Scheduling with two competing agents to minimize total weighted earliness

Abstract

We study a single machine scheduling problem with two competing agents and earliness measures. Given a common deadline for all the jobs of both agents, the objective function is minimizing the total weighted earliness of the first agent, subject to an upper bound on the maximum earliness of the jobs of the second agent. This problem was recently proved to be NP-hard, leaving the question of the complexity class open. We introduce a pseudo-polynomial dynamic programming algorithm, implying that the problem is NP-hard in the ordinary sense. An extensive numerical study indicates that the dynamic programming is very effective for solving medium size instances. We also propose an efficient heuristic, which is shown numerically to produce very close-to-optimal schedules. The dynamic programming algorithm is extended to any (given) number of agents, proving NP-hardness in the ordinary sense of the general multi-agent setting. Finally, we study the inverse problem of minimizing the maximum earliness of one agent subject to an upper bound on the maximum total weighted earliness of the second agent. We introduce a pseudo-polynomial dynamic programming algorithm, a simple greedy-type heuristic and a lower bound. Our numerical tests verify that the heuristic produces very small optimality gaps.

Appendices

Appendix 1: Proofs of Properties 1–4

Proof of Property 1

Consider an optimal schedule, say q, in which an A-job is scheduled between B-jobs. By moving this A job to be scheduled after all the B-jobs, we obtain a new feasible schedule $q^{\prime }$, with a strictly smaller total weighted earliness of the A-jobs. This contradicts the optimality of schedule q. $\square$

Proof of Property 2

Consider an optimal schedule q consisting of a single block of the B-jobs (Property 1), in which the first B-job (say k) is not the largest, and the largest B-job (say l) is scheduled later. Assume that this block starts at time t. Clearly, the maximum earliness of the B-jobs is given by $D-(t+p_{k})$. Create a new schedule $q^{\prime }$ by moving job l to be first and start at t. $q^{\prime }$ is clearly feasible (the maximum earliness becomes smaller: $D-(t+p_{l}))$, with no impact on the total weighted earliness of the A-jobs, implying that $q^{\prime }$ is optimal as well. $\square$

Proof of Property 3

The two-block claim follows Property 1. The order of the jobs within each block is easily proved by a standard pair-wise interchange argument. $\square$

Proof of Property 4

Consider an optimal schedule q with $t>t_{min}+\max \left\{ p_{max}^{A},p_{max}^{B}\right\}$. Create a new schedule $q^{\prime }$ by replacing the last A-job scheduled prior to the B-block with the B-block. It is clear that $q^{\prime }$ is feasible since the new t value is larger than the original (and in particular not smaller than $t_{min})$. Moreover, the total weighted earliness of the A-jobs in $q^{\prime }$ is strictly smaller than that of q, contradicting the optimality of q. $\square$

Fig. 2

The optimal solution of the two-agent problem solved in Example 1

Fig. 3

a The optimal solution of the two-agent problem solved in Example 2, b the schedule obtained by Heuristic 2 in Example 2, c the lower bound obtained in Example 2 based on a preemptive schedule

Appendix 2: numerical examples

Example 1

(for problem $1/ {d_{j}=D (deadline)}/ {\sum \nolimits _{j=1}^{n^{A}} {w_{j}^{A}E_{j}^{A}} :E_{max}^{B}})$:

Consider a 2-agent problem, where each agent needs to process 7 jobs. The job processing times of Agent A are: $p_{1}^{A}=1,\, p_{2}^{A}=2,\, p_{3}^{A}=7,\, p_{4}^{A}=8,\, p_{5}^{A}=4,\, p_{6}^{A}=5,\, p_{7}^{A}=6.$ The weights of the jobs of Agent A are: $w_{1}^{A}=8,\, w_{2}^{A}=2,\, w_{3}^{A}=7,\, w_{4}^{A}=1,\, w_{5}^{A}=3,\, w_{6}^{A}=4,\, w_{7}^{A}=2.$ The job processing times of Agent B are: $p_{1}^{B}=8,\, p_{2}^{B}=2,\, p_{3}^{B}=1,\, p_{4}^{B}=3,\, p_{5}^{B}=7,\, p_{6}^{B}=2,\, p_{7}^{B}=4.$ The upper bound on the maximum earliness of Agent B is $U=33$. It follows that $P^{A}=33,\, P^{B}=27$, and $D=P^{A}+P^{B}=60$. The smallest possible starting time of the B-block is $t_{min}=D-U-p_{max}^{B}=19$. The largest possible starting time of the B-block is $t_{min}+max \left\{ p_{max}^{A},p_{max}^{B} \right\} =27$. The optimal solution obtained by the DP given in Sect. 3 (see Fig. 2) consists of the following job sequence: Job 4 of Agent A is first (starts at zero and is completed at 8); Job 7 of Agent A (completed at 14); Job 6 of Agent A (completed at 19); The block of all the B-jobs (in the interval [19, 46]); Job 5 of Agent A (completed at 50); Job 3 of Agent A (completed at 57); Job 2 of Agent A (completed at 59); Job 1 of Agent A (completed at $D=60)$. The total cost (weighted earliness of Agent A) is 361.

