Generalized Assignment via Submodular Optimization with Reserved Capacity

TL;DR

This paper introduces a novel approximation algorithm for a group-structured variant of the generalized assignment problem, leveraging submodular optimization with reserved capacity to improve feasible solutions.

Contribution

It presents the first $rac{1}{6}$-approximation algorithm for Group GAP with size constraints, using a new submodular function derived from LP formulations.

Findings

Achieves a $rac{1}{6}$-approximation for Group GAP with size constraints.

Develops a submodular maximization algorithm with a knapsack constraint and reserved capacity.

Introduces a new approach potentially applicable to other group assignment problems.

Abstract

We study a variant of the \emph{generalized assignment problem} ({\sf GAP}) with group constraints. An instance of {\sf Group GAP} is a set $I$ of items, partitioned into $L$ groups, and a set of $m$ uniform (unit-sized) bins. Each item $i \in I$ has a size $s_i >0$, and a profit $p_{i,j} \geq 0$ if packed in bin $j$. A group of items is \emph{satisfied} if all of its items are packed. The goal is to find a feasible packing of a subset of the items in the bins such that the total profit from satisfied groups is maximized. We point to central applications of {\sf Group GAP} in Video-on-Demand services, mobile Device-to-Device network caching and base station cooperation in 5G networks. Our main result is a $\frac{1}{6}$-approximation algorithm for {\sf Group GAP} instances where the total size of each group is at most $\frac{m}{2}$. At the heart of our algorithm lies an interesting derivation of a submodular function from the classic LP formulation of {\sf GAP}, which facilitates the construction of a high profit solution utilizing at most half the total bin capacity, while the other half is \emph{reserved} for later use. In particular, we give an algorithm for submodular maximization subject to a knapsack constraint, which finds a solution of profit at least $\frac{1}{3}$ of the optimum, using at most half the knapsack capacity, under mild restrictions on element sizes. Our novel approach of submodular optimization subject to a knapsack \emph{with reserved capacity} constraint may find applications in solving other group assignment problems.

Equations

$$f(S)+f(T)\ge f(S\cup T)+f(S\cap T).$$

$$f(S\cup \{u\})-f(S)\le f(T\cup \{u\})-f(T),$$

$$LP(S)\text{:}$$

$$\sum _{j\in M}{x}_{i,j}\le 1$$

$$\sum _{i\in I}{x}_{i,j}\cdot s_i\le 1$$

$$x_{i,j}=0$$

$$x_{i,j}\ge 0$$

$$M(S)\text{:}$$

$$\sum _{j\in M}{y}_{i,j}\le {\hat{s}}_i$$

$$\sum _{i\in I}{y}_{i,j}\le N$$

$$y_{i,j}=0$$

$$y_{i,j}\ge 0$$

$$h(p_A,p_B,S_A,S_B)=p_A+(1-p_A-p_B)\left(1-e^{-\frac{\frac{1}{2}-S_A}{1-S_A-S_B}}\right)-\frac{p_A+p_B}{k}.$$

$$\max \left(h(p_1,p_2,S_1,S_2),h(p_2,p_1,S_2,S_1)\right)\ge \frac{1}{3}.$$

$$\begin{array}[]{c}f_{U}(T)\geq(1-e^{-1/2})f_{U}(L)\geq f_{U}(L)/3,\end{array}$$

$$\begin{array}[]{c}f_{U}(\{i\})\geq(1-e^{-1/2})f_{U}(L)\geq f_{U}(L)/3.\end{array}$$

$$f(B\cup T)\ge f(B_j)+f_U(T)\ge \frac{1}{3}(f(U)+f_U(L))=\frac{1}{3}\text{mbox}OPT.$$

$$f(B\cup T)\ge f(B)\ge f(B_j)+f_{B_j}(\{i\})\ge f(B_j)+f_U(\{i\})\ge \frac{1}{3}\text{mbox}OPT.$$

$$f_{P_k}(T_j)+f_{P_k}(\{i_j\})\ge (1-e^{-\frac{\frac{1}{2}-S_j}{1-S_1-S_2}})f_{P_k}(L).$$

$$f_{B_j}(T_j)\ge f_{P_k}(T_j)\ge (1-e^{-\frac{\frac{1}{2}-S_j}{1-S_1-S_2}})f_{P_k}(L)-\frac{1}{k}f(P_k).$$

$$\begin{array}[]{ll}f(S_{e}\cup T)&=f(B_{j}\cup T_{j})\geq f(B_{j})+(1-e^{-\frac{\frac{1}{2}-S_{j}}{1-S_{1}-S_{2}}})f_{P_{k}}(L)-\frac{1}{k}f(P_{k})\\ \\ &\geq\displaystyle{\mbox{OPT}\left(p_{j}+(1-p_{1}+p_{2})(1-\exp(-\frac{\frac{1}{2}-S_{j}}{1-S_{1}-S_{2}}))-\frac{p_{1}+p_{2}}{k}\right).}\end{array}$$

$$V_1=\frac{p+d}{2}+(1-p)(1-r)-\frac{p}{k}$$

$$V_2=\frac{p-d}{2}+(1-p)\left(1-\frac{e^{-1}}{r}\right)-\frac{p}{k}$$

$$g_1(x^{\ast })=1-\frac{p}{2}-\frac{\sqrt{d^2+4e^{-1}(1-p{)}^2}}{2}-\frac{p}{k}\ge 1-\frac{p}{2}-\frac{p}{k}-\frac{\sqrt{p^2+4e^{-1}(1-p{)}^2}}{2}.$$

$$g_1(x^{\ast })\ge a(p)\ge \min \left(a(0),a\left(\frac{1}{3}\right)\right)=\min \left(1-e^{-0.5},\frac{5}{6}-\frac{1}{3k}-\frac{\sqrt{1+16e^{-1}}}{6}\right)\ge \frac{1}{3}$$

$$g_1(x^{\ast })=1-\frac{p}{2}-\frac{\sqrt{d^2+4e^{-1}(1-p{)}^2}}{2}-\frac{p}{k}\ge 1-\frac{p}{2}-\frac{p}{k}-\frac{\sqrt{(\frac{2}{3}-p{)}^2+4e^{-1}(1-p{)}^2}}{2}$$

$$b^{\prime \prime }(z)=-\frac{2e^{-1}}{9{\left(4e^{-1}(1-z{)}^2+(\frac{3}{2}-z{)}^2\right)}^{\frac{3}{2}}}<0.$$

$$p_j(U_j^{\prime })+p_{{\ell }_1}(U_{{\ell }_1}^{\prime })+p_{{\ell }_2}(U_{{\ell }_2}^{\prime })\ge p_j(A_j)+p_{{\ell }_1}(U_{{\ell }_1}).$$

$$p_j(U_j^{\prime })+p_{{\ell }_1}(U_{{\ell }_1}^{\prime })+p_{{\ell }_2}(U_{{\ell }_2}^{\prime })\ge p_j(B_j)+p_{{\ell }_2}(U_{{\ell }_2}).$$

$$p_j(U_j^{\prime })+p_{{\ell }_1}(U_{{\ell }_1}^{\prime })+p_{{\ell }_2}(U_{{\ell }_2}^{\prime })\ge \frac{1}{2}\cdot \left(p_j(U_j)+p_{{\ell }_1}(U_{{\ell }_1})+p_{{\ell }_2}(U_{{\ell }_2})\right).$$

$$\begin{array}[]{ll}c&>a\cdot\displaystyle{\frac{2-2r-\frac{1}{2}}{\frac{1}{2}-r}=a\cdot\frac{3-4r}{1-2r}}\\ \\ &=\displaystyle{a\cdot\left(\frac{2-4r}{1-2r}+\frac{1}{1-2r}\right)=a\cdot\left(2+\frac{1}{1-2r}\right)>2a},\end{array}$$

$$\sum _{i\in U_j}{s}_i$$

$$=\sum _{i\in U_{j,1}}{s}_i+\sum _{r=1}^{k_j-1}\sum _{\{i|(i,(j,r))\in E\}}{x}_{(i,(j,r))}^{\prime }{s}_{j,r}^{\min }$$

$$\le \sum _{i\in U_{j,1}}{s}_i+\sum _{r=1}^{k_j-1}\sum _{\{i|(i,(j,r))\in E\}}{x}_{(i,(j,r))}^{\prime }{s}_i\le \sum _{i\in U_{j,1}}{s}_i+\sum _{i\in S}{x}_{i,j}{s}_i$$

$$\le 1+\sum _{i\in U_{j,1}}{s}_i$$

$$\begin{array}[]{ll}g_{T}(S^{*})&\leq g_{T}(S_{r})+\sum_{i\in S^{*}\setminus S_{r}}\left(g_{T}(S_{r}\cup\{i\})-g_{T}(S_{r})\right)\\ \\ &\leq g_{T}(S_{r})+m^{*}\eta_{r+1}\leq\sum_{j=1}^{m_{r}}p_{j}+m^{*}p_{m_{r}+1}\end{array}$$

Full text

Generalized Assignment via Submodular Optimization with Reserved Capacity

Ariel Kulik

  

Kanthi Sarpatwar

  

Baruch Schieber

  

Hadas Shachnai

Our main result is a $\frac{1}{6}$-approximation algorithm for Group GAP instances where the total size of each group is at most $\frac{m}{2}$. At the heart of our algorithm lies an interesting derivation of a submodular function from the classic LP formulation of GAP, which facilitates the construction of a high profit solution utilizing at most half the total bin capacity, while the other half is reserved for later use. In particular, we give an algorithm for submodular maximization subject to a knapsack constraint, which finds a solution of profit at least $\frac{1}{3}$ of the optimum, using at most half the knapsack capacity, under mild restrictions on element sizes. Our novel approach of submodular optimization subject to a knapsack with reserved capacity constraint may find applications in solving other group assignment problems.

