Note: This page has been translated by MathWorks. Please click here

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

This example shows how to convert a problem from mathematical form into Optimization Toolbox™ solver syntax. While the problem is a linear program, the techniques apply to all solvers.

The variables and expressions in the problem represent a model of operating a chemical plant, from an example in Edgar and Himmelblau [1]. There are two videos that describe the problem.

Optimization Modeling 1 shows the problem in pictorial form. It shows how to generate the mathematical expressions of Model Description from the picture.

Optimization Modeling 2 describes how to convert these mathematical expressions into Optimization Toolbox solver syntax. This video shows how to solve the problem, and how to interpret the results.

The remainder of this example is concerned solely with transforming the problem to solver syntax. The example closely follows the video Optimization Modeling 2. The main difference between the video and the example is that this example shows how to use named variables, or index variables, which are similar to hash keys. This difference is in Combine Variables Into One Vector.

The video Optimization Modeling 1 suggests that one way to convert a problem into mathematical form is to:

Get an overall idea of the problem

Identify the goal (maximizing or minimizing something)

Identify (name) variables

Identify constraints

Determine which variables you can control

Specify all quantities in mathematical notation

Check the model for completeness and correctness

For the meaning of the variables in this section, see the video Optimization Modeling 1.

The optimization problem is to minimize the objective function, subject to all the other expressions as constraints.

The objective function is:

`0.002614 HPS + 0.0239 PP + 0.009825 EP`

.

The constraints are:

`2500`

≤ `P1`

≤ `6250`

`I1`

≤ `192,000`

`C`

≤ `62,000`

`I1 - HE1`

≤ `132,000`

`I1 = LE1 + HE1 + C`

```
1359.8
I1 = 1267.8 HE1 + 1251.4 LE1 + 192 C + 3413 P1
```

`3000`

≤ `P2`

≤ `9000`

`I2`

≤ `244,000`

`LE2`

≤ `142,000`

`I2 = LE2 + HE2`

```
1359.8
I2 = 1267.8 HE2 + 1251.4 LE2 + 3413 P2
```

```
HPS
= I1 + I2 + BF1
```

```
HPS = C + MPS +
LPS
```

`LPS = LE1 + LE2 + BF2`

`MPS = HE1 + HE2 + BF1 - BF2`

```
P1
+ P2 + PP
```

≥ `24,550`

```
EP
+ PP
```

≥ `12,000`

`MPS`

≥ `271,536`

`LPS`

≥ `100,623`

All variables are positive.

To solve the optimization problem, take the following steps.

The steps are also shown in the video Optimization Modeling 2.

To find the appropriate solver for this problem, consult the Optimization Decision Table. The table
asks you to categorize your problem by type of objective function
and types of constraints. For this problem, the objective function
is linear, and the constraints are linear. The decision table recommends
using the `linprog`

solver.

As you see in Problems Handled by Optimization Toolbox Functions or
the `linprog`

function reference
page, the `linprog`

solver solves problems of the
form

$$\underset{x}{\mathrm{min}}{f}^{T}x\text{suchthat}\{\begin{array}{c}A\cdot x\le b,\\ Aeq\cdot x=beq,\\ lb\le x\le ub.\end{array}$$ | (1-1) |

*f*means a row vector of constants^{T}x*f*multiplying a column vector of variables*x*. In other words,*f*=^{T}x*f*(1)*x*(1) +*f*(2)*x*(2) + ... +*f*(*n*)*x*(*n*),where

*n*is the length of*f*.*A x*≤*b*represents linear inequalities.*A*is a*k*-by-*n*matrix, where*k*is the number of inequalities and*n*is the number of variables (size of*x*).*b*is a vector of length*k*. For more information, see Linear Inequality Constraints.*Aeq x*=*beq*represents linear equalities.*Aeq*is an*m*-by-*n*matrix, where*m*is the number of equalities and*n*is the number of variables (size of*x*).*beq*is a vector of length*m*. For more information, see Linear Equality Constraints.*lb*≤*x*≤*ub*means each element in the vector*x*must be greater than the corresponding element of*lb*, and must be smaller than the corresponding element of*ub*. For more information, see Bound Constraints.

The syntax of the `linprog`

solver,
as shown in its function reference page, is

[x fval] = linprog(f,A,b,Aeq,beq,lb,ub);

The inputs to the `linprog`

solver are the
matrices and vectors in Equation 1-1.

There are 16 variables in the equations of Model Description. Put these variables into one vector. The
name of the vector of variables is *x* in Equation 1-1. Decide on an order,
and construct the components of *x* out of the variables.

The following code constructs the vector using a cell array of names for the variables.

variables = {'I1','I2','HE1','HE2','LE1','LE2','C','BF1',... 'BF2','HPS','MPS','LPS','P1','P2','PP','EP'}; N = length(variables); % create variables for indexing for v = 1:N eval([variables{v},' = ', num2str(v),';']); end

Executing these commands creates the following named variables in your workspace:

These named variables represent index numbers for the components
of *x*. You do not have to create named variables.
The video Optimization
Modeling 2 shows how to solve the problem simply using the
index numbers of the components of *x*.

There are four variables with lower bounds, and six with upper bounds in the equations of Model Description. The lower bounds:

`P1`

≥ `2500`

`P2`

≥ `3000`

`MPS`

≥ `271,536`

`LPS`

≥ `100,623`

.

Also, all the variables are positive, which means they have a lower bound of zero.

