A mathematical modelling approach for managing sudden disturbances in a three-tier manufacturing supply chain

Abstract

This paper aims to develop a recovery planning approach in a three-tier manufacturing supply chain, which has a single supplier, manufacturer, and retailer under an imperfect production environment, in which we consider three types of sudden disturbances: demand fluctuation, and disruptions to production and raw material supply, which are not known in advance. Firstly, a mathematical model is developed for generating an ideal plan under imperfect production for a finite planning horizon while maximizing total profit, and then we re-formulate the model to generate the recovery plan after happening of each sudden disturbance. Considering the high commercial cost and computational intensity and complexity of this problem, we propose an efficient heuristic, to obtain a recovery plan, for each disturbance type, for a finite future period, after the occurrence of a disturbance. The heuristic solutions are compared with a standard solution technique for a considerable number of random test instances, which demonstrates the trustworthy performance of the developed heuristics. We also develop another heuristic for managing the combined effects of multiple sudden disturbances in a period. Finally, a simulation approach is proposed to investigate the effects of different types of disturbance events generated randomly. We present several numerical examples and random experiments to explicate the benefits of our developed approaches. Results reveal that in the event of sudden disturbances, the proposed mathematical and heuristic approaches are capable of generating recovery plans accurately and consistently.

Appendices

Appendices

1.1 Appendix 1: Modeling for sudden production disruption

Appendix 1 formulates the mathematical model for generating the recovery plan after a sudden production disruption. The model considers the costs of production, delivery, holding, raw materials, backorder, and lost sales, and determines the revenue from the selling price. Finally, the model is formulated as a constrained mathematical programming problem in which the total profit to be maximized is subject to constraints from capacity, demand, delivery, and inventory.

$$ {\text{Total}}\,{\text{production}}\,{\text{cost}} = \frac{{C_{p} }}{r}\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(54)

$$ {\text{Total}}\,{\text{rejection}}\,{\text{cost}} = C_{R} \left( {\frac{1}{r} - 1} \right)\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(55)

$$ {\text{Total}}\,{\text{inspection}}\,{\text{cost}} = \frac{{C_{I} C_{p} }}{r}\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(56)

$$ {\text{Cost}}\,{\text{of}}\,{\text{interest}}\,{\text{and}}\,{\text{depreciation}} = naA^{ - b} r^{c} $$

(57)

$$ {\text{Total}}\,{\text{raw}}\,{\text{material}}\,{\text{cost}} = \mathop \sum \limits_{i = 1}^{n} C_{r} Z_{i} = \frac{{NC_{r} }}{r}\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(58)

$$ {\text{Raw}}\,{\text{material}}\,{\text{holding}}\,{\text{cost}} = \frac{1}{2r}H_{1} N\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(59)

$$ {\text{Total}}\,{\text{delivery}}\,{\text{cost}} = \mathop \sum \limits_{i = 1}^{n} C_{d} Y_{i} $$

(60)

$$ {\text{Total}}\,{\text{ending}}\,{\text{inventory}}\,{\text{holding}}\,{\text{cost}} = H_{2} \mathop \sum \limits_{i = 1}^{n} e_{i} $$

(61)

$$ {\text{Backorder}}\,{\text{cost}} = B\mathop \sum \limits_{i = 2}^{n} \left( {i - 1} \right)(X_{i} - AP_{i} ) $$

(62)

$$ {\text{Lost}}\,{\text{sales}}\,{\text{cost}} = L\left( {\mathop \sum \limits_{i = 1}^{n} AP_{i} - \mathop \sum \limits_{i = 1}^{n} X_{i} } \right) $$

(63)

$$ {\text{Total}}\,{\text{revenue}} = S\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(64)

Final mathematical model

Total profit = total revenue − total costs, is the objective function and to be maximized, which is obtained using Eqs. (54)–(64) and subject to constraints (65)–(73).

$$ e_{i} \ge E_{i} ;\quad \forall i\,\left[ {{\text{Constraint}}\,{\text{of}}\,{\text{ending}}\,{\text{inventory}}} \right] $$

