Approximation Algorithms for Generalized Multidimensional Knapsack

Our two main contributions are $(i)$ a $(2+\varepsilon)$ approximation algorithm for the $(2,d)$ KS problem $(ii)$ define a new problem called Vector-Max-GAP and obtain a PTAS for it.

To obtain the approximation algorithm for $(2,d)$ KS, we first obtain a structural result using corridor decomposition, which shows that if we consider an optimal packing and remove either vertical items (height$\gg$width) or horizontal items (width$\gg$height), we can restructure the remaining packing to possess a ‘nice’ structure that can be efficiently searched for.

In [9], which deals with the 2-D geometric knapsack problem, the authors use the PTAS for the maximum generalized assignment problem (Max-GAP) to efficiently search for such a nice structure. But due to the presence of vector constraints, we needed a generalization of the Max-GAP problem which we call the Vector-Max-GAP problem and obtain a PTAS for it to obtain the desired approximation algorithm for $(2,d)$ KS.

A well-studied problem closely related to the knapsack problem is bin packing. A natural generalization of the bin packing problem is the 2-D geometric bin packing (2GBP) problem where items are rectangles and bins are unit squares. The current best asymptotic approximation ratio of $(1+\ln 1.5+\varepsilon)$ for 2GBP is due to [4]. Another generalization of bin packing is the vector bin packing problem (VBP) where each item has multiple constraints. For the current best approximation ratios of VBP, we refer the reader to [3].

There exist many other variants of the discussed packing problems such as strip packing (see [8, 10]), maximum independent set of rectangles (see [18]), storage allocation problems [22], etc. For an extensive survey of related packing problems, see [5, 16].

In Section 2, we study the Vector-Max-GAP problem and design a PTAS for it. Section 3 describes the $(2+\varepsilon)$ approximation algorithm for the $(2,d)$ Knapsack problem. In Sections 3.1 and 3.2, we use a structural result involving corridor decomposition and reduce our problem to a container packing problem. In Sections 3.3 and 3.4 we will discuss how to model the container packing problem as an instance of the Vector-Max-GAP problem. Finally, in Section 3.5, we put everything together to obtain the $(2+\varepsilon)$ approximation algorithm for the $(2,d)$ Knapsack problem.

Consider the case where item sizes and weights are integers. W.l.o.g., we can assume that the capacities of machines, $M_{j}$ $(j\in[k])$ and weight constraints, $W_{q}$ $(q\in[d])$, are also integers (otherwise, we can round them down to the closest integers).

Arbitrarily order the items and number them from $1$ onwards. Let $\mathrm{VAL}(n,\overrightarrow{M},\overrightarrow{W})$ be the maximum value obtainable by assigning a subset of the first $n$ items to $k$ machines with capacities given by $\overrightarrow{M}$ respecting the global weight constraint $\overrightarrow{W}$. We can express $\mathrm{VAL}(n,\overrightarrow{M},\overrightarrow{W})$ as a recurrence.

$\mathrm{VAL}(n,\overrightarrow{M},\overrightarrow{W})$ can be computed using dynamic programming. We can find the subset of items that gives this much value and it is also easy to ensure that no item assigned to a machine has value 0 in that machine. There are $n\prod_{j=1}^{k}(M_{j}+1)\prod_{q=1}^{d}(W_{q}+1)$ items in the state space and each iteration takes $\Theta(d+k)$ time. Therefore, time taken by the dynamic programming solution is $\Theta\left(n(d+k)\prod_{j=1}^{k}(M_{j}+1)\prod_{q=1}^{d}(W_{q}+1)\right)$.

