Outer-approximation (OA): Difference between revisions

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== Example ==
== Example ==
Minimize <math display=block> f(x)= y_{1} +y_{2} + \big(x_{1}\big)^{2} +\big(x_{2}\big)^{2} </math>
''Minimize'' <math display=block> f(x)= y_{1} +y_{2} + \big(x_{1}\big)^{2} +\big(x_{2}\big)^{2} </math>
Subject to <math display=block>\big(x_{1}-2\big)^{2}-x_{2} \leq 0</math>
''Subject to'' <math display=block>\big(x_{1}-2\big)^{2}-x_{2} \leq 0</math>
<math display=block>x_{1}-2y_{1} \geq 0</math>
<math display=block>x_{1}-2y_{1} \geq 0</math>
<math display=block>x_{1}-x_{2}-3 \big(1-y_{1}\big) \geq 0</math>
<math display=block>x_{1}-x_{2}-3 \big(1-y_{1}\big) \geq 0</math>
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<math display=block>0 \leq x_{2}\leq 4</math>
<math display=block>0 \leq x_{2}\leq 4</math>
<math display=block>y_{1},y_{2} \in  \big\{0,1\big\} </math>
<math display=block>y_{1},y_{2} \in  \big\{0,1\big\} </math>
'''Solution'''<br>
''Step 1a:''  Start from
<math display=inline>y_{1}=y_{2}=1</math> and solve the NLP below: <br>
''Minimize'' <math display=block> f= 2+ \big(x_{1}\big)^{2} +\big(x_{2}\big)^{2} </math>
''Subject to'' <math display=block>\big(x_{1}-2\big)^{2}-x_{2} \leq 0</math>
<math display=block>x_{1}-2 \geq 0</math>
<math display=block>x_{1}-x_{2} \geq 0</math>
<math display=block>x_{1} \geq 0</math>
<math display=block>x_{2}-1 \geq 0</math>
<math display=block>x_{1}+x_{2} \geq 3</math>
<math display=block>0 \leq x_{1}\leq 4</math>
<math display=block>0 \leq x_{2}\leq 4</math>
''Solution:''<math display=inline>x_{1}=2, x_{2}=1, Upper Bound=7</math> and solve the NLP below: <br>


==Conclusion==
==Conclusion==


==References==
==References==

Revision as of 06:25, 26 November 2021

Author: Yousef Aloufi (CHEME 6800 Fall 2021)

Introduction

Theory

Example

Minimize

Subject to
Solution
Step 1a: Start from and solve the NLP below:
Minimize
Subject to
Solution: and solve the NLP below:

Conclusion

References