Create the lower bound vector `lb`

as a vector
of `0`

, then add the four other lower bounds.

lb = zeros(size(variables)); lb([P1,P2,MPS,LPS]) = ... [2500,3000,271536,100623];

The variables with upper bounds are:

`P1`

≤ `6250`

`P2`

≤ `9000`

`I1`

≤ `192,000`

`I2`

≤ `244,000`

`C`

≤ `62,000`

`LE2`

≤ `142000`

.

Create the upper bound vector as a vector of `Inf`

,
then add the six upper bounds.

ub = Inf(size(variables)); ub([P1,P2,I1,I2,C,LE2]) = ... [6250,9000,192000,244000,62000,142000];

There are three linear inequalities in the equations of Model Description:

`I1 - HE1`

≤ `132,000`

`EP + PP`

≥ `12,000`

`P1 + P2 + PP`

≥ `24,550`

.

In order to have the equations in the form *A
x*≤*b*, put
all the variables on the left side of the inequality. All these equations
already have that form. Ensure that each inequality is in "less
than" form by multiplying through by –1 wherever appropriate:

`I1 - HE1`

≤ `132,000`

`-EP - PP`

≤ `-12,000`

`-P1 - P2 - PP`

≤ `-24,550`

.

In your MATLAB^{®} workspace, create the `A`

matrix
as a 3-by-16 zero matrix, corresponding to 3 linear inequalities in
16 variables. Create the `b`

vector with three components.

A = zeros(3,16); A(1,I1) = 1; A(1,HE1) = -1; b(1) = 132000; A(2,EP) = -1; A(2,PP) = -1; b(2) = -12000; A(3,[P1,P2,PP]) = [-1,-1,-1]; b(3) = -24550;

There are eight linear equations in the equations of Model Description:

`I2 = LE2 + HE2`

```
LPS
= LE1 + LE2 + BF2
```

```
HPS = I1 + I2
+ BF1
```

`HPS = C + MPS + LPS`

`I1 = LE1 + HE1 + C`

```
MPS
= HE1 + HE2 + BF1 - BF2
```

```
1359.8 I1
= 1267.8 HE1 + 1251.4 LE1 + 192 C + 3413 P1
```

```
1359.8
I2 = 1267.8 HE2 + 1251.4 LE2 + 3413 P2
```

.

In order to have the equations in the form *Aeq
x*=*beq*, put all the
variables on one side of the equation. The equations become:

`LE2 + HE2 - I2 = 0`

```
LE1
+ LE2 + BF2 - LPS = 0
```

```
I1 + I2 +
BF1 - HPS = 0
```

```
C + MPS + LPS - HPS
= 0
```

`LE1 + HE1 + C - I1 = 0`

`HE1 + HE2 + BF1 - BF2 - MPS = 0`

```
1267.8
HE1 + 1251.4 LE1 + 192 C + 3413 P1 - 1359.8 I1 = 0
```

```
1267.8
HE2 + 1251.4 LE2 + 3413 P2 - 1359.8 I2 = 0
```

.

Now write the `Aeq`

matrix and `beq`

vector
corresponding to these equations. In your MATLAB workspace, create
the `Aeq`

matrix as an 8-by-16 zero matrix, corresponding
to 8 linear equations in 16 variables. Create the `beq`

vector
with eight components, all zero.

Aeq = zeros(8,16); beq = zeros(8,1); Aeq(1,[LE2,HE2,I2]) = [1,1,-1]; Aeq(2,[LE1,LE2,BF2,LPS]) = [1,1,1,-1]; Aeq(3,[I1,I2,BF1,HPS]) = [1,1,1,-1]; Aeq(4,[C,MPS,LPS,HPS]) = [1,1,1,-1]; Aeq(5,[LE1,HE1,C,I1]) = [1,1,1,-1]; Aeq(6,[HE1,HE2,BF1,BF2,MPS]) = [1,1,1,-1,-1]; Aeq(7,[HE1,LE1,C,P1,I1]) = [1267.8,1251.4,192,3413,-1359.8]; Aeq(8,[HE2,LE2,P2,I2]) = [1267.8,1251.4,3413,-1359.8];

The objective function is

*f ^{T}x* =

```
0.002614
HPS + 0.0239 PP + 0.009825 EP
```

.Write this expression as a vector `f`

of multipliers
of the *x* vector:

f = zeros(size(variables)); f([HPS PP EP]) = [0.002614 0.0239 0.009825];

You now have inputs required by the `linprog`

solver.
Call the solver and print the outputs in formatted form:

options = optimoptions('linprog','Algorithm','dual-simplex'); [x fval] = linprog(f,A,b,Aeq,beq,lb,ub,options); for d = 1:N fprintf('%12.2f \t%s\n',x(d),variables{d}) end fval

The result:

Optimal solution found. 136328.74 I1 244000.00 I2 128159.00 HE1 143377.00 HE2 0.00 LE1 100623.00 LE2 8169.74 C 0.00 BF1 0.00 BF2 380328.74 HPS 271536.00 MPS 100623.00 LPS 6250.00 P1 7060.71 P2 11239.29 PP 760.71 EP fval = 1.2703e+03

The `fval`

output gives the smallest value
of the objective function at any feasible point.

The solution vector `x`

is the point where
the objective function has the smallest value. Notice that:

`BF1`

,`BF2`

, and`LE1`

are`0`

, their lower bounds.`I2`

is`244,000`

, its upper bound.The nonzero components of the

`f`

vector are`HPS`

—`380,328.74`

`PP`

—`11,239.29`

`EP`

—`760.71`

The video Optimization Modeling 2 gives interpretations of these characteristics in terms of the original problem.

[1] Edgar, Thomas F., and David M. Himmelblau. *Optimization
of Chemical Processes.* McGraw-Hill, New York, 1988.

Was this topic helpful?