Let $\mu_{j}=\varepsilon M_{j}/n$ and $\delta_{q}=\varepsilon W_{q}/n$ for all $q\in[d]$ and $j\in[k]$. Let $J^{*}$ be the optimal solution to $(I,\overrightarrow{\mathrm{val}},\overrightarrow{s},\overrightarrow{w},\overrightarrow{M},\overrightarrow{W})$. Let $\widehat{J}$ be the optimal solution to $(I,\overrightarrow{\mathrm{val}},\overrightarrow{s^{\prime}},\overrightarrow{w^{\prime}},\overrightarrow{M^{\prime}},\overrightarrow{W^{\prime}})$. By Lemma 1, $\mathrm{val}(\widehat{J})\geq\mathrm{val}(J^{*})$. By Lemma 2, $\widehat{J}$ is feasible for $(\overrightarrow{s},\overrightarrow{w},(1+\varepsilon)\overrightarrow{M},(1+\varepsilon)\overrightarrow{W})$. Also, observe that $|M_{j}^{\prime}|\leq n+M_{j}/\mu_{j}=n(1+1/\varepsilon)$ is a polynomial in $n$. Similarly, $|W_{q}^{\prime}|\leq n(1+1/\varepsilon)$ and hence the optimal solution to $(I,\overrightarrow{\mathrm{val}},\overrightarrow{s^{\prime}},\overrightarrow{w^{\prime}},\overrightarrow{M^{\prime}},\overrightarrow{W^{\prime}})$ can be obtained using the dynamic-programming algorithm in polynomial time. Therefore, the optimal solution to $(I,\overrightarrow{\mathrm{val}},\overrightarrow{s^{\prime}},\overrightarrow{w^{\prime}},\overrightarrow{M^{\prime}},\overrightarrow{W^{\prime}})$ can be obtained using the dynamic-programming algorithm in time ${\displaystyle\Theta\left((d+k)n^{d+k+1}/\varepsilon^{d+k}\right)}$.

Let us define a subroutine $\operatorname{assign-res-aug}_{\varepsilon}(I,\overrightarrow{\mathrm{val}},\overrightarrow{s},\overrightarrow{w},\overrightarrow{M},\overrightarrow{W})$ which takes as input set $I$ with associated values, $\overrightarrow{\mathrm{val}}$, and gives as output the optimal feasible solution to $(\overrightarrow{s},\overrightarrow{w},(1+\varepsilon)\overrightarrow{M},(1+\varepsilon)\overrightarrow{W})$.

Consider a set $I$ of items where each item has length $s(i)$ and profit $p(i)$ (in this subsection, $I$ is an instance of the knapsack problem instead of the Vector-Max-GAP problem).

Suppose for all $i\in I,s(i)\in(0,\varepsilon]$ and $s(I)\leq 1+\delta$. We’ll show that there exists an $R\subseteq I$ such that $s(I-R)\leq 1$ and $p(R)<(\delta+\varepsilon)p(I)$. We call this technique of removing a low-profit subset from $I$ so that it fits in a bin of length 1 trimming.

Arbitrarily order the items and arrange them linearly in a bin of size $1+\delta$. Let $k=\left\lfloor 1/(\delta+\varepsilon)\right\rfloor$. Create $k+1$ intervals of length $\delta$ and $k$ intervals of length $\varepsilon$. Place $\delta$-intervals and $\varepsilon$-intervals alternately. They will fit in the bin because $(k+1)\delta+k\varepsilon=\delta+(\delta+\varepsilon)\left\lfloor 1/(\delta+\varepsilon)\right\rfloor\leq 1+\delta$

Number the $\delta$-intervals from $0$ to $k$ and let $S_{i}$ be the set of items intersecting the $i^{\textrm{th}}$ $\delta$-interval. Note that, all $S_{i}$ are mutually disjoint. Let $i^{*}=\operatorname*{argmin}_{i=0}^{k}p(S_{i})$.

Removing $S_{i^{*}}$ will create an empty interval of length $\delta$ in the bin, and the remaining items can be shifted so that they fit in a bin of length 1.

Consider a Vector-Max-GAP instance $(I,\overrightarrow{\mathrm{val}},\overrightarrow{s},\overrightarrow{w},\overrightarrow{M},\overrightarrow{W})$. Item $i$ is said to be $\varepsilon$-small for this instance iff $\overrightarrow{w}(i)\leq\varepsilon\overrightarrow{W}$ and for all $j\in[k],(s_{j}(i)\leq\varepsilon M_{j}\textit{ or }\mathrm{val}_{j}(i)=0)$. A set $I$ of items is said to be $\varepsilon$-small iff each item in $I$ is $\varepsilon$-small.

