Sequencing situations with Just-in-Time arrival, and related games

Abstract

In this paper sequencing situations with Just-in-Time (JiT) arrival are introduced. This new type of one-machine sequencing situations assumes that a job is available to be handled by the machine as soon as its predecessor is finished. A basic predecessor dependent set-up time is incorporated in the model. Sequencing situations with JiT arrival are first analyzed from an operations research perspective: for a subclass an algorithm is provided to obtain an optimal order. Secondly, we analyze the allocation problem of the minimal joint cost from a game theoretic perspective. A corresponding sequencing game is defined followed by an analysis of a context-specific rule that leads to core elements of this game.

Appendix: Proofs

Proof of Theorem 2.1

Assume $s_0=s^h$ and first consider the case where $\max _{i\in N} s_i = s^l$. This implies that $|N^H_h|=|N^L_h|=0$. For $\sigma \in \Pi (N)$ we obtain

$$\begin{aligned} \gamma _N(\sigma )=\sum _{i\in N} s^l\alpha _i + (s_0-s^l)\alpha _{\sigma (1)}=\sum _{i\in N} s^l\alpha _i + (s^h-s^l)\alpha _{\sigma (1)}. \end{aligned}$$

Assume there exists an order $\sigma '\in \Pi (N)$ such that $|M^{hH}(\sigma ')|=0$, and take such a $\sigma '\in \Pi (N)$. As $s_0=s^h$, we obtain that $\alpha _{\sigma '(1)}=\alpha ^L$ and therefore $\sigma '(1)\in N^L_l$. Hence, for all $\sigma \in \Pi (N)$

$$\begin{aligned} \gamma _N(\sigma ') = \sum _{i\in N} s^l\alpha _i + (s^h-s^l)\alpha ^L\le \gamma _N(\sigma ), \end{aligned}$$

so $\sigma '$ is optimal.

Now assume there exists an order $\sigma ''\in \Pi (N)$ such that $|M^{lL}(\sigma '')|=0$ and $|M^{hH}(\sigma '')|>0$, and take such a $\sigma ''\in \Pi (N)$. Then either $|N^L_l|=0$, which means that $N=N^H_l$ and every order is optimal, or $|N^L_l|=1$ with $\sigma '(1)\in N^L_l$. In the last case $\gamma _N(\sigma '')=\sum _{i\in N} s^l\alpha _i + (s_0-s^l)\alpha ^L\le \gamma _N(\sigma )$ for all $\sigma \in \Pi (N)$, and $\sigma ''$ is optimal.

Now consider the case where $\max _{i\in N} s_i = s^h$. Take an arbitrary $\sigma \in \Pi (N)$. Take $B,D\in \mathbb {N}$ such that $B=|M^{hH}(\sigma )|$ and $D=|M^{lL}(\sigma )|$. Note that by Eqs. (2) and (4) we have

$$\begin{aligned} B-D=|M^{hH}(\sigma )|-|M^{lL}(\sigma )|=|N_h|-|N^L|+\mathbb {1}_{[{s_{\sigma (|N|)}=s^l}] }. \end{aligned}$$

By (3) and (4) it holds that

$$\begin{aligned} \gamma _N(\sigma )&= |M^{hH}(\sigma )|s^h\alpha ^H+|M^{lH}(\sigma )|s^l\alpha ^H+|M^{hL}(\sigma )|s^h\alpha ^L+|M^{lL}(\sigma )|s^l\alpha ^L\\&= Bs^h\alpha ^H+(|N_l|-\mathbb {1}_{[{s_{\sigma (|N|)}=s^l}] }-D)s^l\alpha ^H\\&+ (|N_h|+\mathbb {1}_{[{s_{\sigma (|N|)}=s^l}] }-B)s^h\alpha ^L+Ds^l\alpha ^L\\&\ge (B-\min \{B,D\})s^h\alpha ^H+(|N_l|-\mathbb {1}_{[{s_{\sigma (|N|)}=s^l}] }-D+\min \{B,D\})s^l\alpha ^H\\&+ (|N_h|+\mathbb {1}_{[{s_{\sigma (|N|)}=s^l}] }-B+\min \{B,D\})s^h\alpha ^L+(D-\min \{B,D\})s^l\alpha ^L\\&= \max \{0,|N_h|-|N^L|+\mathbb {1}_{[{s_{\sigma (|N|)}=s^l}] }\}s^h\alpha ^H\\&+ \min \{|N_l|-\mathbb {1}_{[{s_{\sigma (|N|)}=s^l}] },|N^H|\}s^l\alpha ^H\\&+ \min \{|N^L|,|N_h|+\mathbb {1}_{[{s_{\sigma (|N|)}=s^l}] }\}s^h\alpha ^L\\&+ \max \{|N^L|-|N_h|-\mathbb {1}_{[{s_{\sigma (|N|)}=s^l}] },0\}s^l\alpha ^L\\&\ge \max \{0,|N_h|-|N^L|\}s^h\alpha ^H+\min \{|N_l|,|N^H|\}s^l\alpha ^H\\&+ \min \{|N^L|,|N_h|\}s^h\alpha ^L+\max \{|N^L|-|N_h|,0\}s^l\alpha ^L, \end{aligned}$$