(65)

$$ b_{i} \ge B_{i} ;\quad \forall i\left[ {{\text{Constraint}}\,{\text{of}}\,{\text{beginning}}\,{\text{inventory}}} \right] $$

(66)

$$ X_{1} \le r\left( {P - T_{dp} *P} \right)\,\left[ {{\text{Limitation}}\,{\text{of}}\,{\text{production}}\,{\text{quantity}}\,{\text{in}}\,{\text{first}}\,{\text{period}}} \right] $$

(67)

$$ X_{i} \le rP;\quad \forall i \ne 1\left[ {{\text{Limitation}}\,{\text{of}}\,{\text{production}}\,{\text{quantity}}\,{\text{in}}\,{\text{each}}\,{\text{period}}} \right] $$

(68)

$$ X_{i} \ge AP_{i} ;\quad \forall i \ne 1\,\left[ {{\text{Constraint}}\,{\text{for}}\,{\text{production}}\,{\text{in}}\,{\text{recovery}}\,{\text{plan}}} \right] $$

(69)

$$ \mathop \sum \limits_{i = 1}^{n} X_{i} \le \mathop \sum \limits_{i = 1}^{n} D_{i} + b_{n + 1} - b_{1} \,\left[ {{\text{Limitation}}\,{\text{of}}\,{\text{total}}\,{\text{production}}\,{\text{quantity}}} \right] $$

(70)

$$ \mathop \sum \limits_{i = 1}^{n} AP_{i} - \mathop \sum \limits_{i = 1}^{n} X_{i} \ge 0\,\left[ {{\text{Constraint}}\,{\text{of}}\,{\text{lost}}\,{\text{sales}}\,{\text{quantity}}} \right] $$

(71)

$$ \mathop \sum \limits_{i = 1}^{n} Y_{i} \le \mathop \sum \limits_{i = 1}^{n} D_{i} \,\left[ {{\text{Limitation}}\,{\text{of}}\,{\text{total}}\,{\text{delivery}}} \right] $$

(72)

$$ X_{i} ,Y_{i} ,Z_{i} \ge 0\quad {\text{and}}\quad {\text{integer}};\quad \forall i\left[ {{\text{Non-negativity}}\,{\text{constraint}}} \right] $$

(73)

1.2 Appendix 2: Modeling for sudden supply disruption

In this appendix, a constrained mathematical programing model is formulated for generating the recovery plan after a sudden supply disruption in which the total profit is maximized subject to the constraints from capacity, demand, delivery, and inventory.

$$ {\text{Total}}\,{\text{production}}\,{\text{cost}} = \frac{{C_{p} }}{r}\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(74)

$$ {\text{Total}}\,{\text{rejection}}\,{\text{cost}} = C_{R} \left( {\frac{1}{r} - 1} \right)\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(75)

$$ {\text{Total}}\,{\text{inspection}}\,{\text{cost}} = \frac{{C_{I} C_{p} }}{r}\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(76)

$$ {\text{Cost}}\,{\text{of}}\,{\text{interest}}\,{\text{and}}\,{\text{depreciation}} = naA^{ - b} r^{c} $$

(77)

$$ {\text{Total}}\,{\text{raw}}\,{\text{material}}\,{\text{cost}} = \mathop \sum \limits_{i = 1}^{n} C_{r} Z_{i} = \frac{{NC_{r} }}{r}\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(78)

$$ {\text{Raw}}\,{\text{material}}\,{\text{holding}}\,{\text{cost}} = \frac{1}{2r}H_{1} N\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(79)

$$ {\text{Total}}\,{\text{delivery}}\,{\text{cost}} = \mathop \sum \limits_{i = 1}^{n} C_{d} Y_{i} $$

(80)

$$ {\text{Total}}\,{\text{ending}}\,{\text{inventory}}\,{\text{holding}}\,{\text{cost}} = H_{2} \mathop \sum \limits_{i = 1}^{n} e_{i} $$

