Multivariate convex optimization problem involving logarithms 2 The Next CEO of Stack...
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Multivariate convex optimization problem involving logarithms 2
The Next CEO of Stack OverflowMultivariate convex optimization problem involving logarithmsa problem on optimization having a good lookingThe most efficient algorithm to solve the following problemConstraint to unconstraint optimization problem by subsitutionfunction induced by optimizationIs the optimum of this problem unique?Cross-entropy minimization - equivalent unconstrained optimization problemExtension of convex function to boundaryTransformation of Optimization Problem (LP/QP/MIP)?What is the advantage of adding $log$ Barrier to solve a Linear program?Multivariate convex optimization problem involving logarithms
$begingroup$
This is an extension to previous question in Multivariate convex optimization problem involving logarithms. Thanks a lot to David M. for the answer to there. Now, I like to extend the question a little more to include solutions with $a_i=0$ or $b_i=0$.
$$min_{a, b} sum_{i=1}^K b_i I(a_i>0, b_i>0)f(frac{a_i}{b_i}) $$
s.t.
$$ f(x) = (1+x) log(1+x) -log(x) - (1+x) log(2)$$
$$ sum_{i=1}^K a_i = 1.$$
$$ sum_{i=1}^K b_i = 1.$$
$$ a,bgeq 0. $$
$I(a_i>0, b_i>0)$ is indicator function i.e. it is 1 when ($a_i>0$ and $b_i>0$) and $0$ otherwise.
I believe the answer to the question is the set of (a,b) s.t. $sum_{i=1}^K a_i = sum_{i=1}^K b_i = 1$ and $a_i=b_i$ for all $i$. I just wanted to make sure if I am correct.
optimization convex-analysis
asked Mar 18 at 3:24
kkokko
186
$endgroup$
add a comment |
$begingroup$
This is an extension to previous question in Multivariate convex optimization problem involving logarithms. Thanks a lot to David M. for the answer to there. Now, I like to extend the question a little more to include solutions with $a_i=0$ or $b_i=0$.
$$min_{a, b} sum_{i=1}^K b_i I(a_i>0, b_i>0)f(frac{a_i}{b_i}) $$
s.t.
$$ f(x) = (1+x) log(1+x) -log(x) - (1+x) log(2)$$
$$ sum_{i=1}^K a_i = 1.$$
$$ sum_{i=1}^K b_i = 1.$$
$$ a,bgeq 0. $$
$I(a_i>0, b_i>0)$ is indicator function i.e. it is 1 when ($a_i>0$ and $b_i>0$) and $0$ otherwise.
I believe the answer to the question is the set of (a,b) s.t. $sum_{i=1}^K a_i = sum_{i=1}^K b_i = 1$ and $a_i=b_i$ for all $i$. I just wanted to make sure if I am correct.
optimization convex-analysis
asked Mar 18 at 3:24
kkokko
186
$endgroup$
add a comment |
$begingroup$
This is an extension to previous question in Multivariate convex optimization problem involving logarithms. Thanks a lot to David M. for the answer to there. Now, I like to extend the question a little more to include solutions with $a_i=0$ or $b_i=0$.
$$min_{a, b} sum_{i=1}^K b_i I(a_i>0, b_i>0)f(frac{a_i}{b_i}) $$
s.t.
$$ f(x) = (1+x) log(1+x) -log(x) - (1+x) log(2)$$
$$ sum_{i=1}^K a_i = 1.$$
$$ sum_{i=1}^K b_i = 1.$$
$$ a,bgeq 0. $$
$I(a_i>0, b_i>0)$ is indicator function i.e. it is 1 when ($a_i>0$ and $b_i>0$) and $0$ otherwise.
I believe the answer to the question is the set of (a,b) s.t. $sum_{i=1}^K a_i = sum_{i=1}^K b_i = 1$ and $a_i=b_i$ for all $i$. I just wanted to make sure if I am correct.
optimization convex-analysis
asked Mar 18 at 3:24
kkokko
186
$endgroup$
This is an extension to previous question in Multivariate convex optimization problem involving logarithms. Thanks a lot to David M. for the answer to there. Now, I like to extend the question a little more to include solutions with $a_i=0$ or $b_i=0$.
$$min_{a, b} sum_{i=1}^K b_i I(a_i>0, b_i>0)f(frac{a_i}{b_i}) $$
s.t.
$$ f(x) = (1+x) log(1+x) -log(x) - (1+x) log(2)$$
$$ sum_{i=1}^K a_i = 1.$$
$$ sum_{i=1}^K b_i = 1.$$
$$ a,bgeq 0. $$
$I(a_i>0, b_i>0)$ is indicator function i.e. it is 1 when ($a_i>0$ and $b_i>0$) and $0$ otherwise.
