## Dynamic Control with Discrete Variables

## Main.DiscreteVariables History

Hide minor edits - Show changes to output

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An example problem for Mixed Integer Optimal Control is the Lotka Volterra fishing problem. It finds an fishing strategy over a 12 year horizon to equilibrate the predator ~~as prey fish to a minimum ~~steady~~ ~~state value. The Lotka Volterra equations for a predator-prey system have an additional equation to introduce fishing by man and with constants {`c_0=0.4`} and {`c_1=0.2`}.

to:

An example problem for Mixed Integer Optimal Control is the Lotka Volterra fishing problem. It finds an fishing strategy over a 12 year horizon to equilibrate the predator-prey fish to a sustainable steady-state value. The Lotka Volterra equations for a predator-prey system have an additional equation to introduce fishing by man and with constants {`c_0=0.4`} and {`c_1=0.2`}.

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plt.xlabel('Time')

plt.ylabel('Variables')

plt.legend(loc='best')

plt.show()

(:sourceend:)

(:divend:)

(:toggle hide solution2 button show="Show Python GEKKO Solution":)

(:div id=solution2:)

(:source lang=python:)

import numpy as np

import matplotlib.pyplot as plt

from gekko import GEKKO

m = GEKKO() # create GEKKO model

# Add 0.01 as first step

# 0,0.01,0.1,0.2,0.3,...11.9,12.0)

m.time = np.insert(np.linspace(0,12,121),1,0.01)

# change solver options

m.solver_options = ['minlp_gap_tol 0.001',\

'minlp_maximum_iterations 10000',\

'minlp_max_iter_with_int_sol 100',\

'minlp_branch_method 1',\

'minlp_integer_tol 0.001',\

'minlp_integer_leaves 0',\

'minlp_maximum_iterations 200']

c0 = 0.4

c1 = 0.2

last = m.Param(np.zeros(122))

last.value[-1] = 1

x0 = m.Var(value=0.5,lb=0)

x1 = m.Var(value=0.7,lb=0)

x2 = m.Var(value=0.0,lb=0)

w = m.MV(value=0,lb=0,ub=1,integer=True)

w.STATUS = 1

m.Obj(last*x2)

m.Equations([x0.dt() == x0 - x0*x1 - c0*x0*w,\

x1.dt() == - x1 + x0*x1 - c1*x1*w,\

x2.dt() == (x0-1)**2 + (x1-1)**2])

m.options.IMODE = 6

m.options.NODES = 3

m.options.SOLVER = 1

m.options.MV_TYPE = 0

m.solve()

plt.figure(1)

plt.step(m.time,w.value,'r-',label='w (0/1)')

plt.plot(m.time,x0.value,'b-',label=r'$x_0$')

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

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

plt.ylabel('Variables')

plt.legend(loc='best')

plt.show()

(:sourceend:)

(:divend:)

(:toggle hide solution2 button show="Show Python GEKKO Solution":)

(:div id=solution2:)

(:source lang=python:)

import numpy as np

import matplotlib.pyplot as plt

from gekko import GEKKO

m = GEKKO() # create GEKKO model

# Add 0.01 as first step

# 0,0.01,0.1,0.2,0.3,...11.9,12.0)

m.time = np.insert(np.linspace(0,12,121),1,0.01)

# change solver options

m.solver_options = ['minlp_gap_tol 0.001',\

'minlp_maximum_iterations 10000',\

'minlp_max_iter_with_int_sol 100',\

'minlp_branch_method 1',\

'minlp_integer_tol 0.001',\

'minlp_integer_leaves 0',\

'minlp_maximum_iterations 200']

c0 = 0.4

c1 = 0.2

last = m.Param(np.zeros(122))

last.value[-1] = 1

x0 = m.Var(value=0.5,lb=0)

x1 = m.Var(value=0.7,lb=0)

x2 = m.Var(value=0.0,lb=0)

w = m.MV(value=0,lb=0,ub=1,integer=True)

w.STATUS = 1

m.Obj(last*x2)

m.Equations([x0.dt() == x0 - x0*x1 - c0*x0*w,\

x1.dt() == - x1 + x0*x1 - c1*x1*w,\

x2.dt() == (x0-1)**2 + (x1-1)**2])

m.options.IMODE = 6

m.options.NODES = 3

m.options.SOLVER = 1

m.options.MV_TYPE = 0

m.solve()

plt.figure(1)

plt.step(m.time,w.value,'r-',label='w (0/1)')

plt.plot(m.time,x0.value,'b-',label=r'$x_0$')