Suppose $I$ is $\varepsilon$-small. Let $J\subseteq I$ be a feasible solution to $(I,\overrightarrow{\mathrm{val}},\overrightarrow{s},\overrightarrow{w},(1+\varepsilon)\overrightarrow{M},(1+\varepsilon)\overrightarrow{W})$. Let $J_{j}$ be the items assigned to the $j^{\textrm{th}}$ machine.

For each $j\in[k]$, use trimming on $J_{j}$ for sizes $s_{j}$ and then for each $j\in[d]$, use trimming on $J$ for weights $w_{j}$. In both cases, use $\varepsilon:=\varepsilon$ and $\delta:=\varepsilon$. Let $R$ be the removed items and $J^{\prime}=J-R$ be the remaining items. Total value lost is less than $2\varepsilon(d+1)\mathrm{val}(J)$ and $J^{\prime}$ is feasible for $(\overrightarrow{s},\overrightarrow{w},\overrightarrow{M},\overrightarrow{W})$.

Therefore, any resource-augmented solution $J$ of small items can be transformed to get a feasible solution $J^{\prime}$ of value at most $(1-2(d+1)\varepsilon)\mathrm{val}(J)$.

Let $J^{*}$ be an optimal assignment for $(I,\overrightarrow{\mathrm{val}},\overrightarrow{s},\overrightarrow{w},\overrightarrow{M},\overrightarrow{W})$. Let $J_{j}^{*}\subseteq J^{*}$ be the items assigned to the $j^{\textrm{th}}$ machine.

By Theorem 3, $J^{*}$ can be partitioned into sets $X^{*}$, $Y^{*}$ and $Z^{*}$ such that $|X^{*}|\leq\frac{d+k}{\varepsilon^{2}}$ and $\mathrm{val}(Y^{*})\leq\varepsilon\cdot\mathrm{val}(J^{*})$. Let $\overrightarrow{W}^{*}=\overrightarrow{W}-\overrightarrow{w}(X^{*})$ and $M_{j}^{*}=M_{j}-s_{j}(X^{*}\cap J_{j}^{*})$. Then $Z^{*}$ is $\varepsilon$-small for $(\overrightarrow{\mathrm{val}},\overrightarrow{s},\overrightarrow{w},\overrightarrow{M}^{*},\overrightarrow{W}^{*})$.

For a set $S$, define $\Pi_{k}(S)$ as the set of $k$-partitions of $S$. Then the pseudo code in Algorithm 1 provides a PTAS for the Vector-Max-GAP problem.

Correctness: Since $Z$ is feasible for $(\overrightarrow{s},\overrightarrow{w},\overrightarrow{M}^{\prime},\overrightarrow{W}^{\prime})$, $X\cup Z$ is feasible for $(\overrightarrow{s},\overrightarrow{w},\overrightarrow{M},\overrightarrow{W})$.

Approximation guarantee:
For some iteration of Algorithm 1, $X=X^{*}$ and $X_{j}=X^{*}\cap J_{j}^{*}$. When that happens, $\overrightarrow{W}^{\prime}=\overrightarrow{W}^{*}$ and $\overrightarrow{M}^{\prime}=\overrightarrow{M}^{*}$. Let $\mathrm{val}^{*}$ be the maximum value $\varepsilon$-small assignment of items to the machines with capacities given by $\overrightarrow{M}^{\prime}$ and over all weight constraints $\overrightarrow{W}^{\prime}$. Therefore, $\mathrm{val}^{*}\geq\mathrm{val}(Z^{*})$.

To try to find an optimal assignment of small items, we’ll forbid non-small items to be assigned to a machine. To do this, if for item $i$, $s_{j}(i)>\varepsilon M_{j}^{\prime}$, set $\mathrm{val}_{j}(i)$ to 0. Using our resource-augmented Vector-Max-GAP algorithm, we get $\mathrm{val}(Z^{\prime})\geq\mathrm{val}^{*}$. By the property of trimming, $\mathrm{val}(Z)\geq(1-2(d+1)\varepsilon)\mathrm{val}(Z^{\prime})$.

