heuristic solution approach for the capacitated lot …...researches focus on improving the mip...

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Heuristic solution approach for the capacitated lot-sizing and scheduling

problem with sequence-dependent setup times and retunrs

Auteur : Hoang Ngoc Tuan,

Promoteur(s) : Arda, Yasemin

Faculté : HEC-Ecole de gestion de l'ULg

Diplôme : Master en ingénieur de gestion, à finalité spécialisée en Supply Chain Management and

Business Analytics

Année académique : 2015-2016

URI/URL : http://hdl.handle.net/2268.2/1888

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HEURISTIC SOLUTION APPROACH FOR THE

CAPACITATED LOT-SIZING AND SCHEDULING

PROBLEM WITH SEQUENCE-DEPENDENT SETUP

TIMES AND RETURNS

Jury:

Dissertation by

Promoter:

Ngoc Tuan HOANG

Yasemin ARDA

For a Master Degree

Readers:

in Business Engineering

Yves CRAMA

Supply Chain Management & Business

Analytics

Mahmood REZAEI

SADRABADI

Academic year 2015/2016

ACKNOWLEDGMENT

I would like to convey my heartfelt gratitude and sincere appreciation to all people who

have helped and inspired me during my process. This thesis would not have been possible

without those supports from many people.

First of all, I would like to thank to my promoter, Prof. Yasemin Arda, for her endless

guidance during my research. I would like to express my gratitude to other professors and

researchers in the Supply Chain Management specialization. This thesis would not have

been possible without their extraordinary support.

My warmest thanks also go to Ms. Marianna Sneaker, Mr. Gunther Vranken and all

staffs in HEC International Cooperation Department for supporting me in two years of my

master program.

Last but not least, I would like to send my great to my family and friends who have

been standing by my side, supporting me in the study process.

1

TABLE OF CONTENTS

ABTRACT ...................................................................................................................................... 3

CHAPTER 1: INTRODUCTION AND LITERATURE REVIEW .......................................... 5

1.1. The Capacitated Lot-sizing and Scheduling Problem ........................................................... 5

1.2. The Capacitated Lot-sizing and Scheduling Problem with Sequence-Dependent Setup

Costs and Time ............................................................................................................................. 7

1.3. Remanufacturing .................................................................................................................... 8

1.4. Articles directly relating to the thesis .................................................................................... 9

CHAPTER 2: PROBLEM DEFINITION AND MATHEMATICAL MODEL .................... 11

2.1. Mathematical model 1: Lot-Sizing and Scheduling Problem with Sequence-Dependent

Setup Times ................................................................................................................................ 11

2.2. Mathematical model 2: Lot-Sizing and Scheduling Problem with Sequence-Dependent

Setup Times and Returns ............................................................................................................ 15

2.3. Test the correctness of mathematical model 2 ..................................................................... 18

2.4. Development of lower bound .............................................................................................. 21

CHAPTER 3: DEVELOPMENT OF HEURISTICS ................................................................ 23

3.1. Heuristic 1: Fix and Relax ................................................................................................... 23

3.1. Heuristic 2 ............................................................................................................................ 25

CHAPTER 4: NUMERICAL EXPERIMENTS ........................................................................ 31

1.1. Development of instances ................................................................................................ 31

1.2. Computational results ...................................................................................................... 34

Optimally solving the problem .............................................................................................................34

The lower bound ...................................................................................................................................35

Fix and relax heuristic ..........................................................................................................................35

Heuristic 2 ............................................................................................................................................36

2

The effects of demand patterns .............................................................................................................36

The effect of capacity ...........................................................................................................................37

Comparison between the two heuristics ...............................................................................................37

CHAPTER 5: CONCLUDING REMARKS .............................................................................. 39

REFERENCES ............................................................................................................................. 41

APPENDIX ................................................................................................................................... 43

Computational results ................................................................................................................. 43

Instruction to run the code .......................................................................................................... 53

3

ABTRACT

Lot-sizing and Scheduling Problem has always been extremely important with operating

planner. This research seeks to deal with the Capacitated Lot-sizing and Scheduling Problem with

