Concurrent design of cell formation and scheduling with consideration of duplicate machines and alternative process routings

Abstract

Concurrent design of cell formation and scheduling is an effective method for better implementing cellular manufacturing. To address the integrated cell formation and scheduling problem, a nonlinear mixed integer programming mathematical model is developed in this paper. This newly proposed model features the simultaneous consideration of many design attributes, such as duplicate machines, alternative process routings, reentrant parts and variable cell number. Several linearization techniques are proposed to transform it into a mixed integer linear programming formulation. An improved genetic algorithm (IGA) is developed to solve large-scale problems efficiently. To remove redundancy between two chromosomes, a cell renumbering procedure is applied in IGA. An illustrative example problem is solved and the results show that the integration of cell formation and scheduling can remarkably reduce the flowtime of cellular manufacturing systems. A set of thirteen test problems with various scale is used to further evaluate the performance of IGA. Comparison of the results obtained by IGA with those obtained by Lingo and CPLEX reveals the better effectiveness and efficiency of IGA.

Appendix: Proof of two Propositions

Proposition 1

The absolute term in Constraint (7) can be linearized by the transformation $\left| \sum \limits _{m\in \mathbf{M}_{k+1,p} } \sum \limits _{j\in \mathbf{MS}_m } {X_{k+1,pmjc} } -\sum \limits _{m\in \mathbf{M}_{kp}} \sum \limits _{j\in \mathbf{MS}_m } {X_{kpmjc}} \right| = \alpha _{kpc}^+ +\alpha _{kpc}^-$, under the following set of constraints:

$$\begin{aligned}&\sum _{m\in \mathbf{M}_{k+1,p} } {\sum _{j\in \mathbf{MS}_m } {X_{k+1,pmjc}}} -\sum _{m\in \mathbf{M}_{kp} } {\sum _{j\in \mathbf{MS}_m } {X_{kpmjc} } } \nonumber \\&\qquad =\alpha _{kpc}^+ -\alpha _{kpc}^- \quad \forall p\in \mathbf{P},k\in \mathbf{K}_p \backslash \left\{ {K_p } \right\} \end{aligned}$$

(23)

$$\begin{aligned}&\alpha _{kpc}^+ ,\alpha _{kpc}^- \ge 0 \end{aligned}$$

(24)

Proof

Let $\beta _{k+1,p} =\sum \limits _{c\in \mathbf{C}} \sum \limits _{m\in \mathbf{M}_{k+1,p}} \sum \limits _{j\in \mathbf{MS}_m } {T_{k+1,pm} \cdot X_{k\hbox {+}1,pmjc}}$. According to (9), the following formula can be obtained.

$$\begin{aligned} {F_{K_p ,p} }&{\ge F_{K_p -1,p} +\beta _{K_p ,p} +t_{p,K_p -1,K_p}} \nonumber \\&\ge F_{K_p -2,p} +\beta _{K_p -1,p} +t_{p,K_p -2,K_p -1} \nonumber \\&\qquad +\beta _{K_p ,p} +t_{p,K_p -1,K_p } \nonumber \\&\cdots \nonumber \\&{\ge F_{k,p} + {\sum }_{l=k+1}^{K_p } {\beta _{lp} } +{\sum }_{l=k}^{K_p -1} {t_{p,l,l+1}}} \end{aligned}$$

(25)

Based on (7) and Proposition 1, substituting $t_{p,k,k+1} =\frac{TE_p }{2}\cdot \sum \limits _{c\in \mathbf{C}} {\left( {\alpha _{kpc}^+ +\alpha _{kpc}^- } \right) }$ into the formula (25),

$$\begin{aligned} F_{K_p ,p} \ge F_{k,p} +\sum _{l=k+1}^{K_p } {\beta _{lp} } +\frac{TE_p }{2}\sum _{l=k}^{K_p -1} {\sum _{c\in \mathbf{C}} {\left( {\alpha _{kpc}^+ +\alpha _{kpc}^- } \right) } } \end{aligned}$$

(26)

Consider three different cases:

• (i) $\sum \limits _{m\in \mathbf{M}_{k+1,p} } {\sum \limits _{j\in \mathbf{MS}_m } {X_{k+1,pmjc} } } >\sum \limits _{m\in \mathbf{M}_{kp} } {\sum \limits _{j\in \mathbf{MS}_m } {X_{kpmjc} } }$. According to (13), $\alpha _{kpc}^+ -\alpha _{kpc}^- >0$. Since this is a minimization problem, $\alpha _{kpc}^- =0$ and $\alpha _{kpc}^+ =\sum \limits _{m\in \mathbf{M}_{k+1,p}} {\sum \limits _{j\in \mathbf{MS}_m } {X_{k+1,pmjc} } } -\sum \limits _{m\in \mathbf{M}_{kp} } {\sum \limits _{j\in \mathbf{MS}_m } {X_{kpmjc} } }$ will make the right side of the formula (26) be minimum so that the optimal solution can be hold.

• (ii) $\sum \limits _{m\in \mathbf{M}_{k+1,p} } {\sum \limits _{j\in \mathbf{MS}_m } {X_{k+1,pmjc} } } <\sum \limits _{m\in \mathbf{M}_{kp} } {\sum \limits _{j\in \mathbf{MS}_m } {X_{kpmjc} } }$. According to (13), $\alpha _{kpc}^+ -\alpha _{kpc}^- <0$. $\alpha _{kpc}^+ =0$ and $\alpha _{kpc}^- =\sum \limits _{m\in \mathbf{M}_{kp} } {\sum \limits _{j\in \mathbf{MS}_m } {X_{kpmjc} } } -\sum \limits _{m\in \mathbf{M}_{k+1,p} } {\sum \limits _{j\in \mathbf{MS}_m } {X_{k+1,pmjc} } }$ will make the right side of the formula (26) be minimum so that the optimal solution can be hold.

• (iii) $\sum \limits _{m\in \mathbf{M}_{k+1,p} } {\sum \limits _{j\in \mathbf{MS}_m } {X_{k+1,pmjc} } } =\sum \limits _{m\in \mathbf{M}_{kp} } {\sum \limits _{j\in \mathbf{MS}_m} {X_{kpmjc}}}$. According to (13), $\alpha _{kpc}^+ -\alpha _{kpc}^- =0$. $\alpha _{kpc}^+ =0$ and $\alpha _{kpc}^- =0$ will make the right side of the formula (26) be minimum so that the optimal solution can be hold. $\square$

Proposition 2

Constraint (10) can be replaced by the following constraints so that it can be linearized.

$$\begin{aligned} Z_{kpk^{\prime }p^{\prime }mj} +Z_{k^{\prime }p^{\prime }kpmj}\ge & {} \sum _{c\in \mathbf{C}} {\left( {X_{kpmjc} +X_{k^{\prime }p^{\prime }mjc} } \right) } -1 \end{aligned}$$

(27)

$$\begin{aligned} Z_{kpk^{\prime }p^{\prime }mj} +Z_{k^{\prime }p^{\prime }kpmj}\le & {} \sum _{c\in \mathbf{C}} {X_{kpmjc}} \end{aligned}$$

(28)

$$\begin{aligned} Z_{kpk^{\prime }p^{\prime }mj} +Z_{k^{\prime }p^{\prime }kpmj}\le & {} \sum _{c\in \mathbf{C}} {X_{k^{\prime }p^{\prime }mjc} } \quad \end{aligned}$$

(29)

Proof

Since an operation of a certain part cannot be processed in two cells, for every $c_1 ,c_2 \in \mathbf{C}\left( {c_1 \ne c_2 } \right)$, $X_{kpmjc_1 } \cdot X_{kpmjc_2}$ cannot be equal to 1. Consider two different cases:

• (i) $\sum \limits _{c\in \mathbf{C}} {\left( {X_{kpmjc} \cdot X_{k^{\prime }p^{\prime }mjc} } \right) } =0$, which can be resulted from one of the following three sub-cases.

  • (a) $\exists c_1 ,X_{kpmjc_1 } =1$ and $X_{k^{\prime }p^{\prime }mjc_1 } =0$.