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

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

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----

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(:toggle hide solution button show="Show Python Solution":)

to:

(:toggle hide solution button show="Show APM Python Solution":)

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# retrieve apm

# https://raw.githubusercontent.com/APMonitor/apm_python/master/apm.py

# or

# https://apmonitor.com/wiki/index.php/Main/PythonApp

from apm import

# alternative pip install with 'pip install APMonitor'

# from APMonitor.apm import *

# local APMonitor servers are available for Windows or Linux

# https://apmonitor.com/wiki/index.php/Main/APMonitorServer

# with clients in Python, MATLAB, and Julia

to:

from APMonitor.apm import *

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Attach:download.png [[Attach:thermostat_python_matlab.zip|Optimized Home Furnace Source Files]]

to:

Attach:download.png [[Attach:thermostat_python_matlab.zip|Optimized Home Furnace Source Files (APM MATLAB and APM Python)]]

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----

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Solve

to:

'''Solution'''

The integer problem is solved with [[https://gekko.readthedocs.io/en/latest/|Python GEKKO]]. The option '''integer'''=True is used to switch the variable from continuous to integer form. The APOPT solver is required to solve problem with integer variables but other solvers such as IPOPT can provide a relaxed solution where the integer conditions are not enforced.

The integer problem is solved with [[https://gekko.readthedocs.io/en/latest/|Python GEKKO]]. The option '''integer'''=True is used to switch the variable from continuous to integer form. The APOPT solver is required to solve problem with integer variables but other solvers such as IPOPT can provide a relaxed solution where the integer conditions are not enforced.

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----

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'''Solution'''

%width=550px%Attach:milp_thermostat.png

(:toggle hide python button show="Show Python GEKKO Source":)

(:div id=python:)

(:source lang=python:)

from gekko import GEKKO

import matplotlib.pyplot as plt

m = GEKKO() # create GEKKO model

m.time = [0,1,2,4,8,12,16,20,25,30,35,40,45,50,\

55,60,65,70,75,80,85,90,95,100,105,110,\

115,120,125,130,135,140,145,150]

# change solver options

m.solver_options = ['minlp_gap_tol 1.0e-2',\

'minlp_maximum_iterations 10000',\

'minlp_max_iter_with_int_sol 5000']

m.options.SOLVER = 1

m.options.IMODE = 6

# parameters

Kp = 60

taup = 120

T0 = 40

# create heater binary variable

u = m.MV(integer=True,lb=0,ub=1)

u.dcost = 0.1

u.status = 1

# controlled variable (temperature)

T = m.CV(value=T0)

T.SPHI = 68

T.SPLO = 70

T.STATUS = 1

T.TR_INIT = 0

# first order equation

m.Equation(taup * T.dt() == -(T-T0) + Kp * u)

m.solve() # solve MILP

# plot solution

plt.subplot(2,1,1)

plt.plot(m.time,T,'r-',label='Temperature')

plt.plot([0,m.time[-1]],[T.SPHI,T.SPHI],'k:',label=r'$T_{sphi}$')

plt.plot([0,m.time[-1]],[T.SPLO,T.SPLO],'k:',label=r'$T_{splo}$')

plt.ylabel('T (degF)')

plt.legend()

plt.subplot(2,1,2)

plt.plot(m.time,u,'b.-',label='Furnace (On/Off)')

plt.ylabel('Heat (On/Off)')

plt.xlabel('Time (min)')

plt.legend()

plt.show()

(:sourceend:)

(:divend:)

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Solve integer problem with [[https://gekko.readthedocs.io/en/latest/|Python GEKKO]]. Set '''integer'''=True to switch the variable from continuous to integer form. The APOPT solver is required to solve problem with integer variables but other solvers such as IPOPT can provide a relaxed solution where the integer conditions are not enforced.