Sequence-Dependent Setup Times and Returns by heuristic approach. The problem formulization

is inspired by the real working environment of a bottling plant which produces different types of

drink. In the research, two heuristics will be presented, tested and compared to each other. The

first heuristic follows the fix and relax heuristic – one of the efficient heuristics to deal with MIP

problem. The later heuristic tried to break the original problem into two sub-problems in an

attempt to reduce the number of integer variables. In the second heuristic, the problem is

reformulated to a basic Capacitated Lot-sizing and Scheduling Problem and a Travelling

Salesman Problem. Both heuristics will be coded and tested with different types of instances.

Overall, both of them show promising results which could be the motivation to keep following

these heuristic directions in the future.

Key words: Fix and Relax heuristic, Lot-sizing and Scheduling, Travelling Salesman,

Sequence-Dependent Setup.

5

CHAPTER 1: INTRODUCTION AND LITERATURE REVIEW

1.1. The Capacitated Lot-sizing and Scheduling Problem

―Smooth and cost-efficient running of a factory often depends on its manager’s ability to

select appropriate lot-sizes and production schedules‖ (Gupta & Magnusson, 2005). For this

reason, the Lot-sizing and Scheduling Problem has been intensively studied in a large number of

articles. Since the seminal paper of Wagner and Whitin (1958), a considerable amount of

investigation has been done in order develop and deal with various important features of this

problem.

The lot-sizing problem is to find production orders or lots in order to satisfy customer demand

while minimizing production costs, inventory holding costs and setup costs at the same time.

Most of researches consider a finite planning horizon which is divided into discrete time periods.

―Lot-sizing problem is classified into either big bucket or small bucket‖ (Eppen & Martin, 1987).

In small bucket (also called ―small time window‖ in some researches), the planning horizon is

broken into small time periods. At most one type of products can be produced each time period.

Thus, when a setup is performed in the beginning of a period, the system has to remain at this

setup status during the whole period. The small bucket is sometimes called ―The Discrete Lot-

sizing and Scheduling Problem‖, and has been studied extensively in literature. On the other

hand, the Capacitated Lot-sizing and Scheduling Problem (CLSP) is known as the big bucket

(sometimes called ―large time window‖). This problem considers larger time periods which allow

multi products manufacturing. In other words, more than one product can be produced in the

same period in the big bucket problem. Detail review of big bucket and small bucket can be

found in the article Lot-sizing and scheduling – Survey and extensions (Drexl & Kimms, 1997).

In this research, we will focus on the Capacitated Lot-sizing and Scheduling Problem.

Generally, the most basic model of the CLSP can be formulated as the following Mixed Integer

Programming (MIP) model.

We have a single machine which has to produce N products in order to satisfy deterministic

demand in T time periods. Let the products indexed by i = 1,2,…,N and time periods by t =

1,2,…,T. The following notations are used for the parameters:

6

Deterministic demand (in unit) for product i in period t

Available capacity (in time unit) in period t

Capacity needed to produce one unit of product i in period t

Unit inventory holding cost of product i in period t

Setup cost for product i in period t

Decision variables:

Production quantity for product i in period t

Inventory level of product i at the end of period t

{

Objective function:

∑∑

Subject to:

∑

{ }

7

The objective function minimizes the sum of inventory holding costs and setup costs.

Constraint 1 ensures the inventory balance. Constraint 2 makes sure that product i is produced if

and only if the machine is setup for that. Constraint 3 is the capacity constraint, makes sure that

we do not use more than the available capacity. Constraint 4 & 5 express the types of variables.

Exact methods to solve the CLSP have been intensively studied. A great number of

researches have been carried out in order to deal with specific problem such as Belvaux and

Wolsey (2001),Belvaux and Wolsey (2000) or Barany, Van Roy, and Wolsey (1984). These

researches focus on improving the MIP formulation in order to solve the problem with

commercial solvers like CPLEX or XPRESS-MP. Generally, using these solvers, the problem is

often solved by the standard Branch and Bound procedure.

