The single machine serial batch scheduling problems with rejection

Abstract

We consider several serial batch scheduling problems with rejection. Each job is either processed on a single serial batching machine or rejected by paying a penalty. We analyze two models with rejection. The first model is to minimize the sum of the scheduling cost of the accepted jobs and the total penalty of the rejected jobs, where the scheduling costs are the total completion time, the makespan, the maximum lateness and the weighted number of tardy jobs, respectively. For the former two problems, we propose two polynomial time algorithms to solve them. For the last two problems, we derive efficient dynamic programming algorithms. The second model is to minimize the makespan, given an upper bound on the total rejection cost, we present a fully polynomial time approximation scheme.

Appendix 1: Numerical example

The following numerical example illustrates Algorithm DP1. (see (Shabtay 2014)). The job processing times and rejection costs are given in below Table, where s = 2.

j	1	2	3	4	5
p j	3	5	6	8	11
e j	15	17	20	28	24

Next, we apply Algorithm DP1 to solve the example.

First, we compute the case h = 1.

$$ \begin{aligned} & j = 5,F\left( {5,0,0,0} \right) = e_{5} = 24,F\left( {5,1,1,1} \right) = p_{5} + s = 13. \\ & j = 4,F\left( {4,0,0,0} \right) = e_{4} + e_{5} = 52, \\ & F\left( {4,1,1,1} \right) = { \hbox{min} }\left\{ {F\left( {5,0,0,0} \right) + \left( {s + p_{4} } \right),F\left( {5,1,1,1} \right) + e_{4} } \right\} = {\text{min }}\left\{ {24 + 10,13 + 28} \right\} = 34. \\ & F\left( {4,2,1,1} \right) = { \hbox{min} }\left\{ {F\left( {5,1,1,1} \right) + 2\left( {s + p_{4} } \right),F\left( {5,2,1,1} \right) + e_{4} } \right\} = { \hbox{min} } \left\{ {13 + 20, \infty } \right\} = 33. \\ & j = 3,F\left( {3, \, 0, \, 0, \, 0} \right) = e_{3} + e_{4} + e_{5} = 72, \\ & F\left( {3,1,1,1} \right) = { \hbox{min} }\left\{ {F\left( {4,0,0,0} \right) + s + p_{3} ,F\left( {4,1,1,1} \right) + e_{3} } \right\} = { \hbox{min} }\left\{ {52 + 8, 34 + 20} \right\} = 54, \\ & F\left( {3,2,1,1} \right) = { \hbox{min} }\left\{ {F\left( {4,1,1,1} \right) + 2\left( {s + p_{3} } \right),F\left( {4,2,1, 1} \right) + e_{3} } \right\} = { \hbox{min} } \left\{ {34 + 16, \, 33 + 20} \right\} = 50, \\ & F\left( {3,3,1,1} \right) = { \hbox{min} } \left\{ {F\left( {4,2,1,1} \right) + 3\left( {s + p_{3} } \right),F\left( {4,3,1,1} \right) + e_{3} } \right\} = { \hbox{min} } \left\{ {33 + 24, \, \infty } \right\} = 57. \\ & j = 2,F\left( {2,0,0,0} \right) = e_{2} + e_{3} + e_{4} + e_{5} = 89, \\ & F(2,1,1,1) = \hbox{min} \{ F(3,0,0,0) + s + p_{2} ,F(3,1,1,1) + e_{2} \} = \hbox{min} \{ 72 + 7,54 + 17\} = 71 \\ & F(2,2,1,1) = \hbox{min} \{ F(3,1,1,1) + 2(s + p_{2} ),F(3,2,1,1) + e_{2} \} = \hbox{min} \{ 54 + 14,50 + 17\} = 67 \\ & F(2,3,1,1) = \hbox{min} \{ F(3,2,1,1) + 3(s + p_{2} ),F(3,3,1,1) + e_{2} \} = \hbox{min} \{ 50 + 21,57 + 17\} = 71 \\ & F\left( {2,4,1,1} \right) = { \hbox{min} }\left\{ {F\left( {3,3,1,1} \right) + 4\left( {s + p_{2} } \right),F\left( {3,4,1,1} \right) + e_{2} } \right\} = { \hbox{min} } \left\{ {57 + 28, \, \infty } \right\} = 85. \\ & j = 1,F\left( {1,0,0,0} \right) = e_{1} + e_{2} + e_{3} + e_{4} + e_{5} = 104, \\ & F\left( {1,1,1,1} \right) = { \hbox{min} }\left\{ {F\left( {2,0,0,0} \right) + s + p_{1} , F\left( {2, 1, 1, 1} \right) + e_{1} } \right\} = { \hbox{min} }\left\{ {89 + 5, \, 71 + 15} \right\} = 86, \\ & F\left( {1,2,1,1} \right) = { \hbox{min} } \left\{ {F\left( {2,1,1,1} \right) + 2\left( {s + p_{1} } \right), F\left( {2,2,1,1} \right) + e_{1} } \right\} = { \hbox{min} }\left\{ {71 + 10, \, 67 + 15} \right\} = 81, \\ & F\left( {1,3,1,1} \right) = { \hbox{min} } \left\{ {F\left( {2,2,1,1} \right) + 3\left( {s + p_{1} } \right),F\left( {2,3,1,1} \right) + e_{1} } \right\} = { \hbox{min} } \left\{ {67 + 15, \, 71 + 15} \right\} = 82, \\ & F(1,4,1,1) = \hbox{min} \{ F(2,3,1,1) + 4(s + p_{1} ),F(2,4,1,1) + e_{1} \} = \hbox{min} \{ 71 + 20,85 + 15\} = 91 \\ & F\left( {1,5,1,1} \right) = { \hbox{min} }\left\{ {F\left( {2,4,1,1} \right) + 5\left( {s + p_{1} } \right),F\left( {2,5,1,1} \right) + e_{1} } \right\} = { \hbox{min} }\left\{ {85 + 25, \infty } \right\} = 110. \\ \end{aligned} $$

Then, for the case h = 2, 3, 4, 5, we can similarly compute the function value. Finally, the optimal value is min {F(1, i, h, h)| 0 ≤ i ≤ n, 0 ≤ h ≤ i.} = F(1, 2, 1, 1) = 81. The corresponding optimal schedule can be found by backtracking. The set of accepted jobs is $\overline{R} = \{ J_{1} ,J_{4} \}$ and each job is processed in a single batch such that there are two batches: {J ₁} and {J ₄}; The set of rejected jobs is R = {J ₂, J ₃, J ₅}.