From Cornell University Computational Optimization Open Textbook - Optimization Wiki
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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> |
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| | ''MILP Solution: ''<math display=inline>x_{1}=2, x_{2}=1,y_{1}=1, y_{2}=0</math>, Lower Bound = 6 <br> |
| | Lower Bound<Upper Bound, Integer Cut:<math display=inline>y_{1}- y_{2}\leq 0</math> |
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| ==Conclusion== | | ==Conclusion== |
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| ==References== | | ==References== |
Revision as of 06:32, 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: 
, Upper Bound = 7
Step 1a: Solve the MILP master problem with OA for
:
![{\displaystyle f{\big (}x{\big )}={\big (}x_{1}{\big )}^{2}+{\big (}x_{2}{\big )}^{2},~~\bigtriangledown f{\big (}x{\big )}=[2x_{1}~~~~2x_{1}]^{T}~~for~~x^{*}=[2~~~~1]^{T}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3aa00d3f49988a196d93388e7ac41da58fc00066)
![{\displaystyle f{\big (}x^{*}{\big )}+\bigtriangledown f{\big (}x^{*}{\big )}^{T}{\big (}x-x^{*}{\big )}=5+[4~~~~2]{\begin{bmatrix}x_{1}-2\\x_{2}-1\end{bmatrix}}=5+4{\big (}x_{1}-2{\big )}+2{\big (}x_{2}-1{\big )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29222aefe55bdb63f346ae1a3c21c27fe7efa013)
![{\displaystyle g{\big (}x{\big )}={\big (}x_{1}-2{\big )}^{2}-x_{2},~~\bigtriangledown g{\big (}x{\big )}=[2x_{1}-4~~~~-1]^{T}~~for~~x^{*}=[2~~~~1]^{T}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f025d4d61d63efd6a87b01d8e7848abc520f2bfa)
![{\displaystyle g{\big (}x^{*}{\big )}+\bigtriangledown g{\big (}x^{*}{\big )}^{T}{\big (}x-x^{*}{\big )}=-1+[0~~~~-1]{\begin{bmatrix}x_{1}-2\\x_{2}-1\end{bmatrix}}=-x_{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52bbf7d8d87d53306f2542ef0f6ff2b8a75ce4a9)
Minimize

Subject to 










MILP Solution:
, Lower Bound = 6
Lower Bound<Upper Bound, Integer Cut:
Conclusion
References