1.2. The Capacitated Lot-sizing and Scheduling Problem with Sequence-Dependent

Setup Costs and Times

One weakness of CLSP is that it does not sequence products within a period. Instead, setup

occurs whenever a product is produced. Thus, even if the last product of a period is the first

product of the successive period, setup occurs when there is indeed no change in the machine

configuration. The lack of sequence creates a gap between calculated schedule and real working

environment, which sometimes makes the results of the CLSP become infeasible in real world.

For manufacturing environments in which reconfiguring the system takes a considerable amount

of time and money, the need to determine the sequence within a period is even more important.

Therefore, it’s necessary to include the Sequence-Dependent Setup (SDS) into consideration.

In the CLSP with SDS, setup does not occur whenever a product is produced. Indeed, a

changeover occurs when the system switches from a product to another. The costs of a

changeover depend on the pair of products. With SDS, the problem is able to result a sequence of

products for each period. However, SDS also makes the problem become more complex and

difficult to solve.

―Despite CLPS being intensively studied, modeling setup carryover in CLSP has not received

much attention due to model complexity and computational difficulty‖ (Sox & Gao, 1999).

Different researches have tried to reformulate the structure of setup in order to reduce the

complexity of the problem. Gopalakrishnan, Miller, and Schmidt (1995) developed a modeling

8

framework for the CLSP with sequence and product independent setup times and setup carryover.

The authors went to the conclusion that the model complexity can be reduced by looking for

alternative method to model setup changeover. Gopalakrishnan (2000) presented a modified

framework for modeling setup carryover in CLSP which incorporates sequence independent and

product dependent setup times and costs. Sox and Gao (1999) presented mixed integer linear

programming model to CLSP with sequence independent setup costs and no setup times. With a

different approach, Kang, Malik, and Thomas (1999) broke the entire schedule into smaller

segments called split-sequences. With the number of split-sequences being a parameter,

multiple runs with different need to be performed in order to find the optimal value.

1.3. Remanufacturing

―Remanufacturing is the practice of value added recovery in which used products are

collected, processed, and resold” (Rogers and Tibben-Lembke 1999, Guide 2000, Parkinson and

Thompson 2003). In remanufacturing process, used products are returned to manufacturers,

before directly entering the production process as a type of material. The main purpose of

remanufacturing is to save resources as well as reduce the cost. Besides, for some types of

products, remanufacturing also contributes to reducing environmental impacts.

Inventory system with remanufacturing

The graph above shows how remanufacturing system works. Used products, after being

collected from customer, will be stocked in inventory. These return products will enter directly

the production system as a type of material to produce new product. Generally, the system is not

required to reconfigure for remanufacturing. Majority of the researches relating to return products

9

assume that returns are deterministic in advance. Some of studies, however, deal with uncertain

returns in an attempt to simulate the real working environment.

A great number of researches have been carried out for the lot-sizing and scheduling problem

with returns. Teunter, Bayindir, and Den Heuvel (2006) presented a dynamic lot-sizing with

returns and remanufacturing. The authors tested the modified versions of Silver Meal, Least Unit

Cost and Part Period Balancing heuristics. Golany, Yang, and Yu (2001) presented the problem

without restrictive assumptions on the number of returns. They also assumed that returns can be

disposed. Beltrán and Krass (2002) studied the problem in which returns can be directly reused.

Overall, different researches have different assumptions on how the return system works. In this

thesis, the focus will be on the most basic remanufacturing production line in which returns are

deterministic and disposal is not an option.

Generally, Capacitated Lot-sizing and Scheduling Problem with Sequence-Dependent Setup

Times and Returns is close to the case of a drink manufacturer in real working environment. In

this case, different products are different types of drinks, while glass bottles can be considered

returns. Whenever the bottling line changes from a type of drink to another, a changeover process

needs to be executed.