1 Introduction

With the rapid adoption of cloud computing, wireless networks, and other modern platforms, resource allocation problems of various flavors have regained importance. One classic example is the generalized assignment problem (GAP). We are given a set of $n$ items and $m$ bins, $[m]=\{1,2,\ldots,m\}$. Each item $i\in[n]$ has a size $s_{i,j}>0$ and a profit $p_{i,j}\geq 0$ when packed into bin $j\in[m]$. The goal is to feasibly pack in the bins a subset of the items of maximum total profit. GAP has been widely studied, with applications ranging from grouping and loading for manufacturing systems to land use optimization in regional planning (see, e.g., [2, 12]). In discrete optimization, GAP has received considerable attention also as a special case of the separable assignment problem and submodular maximization (see, e.g., [27, 16, 5, 6]). We consider a variant of GAP with group constraints. An instance of Group GAP consists of a set $I=\{1,2,\ldots,n\}$ of $n$ items and $m$ uniform (unit-sized) bins $M=\{1,\ldots,m\}$. Each item $i\in I$ has a size $s_{i}>0$ and a profit $p_{i,j}\geq 0$ when assigned to bin $1\leq j\leq m$. The items in $I$ are partitioned into $L\geq 1$ groups, ${\cal G}=\{G_{1},\ldots,G_{L}\}$. Given an assignment of items to bins, we say that a group is satisfied if all of its items are assigned. The goal is to find a feasible assignment of a subset of the items to bins such that the total profit from satisfied groups is maximized. Formally, a feasible assignment is a tuple $(U_{1},\ldots,U_{m})$, such that $U_{j}\cap U_{k}=\emptyset$ for all $1\leq j<k\leq m$, $U_{j}\subseteq I$ and $\sum_{i\in U_{j}}s_{i}\leq 1$, for all $1\leq j\leq m$. Let $I(U)=\cup_{j}U_{j}$. Then $G_{\ell}\in{\cal G}$ is satisfied if $G_{\ell}\subseteq I(U)$. Let ${\cal G}_{s}=\{G_{\ell_{1}},\ldots G_{\ell_{t}}\}$ be the set of satisfied groups and $I({\cal G}_{s})=\cup_{G_{\ell}\in{\cal G}_{s}}G_{\ell}$. Then we seek an assignment $(U_{1},\ldots,U_{m})$ for which $\sum_{j=1}^{m}\sum_{i\in U_{j}\cap I({\cal G}_{s})}p_{i,j}$ is maximized.

The following scenario suggests a natural application for Group GAP. Consider a Video-on-Demand (VoD) service where each video is given as a collection of segments. The system has a set of $m$ servers of uniform capacity distributed over multiple locations. To obtain revenue from a video the system must store all of its segments (possibly on different servers). The revenue from a specific video also depends on the servers which store the segments. This is due to the content delivery costs resulting from the distance between the servers and the predicted location of the video audience. The objective of the VoD service provider is to select a subset of segments and an allocation of these segments to servers so as to maximize the total revenue. In the Appendix we describe central applications of Group GAP in mobile Device-to-Device network caching and in base station cooperation in 5G networks.

1.1 Prior Work

We note that a Group GAP instance in which each group consists of a single item yields an instance of classic GAP where each item takes a single size across the bins, and all the bins have identical capacities. GAP is known to be APX-hard already in this case, even if there are only two possible item sizes, and each item can take one of two possible profits [10]. Thus, most of the previous research focused on obtaining efficient approximate solutions.111Given an algorithm ${\cal A}$, let ${\cal A}(I),OPT(I)$ denote the profit of the solution output by ${\cal A}$ and by an optimal solution for a problem instance $I$, respectively. For $\rho\in(0,1]$, we say that ${\cal A}$ is a $\rho$-approximation algorithm if, for any instance $I$, $\frac{{\cal A}(I)}{OPT(I)}\geq\rho$. Fleischer et al. [16] obtained a $(1-e^{-1})$-approximation for GAP, as a special case of the separable assignment problem. Feige and Vondrák [14] obtained the current best known ratio of $1-e^{-1}+\varepsilon$, for some absolute constant $\varepsilon>0$.

Chen and Zhang [11] studied the problem of group packing of items into multiple knapsacks (GMKP), a special case of Group GAP where the profit of each item is the same across the bins. Let GMKP$(\delta)$ be the restriction of GMPK to instances in which the total size of items in each group is at most $\delta m$ (that is, a factor $\delta$ of the total capacity of all bins). For $\delta>\frac{2}{3}$, the paper [11] rules out the existence of a constant factor approximation for GMKP$(\delta)$, unless P= NP. For $\frac{1}{3}<\delta\leq\frac{2}{3}$, the authors show that there is no $\left(\frac{1}{2}+\varepsilon\right)$-approximation for GMKP$(\delta)$, unless P= NP, and derive a nearly matching $\left(\frac{1}{2}-\varepsilon\right)$-approximation, for any $\varepsilon>0$. The paper presents also approximation algorithms and hardness results for other special cases of GMPK.

There has been earlier work also on variants of Group GAP with the added constraint that in any feasible assignment there is at most one item from $G_{\ell}$ in bin $j$, for any $1\leq\ell\leq L$, $1\leq j\leq m$. Adany et al. [1] considered this problem, called all-or-nothing GAP (AGAP). They presented a $(\frac{1}{19}-\varepsilon)$-approximation algorithm for general instances, and a $(\frac{1}{3}-\varepsilon)$-approximation for the special case where the profit of an item is identical across the bins, called the group packing (GP) problem. Sarpatwar et al. [24] consider a more general setting for AGAP, where each group of items is associated with a time window in which it can be packed. The paper shows that this variant of the problem, called $\chi$-AGAP, admits an $\Omega(1)$-approximation, assuming the time windows are large enough relative to group sizes. Specifically, for a group $G_{\ell}$ having a time window of $m$ slots (= $m$ bins), it is assumed that $s(G_{\ell})\leq\frac{m}{20}$.

1.1.1 Submodular Maximization

Given a finite set $\Omega$, a function $f:2^{\Omega}\rightarrow\rm I\!R$ is submodular if for every $S,T\subseteq\Omega$ we have

[TABLE]

An equivalent definition of submodularity refers to its diminishing returns:

[TABLE]

for any $T\subseteq S\subseteq\Omega$, and $u\in\Omega\setminus S.$ A set function $f$ is monotone if for every $S\subseteq T\subseteq\Omega$ it holds that $f(S)\leq f(T)$. Submodular functions arise naturally in a wide variety of optimization problems, ranging from coverage problems and graph cut problems to welfare problems (see [7] for a survey on submodular functions). Submodular optimization under various constraints has been widely studied in the past four decades (see, e.g., [26, 8, 15] and [7] and the references therein).

The problem of maximizing a monotone submodular function subject to a knapsack constraint is defined as follows. We are given an oracle to a monotone, non-negative submodular function $f:2^{\Omega}\rightarrow\mathbb{R}$. Each element $i\in\Omega$ is associated with a size $s_{i}\geq 0$. We are also given a capacity $B>0$. The objective is to find a subset $S\subseteq\Omega$ such that $\sum_{i\in S}s_{i}\leq B$ and $f(S)$ is maximized. The best known result is a $(1-e^{-1})$-approximation algorithm due to Sviridenko [26]. The ratio of $(1-e^{-1})$ cannot be improved even when $f$ is a coverage function and element sizes are uniform, unless P= NP [13]. A matching lower bound of $(1-e^{-1})$ is known also for the oracle model with no complexity assumption [22].

