Main

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!!!! Example Problem
~~!!!!~~ [[https://apmonitor.com/do/index.php/Main/DynamicOptimizationBenchmarks|Dynamic Optimization with Intermediates]]

!!!! Example 4

* Batch reactor with consecutive reactions A->B->C

{$\max_{T(t)} x_2 \left( t_f \right)$}

{$\mathrm{subject \; to}$}

{$\frac{dx_1}{dt}=-k_1 \, x_1^2$}

{$\frac{dx_2}{dt}=k_1 \, x_1^2 - k_2 \, x_2$}

{$k_1 = 4000 \, \exp{\left(-\frac{2500}{T}\right)}$}

{$k_2 = 6.2e5 \, \exp{\left(-\frac{5000}{T}\right)}$}

{$x(0) = [1 \; 0]^T$}

{$298 \le T \le 398$}

{$t_f=1$}

!!!! Solution to [[https://apmonitor.com/do/index.php/Main/DynamicOptimizationBenchmarks|Dynamic Optimization with Intermediates]]

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(:toggle hide gekko4 button show="Show GEKKO (Python) Code":)

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(:source lang=python:)

import numpy as np

import matplotlib.pyplot as plt

from gekko import GEKKO

m = GEKKO()

nt = 101

m.time = np.linspace(0,1,nt)

# Parameters

T = m.MV(value=362,ub=398,lb=298)

T.STATUS = 1

T.DCOST = 0

# Variables

x1 = m.Var(value=1)

x2 = m.Var(value=0)

p = np.zeros(nt)

p[-1] = 1.0

final = m.Param(value=p)

# Intermediates

k1 = m.Intermediate(4000*m.exp(-2500/T))

k2 = m.Intermediate(6.2e5*m.exp(-5000/T))

# Equations

m.Equation(x1.dt()==-k1*x1**2)

m.Equation(x2.dt()==k1*x1**2 - k2*x2)

# Objective Function

m.Obj(-x2*final)

m.options.IMODE = 6

m.solve()

print('Objective: ' + str(x2[-1]))

plt.figure(1)

plt.subplot(2,1,1)

plt.plot(m.time,x1.value,'k:',LineWidth=2,label=r'$x_1$')

plt.plot(m.time,x2.value,'b-',LineWidth=2,label=r'$x_2$')

plt.ylabel('Value')

plt.legend(loc='best')

plt.subplot(2,1,2)

plt.plot(m.time,T.value,'r--',LineWidth=2,label=r'$T$')

plt.legend(loc='best')

plt.xlabel('Time')

plt.ylabel('Value')

plt.show()

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~~!! Intermediate Variables and Equations~~

!!! Limitations

%blue%A%red%P%black%Monitor is designed to encourage the user to construct well-posed models for numerical solution. One limitation that may be encountered is the 100 variable limit in each equation. Excessive use of intermediate variables may lead to the violation of this limit. If this limit is encountered, the user can remediate this problem by converting an intermediate variable to a regular implicit variable.

## Intermediates

## Main.Intermediates History

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Intermediate variables are declared in the ''Intermediates ... End Intermediates'' section of the model file. The intermediate variables may be defined in one section or in multiple declarations throughout the model. Intermediate variables are parsed sequentially, from top to bottom. To avoid inadvertent overwrites, intermediate variable can be defined once. In the case of intermediate variables, the order of declaration is critical. If ~~a variable~~ is used before ~~it's~~ definition, ~~it will contain a default value of 1~~.

to:

Intermediate variables are declared in the ''Intermediates ... End Intermediates'' section of the model file. The intermediate variables may be defined in one section or in multiple declarations throughout the model. Intermediate variables are parsed sequentially, from top to bottom. To avoid inadvertent overwrites, intermediate variable can be defined once. In the case of intermediate variables, the order of declaration is critical. If an intermediate is used before the definition, an error reports that there is an uninitialized value.

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The intermediate variables are processed before the implicit equation residuals, every time the solver requests model information. As opposed to implicitly calculated variables, the ~~explicit variables ~~are calculated ~~once~~ and substituted into other ~~explicit~~ or implicit equations.

to:

The intermediate variables are processed before the implicit equation residuals, every time the solver requests model information. As opposed to implicitly calculated variables, the intermediates are calculated repeatedly and substituted into other intermediate or implicit equations.

