how to linearize or convexify the term $ye^{-1/x}$ if $0<1/xll1$ [on hold]












-1














how to linearize or convexify the term $ye^{-1/x}$ if $0<1/xll1$



if $0<1/xll1$;



then $e^{-1/x}=1-1/x$;



then $ye^{-1/x}=y(1-1/x)$;



the obtained term y(1-1/x) still nonlinear and nonconvex






convex-optimization linearization







share|cite|improve this question









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Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as unclear what you're asking by Did, mrtaurho, José Carlos Santos, amWhy, Andrei Jan 4 at 19:37


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    – José Carlos Santos
    Jan 4 at 10:22










  • Okay, so I see that you made the following logic $$ frac{1}{x} approx 0 Rightarrow exp{left(-frac{1}{x}right)} approx 1-frac{1}{x} $$ So far, it seems fine. Then you just multiply it with $y$. What's the next thing that's causing trouble to you?
    – Matti P.
    Jan 4 at 10:23












  • Thanks for modifying it. Actually, in the next step, I want to linearize or convexify the obtained term y*(1-1/x) to be used in a MILP model.
    – Da Xu
    Jan 4 at 10:30












  • There's no way to linearize or convexify this. At best, substituting $z=e^{-1/x}$ gives you a bilinear term $yz$. There's just nothing to be done about that. Approximation is your only recourse, and even then there's no universal approach, it will depend on the context of your entire model.
    – Michael Grant
    Jan 4 at 15:01










  • Thanks very much. I will check my model, and find other ways.
    – Da Xu
    Jan 5 at 10:33
















-1














how to linearize or convexify the term $ye^{-1/x}$ if $0<1/xll1$



if $0<1/xll1$;



then $e^{-1/x}=1-1/x$;



then $ye^{-1/x}=y(1-1/x)$;



the obtained term y(1-1/x) still nonlinear and nonconvex






convex-optimization linearization







share|cite|improve this question









New contributor




Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











put on hold as unclear what you're asking by Did, mrtaurho, José Carlos Santos, amWhy, Andrei Jan 4 at 19:37


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    – José Carlos Santos
    Jan 4 at 10:22










  • Okay, so I see that you made the following logic $$ frac{1}{x} approx 0 Rightarrow exp{left(-frac{1}{x}right)} approx 1-frac{1}{x} $$ So far, it seems fine. Then you just multiply it with $y$. What's the next thing that's causing trouble to you?
    – Matti P.
    Jan 4 at 10:23












  • Thanks for modifying it. Actually, in the next step, I want to linearize or convexify the obtained term y*(1-1/x) to be used in a MILP model.
    – Da Xu
    Jan 4 at 10:30












  • There's no way to linearize or convexify this. At best, substituting $z=e^{-1/x}$ gives you a bilinear term $yz$. There's just nothing to be done about that. Approximation is your only recourse, and even then there's no universal approach, it will depend on the context of your entire model.
    – Michael Grant
    Jan 4 at 15:01










  • Thanks very much. I will check my model, and find other ways.
    – Da Xu
    Jan 5 at 10:33














-1












-1








-1







how to linearize or convexify the term $ye^{-1/x}$ if $0<1/xll1$



if $0<1/xll1$;



then $e^{-1/x}=1-1/x$;



then $ye^{-1/x}=y(1-1/x)$;



the obtained term y(1-1/x) still nonlinear and nonconvex






convex-optimization linearization







share|cite|improve this question









New contributor




Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











how to linearize or convexify the term $ye^{-1/x}$ if $0<1/xll1$



if $0<1/xll1$;



then $e^{-1/x}=1-1/x$;



then $ye^{-1/x}=y(1-1/x)$;



the obtained term y(1-1/x) still nonlinear and nonconvex






convex-optimization linearization




convex-optimization linearization






share|cite|improve this question









New contributor




Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 11:01









Did

246k23221456




246k23221456






New contributor




Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Jan 4 at 10:07









Da XuDa Xu

12




12




New contributor




Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as unclear what you're asking by Did, mrtaurho, José Carlos Santos, amWhy, Andrei Jan 4 at 19:37


