Bin Packing Problem

Outline

• Explanation of original bin packing problem
• On-line algorithms for original bin packing problem
• Definition of extensible bin packing problem
• Brief information about some real world applications related to extensible bin packing problem
• A FPTAAS for extensible bin backing problem
• Conclusion

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Original Bin Packing Problem

Here size(S) = ∑_{a∈ S} size(a) and profit(S) = ∑_{a∈ S} profit(a).

In original bin packing problem, we are given a list of n items L={a₁,a₂,…,a_n} and infinite number of bins with capacity C.

The size of i-th item a_i is s(a_i), where 0 < s(a_i) ≤ C.

The problem is to pack the items in minimum number of bins under the constraint that the total size of items in each bin cannot exceed the capacity of the bin.

Online Algorithms for Original Bin Packing Problem

On-line bin packing algorithms are the algorithms which packs each item one by one without any knowledge of the next items. Thus, approximation factor of these algorithms are higher than offline algorithms in which size and number of items in the list are known before starting to pack them.

Definitions

The following definitions will be used in this section when we talk about on-line algorithms for original bin packing problem.

• There are two types of nonempty bins. Open bins are available for packing additional items whereas closed bins are unavailable for packing. Closed bins cannot be re-opened.
• B_j denotes the items packed in j-th bin.
• |B_j| denotes the number of items in j-th bin.
• The size of a bin is computed as following: c(B_j) = ∑_{ai ∈ Bj} s(a_i)

Next-Fit (NF)

1. Pack the first item into the first bin
2. For each successive item
   1. If it fits in the bin that contains the last packed item, pack in that bin.
   2. Otherwise, close this bin and pack the item in an empty bin.

==Extensible Bin Packing Problem == Extensible bin packing problem is a variation of original bin packing problem. In this problem, number of bins is given as an input. The capacity of each bin is 1 and bins may be extended to hold more than their capacities. The cost of a bin is 1 if it is not extended, and the size of items in it if it is extended. The goal of this problem is to pack a set of items of given sizes into the specified number of bins in a way to minimize the total cost. In this tutorial, a fully asymptotic approximation scheme will be introduced for extensible bin packing problem.

Extensible bin packing has a number of important real world applications such as:

A FTPAAS for Extensible Bin Packing Problem

Introduction

Input

• x_i for i=1,2,3,...,n: size of i-th item.
  
• m: number of bins

3.2 Notations

• b_j for j=1,2,3,...,m: j-th bin
  
• l(b_j): level of b_j (total size of items packed into b_j)
  
• c(b_j): cost of b_j = max(1,l(b_j))

3.3 Goal

minimize ∑_j=1^m c(b_j)

Algorithm

We present an algorithm A(I,ε), which takes a problem instance and a parameter ε and produces an approximation guaranteeing that

A(I,ε) ≤ (1 + ε)OPT(I) + O(ε^-1\log{ε^-1}) in time bounded by a polynomial in n and ε^-1 We can make some assumptions without loss of generality.

First Assumption

There are no small items. In other words
x_i > ε/(1+ε) for i=1,2,…,n

Second Assumption

There are no big items. In other words
x_i < 1 for \(i=1,2,…,n

Third Assumption

Total size of all items cannot exceed 2m
∑_x=1ⁿ x_i < 2m

Fourth Assumption

The level of a bin cannot exceed 3.
l(b_j) < 3 for j=1,2,…,m --- For convenience, we will show that
A(I,ε) ≤ (1 + ε)²OPT(I) + O(ε^-1log{ε^-1})
We can obtain original approximation guarantee by a different choice of ε

We will round the sizes of items up to s₁, s₂, ..., s_N where N is the smallest value for the following inequality
⌊(1+ε)^j/ε⌋ε² ≥ 1
According to the inequality above, N approximately equals to ε^-1lnε{-1}
We define the value of s₁,s₂,…,s_N according to the following equation
s_j = ⌊(1+ε)^j/ε⌋ε² for \1lej<N
s_j = 1 for j=N
Thus,

1. s₁ < s₂ < … < s_N
2. s_N = 1. (Recall that we assume that the maximum size of an item is 1.)
3. If s_j-1 <x_i ≤ s_j then round x_i to s_j.

Increase ratio of items of size less than s₁ s₁/(ε/(1+ε)) ≤ (1+ε)²
Increase ratio of items of size larger than s₁ s_j+1/s_j ≤ (1+ε)²
Rounding increases the cost of the optimum solution by a faxtor of at most (1+ε)²)</b> <br> \(n_j: number of items of size s_j after rounding.
<b>configuration: packing a bin to a level of at most 3 with sizes chosen from s_j M: number of configurations C_ij: number of items of size s_j in the i-th configuration max(1,∑_j=1^N C_ijs_j: cost of packing a bin according to i-th configuration

Ci	zi	s1	s2	s3	sN
1	z1	C11	C12	C13	C1N
2	z2	C11	C22	C23	C2N
3	z3	C11	C32	C33	C3N
4	z4	C11	C42	C43	C4N
5	z5	C11	C52	C53	C5N
M	zM	C11	CM2	CM3	CMN

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Integer Program

z_i: number of times we use i-th configuration
Objective function: min ∑_i=1^M z_i.max(1,∑_j=1^N C_ijs_j)
subject to
∑_i=1^M z_iC_ij ≥ n_j for j=1,2,…,N
∑_i=1^M z_i ≥ m
z_i ∈ {0,1,2,ldots}

When we relax this ILP to LP by allowing z_i to be any nonnegative value.
number of inequalities in LP = N+1

• There will be a feasible solution with at most N+1 nonnegative non-integer values.
• Round LP to ILP increases the solution 3(N+1) because each coefficient in the objective function is at most 3. (Recall that l(b_j) < 3 for j=1,2,…,m

3(N+1) = O(e^-1logε^-1)