I believe the answer to the question is the set of (a,b) s.t. $sum_{i=1}^K a_i = sum_{i=1}^K b_i = 1$ and $a_i=b_i$ for all $i$. I just wanted to make sure if I am correct.
optimization convex-analysis
optimization convex-analysis
asked Mar 18 at 3:24
kkokko
186
asked Mar 18 at 3:24
kkokko
186
asked Mar 18 at 3:24
kkokko
186
asked Mar 18 at 3:24
kkokko
186
asked Mar 18 at 3:24
kkokko
186
186
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Yes I think you're correct. Note that your function $f$ satisfies $f(t)geq0$ for all $t>0$, with $f(t)=0$ if and only if $t=1$. Hence, your objective function (let's call it $g$) satisfies $g(a,b)geq0$ for all $a,b>0$, with $g(a,b)=0$ if and only if $aequiv{b}$.
You can just eliminate all the edge cases explicitly:
- If $a_i=0$ and $b_i>0$, then $b_icdot f(a_i/b_i)=b_icdot f(0)$, which isn't defined. However, $lim_{tto0}f(t)=infty$, so the most sensible thing to do it interpret $f(a_i/b_i)=f(0)=infty$, which obviously isn't the minimizer.
- If $a_i>0$ and $b_i=0$, then $f(a_i/b_i)=f(a_i/0)$ which isn't defined. If we again do the sensible thing, we will define the objective function so that
$$
0cdot f(a_i/0)=lim_{tto0}tcdot fbig(frac{a_i}{t}big)=infty,
$$
again obviously not the minimizer. - Finally, if $a_i=b_i=0$, then $b_icdot f(a_i/b_i)=0cdot f(0/0)$. Again, doing the sensible thing,
$$
0cdot f(0/0)=lim_{tto0}tcdot{fbig(frac{t}{t}big)}=f(1)cdotlim_{tto0}t=0.
$$
Hence, the optimal set is indeed ${a,bgeq0;|;aequiv{b},{1}^text{T}a=1}$.
answered Mar 18 at 17:11
David M.David M.
2,188421
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
Yes I think you're correct. Note that your function $f$ satisfies $f(t)geq0$ for all $t>0$, with $f(t)=0$ if and only if $t=1$. Hence, your objective function (let's call it $g$) satisfies $g(a,b)geq0$ for all $a,b>0$, with $g(a,b)=0$ if and only if $aequiv{b}$.
You can just eliminate all the edge cases explicitly:
- If $a_i=0$ and $b_i>0$, then $b_icdot f(a_i/b_i)=b_icdot f(0)$, which isn't defined. However, $lim_{tto0}f(t)=infty$, so the most sensible thing to do it interpret $f(a_i/b_i)=f(0)=infty$, which obviously isn't the minimizer.
- If $a_i>0$ and $b_i=0$, then $f(a_i/b_i)=f(a_i/0)$ which isn't defined. If we again do the sensible thing, we will define the objective function so that
$$
0cdot f(a_i/0)=lim_{tto0}tcdot fbig(frac{a_i}{t}big)=infty,
$$
again obviously not the minimizer. - Finally, if $a_i=b_i=0$, then $b_icdot f(a_i/b_i)=0cdot f(0/0)$. Again, doing the sensible thing,
$$
0cdot f(0/0)=lim_{tto0}tcdot{fbig(frac{t}{t}big)}=f(1)cdotlim_{tto0}t=0.
$$
Hence, the optimal set is indeed ${a,bgeq0;|;aequiv{b},{1}^text{T}a=1}$.
answered Mar 18 at 17:11
David M.David M.
2,188421
$endgroup$
add a comment |
$begingroup$
Yes I think you're correct. Note that your function $f$ satisfies $f(t)geq0$ for all $t>0$, with $f(t)=0$ if and only if $t=1$. Hence, your objective function (let's call it $g$) satisfies $g(a,b)geq0$ for all $a,b>0$, with $g(a,b)=0$ if and only if $aequiv{b}$.
