Two-stage three-machine assembly scheduling problem with sum-of-processing-times-based learning effect

Abstract

Researchers claim that the processing of most products can be formulated as a two-stage assembly scheduling model. The literature states that cumulative learning experience is neglected in solving two-stage assembly scheduling problems. The sum-of-processing-times-based learning effect means that the actual processing time of a job becomes shorter when it is scheduled later, which depends on the sum of processing time of the jobs already processed. Motivated by this observation, we investigate a novel two-stage assembly scheduling with three machines and sum-of-processing-times-based learning effect to minimize the makespan criterion, where two machines operate at the first stage and an assembly machine operates at the second stage. To solve this NP-hard problem, a branch-and-bound method incorporating with ten dominance properties and a lower bound procedure is first derived to obtain an optimal solution. Three heuristics based on Johnson’s rule with and without improvement are then applied separately to a genetic algorithm and a cloud theory-based simulated annealing algorithm, which are further modified with an interchange pairwise method for finding near-optimal solutions. Finally, the numerical results obtained using all proposed algorithms are reported and evaluated.

Appendix

The following is the detailed proof of Property 1. According to the definition in Eq. (1), we have

$$ C_{1i} \left( \sigma \right) = t_{1} + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } , $$

$$ C_{2i} \left( \sigma \right) = t_{2} + b_{i} \left( {1 - \frac{{{\text{PT}}_{b} }}{{T_{b} }}} \right)^{\beta } , $$

$$ C_{3i} \left( \sigma \right) = \hbox{max} \left\{ {C_{1i} \left( \sigma \right), C_{2i} \left( \sigma \right),t_{3} } \right\} + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta } , $$

$$ C_{1j} \left( \sigma \right) = t_{1} + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} + a_{i} }}{{T_{a} }}} \right)^{\beta } , $$

$$ C_{2j} \left( \sigma \right) = t_{2} + b_{i} \left( {1 - \frac{{{\text{PT}}_{b} }}{{T_{b} }}} \right)^{\beta } + b_{j} \left( {1 - \frac{{{\text{PT}}_{b} + b_{i} }}{{T_{b} }}} \right)^{\beta } , $$

$$ \begin{aligned} C_{3j} \left( \sigma \right) & = \hbox{max} \left\{ {C_{1j} \left( \sigma \right), C_{2j} \left( \sigma \right), { \hbox{max} }\left\{ {C_{1i} \left( \sigma \right),C_{2i} \left( \sigma \right),t_{3} } \right\} + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta }}\right\} + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} + c_{i} }}{{T_{c} }}} \right)^{\beta } \\ & = { \hbox{max} }\left\{ {t_{1} + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} + a_{i} }}{{T_{a} }}} \right)^{\beta } ,} \right. \\ & \quad \quad \left. {t_{2} + b_{i} \left( {1 - \frac{{{\text{PT}}_{b} }}{{T_{b} }}} \right)^{\beta } + b_{j} \left( {1 - \frac{{{\text{PT}}_{b} + b_{i} }}{{T_{b} }}} \right)^{\beta } , { \hbox{max} }\left\{ {C_{1i} \left( \sigma \right),C_{2i} \left( \sigma \right),t_{3} } \right\} + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta } } \right\} \\ & \quad + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} + c_{i} }}{{T_{c} }}} \right)^{\beta } , \\ \end{aligned} $$

$$ C_{1j} \left( {\sigma^{\prime}} \right) = t_{1} + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } , $$

$$ C_{2j} \left( {\sigma^{\prime}} \right) = t_{2} + b_{j} \left( {1 - \frac{{{\text{PT}}_{b} }}{{T_{b} }}} \right)^{\beta } , $$

$$ C_{3j} \left( {\sigma^{\prime}} \right) = \hbox{max} \left\{ {C_{1j} \left( {\sigma^{\prime}} \right), C_{2j} \left( {\sigma^{\prime}} \right),t_{3} } \right\} + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta } . $$

$$ C_{1i} \left( {\sigma^{\prime}} \right) = t_{1} + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} + a_{j} }}{{T_{a} }}} \right)^{\beta } , $$

$$ C_{2i} \left( {\sigma^{\prime}} \right) = t_{2} + b_{j} \left( {1 - \frac{{{\text{PT}}_{b} }}{{T_{b} }}} \right)^{\beta } + b_{i} \left( {1 - \frac{{{\text{PT}}_{b} + b_{j} }}{{T_{b} }}} \right)^{\beta } , $$

$$ \begin{aligned} C_{3i} \left( {\sigma^{\prime}} \right) & = \hbox{max} \left\{ {C_{1i} \left( {\sigma^{\prime}} \right), C_{2i} \left( {\sigma^{\prime}} \right), { \hbox{max} }\left\{ {C_{1j} \left( {\sigma^{\prime}} \right),C_{2j} \left( {\sigma^{\prime}} \right),t_{3} } \right\} + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta } } \right\} \\ & \quad + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} + c_{j} }}{{T_{c} }}} \right)^{\beta } {\text{ = max}}\left\{ {t_{1} + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} + a_{j} }}{{T_{a} }}} \right)^{\beta } ,} \right. \\ & \left. {\quad \quad t_{2} + b_{j} \left( {1 - \frac{{{\text{PT}}_{b} }}{{T_{b} }}} \right)^{\beta } + b_{i} \left( {1 - \frac{{{\text{PT}}_{b} + b_{j} }}{{T_{b} }}} \right)^{\beta } , { \hbox{max} }\left\{ {C_{1i} \left( {\sigma^{\prime}} \right),C_{2i} \left( {\sigma^{\prime}} \right),t_{3} } \right\} + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta } } \right\} \\ & \quad + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} + c_{j} }}{{T_{c} }}} \right)^{\beta } . \\ \end{aligned} $$

We will show that $C_{3j} \left( \sigma \right) \le C_{3i} \left( {\sigma^{\prime}} \right)$ or $C_{3i} \left( {\sigma^{\prime}} \right) - C_{3j} \left( \sigma \right) \ge 0$.