1.2 Contribution and Techniques

Our main result is a $\frac{1}{6}$-approximation algorithm for Group GAP instances where the total size of each group is at most $\frac{m}{2}$. We note that when group sizes can be arbitrary in $(0,m]$, Group GAP cannot be approximated within any bounded ratio, even if item profits are identical across the bins, and $m=2$, unless P= NP. Indeed, in this case, deciding whether a single group of items of total size $2$ and total profit $1$ can be packed in the bins yields an instance of Partition, which is NP-complete [17]. Furthermore, even if group sizes are restricted to be no greater than $\delta m$, for some $\delta>\frac{2}{3}$, then Group GAP still cannot be approximated within a constant factor, as it generalizes GMKP$(\delta)$, for which the paper [11] shows hardness of approximation. Similarly, as we consider in this paper a generalization of GMKP$\left(\frac{1}{2}\right)$, it follows from [11] that our problem cannot be approximated within ratio better than $\frac{1}{2}$.

In solving Group GAP we combine the framework of Adany et al. [1] with the rounding technique of Shmoys and Tardos [25]. The framework of [1] uses submodular maximization to select a collection of groups for the solution. It then finds a feasible assignment for the selected groups.

At the heart of our algorithm lies an interesting derivation of a submodular function from the classic LP formulation of GAP, which facilitates the construction of a high profit solution utilizing at most half the total bin capacity. In particular, we give an algorithm for submodular maximization subject to a knapsack constraint, which finds a solution occupying at most half the knapsack capacity, while the other half is reserved for later use.222We assume throughout the discussion that the size of each element is at most half the knapsack capacity. We show that this algorithm achieves an approximation ratio of $\frac{1}{3}$ relative to an optimal solution that may use the whole knapsack capacity. We note that this ratio is tight. Indeed, it is easy to construct an instance for which the best solution with half the knapsack capacity has only $\frac{1}{3}$ the profit of the optimal solution with full knapsack capacity. We also note that a naive application of the algorithm of Sviridenko [26] with half the knapsack capacity will only guarantee a $\frac{1-e^{-1}}{3}\approx\frac{1}{4.7}$-approximation.

To obtain an integral solution, given a fractional assignment of the selected groups, we apply the rounding technique of Shmoys and Tardos [25], followed by a filling phase. We show that if the total size of the items in the selected groups is at most $\frac{m}{2}$, the rounding procedure yields a feasible assignment of the selected groups, whose profit is at least half the value of the submodular function. Our novel approach of submodular optimization subject to a knapsack with reserved capacity constraint may find applications in solving other group assignment problems.

2 Approximation Algorithm

In this section we present an approximation algorithm for Group GAP. We first introduce several definitions and tools that will be used as building blocks of our algorithm.

2.1 Basic Definitions and Tools

2.1.1 The Submodular Relaxation

For simplicity, we assume that all the numbers are rational. For a subset of elements $I^{\prime}\subseteq I$, let $s(I^{\prime})=\sum_{i\in I^{\prime}}s_{i}$ be the total size of the elements in $I^{\prime}$. We assume throughout the discussion that every $G_{\ell}\in{\cal G}$ satisfies $s(G_{\ell})\leq\frac{m}{2}$. We define below a function ${\phi}:2^{I}\rightarrow{\rm I\!R}^{+}\cup\{0\}$. Let $x_{i,j}\in\{0,1\}$ be an indicator for the assignment of item $i$ to bin $j$, for $1\leq i\leq n$, $1\leq j\leq m$. The following is a linear program associated with a subset $S\subseteq I$, in which $x_{i,j}\geq 0$ for all $i,j$.

[TABLE]

Note that, by the above constraints, all solutions for the LP have the same dimension, regardless of the size of $S$. We define ${\phi}(S)$ as the optimal value of $\mbox{LP}(S)$.

We denote the profit of a solution $x$ for the linear program by $p\cdot x=\sum_{i\in I,j\in M}x_{i,j}\cdot p_{i,j}$. In Section 3 we prove the next result.

Theorem 2.1**.**

The function $\phi$ is submodular.

We note that $\phi$ is also monotone and non-negative. We use $\phi$ to define the group function $\psi:2^{\cal G}\rightarrow{\rm I\!R}^{+}\cup\{0\}$. For any $G^{*}\subseteq{\cal G}$ let $I(G^{*})=\bigcup_{G_{\ell}\in G^{*}}G_{\ell}$ and $\psi(G^{*})=\phi(I(G^{*}))$. As $\phi$ is submodular, monotone and non-negative, it is easy to see that $\psi$ is submodular, monotone and non-negative as well. We optimize $\psi$ subject to a knapsack (budget) constraint, using the next general result.

Theorem 2.2**.**

[Submodular optimization with reserved capacity]*

Let ${\Omega}=\{1,\ldots,n\}$ be a ground set, and $m\geq 0$ a knapsack capacity. Each $i\in{\Omega}$ is associated with non-negative size $s_{i}\leq\frac{m}{2}$. Let $f:2^{\Omega}\rightarrow{\rm I\!R}^{+}\cup\{0\}$ be a non-negative monotone submodular function, and $\mbox{OPT}=\max\{f(S)|S\subseteq{\Omega},\sum_{i\in S}s_{i}\leq m\}$. Then Algorithm 2 (in Section 4) finds in polynomial time333The explicit representation of a submodular function might be exponential in the size of its ground set. Thus, it is standard practice to assume that the function is accessed via a value oracle. Then the number of operations and oracle calls is polynomial in the size of ${\Omega}$ and the maximum length of the representation of $f(S)$. a subset $S\subseteq{\Omega}$ satisfying $f(S)\geq\frac{\mbox{OPT}}{3}$ and $\sum_{i\in S}s_{i}\leq\frac{m}{2}$.*

The proof of Theorem 2.2 is given in Section 4.

2.1.2 Solution Types

Our algorithm uses a few types of intermediate solutions for Group GAP, as defined below. Given $G^{*}\subseteq{\cal G}$, we say that a solution $x$ for $LP(I(G^{*}))$ is a fractional solution. Let $U=(U_{1},\ldots,U_{m})$ be an assignment of elements to bins, where $U_{j}$ is the set of elements assigned to bin $j$. Then $I(U)=\bigcup_{j=1}^{m}U_{j}$ is the subset of elements packed in the bins. We say that $U$ is feasible if for each bin $1\leq j\leq m$ we have $s(U_{j})\leq 1$. We say that $U$ is almost feasible if for each bin $1\leq j\leq m$ there is an element $u_{j}^{*}$ such that $s(U_{j}\setminus\{u_{j}^{*}\})\leq 1$. We also define the profit of an assignment as $p(U)=\sum_{1\leq j\leq m}\sum_{i\in U_{j}}p_{i,j}$.

Our algorithm first obtains a fractional solution, which is then converted to an almost feasible solution. Finally, the algorithm converts this solution to a feasible one. We now state the results used in these conversion steps.

Theorem 2.3**.**

Given $G^{*}\subseteq{\cal G}$, such that $s(I(G^{*}))\leq m$, and a fractional solution $x$ for $LP(I(G^{*}))$, it is possible to construct in polynomial time an almost feasible assignment $U$ such that $p(U)\geq p\cdot x$, and $I(U)=I(G^{*})$.

The theorem easily follows by applying a rounding technique of [25] to a fractional solution in which every element in $I(G^{*})$ is fully assigned (fractionally, in multiple bins). We note that such a solution always exists, since $s(I(G^{*}))\leq m$. We give the proof in the Appendix. To convert an almost feasible solution to a feasible one we use the following result (we give the proof in Section 5).

Theorem 2.4**.**

Let $U=(U_{1},\ldots,U_{m})$ be an almost feasible assignment such that $s(I(U))\leq\frac{m}{2}$, then $U$ can be converted in polynomial time to a feasible assignment $U^{\prime}$, with $I(U^{\prime})=I(U)$ and $p(U^{\prime})\geq\frac{1}{2}p(U)$.

2.2 The Algorithm

Our approximation algorithm for Group GAP follows easily from the tools presented in Section 2.1. Initially, we solve the problem of maximizing a submodular function subject to a knapsack with reserved capacity constraint for the set function $\psi$. Then we solve the linear program and convert the solution to a feasible assignment. We give the pseudocode in Algorithm 1.

Theorem 2.5**.**

Algorithm 1 is a polynomial time $\frac{1}{6}$-approximation algorithm for Group GAP when the total size of a single group is bounded by $\frac{m}{2}$; that is, $\forall G_{\ell}\in{\cal G}:\sum_{i\in G_{\ell}}s_{i}\leq\frac{m}{2}$.

Proof 2.6**.**

It is easy to see that the algorithm runs in polynomial time. By Theorem 2.2, we have that $\psi(S^{*})\geq\mbox{OPT}/3$, where OPT is the value of the optimal solution for the original instance.