1.4. Articles directly relating to the thesis

The Capacitated Lot-sizing and Scheduling Problem with Sequence-Dependent Setups has

been largely studied in the literature. Various versions of the problem have been developed and

studied. Almada-Lobo, Klabjan, Antónia carravilla, and Oliveira (2007) presented a mixed

integer programming formulation with setup times and costs. The authors also proposed the

Compact Formulation – an alternative equivalent formulation to the first formulation. With the

new formulation, they developed a five-step heuristic which is effective both in finding a feasible

solution and in producing good solution. This thesis has followed the same structure of setups and

parameters with the article. However, there is some difference in the binary variables. The article

used two decision binary variables to control the sequence dependent setup, while only one will

be used in the thesis. The difference is due to the different approaches to solve the problem.

The idea of dividing period into sub-periods is borrowed from the research of Mohammadi,

Ghomi, Karimi, and Torabi (2010). In the research, they developed and solved the multi-product

10

multi-level capacitated lot-sizing problem with sequence-dependent setups. Although the

formulation of the problem is much different in compare with the thesis, the binary variable idea

can be applied correctly to the problem in the thesis. The fix and relax heuristic proposed in the

thesis also has the same spirit with the heuristic proposed in the article. However, different

parameters will be setup and tested in the thesis. Besides, the thesis will also propose another

heuristic which is totally different from the fix and relax idea, in order to compare with the fix

and relax heuristic.

With the structure of returns and remanufacturing, the thesis relies on the research of Teunter

et al. (2006). This research focused on dynamic lot-sizing with product returns and

remanufacturing. The authors proposed a simple formulation of lot-sizing problem with returns,

in order to test the performance of some well-known heuristics including Silver Meal, Least Unit

Cost and Part Period Balancing. The structure of demand patterns are kept the same in the thesis.

Overall, in this thesis, the author does not try to reformulate the problem in a different way.

Instead, the effort is put on solving the already proposed problem with a different approach. The

main purpose of this thesis, rather than trying to find the best solution, is to generate, test and

compare different ideas of heuristic solution approach.

11

CHAPTER 2: PROBLEM DEFINITION AND MATHEMATICAL MODEL

In this chapter, two models of the Single machine Lot-Sizing Problem with Sequence-

Dependent Setup Times will be presented. The first one is a basic model of the original problem,

while the later takes into account the Returns. Before creating the mathematical models,

following assumptions need to be made:

There is a single machine which has to produce N products in T periods of time.

The machine is constrained in capacity of time units.

Demands and Returns are predetermined for all T time periods.

Backorders are allowed in the problem.

A time period is divided into sub-periods in which one and only one product is

produced. The number of sub-periods in each period never exceeds the number of

products, since it does not make sense to produce a product more than once in a period.

For example, assume that we have 3 product named A, B, and C. If we produce these

products in the sequence ABCA, meaning product A is produced in 2 sub-periods. This

sequence is obviously more costly than the ABC in which we can save the switch cost

from C to A by producing all required quantity of A in the first sub-period. Therefore,

there will be N sub-periods in each time period.

The triangle inequality needs to be assumed for the switch cost and switch time. It’s

never faster to switch over from one product to another by passing through a third

product. In other words, a direct changeover is always required in the model.

At the beginning of the planning horizon, the machine is configured at a product

which is called the starting setup configuration.

2.1. Mathematical model 1: Lot-Sizing and Scheduling Problem with Sequence-

Dependent Setup Times

12

Indices:

i, j, k

t

s

Product types

Period

Sub - period

Parameters:

N Number of products

T Number of periods

Demand of product i in period t

Unit inventory holding cost of product i in period t

Cost to produce one unit of product i in period t

Time required to produce one unit of product i in period t

Capacity (in time unit) of period t

Time required to switch from product i to product j in period t

Cost required to switch from product i to product j in period t

Unit backorder cost in one period for one unit of product i

The starting configuration of the machine

Decision variables:

Number of product i being produced in sub-period s of period t

13

Inventory level of product i at the end of period t

Backorder level of product i at the end of period t

{

Objective function:

∑∑∑

∑∑

∑∑∑∑

Subject to:

∑

∑∑

∑∑∑

(4)

∑

∑

∑

∑

∑

14

∑

{

⁄ ∑

}

∑∑

∑∑

{ }

In this model, the objective function minimizes the sum of the production costs, the storage

and backorder costs, and the sequence-dependent setup costs. Equation 1 is the inventory balance

constraint which shows the relations of inventory level, backorder level, quantity produced and

demand. Equation 2 makes sure that there will be no backorder at the end of the planning horizon.