Since $t_{3} > t_{1} + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta }$,$t_{3} > t_{2} + b_{j} \left( {1 - \frac{{{\text{PT}}_{b} }}{{T_{b} }}} \right)^{\beta }$, and $a_{j} > a_{i}$, $b_{j} > b_{i} , then\,also t_{3} > t_{1} + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } and\, t_{3} > t_{2} + b_{i} \left( {1 - \frac{{{\text{PT}}_{b} }}{{T_{b} }}} \right)^{\beta } . It\,is\,easily\,seen\,that\,C_{3i} \left( {\sigma^{\prime}} \right) - C_{3j} \left( \sigma \right) can\,be\,simplified\,as\,below.$

$$ \begin{aligned} C_{3i} \left( {\sigma^{\prime}} \right) - C_{3j} \left( \sigma \right) & = \hbox{max} \left\{ {C_{1i} \left( {\sigma^{\prime}} \right), C_{2i} \left( {\sigma^{\prime}} \right), t_{3} + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta } } \right\} + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} + c_{j} }}{{T_{c} }}} \right)^{\beta } \\ & \quad - \hbox{max} \left\{ {C_{1j} \left( \sigma \right), C_{2j} \left( \sigma \right), t_{3} + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta } } \right\} + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} + c_{i} }}{{T_{c} }}} \right)^{\beta } . \\ \end{aligned} $$

(2)

The given condition that

$$ \begin{aligned} C_{1i} \left( {\sigma^{\prime}} \right) & = t_{1} + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} + a_{j} }}{{T_{a} }}} \right)^{\beta } \\ & > \hbox{max} \left\{ { C_{2i} \left( {\sigma^{\prime}} \right), t_{3} + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta } } \right\} \\ & \quad \quad {\text{implies}}\;{\text{that}}\,C_{3i} \left( {\sigma^{\prime}} \right) = C_{1i} \left( {\sigma^{\prime}} \right) + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} + c_{j} }}{{T_{c} }}} \right)^{\beta } . \\ \end{aligned} $$

(3)

The given condition that

$$ \begin{aligned} C_{1j} \left( \sigma \right) & = t_{1} + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} + a_{i} }}{{T_{a} }}} \right)^{\beta } \\ & > { \hbox{max} }\left\{ {C_{2j} \left( \sigma \right), t_{3} + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} }}{{T_{c} }}} \right)^{\beta } } \right\} \\ & \quad \quad {\text{implies}}\;{\text{that}}C_{3j} \left( \sigma \right) = C_{1j} \left( \sigma \right) + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} + c_{i} }}{{T_{c} }}} \right)^{\beta }. \\ \end{aligned} $$

(4)

Now, from Eqs. (2), (3), and (4), $\begin{aligned} C_{3i} \left( {\sigma^{\prime}} \right) - C_{3j} \left( \sigma \right) & = \left[ {C_{1i} \left( {\sigma^{\prime}} \right) + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} + c_{j} }}{{T_{c} }}} \right)^{\beta } } \right] - \left[ { C_{1j} \left( \sigma \right) + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} + c_{i} }}{{T_{c} }}} \right)^{\beta } } \right] \\ & = \left[ {t_{1} + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} + a_{j} }}{{T_{a} }}} \right)^{\beta } + c_{i} \left( {1 - \frac{{{\text{PT}}_{c} + c_{j} }}{{T_{c} }}} \right)^{\beta } } \right] \\ & \quad - \left[ {t_{1} + a_{i} \left( {1 - \frac{{{\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } + a_{j} \left( {1 - \frac{{{\text{PT}}_{a} + a_{i} }}{{T_{a} }}} \right)^{\beta } + c_{j} \left( {1 - \frac{{{\text{PT}}_{c} + c_{i} }}{{T_{c} }}} \right)^{\beta } } \right]. \\ \end{aligned}$

After a simple algebraic,

$$ \begin{aligned} C_{3i} \left( {\sigma^{\prime}} \right) - C_{3j} \left( \sigma \right) & = \left\{ {a_{j} \left[ {\left( {\frac{{T_{a} - {\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } - \left( {\frac{{T_{a} - {\text{PT}}_{a} - a_{i} }}{{T_{a} }}} \right)} \right]^{\beta } } \right. \\ & \quad - \left. {a_{j} \left[ {\left( {\frac{{T_{a} - {\text{PT}}_{a} }}{{T_{a} }}} \right)^{\beta } - \left( {\frac{{T_{a} - {\text{PT}}_{a} - a_{j} }}{{T_{a} }}} \right)^{\beta } } \right]} \right\} \\ & \quad + \left[ { c_{i} \left( {\frac{{T_{c} - {\text{PT}}_{c} - c_{j} }}{{T_{c} }}} \right)^{\beta } - c_{j} \left( {\frac{{T_{c} - {\text{PT}}_{c} - c_{i} }}{{T_{c} }}} \right)^{\beta } } \right]. \\ \end{aligned} $$

(5)

The second term of Eq. (5) is greater than 0, because $c_{i} > c_{j}$ and β > 0, whereas the first term greater than or equal to 0 can easily be obtained by the argument of the proof of Theorem 1 in Koulamas and Kyparisis (2007a, b). Therefore, $C_{3i} \left( {\sigma^{\prime}} \right) - C_{3j} \left( \sigma \right) \ge 0$, as required.