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






put on hold as unclear what you're asking by Did, mrtaurho, José Carlos Santos, amWhy, Andrei Jan 4 at 19:37


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    – José Carlos Santos
    Jan 4 at 10:22










  • Okay, so I see that you made the following logic $$ frac{1}{x} approx 0 Rightarrow exp{left(-frac{1}{x}right)} approx 1-frac{1}{x} $$ So far, it seems fine. Then you just multiply it with $y$. What's the next thing that's causing trouble to you?
    – Matti P.
    Jan 4 at 10:23












  • Thanks for modifying it. Actually, in the next step, I want to linearize or convexify the obtained term y*(1-1/x) to be used in a MILP model.
    – Da Xu
    Jan 4 at 10:30












  • There's no way to linearize or convexify this. At best, substituting $z=e^{-1/x}$ gives you a bilinear term $yz$. There's just nothing to be done about that. Approximation is your only recourse, and even then there's no universal approach, it will depend on the context of your entire model.
    – Michael Grant
    Jan 4 at 15:01










  • Thanks very much. I will check my model, and find other ways.
    – Da Xu
    Jan 5 at 10:33


















  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    – José Carlos Santos
    Jan 4 at 10:22










  • Okay, so I see that you made the following logic $$ frac{1}{x} approx 0 Rightarrow exp{left(-frac{1}{x}right)} approx 1-frac{1}{x} $$ So far, it seems fine. Then you just multiply it with $y$. What's the next thing that's causing trouble to you?
    – Matti P.
    Jan 4 at 10:23












  • Thanks for modifying it. Actually, in the next step, I want to linearize or convexify the obtained term y*(1-1/x) to be used in a MILP model.
    – Da Xu
    Jan 4 at 10:30












  • There's no way to linearize or convexify this. At best, substituting $z=e^{-1/x}$ gives you a bilinear term $yz$. There's just nothing to be done about that. Approximation is your only recourse, and even then there's no universal approach, it will depend on the context of your entire model.
    – Michael Grant
    Jan 4 at 15:01










  • Thanks very much. I will check my model, and find other ways.
    – Da Xu
    Jan 5 at 10:33
















Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Jan 4 at 10:22




Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
Jan 4 at 10:22












Okay, so I see that you made the following logic $$ frac{1}{x} approx 0 Rightarrow exp{left(-frac{1}{x}right)} approx 1-frac{1}{x} $$ So far, it seems fine. Then you just multiply it with $y$. What's the next thing that's causing trouble to you?
– Matti P.
Jan 4 at 10:23






Okay, so I see that you made the following logic $$ frac{1}{x} approx 0 Rightarrow exp{left(-frac{1}{x}right)} approx 1-frac{1}{x} $$ So far, it seems fine. Then you just multiply it with $y$. What's the next thing that's causing trouble to you?
– Matti P.
Jan 4 at 10:23














Thanks for modifying it. Actually, in the next step, I want to linearize or convexify the obtained term y*(1-1/x) to be used in a MILP model.
– Da Xu
Jan 4 at 10:30






Thanks for modifying it. Actually, in the next step, I want to linearize or convexify the obtained term y*(1-1/x) to be used in a MILP model.
– Da Xu
Jan 4 at 10:30














There's no way to linearize or convexify this. At best, substituting $z=e^{-1/x}$ gives you a bilinear term $yz$. There's just nothing to be done about that. Approximation is your only recourse, and even then there's no universal approach, it will depend on the context of your entire model.
– Michael Grant
Jan 4 at 15:01




There's no way to linearize or convexify this. At best, substituting $z=e^{-1/x}$ gives you a bilinear term $yz$. There's just nothing to be done about that. Approximation is your only recourse, and even then there's no universal approach, it will depend on the context of your entire model.
– Michael Grant
Jan 4 at 15:01












Thanks very much. I will check my model, and find other ways.
– Da Xu
Jan 5 at 10:33




Thanks very much. I will check my model, and find other ways.
– Da Xu
Jan 5 at 10:33










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