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I have an optimization problem with the following objective function.

$\max_{a^{l}_{n,k} b^l_{n,k}} \sum_{n=1}^{\overline{N_l}} b_{k,n} \frac{C_1}{C_2} \log_2 \bigg(1 +\frac{a_{k,n} h_{k,n}} {b_{n,k} c_3}\bigg) $

The constraints are linear.

The objective is concave, if I keep all the constants as 1, for simplicity the objective function is:

$f = b \cdot \log_2(1+a/b)$

which is concave right, or does it depends on the actual values of the constants?

Also if I add another parameter in the denominator of the log term as:

$\max_{a^{l}_{n,k} b^l_{n,k}} \sum_{n=1}^{\overline{N_l}} b_{k,n} \frac{C_1}{C_2} \log_2 \bigg(1 +\frac{a_{k,n} h_{k,n}} {b_{n,k} c_3 + X}\bigg) $ does it still remain concave.

$f= b⋅ \log (1+\frac{a}{b+1})$ is not concave right? Or does it depends on the value of X and other constants(which I kept one)?

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$b\log(1+\frac{a}{b})$ is concave (it is the perspective of the concave function $\log(1+a)$. The function $b\log(1+\frac{a}{1+b})$ is not concave (You can compute Hessian to prove if you want, or pick two suitable points and check if the point in between has a lower function value than the average of the function values at the end points. Test the two end points $(a ,b)=(0.01,0.01)$ and $(a, b) = (1,1)$ for instance). I just plotted the function and saw that it is neither convex nor concave. Try simple things first...)

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  • $\begingroup$ So sir, the second objective is non-convex? No-matter what the values of my constants are according to the domain? $\endgroup$ – cody Feb 7 '14 at 6:56
  • $\begingroup$ It is neither convex nor concave. Most likely regardless of any constants or scalings. $\endgroup$ – Johan Löfberg Feb 7 '14 at 8:28
  • $\begingroup$ Thats all I wanted to hear sir..:D THANK YOU!! THANKS A LOT!! $\endgroup$ – cody Feb 7 '14 at 9:51
  • $\begingroup$ Sir, in the 1st function, if $a = b \times C$, i.e. $a$ is a function w.r.t. $b$. Whether this correlation changes the convexity of the function? Is the function still negative semi-definite? $\endgroup$ – cody Jun 10 '14 at 16:50
  • $\begingroup$ Then you have $b\log(1+C)$ which trivially is convex (and concave) assuming $C$ is constant $\endgroup$ – Johan Löfberg Jun 10 '14 at 18:03

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