Example 2

(for problem $1 /{d_{j}=D \,(deadline)}/ {E_{max}^{B}: \sum \nolimits _{j=1}^{n^{A}} {w_{j}^{A}E_{j}^{A}} })$:

We use the same data of Example 1 (processing times of both agents, and weights of Agent A). As above, $P^{A}=33, P^{B}=27$, $D=P^{A}+P^{B}=60$. The upper bound on the total weighted earliness of Agent A is $U=310$. The interval of a feasible t-values is [0, 33]. The optimal solution obtained by the DP given in Sect. 6 (see Fig. 3a) consists of the following job sequence: Job 4, Job 7, Job 2, the B-block, Job 5, Job 6, Job 3, Job 1. $t^{opt}=16$. The optimal maximal earliness of Agent B is $E_{max}^{B}=D-t^{opt}-p_{max}^{B}=60-16-8=36$.

We now solve the problem using Heuristic 2. The schedule obtained by the heuristic (see Fig. 3b) consists of the following job sequence: Job 4, Job 7, the B-block, Job 5, Job 6, Job 3, Job 2, Job 1. $t^{heur}=14$. Thus, the maximal earliness of Agent B obtained by Heuristic 2 is $E_{max}^{B}=D-p_{max}^{B}=60-14-8=38$. Note that the total weighted earliness of Agent A is 252 (smaller than $U=310)$.

Finally, we calculate the lower bound obtained by the following preempted schedule (see Fig. 3c): Job 4, Job 7, Job 5 (starts at time 14 and is preempted at time 17 by the B-block), the B-block (which starts at time 17 and is completed at time 44), the last unit of time of Job 5, Job 6, Job 3, Job 2, Job 1. $t^{LB}=17$. The lower bound on the maximal earliness of Agent B is $E_{max}^{B}=D-t^{LB}-p_{max}^{B}=60-17-8=35$.

In summary: the optimal $E_{max}^{B}$ in Example 2 (obtained by the DP) is 36, the heuristic value is 38, and the lower bound is 35.

Appendix 3: a dynamic programming for the case of $m\,B$-agents

We now assume that each Agent $B_{i}$ has a set $J^{B_{i}}$ of $n^{B_{i}}$ jobs, $i=1,\ldots ,m$. The above definitions of $p_{j}^{B_{i}}$, $P^{B_{i}}$, $p_{max}^{B_{i}}, D$, $U_{i}$ and $t_{i}$ remain unchanged, $i=1,\ldots ,m$. Clearly, Properties 1–3 remain valid. Similar to the case of $m=2$, the number of B-blocks may reduce when two (or more) B-blocks are processed continuously with no A-jobs among them. Again, without loss of generality we assume $t_{1}\le t_{2}\le \cdots \le t_{m}$. A lower bound on $t_{i}$ is given by: $t_{min}^{B_{i}}=D-U_{i}-p_{max}^{B_{i}}$. An upper bound is $t_{min}^{B_{i}}+\max \left\{ p_{max}^{A},p_{max}^{B_{i}}\right\}$, $i=1,\ldots ,m$. The extension of Property 4 to the setting of m agents is that an optimal schedule exists such that the starting times of the $B_{i}$-blocks are bounded by: $t_{min}^{B_{i}}\le t_{i}\le t_{min}^{B_{i}}+\max \left\{ p_{max}^{A},p_{max}^{B_{i}}\right\}$, $i=1,\ldots ,m$.

The A-jobs are scheduled in (at most) $m+1$ blocks. We sort the jobs in a non-decreasing order of $p_{j}^{A}/w_{j}^{A}$. The updated state variables are the following:

j :

The number of $A-$jobs already scheduled;

$e_{i}$ :

The total processing time of the A-jobs scheduled after block $B_{i}$ and prior to block $B_{i+1}$, $i=1,\ldots ,m-1$

$e_{m}$ :

The total processing time of the A-jobs scheduled after block $B_{m}$.

$P_{j}^{A}-\sum \nolimits _{i=1}^m e_{i}$ is the total processing time of the A-jobs scheduled prior to block $B_{1}$.