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Attach:download.png [[Attach:discrete_optimization_source.zip|Discrete Optimization Source ~~Files~~]]

to:

Attach:download.png [[Attach:discrete_optimization_source.zip|Discrete Optimization Source (MATLAB and Python)]]

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(:source lang=python:)

from gekko import GEKKO

m = GEKKO() # create GEKKO model

# create integer variables

x1 = m.Var(integer=True,lb=-1,ub=1)

x2 = m.Var(integer=True,lb=-1,ub=2)

m.Obj(4*x1**2-4*x2*x1**2+x2**2+x1**2-x1+1)

m.options.SOLVER = 1 # APOPT solver

m.solve()

print('x1: ' + str(x1.value[0]))

print('x2: ' + str(x2.value[0]))

(:sourceend:)

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(:source lang=python:)

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(:sourceend:)

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{$\min_{x,w} x_2\left(t_f\right$}

to:

{$\min_{x,w} x_2 \left( t_f \right)$}

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{$\quad \quad \frac{dx_~~0~~}{dt} = \left(x_0-1\right)^2 + \left(x_1-1\right)^2$}

to:

{$\quad \quad \frac{dx_2}{dt} = \left(x_0-1\right)^2 + \left(x_1-1\right)^2$}

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to:

{$\quad \quad w(t) \in {{0,1}}$}

The optimal solution has a final objective of 1.344.

%width=550px%Attach:lotka_volterra_fishing.png

(:toggle hide solution button show="Show Python Solution":)

(:div id=solution:)

import numpy as np

import matplotlib.pyplot as plt

# retrieve apm.py from

# https://raw.githubusercontent.com/APMonitor/apm_python/master/apm.py

# or

# https://apmonitor.com/wiki/index.php/Main/PythonApp

from apm import *

# alternative pip install with 'pip install APMonitor'

# from APMonitor.apm import *

# local APMonitor servers are available for Windows or Linux

# https://apmonitor.com/wiki/index.php/Main/APMonitorServer

# with clients in Python, MATLAB, and Julia

# write model

model = '''

! apopt MINLP solver options (see apopt.com)

File apopt.opt

minlp_maximum_iterations 1000 ! minlp iterations

minlp_max_iter_with_int_sol 50 ! minlp iterations if integer solution is found

minlp_as_nlp 0 ! treat minlp as nlp

nlp_maximum_iterations 200 ! nlp sub-problem max iterations

minlp_branch_method 1 ! 1 = depth first, 2 = breadth first

minlp_gap_tol 0.001 ! covergence tolerance

minlp_integer_tol 0.001 ! maximum deviation from whole number to be considered an integer

minlp_integer_leaves 0 ! create soft (1) integer leaves or hard (2) integer leaves with branching

End File

Constants

c0 = 0.4

c1 = 0.2

Parameters

last

Variables

x0 = 0.5 , >= 0

x1 = 0.7 , >= 0

x2 = 0.0 , >= 0

int_w = 0 , >= 0 , <= 1

Intermediates

w = int_w

Equations

minimize last * x2

$x0 = x0 - x0*x1 - c0*x0*w

$x1 = - x1 + x0*x1 - c1*x1*w

$x2 = (x0-1)^2 + (x1-1)^2

'''

fid = open('lotka_volterra.apm','w')

fid.write(model)

fid.close()

# write data file

time = np.linspace(0,12,121)

time = np.insert(time, 1, 0.01)

last = np.zeros(122)

last[-1] = 1.0

data = np.vstack((time,last))

np.savetxt('data.csv',data.T,delimiter=',',header='time,last',comments='')

# specify server and application name

s = 'https://byu.apmonitor.com'

#s = 'https://127.0.0.1/' # for local APMonitor server

a = 'lotka'

apm(s,a,'clear all')

apm_load(s,a,'lotka_volterra.apm')

csv_load(s,a,'data.csv')

apm_option(s,a,'nlc.imode',6) # Nonlinear control / dynamic optimization

apm_option(s,a,'nlc.nodes',3)

apm_info(s,a,'MV','int_w') # M or MV = Manipulated variable - independent variable over time horizon

apm_option(s,a,'int_w.status',1) # Status: 1=ON, 0=OFF

apm_option(s,a,'int_w.mv_type',0) # MV Type = Zero Order Hold

apm_option(s,a,'nlc.solver',1) # 1 = APOPT

# solve

output = apm(s,a,'solve')

print(output)