where the first inequality follows from the observation that $(s^h-s^l)(\alpha ^H-\alpha ^L)>0$. If $|N_h|-|N^L|\ge 0$, and therefore $|N_l|-|N^H|\le 0$, then the second inequality follows from $(s^h-s^l)\alpha ^H>0$. If $|N_h|-|N^L|< 0$, then the second inequality follows from $(s^h-s^l)\alpha ^L>0$. The first inequality holds with equality if either $B=0$ or $D=0$, and the second inequality holds with equality if $s_{\sigma (|N|)}=s^h$. This shows that every order $\sigma \in \Pi (N)$ with $s_{\sigma (|N|)}=s^h$ and either $|M^{hH}(\sigma )|=0$ or $|M^{lL}(\sigma )|=0$ is optimal.

Next, assume $s_0=s^l$ and first consider the case where $\max _{i\in N} s_i = s^l$. In this case, for any order $\sigma \in \Pi (N)$ we have

$$\begin{aligned} \gamma _N(\sigma )=\sum _{i\in N} s^l\alpha _i. \end{aligned}$$

So, every order is optimal.

Now consider the case where $\max _{i\in N} s_i = s^h$. Take an arbitrary $\sigma \in \Pi (N)$. Take $B,D\in \mathbb {N}$ such that $B=|M^{hH}(\sigma )|$ and $D=|M^{lL}(\sigma )|$. Note that by Eq. (2) and (6) we have

$$\begin{aligned} B-D=|M^{hH}(\sigma )|-|M^{lL}(\sigma )|=|N_h|-|N^L|-\mathbb {1}_{[{s_{\sigma (|N|)}=s^h}] }. \end{aligned}$$

By (5) and (6) it holds that

$$\begin{aligned} \gamma _N(\sigma )&= |M^{hH}(\sigma )|s^h\alpha ^H+|M^{lH}(\sigma )|s^l\alpha ^H+|M^{hL}(\sigma )|s^h\alpha ^L+|M^{lL}(\sigma )|s^l\alpha ^L\\&= Bs^h\alpha ^H+(|N_l|+\mathbb {1}_{[{s_{\sigma (|N|)}=s^h}] }-D)s^l\alpha ^H\\&+ (|N_h|-\mathbb {1}_{[{s_{\sigma (|N|)}=s^h}] }-B)s^h\alpha ^L+Ds^l\alpha ^L\\&\ge (B-\min \{B,D\})s^h\alpha ^H+(|N_l|+\mathbb {1}_{[{s_{\sigma (|N|)}=s^h}] }-D+\min \{B,D\})s^l\alpha ^H\\&+ (|N_h|-\mathbb {1}_{[{s_{\sigma (|N|)}=s^h}] }-B+\min \{B,D\})s^h\alpha ^L+(D-\min \{B,D\})s^l\alpha ^L\\&= \max \{0,|N_h|-|N^L|-\mathbb {1}_{[{s_{\sigma (|N|)}=s^h}] }\}s^h\alpha ^H\\&+ \min \{|N_l|+\mathbb {1}_{[{s_{\sigma (|N|)}=s^h}] },|N^H|\}s^l\alpha ^H\\&+ \min \{|N^L|,|N_h|-\mathbb {1}_{[{s_{\sigma (|N|)}=s^h}] }\}s^h\alpha ^L\\&+ \max \{|N^L|-|N_h|+\mathbb {1}_{[{s_{\sigma (|N|)}=s^h}] },0\}s^l\alpha ^L\\&\ge \max \{0,|N_h|-|N^L|-1\}s^h\alpha ^H+\min \{|N_l|+1,|N^H|\}s^l\alpha ^H\\&+ \min \{|N^L|,|N_h|-1\}s^h\alpha ^L+\max \{|N^L|-|N_h|+1,0\}s^l\alpha ^L, \end{aligned}$$