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!!!! Example ~~4~~

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!!!! Example Problem

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Attach:download.png [[https://apmonitor.com/do/index.php/Main/DynamicOptimizationBenchmarks|Dynamic Optimization with Intermediates]] (see Problem #4)

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!!!!~~ Solution to~~ [[https://apmonitor.com/do/index.php/Main/DynamicOptimizationBenchmarks|Dynamic Optimization with Intermediates]]

to:

!!!! [[https://apmonitor.com/do/index.php/Main/DynamicOptimizationBenchmarks|Dynamic Optimization with Intermediates]]

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!!!! Example 4

* Batch reactor with consecutive reactions A->B->C

{$\max_{T(t)} x_2 \left( t_f \right)$}

{$\mathrm{subject \; to}$}

{$\frac{dx_1}{dt}=-k_1 \, x_1^2$}

{$\frac{dx_2}{dt}=k_1 \, x_1^2 - k_2 \, x_2$}

{$k_1 = 4000 \, \exp{\left(-\frac{2500}{T}\right)}$}

{$k_2 = 6.2e5 \, \exp{\left(-\frac{5000}{T}\right)}$}

{$x(0) = [1 \; 0]^T$}

{$298 \le T \le 398$}

{$t_f=1$}

!!!! Solution to [[https://apmonitor.com/do/index.php/Main/DynamicOptimizationBenchmarks|Dynamic Optimization with Intermediates]]

(:html:)

<iframe width="560" height="315" src="https://www.youtube.com/embed/yFprG0iJQUE" frameborder="0" allowfullscreen></iframe>

(:htmlend:)

(:toggle hide gekko4 button show="Show GEKKO (Python) Code":)

(:div id=gekko4:)

(:source lang=python:)

import numpy as np

import matplotlib.pyplot as plt

from gekko import GEKKO

m = GEKKO()

nt = 101

m.time = np.linspace(0,1,nt)

# Parameters

T = m.MV(value=362,ub=398,lb=298)

T.STATUS = 1

T.DCOST = 0

# Variables

x1 = m.Var(value=1)

x2 = m.Var(value=0)

p = np.zeros(nt)

p[-1] = 1.0

final = m.Param(value=p)

# Intermediates

k1 = m.Intermediate(4000*m.exp(-2500/T))

k2 = m.Intermediate(6.2e5*m.exp(-5000/T))

# Equations

m.Equation(x1.dt()==-k1*x1**2)

m.Equation(x2.dt()==k1*x1**2 - k2*x2)

# Objective Function

m.Obj(-x2*final)

m.options.IMODE = 6

m.solve()

print('Objective: ' + str(x2[-1]))

plt.figure(1)

plt.subplot(2,1,1)

plt.plot(m.time,x1.value,'k:',LineWidth=2,label=r'$x_1$')

plt.plot(m.time,x2.value,'b-',LineWidth=2,label=r'$x_2$')

plt.ylabel('Value')

plt.legend(loc='best')

plt.subplot(2,1,2)

plt.plot(m.time,T.value,'r--',LineWidth=2,label=r'$T$')

plt.legend(loc='best')

plt.xlabel('Time')

plt.ylabel('Value')

plt.show()

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(y+2/x)^(x*z) * (log(tanh(sqrt(y-x+x^2))+3))^2 = 2+sinh(y)+acos(x+y)+asin(x/y)

to:

(y+2/x)^(x*z) * (log(tanh(sqrt(y-x+x^2))+3))^2 = &

2+sinh(y)+acos(x+y)+asin(x/y)

2+sinh(y)+acos(x+y)+asin(x/y)

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!!! Limitations

%blue%A%red%P%black%Monitor is designed to encourage the user to construct well-posed models for numerical solution. One limitation that may be encountered is the 100 variable limit in each equation. Excessive use of intermediate variables may lead to the violation of this limit. If this limit is encountered, the user can remediate this problem by converting an intermediate variable to a regular implicit variable.

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!! Intermediate Variables and Equations

Intermediate variables are useful to decrease the complexity of the model. These variables store temporary calculations with results that are not reported in the final solution reports. In many models, the temporary variables outnumber the regular variables by many factors. This model reduction often aides the solver in finding a solution by reducing the problem size.

