A turnpike approach to solving the linear bottleneck assignment problem

Abstract

The linear bottleneck assignment problem (LBAP), which is a variation of the classical assignment problem (CAP), seeks to minimize the longest completion time rather than the sum of the completion times when a number of jobs are to be assigned to the same number of workers. Several procedures have been proposed in the current literature to convert the LBAP into an equivalent CAP and then apply the Hungarian method to solve it efficiently. However, their applicability is limited because some of the elements in the transformed assignment matrix for the CAP can be too large to be handled by most computer programs. In this paper, we suggest a turnpike approach to alleviate the problem so that the conversion methodology will become more useful in practice. A numerical example is provided to demonstrate the superiority of the new algorithm to the existing ones.

Appendix 1: Proofs of Propositions

1.1 Proof of Proposition 1

Suppose that x ^* = (x ₁₁ ^*, x ₁₂ ^*, …, x _nn ^*) is an optimal solution to (M2). Let S = {(i, j): x _ij ^* = 1, 1 ≤ i, j ≤ n}. Since c _ij  ≥ r _i ∀ (i, j) ∈ S, Z ^* = max {c _ij : (i, j) ∈ S} ≥ max {r _i : (i, j) ∈ S} = max {r _i : 1 ≤ i ≤ n} = r or Z ^* ≥ r. Similarly, c _ij  ≥ c _j ∀ (i, j) ∈ S and Z ^* = max {c _ij : (i, j) ∈ S} ≥ max {c _j : (i, j) ∈ S} = max {c _j : 1 ≤ j ≤ n} = c or Z ^* ≥ c. It follows that Z ^* ≥ max {r, c} = LB or Z ^* ≥ LB, which implies that Z ^* is bounded below by LB.

1.2 Proof of Proposition 4

Call all the elements in [c _ij ] greater than c(n ² − n + 1) inadmissible and replace each of them with an extremely large positive number M. Rank the matrix columns in non-decreasing order of number of admissible elements and label them as [1], [2], …, [n]. Since there are only n − 1 inadmissible elements in [c _ij ], column [j] contains at least j admissible elements, 1 ≤ j ≤ n. Beginning with column [1], match the job with the worker associated with any admissible element in the column and cross out the corresponding row and column. Repeat the same procedure on the continuously reduced matrix until the matching in column [n] is made.

Suppose the job-worker assignments determined in this way are represented by w = (w ₁₁, w ₁₂, …, w _nn ), where w _ij  ∈ {0, 1}, ${\displaystyle \sum_{j=1}^n{w}_{ij}}=1,1\le i\le n$, and ${\displaystyle \sum_{i=1}^n{w}_{ij}}=1,1\le j\le n$. It is trivial to see that w is a feasible solution to (M2). Furthermore, Z ^* ≤ Z = max {c _ij : w _ij  = 1, 1 ≤ i, j ≤ n} ≤ c(n ² − n + 1) = w or Z ^* ≤ w, which implies that Z ^* is bounded above by w.

1.3 Proof of Proposition 7

Assume that x ^* = (x ₁₁ ^*, x ₁₂ ^*, …, x _nn ^*) is an optimal solution to the CAP with [f _ij ]. To prove the proposition, it suffices to show that x ^* also solves the LBAP with [e _ij ]. Suppose otherwise and y = (y ₁₁, y ₁₂, …, y _nn ) is a better feasible solution to the LBAP; that is, $\underset{1\le i,j\le n}{ \max }$ {e _ij : y _ij  = 1} < $\underset{1\le i,j\le n}{ \max }$ {e _ij : x _ij ^* = 1}. Let t′ and t ^* be the unique positive integers such that $\underset{1\le i,j\le n}{ \max }$ {e _ij : y _ij  = 1} = t′ ∈ S(t′) and $\underset{1\le i,j\le n}{ \max }$ {e _ij : x _ij ^* = 1} = t ^* ∈ S(t ^*). Further, let σ′ and θ′ be associated with t′ while σ ^* and θ ^* be associated with t ^*. Given t′ < t ^*, we have f(t′) < f(t ^*) since f(.) is an increasing function. Also, note that θ ^* ≥ f(t′) + [θ′ − f(σ′)], which means that the sum of the largest n elements smaller than f(t ^*) in [f _ij ] is greater than or equal to the sum of f(t′) and the largest n − 1 elements smaller than f(t′).

Based on the above discussion, we have $\begin{array}{l}{\displaystyle \sum_{i=1}^n{\displaystyle \sum_{j=1}^n{f}_{ij}{x}_{ij}^{\ast }}}={\displaystyle \sum_{i=1}^n{\displaystyle \sum_{j=1}^n\left\{{f}_{ij}:{x}_{ij}^{\ast }=1\right\}}}=\underset{1\le i,j\le n}{ \max}\left\{{f}_{ij}:{x_{ij}}^{*}=1\right\}\ \hfill \\ {}+\left[{\displaystyle \sum_{i=1}^n{\displaystyle \sum_{j=1}^n\left\{{f}_{ij}:{x}_{ij}^{\ast }=1\right\}}}\kern.3em -\underset{1\le i,j\le n}{ \max}\left\{{f}_{ij}:{x_{ij}}^{*}=1\right\}\right]=f\left({t}^{*}\right)+\left[{\displaystyle \sum_{i=1}^n{\displaystyle \sum_{j=1}^n\left\{{f}_{ij}:{x}_{ij}^{\ast }=1\right\}}}\kern.3em -\underset{1\le i,j\le n}{ \max}\left\{{f}_{ij}:{x_{ij}}^{*}=1\right\}\right]\ge f\left({t}^{*}\right)+\beta \hfill \\ {}=\left({\theta}^{*}-\beta +1\right)+\beta ={\theta}^{*}+1>{\theta}^{*}\ge f\left(t\prime \right)+\left[\theta \prime -f\left(\sigma \prime \right)\right]=\underset{1\le i,j\le n}{ \max}\left\{{f}_{ij}:{y}_{ij}=1\right\}+\left[\theta \prime -f\left(\sigma \prime \right)\right]\ge \underset{1\le i,j\le n}{ \max}\left\{{f}_{ij}:{y}_{ij}=1\right\}\hfill \\ {}+\left[{\displaystyle \sum_{i=1}^n{\displaystyle \sum_{j=1}^n\left\{{f}_{ij}:{y}_{ij}=1\right\}}}\kern.3em -\underset{1\le i,j\le n}{ \max}\left\{{f}_{ij}:{y}_{ij}=1\right\}\right]={\displaystyle \sum_{i=1}^n\kern5pt {\displaystyle \sum_{j=1}^n\kern5pt \left\{{f}_{ij}:{y}_{ij}=1\right\}}={\displaystyle \sum_{i=1}^n\kern5pt {\displaystyle \sum_{j=1}^n{f}_{ij}{y}_{ij}}}}\hfill \end{array}$ or ${\displaystyle \sum_{i=1}^n{\displaystyle \sum_{j=1}^n{f}_{ij}{x}_{ij}^{\ast }}}\ge {\displaystyle \sum_{i=1}^n{\displaystyle \sum_{j=1}^n{f}_{ij}{y}_{ij}}}$, which implies that y is a better solution to the CAP than x. Since this a contradiction to the assumption that x ^* is an optimal solution, the proof is complete.