Minimize $\int\limits_{0}^{1} c(t)u(t)\ \text dt$ subject to $\int\limits_{0}^{1} u(t)\ \text dt = 1$ with $∫\limits_0^1 \frac{\text dt}{c(t)} < ∞.$

Asked Jan 16 '25 at 12:27

Active Jan 16 '25 at 13:48

Viewed 79 times

I had the final exam of my (convex) optimization course last week and was completely stumped when I saw the following problem. We did not cover much of these types of problems during the course, maybe one example relating to Lagrange relaxation, so I was not able to do much on the problem. I would really like to know how to do these. Here is the problem:

We are presented with the problem (P) to minimize

$$\int\limits_{0}^{1} c(t)u(t)\ \text dt$$

subject to

$$\int\limits_{0}^{1} u(t)\ \text dt = 1.$$

Here $c(t)$ and $u(t)$ are continuous on $[0, 1]$ and $c(t)$ (which is a fixed function) is positive with

$$\int\limits_{0}^{1} \frac{\text dt}{c(t)} < \infty.$$

(a) Determine the Lagrangian dual of (P).

(b) Discuss the duality gap.

edited Jan 16 '25 at 13:48

Alwaz

asked Jan 16 '25 at 12:27

nazorated

Didn't you miss something in your exercise? For a continuous positive function on $[0,1]$, the inverse is also continuous and the last integral is always finite. – Christophe Boilley Jan 16 '25 at 15:43
1

@ChristopheBoilley $c(t)=2t$ would be a function that violates this condition. Inverses are not involved. – Paul Jan 16 '25 at 16:16
@Paul I thought "positive" means $c(t)>0$. In English, I’m used to qualify $c(t)=2t$ as non-negative : https://en.wikipedia.org/wiki/Sign_(mathematics)#Terminology_for_signs – Christophe Boilley Jan 16 '25 at 18:04
So $u(t)$ is not constrained to be positive? – John Barber Jan 17 '25 at 00:01
@JohnBarber, there is no sign restriction on $u(t)$. – nazorated Jan 17 '25 at 14:28

Minimize $\int\limits_{0}^{1} c(t)u(t)\ \text dt$ subject to $\int\limits_{0}^{1} u(t)\ \text dt = 1$ with $∫\limits_0^1 \frac{\text dt}{c(t)} < ∞.$

0 Answers0