Jan 2020 Exam Problem 3 question 3

Jan 2020 Exam Problem 3 question 3

par Tiannan Sha,
Number of replies: 1

Hi,

I don't quite follow this symmetric argument. So we are maximising logx+logy+logz over constraints x,y,z>=0, x<=10, y+z <=40. The solution says the problem is symmetric in y,z and since we know there is a unique optimum, then we must have y=z. I don't understand why we can have y=z.

Thanks,

Tiannan


In reply to Tiannan Sha

Re: Jan 2020 Exam Problem 3 question 3

par Seyed Mohammadhossein Tabatabaee,
Hello,

Thank you for your question. The proportionally fair allocation is unique, which means this maximization problem has a unique optimal solution. Now, let us show by contradiction that in the optimal solution, we must y = z: Assume that at the optimal solution we have y ~=z; we can exchange the values of y and z, and obtain another proportionally fair allocation; this contradicts the uniqueness of the proportionally fair allocation. Hence, we have y = z.

Best,
Hossein