By Theorems 2.3 and 2.4, we are guaranteed to find in Steps 3$-$4 a feasible assignment $U$ of all elements in $I(S^{*})$, such that $p(U)\geq\frac{1}{2}\psi(S^{*})\geq\frac{1}{6}\mbox{OPT}$.

3 Submodularity

In this section we show that the function $\phi$ is submodular. Our proof builds on the useful relation of our problem to maximum weight bipartite matching. Let $G=(A\cup B,E)$ be a bipartite (edge) weighted graph, where $|B|\geq|A|$. Assume that the graph is complete (by adding zero weight edges if needed). For $e\in E$, let $W(e)$ be the weight of edge $e$, and for $F\subseteq E$, let $W(F)=\sum_{e\in F}W(e)$ be the total weight of edges in $F$. For $S\subseteq A$, define $h(S)$ to be the value of the maximum weight matching in $G[S\cup B]$, the graph induced by $S\cup B$. We call $h$ the partial maximum weight matching function. The next result was shown by Bar-Noy and Rabanca [3].

Theorem 3.1**.**

If the edge weights are non-negative then the function $h$ is (monotone) submodular.

We give a simpler proof in the Appendix. We are now ready to prove our main result.

Proof of Theorem 2.1: We first note that since all numbers are rational, for some $N\in{\mathbb{Z}}^{+}$, we can write $s_{i}=\frac{{\hat{s}}_{i}}{N}$, where ${{\hat{s}}_{i}}\in{\mathbb{Z}}^{+}$ for all $i\in I$.444Note that $N$, which may be arbitrarily large, is used just for the proof. Our algorithm does not rely on obtaining a solution (or an explicit formulation) for $M(S)$.

Now, set the capacity of each bin $1\leq j\leq m$ to be $b_{j}=N$, and let $0\leq y_{i,j}\leq{{\hat{s}}_{i}}$ indicate the size of item $i$ assigned to bin $j$. For a subset of items $S\subseteq I$, we now write the following linear program.

[TABLE]

Indeed, Constraint (3) ensures that the total size assigned for item $i$ over the bins is upper bounded by ${{\hat{s}}_{i}}$, and Constraint (4) guarantees that the capacity constraint is satisfied for all the bins $j\in M$. Given a subset of elements $S\subseteq I$, let ${\eta}(S)$ be the value of an optimal solution for $M(S)$.

Now, observe that any feasible solution for $LP(S)$ induces a feasible solution for $M(S)$ of the same value, by setting $y_{i,j}=x_{i,j}\cdot{{\hat{s}}_{i}}$ for all $i\in I$ and $j\in M$. Similarly, a feasible solution for $M(S)$ induces a feasible solution for $LP(S)$ of the same value. Hence, $\phi(S)={\eta}(S)$ for all $S\subseteq I$.

By the above discussion, to prove the theorem it suffices to show that ${\eta}$ is submodular. Given our Group GAP instance, we construct the following bipartite graph $G$. For each item $i\in I$, we define ${{\hat{s}}_{i}}$ vertices, $V_{i}=\{v_{i,1},\ldots,v_{i,{{\hat{s}}_{i}}}\}$. For each bin $j\in M$, we define $N$ vertices $U_{j}=\{u_{j,1},\ldots,u_{j,N}\}$. For any $1\leq i\leq n$ and $1\leq j\leq m$, there are edges $(v_{i,s},u_{j,r})$ of weight $p_{i,j}/{{\hat{s}}_{i}}$, for all $1\leq s\leq{{\hat{s}}_{i}}$, $1\leq r\leq N$. Let $V_{I}=\cup_{i\in I}V_{i}$, $U_{M}=\cup_{j\in M}U_{j}$, and let $E$ be the set of edges. Consider the bipartite graph $G=(V_{I}\cup U_{M},E)$. W.l.o.g we may assume that $|U_{M}|\geq|V_{I}|$; otherwise, we can add new bins $j=m+1,m+2,\ldots$ with corresponding sets of vertices $U_{j}=\{u_{j,1},\ldots,u_{j,N}\}$ and zero weight edges $(v_{i,s},u_{j,r})$ for all $1\leq i\leq n$, $1\leq s\leq{{\hat{s}}_{i}}$, $1\leq r\leq N$.

We note that, given a subset of items $S\subseteq I$, $M(S)$ is the linear programming relaxation of the problem of finding a maximum weight matching in the subgraph $G[V_{S}\cup U_{M}]$, where $V_{S}\subseteq V_{I}$ is the subset of vertices in $G$ that corresponds to $S$. Using standard techniques (see, e.g., [19]), it can be shown that $M(S)$ has an optimal integral solution. Hence, ${\eta}(S)=h(V_{S})$, where $h:2^{V_{I}}\rightarrow{\rm I\!R}^{+}$ is a partial maximum weight matching function in $G$. By Theorem 3.1, $h$ is (monotone) submodular. Hence, ${\eta}$ is also (monotone) submodular. ∎

4 Submodular Optimization with Reserved Capacity

In this section we prove Theorem 2.2. We start with some definitions and notation. Assume we are given a ground set ${\Omega}=\{1,\ldots,n\}$ and capacity $m>0$, where each element $i\in{\Omega}$ is associated with a non-negative size $s_{i}\leq\frac{m}{2}$. For $S\subseteq{\Omega}$, let $s(S)=\sum_{i\in S}s_{i}$. Also, for $S,T\subseteq{\Omega}$ let $f_{S}(T)=f(S\cup T)-f(S)$. We use throughout this section basic properties of monotone submodular functions (see, e.g., [7]).

In the following we give an outline of an algorithm for maximizing a monotone submodular function $f$ subject to a knapsack with reserved capacity constraint. Specifically, assuming that the knapsack capacity is $m$ for some $m>0$, the algorithm solves the problem $\max_{\{S\subseteq{\Omega}:s(S)\leq\frac{m}{2}\}}f(S)$. The algorithm, SubmodularOpt, initially guesses the set of at most six items of highest profits in some optimal solution (for the problem with knapsack capacity $m$), and a subset of these profitable items, whose total size is at most $m/2$. Then the algorithm calls a procedure which applies the Greedy approach as in [26] to find the remaining items in the solution. We give a pseudocode of SubmodularOpt in Algorithm 2. It is important to note that while the algorithm produces a solution of size at most $m/2$, the analysis compares this solution against an optimal solution of size at most $m$.

The next lemma, which plays a key role in our analysis, follows from the technique presented in [26].

Lemma 4.1**.**

Given the knapsack capacity $m>0$, let $0<m^{\prime}\leq m^{*}\leq m$. Let $S^{*}\subseteq{\Omega}$ be a non-empty subset of elements, such that $s(S^{*})\leq m^{*}$. Also, let $g:2^{\Omega}\rightarrow\rm I\!R$ be a monotone submodular function satisfying $g(\emptyset)=0$, and let $S=\textsc{Greedy}(g,m^{\prime})$. Then, there is an element $i^{*}\in S^{*}$ such that $g(S)+g(\{i^{*}\})\geq(1-e^{m^{\prime}/m^{*}})g(S^{*})$.

In the Appendix we prove a more general result (see Lemma A.2).555The statement of Lemma 4.1 is obtained by taking in Lemma A.2 $T=\emptyset$. In [26] the above lemma was applied in conjunction with a guessing phase, used to ensure the three most profitable elements in an optimal solution are selected by the algorithm, thus limiting the value of $g(\{i^{*}\})$ in the special case where $m^{\prime}=m^{*}$.

Attempting to apply a similar approach in solving the problem with reserved capacity, several difficulties arise. The first one is that the most profitable elements in any optimal solution may already exceed the reduced capacity, and therefore cannot be added to the solution. Another difficulty is that even if these elements do fit in the smaller knapsack, one can easily come up with a scenario in which it is better not to include them in the solution.

To overcome these difficulties we use the following main observation. Given $P_{k}$, the set of $k=6$ most profitable elements in an optimal solution666The value $k=6$ is derived from Lemma 4.2. It may be possible to obtain the same approximation ratio using smaller values of $k$, leading to a more efficient algorithm., and a partition of this set into two subsets $B_{1},B_{2}$, each of size at most $m/2$, adding elements to either $B_{1}$ or $B_{2}$ using the greedy procedure leads to a solution of value at least one third of the value of an optimal solution. This observation comes into play in Case 2.2 in the proof of Theorem 2.2. The next technical lemma is used to prove this observation (we give the proof below).