# retrieve solution

y = apm_sol(s,a)

plt.figure(1)

plt.step(y['time'],y['int_w'],'r-',label='w (0/1)')

plt.plot(y['time'],y['x0'],'b-',label=r'$x_0$')

plt.plot(y['time'],y['x1'],'k-',label=r'$x_1$')

plt.plot(y['time'],y['x2'],'g-',label=r'$x_2$')

plt.xlabel('Time')

plt.ylabel('Variables')

plt.legend(loc='best')

plt.show()

(:divend:)

Attach:download.png [[Attach:lotka_volterra_fishing.zip|Download Solution Files in MATLAB and Python]]

The optimal solution has a final objective of 1.344.

%width=550px%Attach:lotka_volterra_fishing.png

(:toggle hide solution button show="Show Python Solution":)

(:div id=solution:)

import numpy as np

import matplotlib.pyplot as plt

# retrieve apm.py from

# https://raw.githubusercontent.com/APMonitor/apm_python/master/apm.py

# or

# https://apmonitor.com/wiki/index.php/Main/PythonApp

from apm import *

# alternative pip install with 'pip install APMonitor'

# from APMonitor.apm import *

# local APMonitor servers are available for Windows or Linux

# https://apmonitor.com/wiki/index.php/Main/APMonitorServer

# with clients in Python, MATLAB, and Julia

# write model

model = '''

! apopt MINLP solver options (see apopt.com)

File apopt.opt

minlp_maximum_iterations 1000 ! minlp iterations

minlp_max_iter_with_int_sol 50 ! minlp iterations if integer solution is found

minlp_as_nlp 0 ! treat minlp as nlp

nlp_maximum_iterations 200 ! nlp sub-problem max iterations

minlp_branch_method 1 ! 1 = depth first, 2 = breadth first

minlp_gap_tol 0.001 ! covergence tolerance

minlp_integer_tol 0.001 ! maximum deviation from whole number to be considered an integer

minlp_integer_leaves 0 ! create soft (1) integer leaves or hard (2) integer leaves with branching

End File

Constants

c0 = 0.4

c1 = 0.2

Parameters

last

Variables

x0 = 0.5 , >= 0

x1 = 0.7 , >= 0

x2 = 0.0 , >= 0

int_w = 0 , >= 0 , <= 1

Intermediates

w = int_w

Equations

minimize last * x2

$x0 = x0 - x0*x1 - c0*x0*w

$x1 = - x1 + x0*x1 - c1*x1*w

$x2 = (x0-1)^2 + (x1-1)^2

'''

fid = open('lotka_volterra.apm','w')

fid.write(model)

fid.close()

# write data file

time = np.linspace(0,12,121)

time = np.insert(time, 1, 0.01)

last = np.zeros(122)

last[-1] = 1.0

data = np.vstack((time,last))

np.savetxt('data.csv',data.T,delimiter=',',header='time,last',comments='')

# specify server and application name

s = 'https://byu.apmonitor.com'

#s = 'https://127.0.0.1/' # for local APMonitor server

a = 'lotka'

apm(s,a,'clear all')

apm_load(s,a,'lotka_volterra.apm')

csv_load(s,a,'data.csv')

apm_option(s,a,'nlc.imode',6) # Nonlinear control / dynamic optimization

apm_option(s,a,'nlc.nodes',3)

apm_info(s,a,'MV','int_w') # M or MV = Manipulated variable - independent variable over time horizon

apm_option(s,a,'int_w.status',1) # Status: 1=ON, 0=OFF

apm_option(s,a,'int_w.mv_type',0) # MV Type = Zero Order Hold

apm_option(s,a,'nlc.solver',1) # 1 = APOPT

# solve

output = apm(s,a,'solve')

print(output)

# retrieve solution

y = apm_sol(s,a)

plt.figure(1)

plt.step(y['time'],y['int_w'],'r-',label='w (0/1)')

plt.plot(y['time'],y['x0'],'b-',label=r'$x_0$')

plt.plot(y['time'],y['x1'],'k-',label=r'$x_1$')

plt.plot(y['time'],y['x2'],'g-',label=r'$x_2$')

plt.xlabel('Time')

plt.ylabel('Variables')

plt.legend(loc='best')

plt.show()

(:divend:)

Attach:download.png [[Attach:lotka_volterra_fishing.zip|Download Solution Files in MATLAB and Python]]

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!!!! Exercise 3

An example problem for Mixed Integer Optimal Control is the Lotka Volterra fishing problem. It finds an fishing strategy over a 12 year horizon to equilibrate the predator as prey fish to a minimum steady state value. The Lotka Volterra equations for a predator-prey system have an additional equation to introduce fishing by man and with constants {`c_0=0.4`} and {`c_1=0.2`}.