where the first inequality follows from the observation that $(s^h-s^l)(\alpha ^H-\alpha ^L)>0$. If $|N_h|-|N^L|\le 0$, then the second inequality follows from $(s^l-s^h)\alpha ^L<0$. If $|N_h|-|N^L|>0$, then the second inequality follows from $(s^l-s^h)\alpha ^H<0$. The first inequality holds with equality if either $B=0$ or $D=0$, and the second inequality holds with equality if $s_{\sigma (|N|)}=s^h$.

Hence for both values of $\sigma _0$ every order $\sigma \in \Pi (N)$ with $s_{\sigma (|N|)}=s^h$ and either $|M^{hH}(\sigma )|=0$ or $|M^{lL}(\sigma )|=0$ is optimal. $\square$

Proof of Theorem 2.2

Let $\tilde{\sigma }\in \Pi (N)$ be an order provided by Algorithm 1. In Step 2 of the algorithm, it is made sure that there is always a player with the highest available set-up time left to place at the last position. Hence, $s_{\tilde{\sigma }(|N|)}=s^h$, unless $N^H_h\cup N^L_h=\emptyset$ which implies that there is in fact only one value for $s_i$ and $s_{\tilde{\sigma }(|N|)}=s^l=\max _{i\in N} s_i$.

We prove that either optimality of $\tilde{\sigma }$ follows directly from Theorem 2.1, i.e., $|M^{hH}(\tilde{\sigma })|=0$ or $|M^{lL}(\tilde{\sigma })|=0$, or $N^H_h\cup N^L_l=\emptyset$. For the latter case we show that $\tilde{\sigma }$ is optimal as well.

Assume that optimality of $\tilde{\sigma }$ does not follow directly from Theorem 2.1, i.e., $|M^{hH}(\tilde{\sigma })|>0$ and $|M^{lL}(\tilde{\sigma })|>0$. Then there exist $p,r\in \{0, \ldots ,|N|-1\}$ such that $(\tilde{\sigma }(p),\tilde{\sigma }(p+1))\in M^{hH}(\tilde{\sigma })$ and $(\tilde{\sigma }(r),\tilde{\sigma }(r+1))\in M^{lL}(\tilde{\sigma })$.

Assume $r<p$. According to the algorithm, job $\tilde{\sigma }(r+1)$ is only placed behind job $\tilde{\sigma }(r)$ if there is no job $j$ with $\alpha _{j}=\alpha ^H$ left that is not yet placed, or there is only one job $j$ with $\alpha _j=\alpha ^H$ left, but this job has to be reserved for the last spot because it is the only remaining job with high set-up time. In the first case, we have a contradiction, since job $\tilde{\sigma }(p+1)$ is not yet placed. The second case also results in a contradiction, since both $s_{\tilde{\sigma }(p)}=s^h$ and $s_{\tilde{\sigma }(|N|)}=s^h$.

Now assume $p<r$. According to the algorithm, job $\tilde{\sigma }(p+1)$ is only placed behind job $\tilde{\sigma }(p)$ if there is no job $j$ with $\alpha _{j}=\alpha ^L$ left that is not yet placed, or there is only one job $j$ with $\alpha _j=\alpha ^L$ left, but this job has to be reserved for the last spot because it is the only remaining job with high set-up time. In the first case, we have a contradiction, since job $\tilde{\sigma }(r+1)$ is not yet placed.