Lemma 4.2**.**

For $k=6$, $p_{A},p_{B},S_{A},S_{B}\geq 0$ define

[TABLE]

Then for $p_{1},p_{2},S_{1},S_{2}$ such that $0\leq p_{1},p_{2}\leq\frac{1}{3}$ and $0\leq S_{1},S_{2}\leq\frac{1}{2}$ it holds that

[TABLE]

Another main tool used in the proof of Theorem 2.2 is a simple partitioning procedure. It shows that $P_{k}$ can either be partitioned into two sets as required in the above observation, or we reach a simple corner case (Case 2.1 in the proof) in which at least one third of the optimal value can be easily attained. For the latter case, we use the following result, due to [20].

Lemma 4.3**.**

Let $g:2^{\Omega}\rightarrow\rm I\!R$ be a non-negative and monotone submodular function. Let $\mbox{OPT}=\max\{g(S)|S\subseteq{\Omega},\sum_{i\in S}s_{i}\leq m\}$, and $S=\textsc{Greedy}(g,m)$. Then either $g(S)\geq(1-e^{-1/2})\mbox{OPT}$, or there is an element $i\in{\Omega}$ such that $g(\{i\})\geq(1-e^{-1/2})OPT$ and $s_{i}\leq m$.

Proof of Theorem 2.2 [Submodular optimization with reserved capacity]: It is easy to see that the running time of the algorithm is polynomial. Let $S\subseteq{\Omega}$, $s(S)\leq m$, $f(S)=\mbox{OPT}$, and $k=6$.

Case 1: We first handle the case where $|S|<k$. Start with $A_{1}=\emptyset$, iterate over the elements of $S$ and add them to $A_{1}$, as long as $s(A_{1})\leq m/2$. If $S\neq A_{1}$, let $j\in S\setminus A_{1}$, and set $A_{2}=\{j\}$ and $A_{3}=S\setminus(A_{1}\cup A_{2})$. Clearly, $s(A_{2})\leq m/2$, and since $s(A_{1}\cup A_{2})>m/2$ and $s(S)\leq m$, we have that $s(A_{3})\leq m/2$. If $S=A_{1}$ set $A_{2}=A_{3}=\emptyset$.

By the submodularity of $f$, we have $f(S)\leq f(A_{1})+f(A_{2})+f(A_{3})$. Hence, for some $r\in\{1,2,3\}$, $f(A_{r})\geq f(S)/3=\mbox{OPT}/3$. We also have that $|A_{r}|\leq 5$; therefore, at some iteration of the algorithm $S_{e}=B=A_{r}$, and following this iteration $f(R)\geq\mbox{OPT}/3$.

Case 2: Assume now that $|S|\geq k$. Let $S=\{i_{1},i_{2},\ldots,i_{\ell}\}$ such that the elements are ordered by their marginal profits: $i_{j}=\operatorname*{arg\,max}_{j\leq r\leq\ell}f_{\{i_{1},\ldots i_{j-1}\}}(\{i_{r}\})$. Set $P_{k}=\{i_{1},i_{2},\ldots,i_{k}\}$.

Consider the following process. Start with $B_{1}=\emptyset$ and $B_{2}=\emptyset$. Iterate over the elements $i\in P_{k}$ in decreasing order by size. For each element $i$, let $t=\operatorname*{arg\,min}_{j=1,2}s(B_{j})$. If $s(B_{t})+s_{i}\leq m/2$ then $B_{t}=B_{t}\cup\{i\}$; otherwise, Stop. We now distinguish between two sub-cases for the termination of the process.

Case 2.1: Suppose that the process terminates due to an element $i$ which cannot be added to any of the sets. Let $B_{1}$ and $B_{2}$ be the sets in this iteration. Also, set $B_{3}=\{i\}$, $U=B_{1}\cup B_{2}\cup B_{3}$, and $L=S\setminus U$. W.l.o.g assume that $s(B_{1})\geq s(B_{2})$. As the process terminated, we have that $s(B_{3})+s(B_{2})=s_{i}+s(B_{2})>m/2$. The sets $B_{1},B_{2},B_{3}$ and $L$ form a partition of $S$, and $s(S)\leq m$. We conclude that $s(B_{1})+s(L)\leq m/2$. Hence, $s(B_{2})+s(L)\leq m/2$, and $s(B_{3})+s(L)\leq m/2$ as well (it is easy to see that $s(B_{3})\leq s(B_{1})$).

By the submodularity of $f$, $f(U)\leq f(B_{1})+f(B_{2})+f(B_{3})$; thus, there is $j\in\{1,2,3\}$ such that $f(B_{j})\geq f(U)/3$. As none of the sets $B_{1},B_{2},B_{3}$ is empty, we have that $|B_{j}|\leq|P_{k}|-2=4$.

Let $T=\textsc{Greedy}(f_{U},m/2-s(B_{j}))$. By Lemma 4.3, either

[TABLE]

or there is $i\in L$ such that

[TABLE]

In the former case, we can consider the iteration in which $S_{e}=U,B=B_{j}$. In this iteration, we have

[TABLE]

In the latter case, we can consider the iteration where $S_{e}=B=B_{j}\cup\{i\}$, and in which

[TABLE]

Case 2.2: The process terminated with $B_{1},B_{2}$ satisfying $B_{1}\cup B_{2}=P_{k}$, and $s(B_{1}),s(B_{2})\leq m/2$.

Let $p_{1}=f(B_{1})/\mbox{OPT}$, $p_{2}=(f(P_{k})-f(B_{1}))/\mbox{OPT}$, $S_{1}=s(B_{1})$, $S_{2}=s(B_{2})$ and $L=S\setminus P_{k}$. If $p_{1}\geq\frac{1}{3}$ (or $p_{2}\geq\frac{1}{3}$ ) we have that in the iteration where $S_{e}=P_{k}$ and $B=B_{1}$ (or $B=B_{2}$) the algorithm finds a solution of value at least $\mbox{OPT}/3$, and the theorem holds. Thus, we may assume that $p_{1},p_{2}\leq\frac{1}{3}$.

For $j=1,2$, let $T_{i}=\textsc{Greedy}(f_{P_{k}},m/2-S_{j})$. Using Lemma 4.1 with $S^{*}=L$, we have that there is $i_{j}\in L$ for which

[TABLE]

By the selection of elements in $P_{k}$, we have $f_{P_{k}}(\{i_{j}\})\leq\frac{1}{k}f(P_{k})$. Thus,

[TABLE]

Hence, in the iteration where $S_{e}=P_{k},B=B_{j}$, we obtain a solution satisfying

[TABLE]

By Lemma 4.2, in one of these iterations we obtain a solution of value at least $\mbox{OPT}/3$, implying the statement of the theorem. ∎

Proof of Lemma 4.2: Let $p_{1},p_{2},S_{1},S_{2}$ be values that satisfy the conditions in the lemma. Denote $p=p_{1}+p_{2}$, $d=p_{1}-p_{2}$ and $r=\exp\left(-\frac{\frac{1}{2}-S_{1}}{1-S_{1}-S_{2}}\right)$. It is easy to see that $\exp\left(-\frac{\frac{1}{2}-S_{2}}{1-S_{1}-S_{2}}\right)=\frac{e^{-1}}{r}$. Define $V_{1}=h(p_{1},p_{2},S_{1},S_{2})$ and $V_{2}=h(p_{2},p_{1},S_{2},S_{1})$. By the definition of $h$ and above definitions we get

[TABLE]

and

[TABLE]

Let $g_{1}(x)=\frac{p+d}{2}+(1-p)(1-x)-\frac{p}{k}$ and $g_{2}(x)=\frac{p-d}{2}+(1-p)\left(1-\frac{e^{-1}}{x}\right)-\frac{p}{k}$. Clearly, $V_{1}=g_{1}(r)$ and $V_{2}=g_{2}(r)$. It is also easy to see that $g_{1}$ is decreasing and $g_{2}$ is increasing (for $x>0$).

Claim 1**.**

It holds that $g_{1}(x^{*})=g_{2}(x^{*})$ where $x^{*}=\frac{d+\sqrt{d^{2}+4e^{-1}(1-p)^{2}}}{2(1-p)}$.

Proof 4.4**.**

By rearranging terms we have $g_{1}(x)=g_{2}(x)$ if and only if $d=(1-p)(x-\frac{e^{-1}}{x})$, which holds for $x>0$ if and only if $0=(1-p)x^{2}-dx-e^{-1}(1-p)$. The latter is a quadratic equation and $x^{*}>0$ is a root.

If $r\geq x^{*}$, since $g_{2}$ is increasing, we have $V_{2}=g_{2}(r)\geq g_{2}(x^{*})=g_{1}(x^{*})$, and if $r\leq x^{*}$, as $g_{1}$ is decreasing, we have $V_{1}=g_{1}(r)\geq g_{1}(x^{*})$. Therefore $\max(V_{1},V_{2})\geq g_{1}(x^{*})$.