(81)

$$ {\text{Backorder}}\,{\text{cost}} = B\mathop \sum \limits_{i = 2}^{n} \left( {i - 1} \right)(X_{i} - AP_{i} ) $$

(82)

$$ {\text{Lost}}\,{\text{sales}}\,{\text{cost}} = L\left( {\mathop \sum \limits_{i = 1}^{n} AP_{i} - \mathop \sum \limits_{i = 1}^{n} X_{i} } \right) $$

(83)

$$ {\text{Total}}\,{\text{revenue}} = S\mathop \sum \limits_{i = 1}^{n} X_{i} $$

(84)

Final mathematical model

Total profit = total revenue − total costs, is the objective function and to be maximized, which is obtained using Eqs. (74)–(84) and subject to constraints (85)–(93).

$$ e_{i} \ge E_{i} ;\quad \forall i\,\left[ {{\text{Constraint}}\,{\text{of}}\,{\text{ending}}\,{\text{inventory}}} \right] $$

(85)

$$ b_{i} \ge B_{i} ;\quad \forall i\left[ {{\text{Constraint}}\,{\text{of}}\,{\text{beginning}}\,{\text{inventory}}} \right] $$

(86)

$$ X_{1} \le r\left( {P - T_{ds} *P} \right)\left[ {{\text{Limitation}}\,{\text{of}}\,{\text{production}}\,{\text{quantity}}\,{\text{in}}\,{\text{first}}\,{\text{period}}} \right] $$

(87)

$$ X_{i} \le rP;\quad \forall i \ne 1\,\left[ {{\text{Limitation}}\,{\text{of}}\,{\text{production}}\,{\text{quantity}}\,{\text{in}}\,{\text{each}}\,{\text{period}}} \right] $$

(88)

$$ X_{i} \ge AP_{i} ; \quad \forall i \ne 1\,\left[ {{\text{Constraint}}\,{\text{for}}\,{\text{production}}\,{\text{in}}\,{\text{recovery}}\,{\text{plan}}} \right] $$

(89)

$$ \mathop \sum \limits_{i = 1}^{n} X_{i} \le \mathop \sum \limits_{i = 1}^{n} D_{i} + b_{n + 1} - b_{1} \,\left[ {{\text{Limitation}}\,{\text{of}}\,{\text{total}}\,{\text{production}}\,{\text{quantity}}} \right] $$

(90)

$$ \mathop \sum \limits_{i = 1}^{n} AP_{i} - \mathop \sum \limits_{i = 1}^{n} X_{i} \ge 0\,\left[ {{\text{Constraint}}\,{\text{of}}\,{\text{lost}}\,{\text{sales}}\,{\text{quantity}}} \right] $$

(91)

$$ \mathop \sum \limits_{i = 1}^{n} Y_{i} \le \mathop \sum \limits_{i = 1}^{n} D_{i} \,\left[ {{\text{Limitation}}\,{\text{of}}\,{\text{total}}\,{\text{delivery}}} \right] $$

(92)

$$ X_{i} ,Y_{i} ,Z_{i} \ge 0\quad {\text{and}}\quad {\text{integer}};\quad \forall i\left[ {{\text{Non-negativity}}\,{\text{constraint}}} \right] $$

(93)

1.3 Appendix 3: Heuristic 2: generating recovery plan for a sudden production disruption

This appendix shows heuristic steps to generate recovery plan for a sudden production disruption.

1.4 Appendix 4: Heuristic 3: generating recovery plan for a sudden supply disruption

This appendix shows heuristic steps to generate recovery plan for a sudden supply disruption.

1.5 Appendix 5: Heuristic 4: for considering combined effect multiple disturbances

A demand fluctuation occurs at the retailer end, a supply disruption at the supplier end, and a production disruption at the manufacturing plant. Multiple disturbances can happen together in a period, in which case their effects must be considered when formulating a recovery plan. We develop a heuristic to deal with multiple disturbances and use random data to develop multiple disturbance scenarios. The steps in Heuristic 4 for managing multiple disturbances in a period are presented below.