The second case can only hold if $r+1=|N|$. For all jobs $i\in \{\tilde{\sigma }(p+1),\ldots ,\tilde{\sigma }(r)\}$ it then must hold that $s_i=s^l$, otherwise job $\tilde{\sigma }(r+1)$ would have been placed at position $p+1$. Furthermore, $\alpha _i=\alpha ^H$ otherwise job $i$ would have been placed at position $p+1$ as this would avoid the combination of $s^h$ and $\alpha ^H$. So, we obtain that $i\in N^H_l$ for all $i\in \{\tilde{\sigma }(p+1),\ldots ,\tilde{\sigma }(r)\}$. Since the algorithm first places the jobs in $N^H_l$ before placing the jobs in $N^H_h$, and $\tilde{\sigma }(|N|)\not \in N^H_h$, we obtain that $N^H_h=\emptyset$ and therefore $\tilde{\sigma }(p)\in N^L_h$. Furthermore, if there existed a job $i\in N^L_l$ then the algorithm would place every job in $N^H_l$ directly behind this job. But since $\tilde{\sigma }(p)\not \in N^L_l$ this implies that $N^L_l=\emptyset$. Hence, the second case only allows players in $N^H_l$ and $N^L_h$, so $N^H_h\cup N^L_l=\emptyset$. If $s_0=s^l$, then the algorithm first places all players in $N^H_l$ and then all players in $N^L_h$, which contradicts $\tilde{\sigma }(s)\in N^L_h$. So, $s_0=s^h$ and the solution provided by the algorithm for this situation (first all players in $N^L_h$ but one, then all players in $N^H_l$ and finally the last player in $N^L_h$) is clearly optimal. $\square$

Proof of Proposition 3.1

Here, we will only prove the first case, the other cases follow from a similar reasoning. First of all, we have

$$\begin{aligned} \sum _{i\in S} \gamma _i(\sigma ^*_{\{i\}})=(|S^H_h|+|S^H_l|)s^h\alpha ^H+ (|S^L_h|+|S^L_l|)s^h\alpha ^L, \end{aligned}$$

for every $S\in 2^N$. $\square$

Take $S\in 2^N$ such that $S^H_h\not =\emptyset$ and $|S^H_h|\ge |S^L_l|$. Since $S^H_h\not =\emptyset$, it holds for every (optimal) order $\tilde{\sigma }_{S}$ provided by Algorithm 1 that $s_{\sigma ^*_{S}(|S|)}=s^h$. Furthermore, since $S^H_h\not =\emptyset$ we have by Proposition 2.1 that either $|M^{hH}(\tilde{\sigma }_S)|=0$ or $|M^{lL}(\tilde{\sigma }_S)|=0$. It must hold that $|M^{lL}(\tilde{\sigma }_S)|=0$, since $|S^H_h|\ge |S^L_l|$ together with (1) and (3) implies that $|M^{hH}(\tilde{\sigma }_S)|\ge |M^{lL}(\tilde{\sigma }_S)|$. So, we have

$$\begin{aligned} \gamma _S(\tilde{\sigma }_S)&= |M^{hH}(\tilde{\sigma }_S)|s^h\alpha ^H+|M^{lH}(\tilde{\sigma }_S)|s^l\alpha ^H+|M^{hL}(\tilde{\sigma }_S)|s^h\alpha ^L+|M^{lL}(\tilde{\sigma }_S)|s^l\alpha ^L\\&= (|S^H_h|-|S^L_l|)s^h\alpha ^H+(|S^H_l|+|S^L_l|)s^l\alpha ^H+(|S^L_h|+|S^L_l|)s^h\alpha ^L, \end{aligned}$$

and we may conclude that

$$\begin{aligned} v^\Psi (S)&= \sum _{i\in S} \gamma _{i}(\tilde{\sigma }_{\{i\}}) - \gamma _S(\tilde{\sigma }_S)\\&= (|S^H_h|+|S^H_l|)s^h\alpha ^H+ (|S^L_h|+|S^L_l|)s^h\alpha ^L,\\&- \left( (|S^H_h|-|S^L_l|)s^h\alpha ^H+(|S^H_l|+|S^L_l|)s^l\alpha ^H+(|S^L_h|+|S^L_l|)s^h\alpha ^L\right) \\&= (|S^H_l|+|S^L_l|)(s^h-s^l)\alpha ^H. \end{aligned}$$