    It means the jth copy of machine type m is assigned to cell $c_{1} (Y_{mjc_1 } =1)$ and kth operation of part p is processed in cell $c_{1}$. Since a machine cannot be assigned to more than one cell,

    $$\begin{aligned} Y_{mjc} =0\quad \forall c\in \mathbf{C}\backslash \left\{ {c_1 } \right\} \end{aligned}$$

    Based on Constraint (4),

    $$\begin{aligned} X_{k^{\prime }p^{\prime }mjc} = 0\quad \forall c\in \mathbf{C}\backslash \left\{ {c_1 } \right\} \end{aligned}$$

    Since an operation of a certain part cannot be processed in two cells,

    $$\begin{aligned} X_{kpmjc} = 0\quad \forall c\in \mathbf{C}\backslash \left\{ {c_1 } \right\} \end{aligned}$$

    Therefore,

    $$\begin{aligned} \sum _{c\in \mathbf{C}} {X_{k^{\prime }p^{\prime }mjc} }= & {} \sum _{c\in \mathbf{C}\backslash \left\{ {c_1 } \right\} } {X_{k^{\prime }p^{\prime }mjc} } +X_{k^{\prime }p^{\prime }mjc_1 } =0\\ \sum _{c\in \mathbf{C}} {X_{kpmjc} }= & {} \sum _{c\in \mathbf{C}\backslash \left\{ {c_1 } \right\} } {X_{kpmjc} } +X_{kpmjc_1 } =1 \end{aligned}$$

    Under Constraint (29), $Z_{kpk^{\prime }p^{\prime }mj} =0$ and $Z_{k^{\prime }p^{\prime }kpmj} =0$.

  • (b) $\exists c_1 ,X_{k^{\prime }p^{\prime }mjc_1 } =1$ and $X_{kpmjc_1 } =0$, which is similar to the last sub-case.

  • (c) $X_{kpmjc} =0$ and $X_{k^{\prime }p^{\prime }mjc} =0$ for every $c\in \mathbf{C}$. Under Constraints (29), $Z_{kpk^{\prime }p^{\prime }mj} =0$ and $Z_{k^{\prime }p^{\prime }kpmj} =0$.

  To sum up, when $\sum \limits _{c\in \mathbf{C}} {\left( {X_{kpmjc} \cdot X_{k^{\prime }p^{\prime }mjc} } \right) } =0$,

  $$\begin{aligned} Z_{kpk^{\prime }p^{\prime }mj} +Z_{k^{\prime }p^{\prime }kpmj} =\sum _{c\in \mathbf{C}} {\left( {X_{kpmjc} \cdot X_{k^{\prime }p^{\prime }mjc} } \right) } =0 \end{aligned}$$

• (ii) $\sum \limits _{c\in \mathbf{C}} {\left( {X_{kpmjc} \cdot X_{k^{\prime }p^{\prime }mjc} } \right) } =1$.

  Such a situation arises only when $\exists c_1 ,X_{kpmjc_1 } =1$ and $X_{k^{\prime }p^{\prime }mjc_1 } =1$. Since an operation of a certain part cannot be processed in two cells,

  $$\begin{aligned} \sum _{c\in \mathbf{C}} {X_{k^{\prime }p^{\prime }mjc} }= & {} \sum _{c\in \mathbf{C}\backslash \left\{ {c_1 } \right\} } {X_{k^{\prime }p^{\prime }mjc} } +X_{k^{\prime }p^{\prime }mjc_1 } =0+1=0 \\ \sum _{c\in \mathbf{C}} {X_{kpmjc} }= & {} \sum _{c\in \mathbf{C}\backslash \left\{ {c_1 } \right\} } {X_{kpmjc} } +X_{kpmjc_1 } =0+1=1 \end{aligned}$$

  Under Constraints (27)–(29),

  $$\begin{aligned} Z_{kpk^{\prime }p^{\prime }mj} +Z_{k^{\prime }p^{\prime }kpmj} =1 \end{aligned}$$

  Therefore, when $\sum \limits _{c\in \mathbf{C}} {\left( {X_{kpmjc} \cdot X_{k^{\prime }p^{\prime }mjc} } \right) } =1$,

  $$\begin{aligned} Z_{kpk^{\prime }p^{\prime }mj} +Z_{k^{\prime }p^{\prime }kpmj} =\sum _{c\in \mathbf{C}} {\left( {X_{kpmjc} \cdot X_{k^{\prime }p^{\prime }mjc} } \right) } =1 \end{aligned}$$