This gives us a $(1-(2d+3)\varepsilon)$-approx solution. The running time can be easily seen to be polynomial as $\operatorname{assign-res-aug}$ runs in polynomial time and the number of iterations of the outer loop in Algorithm 1 is polynomial in $n$ and for one iteration of the outer loop, the inner loop runs at most constant number of times.

Consider a set of items $S$ that are packed in a knapsack. We now state a structural result, inspired by [9], where only a subset of items $S^{\prime}\subseteq S$ is packed into the knapsack. We may lose some profit but the packing has a nice structure which can be searched for, efficiently.

Now, let us go back to our original problem instance $I$. Let $I_{\mathrm{OPT}}$ be the set of items packed into the knapsack in an optimal packing $\mathcal{P}$.

Let us apply Theorem 4 to the set of items $I_{\mathrm{OPT}}$ with $\varepsilon:=\varepsilon_{\mathrm{struct}}$ ($\varepsilon_{\mathrm{struct}}$ will be defined later). Let $I_{\mathrm{OPT}}^{\prime}$ be the resulting analog of $S^{\prime}$ in the theorem (there can be many candidates for $I_{\mathrm{OPT}}^{\prime}$ but let us pick one). Therefore,

In this subsection, we will prove Theorem 4.

The strategy to prove the theorem is to use the corridor decomposition scheme, essentially taken from [1]. First, we assume that we can remove $O_{1/\varepsilon}(1)$ number of items at the cost of zero profit loss from the originally packed items. Under this assumption, we show that we lose at most half profit by our restructuring. Finally, we show how to get rid of this assumption by using shifting argumentations.

Removing Medium items: Let $\varepsilon_{\mathrm{small}},\varepsilon_{\mathrm{big}}$ be two fixed constants such that $\varepsilon_{\mathrm{big}}>\varepsilon_{\mathrm{small}}$. We partition the items in $S$ based on the values of $\varepsilon_{\mathrm{small}}$ and $\varepsilon_{\mathrm{big}}$ as follows:

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  $S_{S}=\left\{i\in S\>:\>w(i)\leq\varepsilon_{\mathrm{small}}\textup{ and }h(i)\leq\varepsilon_{\mathrm{small}}\right\}$

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  $S_{B}=\left\{i\in S\>:\>w(i)>\varepsilon_{\mathrm{big}}\textup{ and }h(i)>\varepsilon_{\mathrm{big}}\right\}$

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  $S_{W}=\left\{i\in S\>:\>w(i)>\varepsilon_{\mathrm{big}}\textup{ and }h(i)\leq\varepsilon_{\mathrm{small}}\right\}$

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  $S_{T}=\left\{i\in S\>:\>w(i)\leq\varepsilon_{\mathrm{small}}\textup{ and }h(i)>\varepsilon_{\mathrm{big}}\right\}$

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  $S_{\mathrm{med}}=\left\{i\in S\>:\>w(i)>\varepsilon_{\mathrm{small}}\textup{ and }w(i)\leq\varepsilon_{\mathrm{big}}\quad\textup{OR}\quad h(i)\leq\varepsilon_{\mathrm{big}}\textup{ and }h(i)>\varepsilon_{\mathrm{small}}\right\}$

We call the items in $S_{S}$ as small items as they are small in both the dimensions. Similarly, we call the items in $S_{B},S_{W},S_{T},S_{\mathrm{med}}$ as big, wide, tall, medium respectively.

By standard arguments, it is possible to choose the constants $\varepsilon_{\mathrm{small}}$ and $\varepsilon_{\mathrm{big}}$ such that the total profit of all the medium items is at most $\varepsilon\cdot p(S)$. Hence, we can discard the items in $S_{\mathrm{med}}$ from $S$ while losing a very small profit. We omit the proof as it is present in many articles on packing (see, for example, [9]).