Knapsack Secretary Through Boosting

We revisit the knapsack-secretary problem (Babaioff et al.; APPROX 2007), a generalization of the classic secretary problem in which items have different sizes and multiple items may be selected if their total size does not exceed the capacity $B$ of a knapsack. Previous works show competitive ratios of $1/(10e)$ (Babaioff et al.), $1/8.06$ (Kesselheim et al.; STOC 2014), and $1/6.65$ (Albers, Khan, and Ladewig; APPROX 2019) for the general problem but no definitive answers for the achievable competitive ratio; the best known impossibility remains $1/e$ as inherited from the classic secretary problem. In an effort to make more qualitative progress, we take an orthogonal approach and give definitive answers for special cases. Our main result is on the $1$-$2$-knapsack secretary problem, the special case in which $B=2$ and all items have sizes $1$ or $2$, arguably the simplest meaningful generalization of the secretary problem towards the knapsack secretary problem. Our algorithm is simple: It $\textit{boosts}$ the value of size-$1$ items by a factor $α>1$ and then uses the size-oblivious approach by Albers, Khan, and Ladewig. We show by a nontrivial analysis that this algorithm achieves a competitive ratio of $1/e$ if and only if $1.40\lesssimα\leq e/(e-1)\approx 1.58$. Towards understanding the general case, we then consider the case when sizes are $1$ and $B$, and $B$ is large. While it remains unclear if $1/e$ can be achieved in that case, we show that algorithms based only on the relative ranks of the item values can achieve precisely a competitive ratio of $1/(e+1)$. To show the impossibility, we use a non-trivial generalization of the factor-revealing linear program for the secretary problem (Buchbinder, Jain, and Singh; IPCO 2010).