Consider two cases:

Case 1: $p\leq\frac{1}{3}$. We have $|d|\leq p$ and

[TABLE]

Let $a(y)=1-\frac{y}{2}-\frac{y}{k}-\frac{\sqrt{y^{2}+4e^{-1}(1-y)^{2}}}{2}$. Clearly, $g_{1}(x^{*})\geq a(p)$. We note that $a^{\prime\prime}(y)=-\frac{2e^{-1}}{(y^{2}+4e^{-1}(1-y)^{2})^{\frac{3}{2}}}$. As $a^{\prime\prime}(y)<0$, $a(y)$ does not have a local minima in $[0,\frac{1}{3}]$. Thus as $p\in[0,\frac{1}{3}]$ we have

[TABLE]

Case 2: $p\geq\frac{1}{3}$, as $\frac{p+d}{2},\frac{p-d}{2}\leq\frac{1}{3}$ we have that $|d|\leq\frac{2}{3}-p$. Therefore,

[TABLE]

Let $b(z)=1-\frac{z}{2}-\frac{z}{k}-\frac{\sqrt{(\frac{2}{3}-z)^{2}+4e^{-1}(1-z)^{2}}}{2}$. Clearly, $g_{1}(x^{*})\geq b(p)$. We note that

[TABLE]

Thus, $b(z)$ does not have a local minima in $[\frac{1}{3},\frac{2}{3}]$. Hence, we have that $g_{1}(x^{*})\geq b(p)\geq\min(b(\frac{1}{3}),b(\frac{2}{3}))\geq\frac{1}{3}$.

In both cases we get $g_{1}(x^{*})\geq\frac{1}{3}$, and as $\max(V_{1},V_{2})\geq g_{1}(x^{*})$, the lemma immediately follows. ∎

5 Filling Phase

In this section we prove Theorem 2.4. Define the size of bin $j$ in assignment $U$ as $s^{U}_{j}=\sum_{i\in U_{j}}s_{i}$. We first divide the bins and items into types. We say that a bin $j$ is full if $s^{U}_{j}>1$, semi-full if $\frac{1}{2}\leq s^{U}_{j}\leq 1$, and semi-vacant if $s^{U}_{j}<\frac{1}{2}$. An item $i\in I(U)$ is big if $s_{i}>\frac{1}{2}$; otherwise, $i$ is small. Clearly, there are no big items in semi-vacant bins.

Informally, we use in the proof several types of resolution steps. Each step takes as input a full bin and possibly one or two semi-vacant bins, and reassigns some of the items into the bins while evicting others. These resolution steps ensure that the new assignment has at least half the profit of the original assignment, the assignment to any bin remains feasible, and only small items are evicted.

We apply the resolution steps repeatedly, but once a bin participated in a resolution step it may not participate in another one. We then prove that as long as there are full bins, one of the steps can be applied. Hence, by applying the resolution steps, we have a new assignment in which all bins are feasible, and the total profit is at least half the profit of the original assignment. To handle the evicted items, we note that as $s(I(U))\leq m/2$ and all the evicted items are small, it is possible to assign the evicted items to bins without violating the capacity constraints.

Proof of Theorem 2.4: For any bin $1\leq j\leq m$ and $A\subseteq I$, let $p_{j}(A)=\sum_{i\in A}p_{i,j}$ be the total profit gained from packing $A$ into $j$. The first step is to resolve the violation of the capacity constraint in each full bin. We do that using four types of resolution steps. Each step takes a full bin and possibly one or two semi-vacant bins, modifies their contents and adds some small elements to a set $V$ of evicted elements, that will be handled later. Throughout the discussion, we consider for a full bin $j$ a partition of the elements in $U_{j}$ into two feasible subsets, given by $\{A_{j},B_{j}\}$. We use the following resolution steps.

Consider a full bin $j$ such that $U_{j}$ has no big elements. If $p_{j}(A_{j})>p_{j}(B_{j})$ then set $U^{\prime}_{j}=A_{j}$ and evict $B_{j}$ ($V:=V\cup B_{j}$); otherwise, set $U^{\prime}_{j}=B_{j}$ and evict $A_{j}$, i.e., $V:=V\cup A_{j}$. In both cases $U^{\prime}_{j}$ is feasible and $p_{j}(U^{\prime}_{j})\geq\frac{1}{2}\cdot p_{j}(U_{j})$. 2. 2.

Now, suppose we have a full bin $j$ such that $U_{j}$ has a single big element, and a semi-vacant bin $\ell$. Let $\{A^{*},B^{*}\}=\{A_{j},B_{j}\}$, such that the big element is in $A^{*}$.

If $p_{j}(A^{*})+p_{\ell}(U_{\ell})>p_{j}(B^{*})$, set $U^{\prime}_{j}=A^{*}$, $U^{\prime}_{\ell}=U_{\ell}$ and evict the elements in $B^{*}$ ($V:=V\cup B^{*}$). We note that in this case $p_{j}(U^{\prime}_{j})+p_{\ell}(U^{\prime}_{\ell})=p_{j}(A^{*})+p_{\ell}(U_{\ell})$.

Otherwise, set $U^{\prime}_{j}=B^{*}$ and $U^{\prime}_{\ell}=A^{*}$, and evict all the elements in $U_{\ell}$, i.e., $V:=V\cup U_{\ell}$. In this case we have $p_{j}(U^{\prime}_{j})+p_{\ell}(U^{\prime}_{\ell})\geq p_{j}(B^{*})$.

Therefore, in both cases have $p_{j}(U^{\prime}_{j})+p_{\ell}(U^{\prime}_{\ell})\geq\frac{1}{2}\cdot\left(p_{j}(U_{j})+p_{\ell}(U_{\ell})\right).$ 3. 3.

Consider a full bin $j$ such that $U_{j}$ has two big elements, and a semi-vacant bin $\ell$, such that one of the big elements has space in bin $\ell$; that is, there is a big element $i^{*}\in U_{j}$ such that $s_{i^{*}}+s^{U}_{\ell}\leq 1$.

Let $\{A^{*},B^{*}\}=\{A_{j},B_{j}\}$ such that $i^{*}\in A^{*}$. We note that there cannot be any other big element in $A^{*}$ other that $i^{*}$.

If $p_{j}(B^{*})+p_{\ell}(U_{\ell})>p_{j}(A^{*})$ set $U^{\prime}_{j}=B^{*}$ and $U^{\prime}_{\ell}=U_{\ell}\cup\{i^{*}\}$ (note that $s^{U^{\prime}}_{\ell}\leq 1$). Also, evict all elements in $A^{*}\setminus\{i^{*}\}$. In this case we have $p_{j}(U^{\prime}_{j})+p_{\ell}(U^{\prime}_{\ell})\geq p_{j}(B^{*})+p_{\ell}(U_{\ell})$.

Otherwise, we set $U^{\prime}_{j}=A^{*}$ and $U^{\prime}_{\ell}=B^{*}$, and evict $U_{\ell}$ ($V:=V\cup U_{\ell}$). In this case we have $p_{j}(U^{\prime}_{j})+p_{\ell}(U^{\prime}_{\ell})\geq p_{j}(A^{*})$.

Thus, in both cases $p_{j}(U^{\prime}_{j})+p_{\ell}(U^{\prime}_{\ell})\geq\frac{1}{2}\cdot\left(p_{j}(U_{j})+p_{\ell}(U_{\ell})\right).$ 4. 4.

Finally, consider a full bin $j$, such that $U_{j}$ has two big elements, and two semi-vacant bins $\ell_{1}$ and $\ell_{2}$. Recall $A_{j},B_{j}$ is a partition of the elements in $U_{j}$ into two feasible subsets.

If $p_{j}(A_{j})+p_{\ell_{1}}(U_{\ell_{1}})>p_{j}(B_{j})+p_{\ell_{2}}(U_{\ell_{2}})$, set $U^{\prime}_{j}=A_{j}$ , $U^{\prime}_{\ell_{1}}=U_{\ell_{1}}$ and $U^{\prime}_{\ell_{2}}=B_{j}$, and evict $U_{\ell_{2}}$. Thus, we have

[TABLE]

Otherwise, set $U^{\prime}_{j}=B_{j}$, $U^{\prime}_{\ell_{1}}=A_{j}$ and $U^{\prime}_{\ell_{2}}=U_{\ell_{2}}$ and evict $U_{\ell_{1}}$. Here

[TABLE]

And finally, we always have

[TABLE]

Each time we execute a step we mark the bins used in this step as resolved, and we do not consider them in the next steps. We first use Steps 1,2, and 3, until none of them can be applied.

Consider the average size of the unresolved bins. When we start, there are $m$ bins of average size no greater than half. Each of Steps 1,2, 3 reduces the size of the unresolved bins by at least one (as a full bin is removed) and reduces the number of bins by at most two. Therefore, the average size of the unresolved bins remains no more than half. Also, marking all the semi-full bins as resolved preserves the property.