{$\min_{x,w} x_2\left(t_f\right$}

{$\mathrm{s.t.} \quad \frac{dx_0}{dt} = x_0 - x_0 x_1 - c_0 x_0 w$}

{$\quad \quad \frac{dx_1}{dt} = -x_1 + x_0 x_1 - c_1 x_1 w$}

{$\quad \quad \frac{dx_0}{dt} = \left(x_0-1\right)^2 + \left(x_1-1\right)^2$}

{$\quad \quad x(0) = (0.5,0.7,0)^T$}

{$\quad \quad x(t) \in {0,1}$}

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* [[https://mintoc.de|Mixed Integer Optimal Control Problems]]

to:

* [[https://mintoc.de|Mixed Integer Optimal Control Problems (mintOC)]]

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* [[~~Mixed Integer Optimal Control Problems|~~https://mintoc.de~~]]~~

* [[CMU-IBM Cyber-Infrastructure for MINLP collaborative site|https://minlp.org]]

* [[CMU-IBM Cyber-Infrastructure for MINLP collaborative site|

to:

* [[https://mintoc.de|Mixed Integer Optimal Control Problems]]

* [[https://minlp.org|CMU-IBM Cyber-Infrastructure for MINLP]]

* [[https://minlp.org|CMU-IBM Cyber-Infrastructure for MINLP]]

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(:htmlend:)

to:

(:htmlend:)

!!!! Additional Resources

* [[Mixed Integer Optimal Control Problems|https://mintoc.de]]

* [[CMU-IBM Cyber-Infrastructure for MINLP collaborative site|https://minlp.org]]

!!!! Additional Resources

* [[Mixed Integer Optimal Control Problems|https://mintoc.de]]

* [[CMU-IBM Cyber-Infrastructure for MINLP collaborative site|https://minlp.org]]

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!!!! Integer Variables

Integer or binary variables are defined in the APMonitor Modeling Language by appending a variable name with '''int'''. An binary decision variable is an integer variable with bounds between 0 and 1.

! Binary decision variable (0 or 1)

Variables

int_b >=0 <=1

The range of upper and lower bounds can be increased or decreased to any range to create a more general integer variable.

! Integer decision variable (-5 to 10)

Variables

int_v >=-5 <=10

Nonlinear programming solvers (such as IPOPT) may not return an integer solution because they are designed for continuous variables. Mixed Integer Nonlinear Programming solvers (such as APOPT) are equipped to solve for binary or integer variables. Select the appropriate [[https://apmonitor.com/wiki/index.php/Main/OptionApmSolver|solver option]] to either find an initial solution without integer variables or an integer solution. It is sometimes desirable to find a non-integer solution first because of the often significant reduction in computation time without the integer variables.

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(:html:)

<iframe width="560" height="315" src="https://www.youtube.com/embed/RPKPjqdiX1c" frameborder="0" allowfullscreen></iframe>

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(:html:)

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!!!! Exercise

to:

!!!! Exercise 1

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!!!! Exercise 2

Design a thermostat for a home that has a furnace with On/Off control. The home is initially unheated in 40 degF weather. When the heater is turned on, the temperature in the home reaches 63% of the steady state value in 120 minutes, starting from 40 degF. If left on indefinitely, the home would eventually reach 100 degF. Optimize the heater response for the first 2 1/2 hours starting from a cold home at 40 degF. The target range for the home is between 68 and 70 degF.

Attach:download.png [[Attach:thermostat_python_matlab.zip|Optimized Home Furnace Source Files]]

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'''Objective:''' Solve a discrete optimization problem with the branch and bound technique. ''Estimated time: ~~2 hours~~.''

to:

'''Objective:''' Solve a discrete optimization problem with the branch and bound technique. ''Estimated time: 1 hour.''

