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










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put on hold as unclear what you're asking by Did, mrtaurho, José Carlos Santos, amWhy, Andrei 5 hours ago


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
    14 hours ago










  • 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.
    14 hours ago












  • 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
    14 hours ago












  • 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
    10 hours ago
















-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










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 5 hours ago


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
    14 hours ago










  • 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.
    14 hours ago












  • 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
    14 hours ago












  • 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
    10 hours ago














-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










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






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 14 hours ago









Did

246k23221455




246k23221455






New contributor




Da Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 15 hours ago









Da 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 5 hours ago


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 5 hours ago


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
    14 hours ago










  • 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.
    14 hours ago












  • 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
    14 hours ago












  • 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
    10 hours ago


















  • 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
    14 hours ago










  • 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.
    14 hours ago












  • 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
    14 hours ago












  • 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
    10 hours ago
















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
14 hours ago




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
14 hours ago












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.
14 hours ago






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.
14 hours ago














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
14 hours ago






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
14 hours ago














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
10 hours ago




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
10 hours ago










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