Let $a$ be the number of unresolved full bins and $c$ be the number of unresolved semi-vacant bins. Due to the average size of bins, we have $a\leq\frac{a+c}{2}$, therefore $a\leq c$. Hence, if we have a full bin, there must be a semi-vacant bin as well. As we used Steps 1,2, 3 to exhaustion, every full bin must have two big elements (no bin in $U$ contains more than two big elements), and none of these big elements can fit into one of the semi-vacant bins.

Denote the minimal size of a semi-vacant unresolved bin by $r$. Then each of the full bins has two big elements of size greater than $1-r$. Hence, we have $c\cdot r+2a\cdot(1-r)<\frac{a+c}{2}$, which leads to $2a(1-r)-\frac{a}{2}<c\left(\frac{1}{2}-r\right)$, and

[TABLE]

implying that we can now run Step 4, until there are no more unresolved full bins.

We use the resolution steps to eliminate all the full bins. Every time we run such a step over a set of bins we lose at most half the profit of the bins participating in the step. As the total size of items in the assignment is bounded by $m/2$, we are guaranteed that it is possible to assign all the evicted elements to some bins. Thus, we are able to resolve the capacity overflow while losing at most half of the profit. ∎

6 Discussion and Future Work

In this paper we presented a $\frac{1}{6}$-approximation algorithm for Group GAP, using a mild restriction on group sizes. A key component in our result is an algorithm for submodular maximization subject to a knapsack constraint, which finds a solution occupying at most half the knapsack capacity, while the other half is reserved for later use. Our results leave several avenues for future work.

As mentioned above, Group GAP with no assumption on group sizes cannot be approximated within any constant factor. Yet, the maximum group size that still allows to obtain a constant ratio can be anywhere in $[\frac{m}{2},\frac{2}{3}m]$. A natural question is whether our results can be applied to instances with larger group sizes. We note that the ratio stated in Theorem 2.5 may not hold already for instances in which group sizes can be at most $\frac{m}{2}(1+\varepsilon)$, for some $\varepsilon>0$. Indeed, for such instances, it may be the case that no set of groups of total size at most $m/2$ is ‘good’ relative to the optimum. The existence of an algorithm that yields a constant ratio for such instances remains open.

While our result for submodular optimization with reserved capacity (Theorem 2.2) gives an optimal approximation ratio for the studied subclass of instances, we believe the result can be extended to other subclasses. In particular, we conjecture that for instances where each item is of size at most $\delta>0$, the approximation ratio approaches $1-e^{-\frac{1}{2}}$ as $\delta\rightarrow 0$. Such result would yield improved the approximation ratio for instances of Group GAP in which the total size of each group is bounded by $\delta m$. We defer this line of work to the full version of the paper.

Lastly, we introduced in the paper the novel approach of submodular optimization subject to a knapsack with reserved capacity constraint. We applied the approach along with a framework similar to the one developed in [1]. It is natural to ask whether the approach can be used to improve the approximation ratio obtained in [1] for all-or-nothing GAP.

Appendix A Appendix

A.1 Applications of Group GAP

As a primary motivation for our study, consider wireless networks (see, e.g., [18, 28]) a carrier supplies a set of data items to a set of users upon demand. Each user has an isolated cache memory of a given size; each item consists of a set of blocks. Time is slotted. The carrier smooths out the network load by caching at time $t=0$ data block within the user caches, and later lets the users share these blocks at time $t\geq 1$ via a D2D (Device-to-Device) network. There is a utility associated with storing a data block of an item in each user cache (depending, e.g., on the location of this user in the network). Thus, viewing each item as a group of blocks, finding a feasible allocation of data blocks in the users caches so as to maximize the total utility yields an instance of Group GAP.

Group GAP has a central application also in 5G networks. Caching at small-cell base stations (SBSs) to meet the overwhelming growth in mobile data demand is expected to be utilized in the next generation (5G) cellular networks [23, 4]. This technique leverages the expected deployment of short-range, low-power, and low-cost SBSs in cellular networks. It is expected that this will enable immersive Virtual Reality (VR) and Augmented Reality (AR) mobile applications that are currently limited to off-line settings [9]. A VR/AR $360^{\circ}$-navigable 3D video presents the user with an interactive immersive visual representation of the remote scene of interest that the user can navigate from any possible 3D viewpoint. To enable such capability in a mobile device with limited computation power and storage, the 3D video is decomposed into different views and all these views must be cached in the SBSs. There is a utility associated with storing a specific view in each SBS (depending, e.g., on the projected popularity of this view, its size relative to the cache size, etc.). Thus, viewing each 3D video as a group of views, finding a feasible allocation of these views to SBSs to maximize the total utility yields an instance of Group GAP.

A.2 Some Proofs

Proof of Theorem 2.3: Let $S=I(G^{*})$. We first note that since $s(S)=s(I(G^{*}))\leq m$, any fractional solution $x$ can be modified to a solution $x^{\prime}$ of at least the same profit, such that for all $i\in S$ it holds that $\sum_{j\in M}x^{\prime}_{i,j}=1$. Thus, we may assume that for any $i\in S$ we have $\sum_{j\in M}x_{i,j}=1$. W.l.o.g., assume the items are ordered by their size; that is, $s_{1}\geq s_{2}\geq\ldots\geq s_{n}$. We use Algorithm 3 to generate a bipartite graph $G=(S,R,E)$ with weight $w_{e}\geq 0$ on any edge $e\in E$, and a fractional matching $x^{\prime}$ in $G$.

We note the following observations about Algorithm 3.

• •

The algorithm returns a fractional matching $x^{\prime}$ which is complete for $S$. For each item $i\in S$ and bin $j\in M$ it holds that $x_{i,j}=\sum_{e=(i,(j,r))\in E}x^{\prime}_{e}$. Therefore, the weight of $x^{\prime}$ is equal to the profit of the solution $x$ for the LP, i.e., $w(x^{\prime})=p\cdot x$.

• •

For any bin $j\in M$ let $k_{j}$ be the maximum number such that $(j,k_{j})\in R$. For any $1\leq r\leq k_{j}$ let $s_{j,r}^{\max}=\max\{s_{i}|(i,(j,r))\in E\}$ and $s_{j,r}^{\min}=\min\{s_{i}|(i,(j,r))\in E\}$. By the construction, it is easy to see that $s_{j,1}^{\max}\geq s_{j,1}^{\min}\geq s_{j,2}^{\max}\geq s_{j,2}^{\min}\geq\ldots s_{j,k_{j}}^{\max}\geq s_{j,k_{k}}^{min}$

• •

For any bin $j\in M$ and $1\leq r<k_{j}$ we have $\sum_{i|(i,(j,r))\in E}x^{\prime}_{(i,(j,r))}=1$.

Using standard results from matching theory (see, e.g., [21]), there is a matching ${\cal M}$ for $G$ which is complete for $S$ and $\sum_{e\in M}w_{e}\geq w(x^{\prime})$. Such a matching ${\cal M}$ can also be found in polynomial time.

We define an assignment based on ${\cal M}$; that is, for every bin $j\in M$ and $1\leq r\leq k_{j}$ set $U_{j,r}=\{i\in S|(i,(j,r))\in{\cal M}\}$. Now, define for every $j\in M$: $U_{j}=\bigcup_{r=1}^{k_{j}}U_{j,r}$ and $U=(U_{1},U_{2},\ldots,U_{m})$. As ${\cal M}$ is complete for $S$, each element $i\in S$ is assigned to a single bin, and each of the sets $U_{j,r}$ contains at most a single element. Now, for every $j\in M$ we have

[TABLE]

The first inequality follows from the definition of $s_{j,r}^{\max}$ and $|U_{j,r}|\leq 1$. The second inequality holds since $s_{j,r}^{\max}\leq s_{j,r-1}^{\min}$. The second equality holds since $\sum_{i|(i,(j,r))\in E}x^{\prime}_{(i,(j,r))}=1$ for $1\leq r<k_{j}$. In the third inequality we use the definition of $s_{j,r}^{\min}$, and the fourth inequality follows from $x_{i,j}=\sum_{e=(i,(j,r))\in E}x^{\prime}_{e}$. The last inequality holds since $x$ is a (feasible) solution for $\mbox{LP}(S)$.

By inequality (5) and since $U_{j,1}$ contains at most a single element, we have that $U$ is an almost feasible assignment. We note that the profit of the assignment $U$ is exactly the weight of the matching ${\cal M}$, which is at least the profit of $x$.