You can just eliminate all the edge cases explicitly:
- If $a_i=0$ and $b_i>0$, then $b_icdot f(a_i/b_i)=b_icdot f(0)$, which isn't defined. However, $lim_{tto0}f(t)=infty$, so the most sensible thing to do it interpret $f(a_i/b_i)=f(0)=infty$, which obviously isn't the minimizer.
- If $a_i>0$ and $b_i=0$, then $f(a_i/b_i)=f(a_i/0)$ which isn't defined. If we again do the sensible thing, we will define the objective function so that
$$
0cdot f(a_i/0)=lim_{tto0}tcdot fbig(frac{a_i}{t}big)=infty,
$$
again obviously not the minimizer. - Finally, if $a_i=b_i=0$, then $b_icdot f(a_i/b_i)=0cdot f(0/0)$. Again, doing the sensible thing,
$$
0cdot f(0/0)=lim_{tto0}tcdot{fbig(frac{t}{t}big)}=f(1)cdotlim_{tto0}t=0.
$$
Hence, the optimal set is indeed ${a,bgeq0;|;aequiv{b},{1}^text{T}a=1}$.
answered Mar 18 at 17:11
David M.David M.
2,188421
$endgroup$
add a comment |
$begingroup$
Yes I think you're correct. Note that your function $f$ satisfies $f(t)geq0$ for all $t>0$, with $f(t)=0$ if and only if $t=1$. Hence, your objective function (let's call it $g$) satisfies $g(a,b)geq0$ for all $a,b>0$, with $g(a,b)=0$ if and only if $aequiv{b}$.
You can just eliminate all the edge cases explicitly:
- If $a_i=0$ and $b_i>0$, then $b_icdot f(a_i/b_i)=b_icdot f(0)$, which isn't defined. However, $lim_{tto0}f(t)=infty$, so the most sensible thing to do it interpret $f(a_i/b_i)=f(0)=infty$, which obviously isn't the minimizer.
- If $a_i>0$ and $b_i=0$, then $f(a_i/b_i)=f(a_i/0)$ which isn't defined. If we again do the sensible thing, we will define the objective function so that
$$
0cdot f(a_i/0)=lim_{tto0}tcdot fbig(frac{a_i}{t}big)=infty,
$$
again obviously not the minimizer. - Finally, if $a_i=b_i=0$, then $b_icdot f(a_i/b_i)=0cdot f(0/0)$. Again, doing the sensible thing,
$$
0cdot f(0/0)=lim_{tto0}tcdot{fbig(frac{t}{t}big)}=f(1)cdotlim_{tto0}t=0.
$$
Hence, the optimal set is indeed ${a,bgeq0;|;aequiv{b},{1}^text{T}a=1}$.
answered Mar 18 at 17:11
David M.David M.
2,188421
$endgroup$
Yes I think you're correct. Note that your function $f$ satisfies $f(t)geq0$ for all $t>0$, with $f(t)=0$ if and only if $t=1$. Hence, your objective function (let's call it $g$) satisfies $g(a,b)geq0$ for all $a,b>0$, with $g(a,b)=0$ if and only if $aequiv{b}$.
You can just eliminate all the edge cases explicitly:
- If $a_i=0$ and $b_i>0$, then $b_icdot f(a_i/b_i)=b_icdot f(0)$, which isn't defined. However, $lim_{tto0}f(t)=infty$, so the most sensible thing to do it interpret $f(a_i/b_i)=f(0)=infty$, which obviously isn't the minimizer.
- If $a_i>0$ and $b_i=0$, then $f(a_i/b_i)=f(a_i/0)$ which isn't defined. If we again do the sensible thing, we will define the objective function so that
$$
0cdot f(a_i/0)=lim_{tto0}tcdot fbig(frac{a_i}{t}big)=infty,
$$
again obviously not the minimizer. - Finally, if $a_i=b_i=0$, then $b_icdot f(a_i/b_i)=0cdot f(0/0)$. Again, doing the sensible thing,
$$
0cdot f(0/0)=lim_{tto0}tcdot{fbig(frac{t}{t}big)}=f(1)cdotlim_{tto0}t=0.
$$
Hence, the optimal set is indeed ${a,bgeq0;|;aequiv{b},{1}^text{T}a=1}$.
answered Mar 18 at 17:11
David M.David M.
2,188421
answered Mar 18 at 17:11
David M.David M.
2,188421
answered Mar 18 at 17:11
David M.David M.
2,188421
answered Mar 18 at 17:11
David M.David M.
2,188421
2,188421
add a comment |
add a comment |
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