Intermediate variables are declared in the ''Intermediates ... End Intermediates'' section of the model file. The intermediate variables may be defined in one section or in multiple declarations throughout the model. Intermediate variables are parsed sequentially, from top to bottom. To avoid inadvertent overwrites, intermediate variable can be defined once. In the case of intermediate variables, the order of declaration is critical. If a variable is used before it's definition, it will contain a default value of 1.

!!! Explicit calculation

The intermediate variables are processed before the implicit equation residuals, every time the solver requests model information. As opposed to implicitly calculated variables, the explicit variables are calculated once and substituted into other explicit or implicit equations.

!!! Clipping

When the intermediate variable is solely a function of parameters (not variables), the value may be clipped. This is accomplished by adding inequalities to the expression, separated by a comma. The inequalities may also be a function of other intermediate or regular variables.

!!! Example

(:table border=1 width=50% align=left bgcolor=#EEEEEE cellspacing=0:)

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! Original model

Model example

Parameters

p = 2

End Parameters

Variables

x

y

z

End Variables

Equations

exp(x*p)=y

z = p*$x + x

(y+2/x)^(x*z) * (log(tanh(sqrt(y-x+x^2))+3))^2 = 2+sinh(y)+acos(x+y)+asin(x/y)

End Equations

End Model

! Model simplified with use of intermediate variables

Model example

Parameters

p = 2

End Parameters

Variables

x

y

z

End Variables

Intermediates

exp_result = exp(x*p)

left = (y+2/x)^(x*z) * (log(tanh(sqrt(y-x+x^2))+3))^2

right = 2+sinh(y)+acos(x+y)+asin(x/y)

End Intermediates

Equations

exp_result=y

z = p*$x + x

left = right

End Equations

End Model

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! Example with intermediate variable clipping

Model example

Parameters

p = 1

End Parameters

Variables

x

End Variables

Intermediates

pi = p*3.1415, <1.5

End Intermediates

Equations

x = 0.5 * pi

End Equations

End Model

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Intermediate variables are useful to decrease the complexity of the model. These variables store temporary calculations with results that are not reported in the final solution reports. In many models, the temporary variables outnumber the regular variables by many factors. This model reduction often aides the solver in finding a solution by reducing the problem size.

Intermediate variables are declared in the ''Intermediates ... End Intermediates'' section of the model file. The intermediate variables may be defined in one section or in multiple declarations throughout the model. Intermediate variables are parsed sequentially, from top to bottom. To avoid inadvertent overwrites, intermediate variable can be defined once. In the case of intermediate variables, the order of declaration is critical. If a variable is used before it's definition, it will contain a default value of 1.

!!! Explicit calculation

The intermediate variables are processed before the implicit equation residuals, every time the solver requests model information. As opposed to implicitly calculated variables, the explicit variables are calculated once and substituted into other explicit or implicit equations.

!!! Clipping

When the intermediate variable is solely a function of parameters (not variables), the value may be clipped. This is accomplished by adding inequalities to the expression, separated by a comma. The inequalities may also be a function of other intermediate or regular variables.

!!! Example

(:table border=1 width=50% align=left bgcolor=#EEEEEE cellspacing=0:)

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! Original model

Model example

Parameters

p = 2

End Parameters

Variables

x

y

z

End Variables

Equations

exp(x*p)=y

z = p*$x + x

(y+2/x)^(x*z) * (log(tanh(sqrt(y-x+x^2))+3))^2 = 2+sinh(y)+acos(x+y)+asin(x/y)

End Equations

End Model

! Model simplified with use of intermediate variables

Model example

Parameters

p = 2

End Parameters

Variables

x

y

z

End Variables

Intermediates

exp_result = exp(x*p)

left = (y+2/x)^(x*z) * (log(tanh(sqrt(y-x+x^2))+3))^2

right = 2+sinh(y)+acos(x+y)+asin(x/y)

End Intermediates

Equations

exp_result=y

z = p*$x + x

left = right

End Equations

End Model

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! Example with intermediate variable clipping

Model example

Parameters

p = 1

End Parameters

Variables

x

End Variables

Intermediates

pi = p*3.1415, <1.5

End Intermediates

Equations

x = 0.5 * pi

End Equations

End Model

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