This suggests the following conversion procedure. Given a solution $x$ for $\mbox{LP}(S)$, use Algorithm 3 to obtain a bipartite graph $G=(S,R,E)$ with weights $w$. Find a maximal weight matching ${\cal M}$ in $G$ which is complete for $S$. Define an assignment $U$ using ${\cal M}$ and return this assignment. By the above discussion, this procedure is polynomial and returns an assignment as required by the theorem. ∎

Proof of Theorem 3.1: Consider two subsets $S\subset T\subset A$. Let $v\in A\setminus T$. To prove submodularity we show that $h(S\cup\{v\})-h(S)\geq h(T\cup\{v\})-h(T)$.

Let $S^{\prime}=S\cup\{v\}$ and $T^{\prime}=T\cup\{v\}$. For a subset $X\subseteq A$ let $M_{X}$ be a matching whose weight achieves $h(X)$. (Note that there may be more than one such matching, we fix one arbitrarily.) For $u\in X$, define $M_{X}(u)$ to be the vertex in $B$ that is matched to $u$. Also, for $Y\subseteq X$ let $M_{X}(Y)$ be the set of matching edges that touch the vertices in $Y$. That is, $M_{X}(Y)=\cup_{y\in Y}\{(y,M_{X}(y))\}$.

To prove the theorem we need the following lemma.

Lemma A.1**.**

There exists a subset $X=\{x_{1}=v,\ldots,x_{k}\}\subseteq S^{\prime}$, for some $k\geq 1$, with the following two properties.

For $i\in[1..k-1]$, $M_{S}(x_{i+1})=M_{T^{\prime}}(x_{i})$ 2. 2.

$M_{T^{\prime}}(x_{k})$* is not matched in $M_{S}$*

Before we prove the lemma, we show how it implies the theorem. Let $E_{1}=M_{S}(X\setminus\{x_{1}\})$, and $E_{2}=M_{T^{\prime}}(X)$.

We claim that $h(S^{\prime})-h(S)\geq W(E_{2})-W(E_{1})$. Consider the set of edges $E_{2}\cup M_{S}(S\setminus X)$. Notice that it is a matching in the subgraph induced by $S^{\prime}\cup B$. Clearly, $h(S^{\prime})\geq W(E_{2}\cup M_{S}(S\setminus X))=W(E_{2})+W(M_{S}(S\setminus X))$. Also, by definition, $h(S)=W(E_{1}\cup M_{S}(S\setminus X))=W(E_{1})+W(M_{S}(S\setminus X))$. Thus, $h(S^{\prime})-h(S)\geq W(E_{2})-W(E_{1})$.

We claim that $h(T^{\prime})-h(T)\leq W(E_{2})-W(E_{1})$. By definition, $h(T^{\prime})=W(E_{2}\cup M_{T^{\prime}}(T\setminus X))=W(E_{2})+W(M_{T^{\prime}}(T\setminus X))$. Consider the set of edges $E_{1}\cup M_{T^{\prime}}(T\setminus X)$. Notice that it is a matching in the subgraph induced by $T\cup B$. Clearly, $h(T)\geq W(E_{1}\cup M_{T^{\prime}}(T\setminus X))=W(E_{1})+W(M_{T^{\prime}}(T\setminus X))$. Thus, $h(T^{\prime})-h(T)\leq W(E_{2})-W(E_{1})$. The proof follows. ∎

Proof of Lemma A.1: We build $X$ iteratively. We start with $X=\{v\}$. If $M_{T^{\prime}}(v)$ is not matched in $M_{S}$, we are done since $X$ satisfies both properties. (The first property is satisfied vacuously.) Otherwise, define $x_{2}$ to be the vertex that satisfies $M_{S}(x_{2})=M_{T^{\prime}}(x_{1}=v)$. Note that $x_{2}\neq x_{1}$. Also, $X=\{x_{1},x_{2}\}$ satisfies the first property by definition. If $M_{T^{\prime}}(x_{2})$ is not matched in $M_{S}$, we are done since $X$ satisfies both properties. Otherwise, define $x_{3}$ to be the vertex that satisfies $M_{S}(x_{3})=M_{T^{\prime}}(x_{2})$. Note that $x_{3}\notin\{x_{1},x_{2}\}$. We continue in the same manner if we are not done. Namely, at stage $i$ we have the set $\{x_{1},\ldots,x_{i}\}$ that satisfies the first property. If $M_{T^{\prime}}(x_{i})$ is not matched in $M_{S}$, we are done. Otherwise, define $x_{i+1}$ to be the vertex that satisfies $M_{S}(x_{i+1})=M_{T^{\prime}}(x_{i})$. Again, $x_{i+1}\notin\{x_{1},\ldots,x_{i}\}$. Since in each stage we add a new vertex from $S$, and $S$ is finite, we are guaranteed to be done at some stage.

∎

A.3 Missing Details in the Proof of Theorem 2.2

Lemma A.2**.**

Let $S^{*}\subseteq{\Omega}$ be a non-empty subset of items, such that $s(S^{*})\leq m^{*}$. Also, let $g:2^{\Omega}\rightarrow\rm I\!R$ be a submodular, non-negative and monotone function such that $g(\emptyset)=0$, and $S=\textsc{Greedy}(g,m^{\prime})$. Then, for any subset $T\subseteq{\Omega}$, there is an item $i^{*}\in S^{*}$ such that $g(S)+g_{T}(\{i^{*}\})\geq(1-e^{m^{\prime}/m^{*}})g_{T}(S^{*})$.

Proof A.3**.**

If $S^{*}\subseteq S$ then the claim trivially holds. Thus, we may assume that $S^{*}\nsubseteq S$. Let $S=\{i_{1},i_{2},\ldots,i_{\ell}\}$, where the elements are in the order they were added to $S$ during the execution of Greedy. Let $S_{r}=\{i_{1},i_{2},\ldots,i_{r}\}$ denote the first $r$ items in $S$, $0\leq r\leq\ell$, where $S_{0}=\emptyset$. We denote by $\theta_{r}=\frac{g_{S_{r-1}}(\{i_{r}\})}{s_{i_{r}}}$ the marginal gain density from adding $\{i_{r}\}$ to $S_{r-1}$. Given a subset $T\subseteq{\Omega}$, we define $\eta_{r}=\max_{i\in S^{*}\setminus S_{r-1}}\frac{g_{T}(S_{r-1}\cup\{i\})-g_{T}(S_{r-1})}{s_{i}}$ for $1\leq r\leq\ell+1$ (since $S^{*}\nsubseteq S$ the maximization is never over an empty set). By the definition of $\theta_{r}$ we have $g(S_{r})=\sum_{j=1}^{r}\theta_{j}s_{i_{j}}$ for $1\leq r\leq\ell$. If there is $1\leq r\leq\ell$ for which $\eta_{r}>\theta_{r}$, set $t=\min\{r|1\leq r\leq\ell,\eta_{r}>\theta_{r}\}-1$, otherwise set $t=\ell$.

Let $i^{*}\in S^{*}\setminus S_{t}$ such that $\eta_{t+1}=\frac{g_{T}(S_{t}\cup\{i^{*}\})-g_{T}(S_{t})}{s_{i^{*}}}$ (one exists by the definition of $\eta_{r}$). If $t<\ell$, by the submodularity of $g$ we have that $\frac{g_{S_{t}}(\{i^{*}\})}{s_{i}}\geq\eta_{t+1}>\theta_{t+1}$; therefore in the iteration of Greedy in which $i_{t+1}$ was added to $S$, it holds that $i^{*}\notin E$ (otherwise it would have been selected instead of $i_{t+1}$), and we can conclude that $s(S_{t})+s_{i^{*}}>m^{\prime}$. If $t=\ell$, we know that by the end of the execution of Greedy $i^{*}\notin E$, therefore $s(S_{t})+s_{i^{*}}>m^{\prime}$ as well.

Define $m_{r}=s(S_{r})$ for $0\leq r\leq t$ and $m_{t+1}=s(S_{t})+s_{i^{*}}$. Also, define $p_{j}=\theta_{r}$ for $1\leq r\leq t$ and $j=m_{r-1}+1,m_{r-1}+2,\ldots,m_{r}$, and $p_{j}=\eta_{t+1}$ for $j=m_{t}+1,m_{t}+2,\ldots,m_{t+1}$. Clearly, by the submodularity of $g$, it holds that $g_{T}(S_{r})\leq g(S_{r})=\sum_{j=1}^{m_{r}}p_{j}$, for $0\leq r\leq t$. Also, it is easy to see that for any $0\leq r\leq t$ we have $\eta_{r+1}\leq p_{m_{r}+1}$.

The following sequence of inequalities follows from the above observation and the submodularity of $g$. For any $0\leq r\leq t$:

[TABLE]

Therefore,

[TABLE]

We now use inequality (4) in [26]:

[TABLE]

and the lemma follows immediately.