Equation 3 indicates the capacity constraint, showing that the total amount of time spending

on production and switchover cannot exceed the given capacity. Equation 4 and 5 force the

machine to start at the starting configuration in the first sub-period of the first period.

Equation 6 and 7 force the machine to switch from one product to another, while equation 9

and 10 makes sure that the machine produces one and only one product in a sub-period.

Equation 8 indicates the relation between variable Y and variable X. It guarantees that

product i is produced if and only if the machine is switched to product i at that sub-period.

15

Equation 11, 12, 13, and 14 represent the type of variables.

2.2. Mathematical model 2: Lot-Sizing and Scheduling Problem with Sequence-

Dependent Setup Times and Returns

This model is indeed a development of the mathematical model 1. In the later model, returns

are added, which leads to some slight changes in the model.

Indices:

i, j, k

t

s

Product types

Period

Sub - period

Parameters:


T Number of periods


Number of product i being returned for remanufacturing at the beginning of period t

Unit inventory holding cost of new product i in period t

Unit inventory holding cost of returned product i in period t

Cost to produce one unit of new product i in period t

Cost to reproduce one unit of product i in period t

Time required to produce or reproduce one unit of product i in period t

16


Time required to switch from product i to product j in period t

Cost required to switch from product i to product j in period t


The starting configuration of the machine

Decision variables:

Number of product i being produced in sub-period s of period t

Number of product i being reproduced in sub-period s of period t

Inventory level of finished-product i at the end of period t

Inventory level of returned-product i at the end of period t


{

Objective function:

∑∑∑

∑∑

∑∑∑∑

17

Subject to:

∑

∑

∑∑

∑∑∑

(4)

∑

∑

∑

∑

∑

∑

{

⁄ ∑

}

∑∑

18

∑∑

{ }

Again, the objective function minimizes the sum of the production costs, the storage and

backorder costs, and the sequence-dependent costs.

The inventory balance constraints need to be written for both finished products and return

products in equation 1.a and 1.b. There is also a slight change in the capacity constraint in

equation 3. The other constraints remain the same in this second model.

From this point, we will work only on this model to solve the problem.

2.3. Test the correctness of mathematical model 2

Before carrying any sort of experiment, it’s important to make sure that we are working on a

correct model. Therefore, a simple instance is provided below in order to show how the model

works. Data of this very first instance is presented in the tables below.

Demand

Return

week 1 week 2 week 3


A 100 100 100

A 30 30 30

B 100 100 100

B 30 30 30

C 100 100 100

C 30 30 30

19

Inventory cost for new products

Inventory cost for return products



A 2 2 2

A 1 1 1

B 2 2 2

B 1 1 1

C 2 2 2

C 1 1 1

Production cost for new products

Production cost for return products



A 10 10 15

A 6 6 9

B 10 10 15

B 6 6 9

C 10 10 15

C 6 6 9

Production time

Switch cost (for all 3 weeks)


A B C

A 20 20 30

A 0 40 40

B 20 20 30

B 40 0 79

C 20 20 30

C 40 40 0

Switch time (for all 3 weeks)

Backorder cost

A B C

A 5

A 0 10 10

B 5

B 10 0 19

C 5

C 10 10 0

Capacity (time unit)

Starting config.


jo A

9000 9000 9000

20

In this instance, the production cost in week 3 is set higher than the other weeks on purpose.

Since it’s costly to produce new products in week 3, we should see an optimal plan which

produces mostly in week 1 & 2 and less in week 3. Similarly, the changeover cost from product B

to product C is higher than the other pairs. Thus, we do not expect to see the machine switching

from B to C in the result.

