MAX, GUBs

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 (Contd)

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



We define the transportation loads (tons/day) going from each MeOH plant to each MeAc 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 = 2X1A + X1B + 5X1C + 3X2A + 0X2B + 8X2C + 11X3A + 6X3B + 15X3C + 7X4A + X4B + 9X4C



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



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


examples

Documents

x1 x2

value of variable x1

value of variable x2

gasification capacity

variable value reduced

cost of coal

nonnegativity x1

objective value