The Bruss–Robertson–Steele inequality

Abstract

The Bruss–Robertson–Steele (BRS) inequality bounds the expected number of items of random size which can be packed into a given suitcase. Remarkably, no independence assumptions are needed on the random sizes, which points to a simple explanation; the inequality is the integrated form of an $\omega$-by-$\omega$ inequality, as this note proves.

1. The basic problem

The Bruss–Robertson–Steele (BRS) inequality was first proved in [2], and later generalized in [3]. The recent survey in [1] gives a fine review.

You have N objects which you would like to take in your suitcase on a flight. The weight of object j is $Z_j$ , but the total weight you are allowed to take on the flight must not exceed $s>0$ . You want to maximise the number of objects that you can take, subject to this constraint. This can be posed as a linear programme:

\begin{equation*} \max_{x \geq 0} \; \sum_i x_i \quad {\rm subject \, to} \quad x_i \leq 1 \ \text{for all } i, \ \sum_i x_i\,Z_i \leq s.\end{equation*}

Strictly, we have to have that each $x_i$ is in $\{0,1\}$ , but this additional constraint will only reduce the value; since we are looking for upper bounds, the gap here will help us. We can write this in canonical matrix form,

\begin{equation*} \max_{x \geq 0} c^\top x \quad {\rm subject \, to}\quad Ax \leq b,\end{equation*}

where $c^\top = (1, \ldots,1)$ , $b^\top = (1, \ldots, 1,s)$ , and

\begin{equation*} A = \begin{pmatrix} I \\[4pt] Z^\top \end{pmatrix}.\end{equation*}

The dual linear programme is

\begin{equation*} \min_{y \geq 0} \; b^\top y \quad{\rm subject \, to}\quad c \leq A^\top y.\end{equation*}

Written out more fully, this is

(1) \begin{equation} \min_{y \geq 0} \; \sum_{j=1}^N y_j + s y_{N+1} \quad {\rm subject \, to}\quad 1 \leq y_j + Z_j y_{N+1} \ \text{for all } j. \end{equation}

The value of the dual problem is the value of the primal problem (e.g. [4, Section 4.2]), and for any dual-feasible y the value $b^\top y$ is an upper bound for the value of the problem. If we write $y_{N+1} = \eta$ for short, the problem in (1) requires

\begin{equation*} \min_{y \geq 0} \; \sum_{j=1}^N y_j + \eta s \quad {\rm subject \, to}\quad 1 \leq y_j + \eta Z_j \ \text{for all } j.\end{equation*}

Obviously, once $\eta>0$ has been chosen, the best dual-feasible choice of $y_1, \ldots, y_N$ will be $y_j = (1-\eta Z_j)^+$ . Thus, for any $\eta>0$ , the value $\Phi^*$ of the problem is bounded above by

\begin{equation*} \Phi(\eta) \equiv \sum_{j=1}^N (1-\eta Z_j)^+ + \eta s,\end{equation*}

which is clearly a convex piecewise-linear function of $\eta$ .

2. The BRS inequality

In the problem studied by Bruss and Robertson, and later in greater generality by Steele, $Z_1, \ldots, Z_N$ are positive random variables, and the distribution function of $Z_j$ is $F_j$ , assumed for convenience to be continuous. In this situation, the value $\Phi^*$ of the suitcase packing problem will of course be random, and the BRS inequality gives an upper bound for $E \Phi^*$ . Let us see how the BRS inequality follows easily from the linear-programming story of the previous section.

Clearly, for any $\eta>0$ we have

(2) \begin{align} \mathbb{E} \Phi^* \leq \mathbb{E} \Phi(\eta) & = \mathbb{E} \sum_{j=1}^N (1-\eta Z_j)^+ + \eta s \nonumber \\ & = \sum_{j=1}^N \int_0^{1/\eta} (1-\eta z)\; F_j(\mathrm{d} z) + \eta s. \end{align}

Now we optimize the bound in (2) by differentiating:

\begin{equation*} 0 = - \sum_{j=1}^N \int_0^{1/\eta} z \; F_j(\mathrm{d} z) + s ,\end{equation*}

which will be satisfied when $\eta = \eta^*$ , the root of

(3) \begin{equation} \sum_{j=1}^N \int_0^{1/\eta}z \; F_j(\mathrm{d} z) = s. \end{equation}

Taking $\eta = \eta^*$ and using (3), the bound in (2) for $\mathbb{E} \Phi^*$ is easily seen to be

\begin{equation*} \mathbb{E}\Phi^* \leq \sum_{j=1}^N \int_0^{1/\eta^*} F_j(\mathrm{d} x),\end{equation*}

which is the BRS inequality.

Remark 1. Even if the implicit equation in (3) cannot be solved explicitly, the bound in (2) can still be applied for any choice of $\eta$ .