Let $f_{t_{1},\ldots {,t}_{m}}\left( j,e_{1},\ldots ,e_{m} \right)$ denote the optimal total weighted earliness of the jobs $j+1, j+2,\ldots ,n^{A}$ (of Agent A), given a total load of $e_{i}$ of the A-jobs scheduled after the $B_{i}$-block, $i=1,\ldots ,m$.

The DP compares $m+1$ options at each iteration: schedule the next A-job as late as possible after block $B_{i},i=1,\ldots m$, or as late as possible prior to block $B_{1}$. The recursion becomes:

$$\begin{aligned}&f_{t_{1},\ldots {,t}_{m}}\left( j,e_{1},\ldots ,e_{m} \right) \\&\quad =min\left\{ {\begin{array}{ll} {\begin{array}{ll} \left\{ {\begin{array}{c} \left( D-t_{1}+\left( P_{j}^{A}-\sum \nolimits _{i=1}^m e_{i} \right) \right) w_{j+1}^{A}+f_{t_{1},{\ldots ,t}_{m}}\left( j+1,e_{1},\ldots ,e_{m} \right) \\ \infty \\ \end{array} } \right\} &{}\quad {\begin{array}{l} if \,P_{j+1}^{A}-\sum \nolimits _{i=1}^m e_{i} \le t_{1}\\ otherwise\\ \end{array} } \\ \left\{ {\begin{array}{ll} \left( D-t_{2}+e_{1} \right) w_{j+1}^{A}+f_{t_{1},{\ldots ,t}_{m}}\left( j+1,e_{1}+p_{j+1}^{A},e_{2},\ldots ,e_{m} \right) \\ \infty \\ \end{array} } \right\} &{}\quad {\begin{array}{c} if \,t_{1}+P^{B_{1}}+e_{1}+p_{j+1}^{A}\le t_{2} \\ otherwise\\ \end{array} } \\ \left\{ {\begin{array}{ll} \left( D-t_{3}+e_{2} \right) w_{j+1}^{A}+f_{t_{1},{\ldots ,t}_{m}}\left( j+1,e_{1},e_{2}+p_{j+1}^{A},\ldots ,e_{m} \right) \\ \infty \\ \end{array} } \right\} &{}\quad {\begin{array}{l} if \,t_{2}+P^{B_{2}}+e_{2}+p_{j+1}^{A}\le t_{3} \\ otherwise\\ \end{array} }\\ \ldots \\ \ldots \\ \ldots \\ \left\{ {\begin{array}{ll} e_{m}w_{j+1}^{A}+f_{t_{1},{\ldots ,t}_{m}}\left( j+1,e_{1},{\ldots {,e}_{m-1},e}_{m}+p_{j+1}^{A} \right) \\ \infty \\ \end{array} } \right\} &{}\quad {\begin{array}{l} if\, t_{m}+P^{B_{m}}+e_{m}+p_{j+1}^{A}\le D \\ otherwise\\ \end{array} } \\ \end{array} }\\ \end{array} } \right\} \end{aligned}$$

The boundary conditions are:

$$\begin{aligned} f_{t_{1},{\ldots ,t}_{m}}=f_{t_{1},{\ldots ,t}_{m}}( n^{A},e_{1},\ldots ,e_{m})=0,\quad e_{1},\ldots ,e_{m}=0,1,\ldots ,P^{A},\quad e_{1}+\ldots +e_{m}\le P^{A}. \end{aligned}$$

The solution is given by: $f_{t_{1},{\ldots ,t}_{m}}\left( 0,0,\ldots ,0 \right)$.

The DP needs to be executed for all relevant combinations of $t_{i}$, which leads to the following:

Theorem 7

For the general setting of $m \, B$-agents, the running time of the DP is: $O\left( \left( n^{A} \right) ^{m+1}{(p_{max}^{A})}^{m}\times \left( max \left\{ p_{max}^{A},p_{max}^{B} \right\} \right) ^{m} \right)$ time, where $p_{max}^{B}=\max \left\{ p_{max}^{B_{i}},i=1,\ldots m\right\}$

Proof

$e_{i}, i=1,\ldots ,m$ is bounded by $P^{A}$, which is bounded by $n^{A}p_{max}^{A}$. Therefore, for given $t_{1},t_{2},\ldots ,t_{m}$, the running time is $O\left( \left( n^{A} \right) ^{m+1}{(p_{max}^{A})}^{m} \right)$. The DP is repeated up to $\left( max \left\{ p_{max}^{A},p_{max}^{B} \right\} \right) ^{m}$ times. Thus, the total running time is:

$$\begin{aligned} O\left( \left( n^{A} \right) ^{m+1}{(p_{max}^{A})}^{m}\times \left( max \left\{ p_{max}^{A},p_{max}^{B} \right\} \right) ^{m} \right) . \end{aligned}$$

$\square$