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to:

One often encounters problems in which manipulated variables or adjustable parameters must be selected from among a set of discrete values. Examples of discrete variables include ON/OFF state, selection of different feed lines, and other variables that are naturally integers. Many dynamic optimization problems are discrete in nature.

At first glance it might seem solving a discrete variable problem would be easier than a continuous problem. After all, for a variable within a given range, a set of discrete values within the range is finite whereas the number of continuous values is infinite. When searching for an optimum, it seems it would be easier to search from a finite set rather than from an infinite set.

This is not the case, however. Solving discrete problems is harder than continuous problems. This is because of the combinatorial explosion that occurs in all but the smallest problems. For example if we have two variables which can each take 10 values, we have 10*10 = 100 possibilities. If we have 10 variables that can each take 10 values, we have 10^10 possibilities. Even with the fastest computer, it would take a long time to evaluate all of these.

Discrete decision variables are those that have only certain levels or quantities that are acceptable at an optimal solution. Examples of discrete variables are binary (e.g. off/on or 0/1), integer (e.g. 4,5,6,7), or general discrete values that are not integer (e.g. 1/4 cm, 1/2 cm, 1 cm).

!!!! Exercise

'''Objective:''' Solve a discrete optimization problem with the branch and bound technique. ''Estimated time: 2 hours.''

* [[Attach:minlp_worksheet.pdf | Branch and Bound Exercise (pdf)]]

Attach:bnb_contour.png

!!!! Solution

Attach:download.png [[Attach:discrete_optimization_source.zip|Discrete Optimization Source Files]]

(:html:)

<iframe width="560" height="315" src="//www.youtube.com/embed/XhqteZIydT0?rel=0" frameborder="0" allowfullscreen></iframe>

(:htmlend:)

At first glance it might seem solving a discrete variable problem would be easier than a continuous problem. After all, for a variable within a given range, a set of discrete values within the range is finite whereas the number of continuous values is infinite. When searching for an optimum, it seems it would be easier to search from a finite set rather than from an infinite set.

This is not the case, however. Solving discrete problems is harder than continuous problems. This is because of the combinatorial explosion that occurs in all but the smallest problems. For example if we have two variables which can each take 10 values, we have 10*10 = 100 possibilities. If we have 10 variables that can each take 10 values, we have 10^10 possibilities. Even with the fastest computer, it would take a long time to evaluate all of these.

Discrete decision variables are those that have only certain levels or quantities that are acceptable at an optimal solution. Examples of discrete variables are binary (e.g. off/on or 0/1), integer (e.g. 4,5,6,7), or general discrete values that are not integer (e.g. 1/4 cm, 1/2 cm, 1 cm).

!!!! Exercise

'''Objective:''' Solve a discrete optimization problem with the branch and bound technique. ''Estimated time: 2 hours.''

* [[Attach:minlp_worksheet.pdf | Branch and Bound Exercise (pdf)]]

Attach:bnb_contour.png

!!!! Solution

Attach:download.png [[Attach:discrete_optimization_source.zip|Discrete Optimization Source Files]]

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<iframe width="560" height="315" src="//www.youtube.com/embed/XhqteZIydT0?rel=0" frameborder="0" allowfullscreen></iframe>

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(:title Dynamic Control with Discrete Variables:)

(:keywords dynamic control, model predictive control, integer, binary, discrete:)

(:description Model predictive control may require binary (On/Off), integer (select number of pumps), or other discrete decision variables:)

Discrete decision variables are those that have only certain levels or quantities that are acceptable at an optimal solution. Examples of discrete variables are binary (e.g. off/on or 0/1), integer (e.g. 4,5,6,7), or general discrete values that are not integer (e.g. 1/4 cm, 1/2 cm, 1 cm).

(:keywords dynamic control, model predictive control, integer, binary, discrete:)

(:description Model predictive control may require binary (On/Off), integer (select number of pumps), or other discrete decision variables:)

Discrete decision variables are those that have only certain levels or quantities that are acceptable at an optimal solution. Examples of discrete variables are binary (e.g. off/on or 0/1), integer (e.g. 4,5,6,7), or general discrete values that are not integer (e.g. 1/4 cm, 1/2 cm, 1 cm).