The model is coded and solved by Gorubi 6.5. The optimal solution of this instance is

presented in the following tables.

Production plan

Week 1 Week 2 Week 3

(jo = A) A C B B A C C A B

New product 70 70 70 140 140 140 0 0 0

Return product 30 30 30 0 29 0 60 31 60

Variable Y - Week 1

Sub-period 1 Sub-period 2 Sub-period 3

A B C A B C A B C

A 1 0 0 0 0 1 0 0 0

B 0 0 0 0 0 0 0 0 0

C 0 0 0 0 0 0 0 1 0

Variable Y - Week 2


A B C A B C A B C

A 0 0 0 0 0 0 0 0 1

B 0 1 0 1 0 0 0 0 0

C 0 0 0 0 0 0 0 0 0

21

Variable Y - Week 3


A B C A B C A B C

A 0 0 0 1 0 0 0 1 0

B 0 0 0 0 0 0 0 0 0

C 0 0 1 0 0 0 0 0 0

Inventory level of new products

Inventory level of return products



A 0 69 0

A 0 1 0

B 0 40 0

B 0 30 0

C 0 40 0

C 0 30 0

From the results, it can be seen that the model works as expected. The machine starts at

product A, which makes sense since the starting configuration is A. We also do not see the

machine switching from product B to product C. The relations between variable XN, XR and

variable Y are correct as expected. Similarly, inventory balance is also checked out in the result.

Therefore, it can be confidently concluded that the model works correctly.

2.4. Development of lower bound

So far, the problem has been formulated successfully. However, it’s too hard to solve

optimally the problem when the numbers of products and time periods are high. (The detail about

optimally computable size of the problem will be presented in the numerical experiment chapter).

Therefore, in order to test the accuracy of heuristics, it’s necessary to develop a lower bound.

It should be noticed that we are working with a Mixed Integer Programming (MIP) problem.

Theoretically, the common way to find lower bound of MIP problem is relaxing all integer

variables. In our problem, variable Y is a binary variable – a special case of integer variable.

Thus, relaxing variable Y seems to be the only choice to get a lower bound.

22

By relaxing variable Y, we have generated a relaxed problem. This relaxed problem indeed is

a Linear Programming problem which will be easier to solve. The optimal value of this relaxed

problem is a lower bound of the original MIP problem. The efficiency of this lower bound will be

tested when carrying out numerical experiments.

23

CHAPTER 3: DEVELOPMENT OF HEURISTICS

3.1. Heuristic 1: Fix and Relax

The main idea behind this heuristic approach is that if we relax part of the problem, we will

get a relaxed problem which is easier to solve optimally. In our problem, the integer variables Y

will be grouped by time periods. Integer variables in a number of periods will be fixed, while

integer variables in other periods will be relaxed in the same iteration. When binary variable Y is

relaxed, it becomes a continuous variable that 0 ≤ Y ≤ 1. Thus, instead of solving the problem all

at once, we will solve the problem part by part until we get all the fixed integer variables.

The only question left is which set of integer variables to fix and which to relax. At this point,

it’s necessary to introduce 2 parameters.

- St: the number of periods which we keep them remaining integer variables in each

iteration.

- F: the number of periods which we fix the integer variable at the results from the previous

iteration.

In iteration k, we will fix (k-1)*F periods, while keeping periods from (k-1)*F +1 to (k-1)*F

+ St remaining integer and relax the rest. Since we fix F periods each iteration, we have to solve

T/F iterations to get the results of the whole problem. By monitoring theses 2 parameters, we are

able to carry out various experiments.

A concrete example is presented below in order to show how these parameters work. Assume

that we have an instance with 10 time periods. In this example, St = 3 periods and F = 2 periods.

Iteration 1 Integer Relaxed

Period 1 2 3 4 5 6 7 8 9 10

Iteration 2 Fixed Integer Relaxed

Period 1 2 3 4 5 6 7 8 9 10

24


Period 1 2 3 4 5 6 7 8 9 10


Period 1 2 3 4 5 6 7 8 9 10

Iteration 5 Fixed Integer

Period 1 2 3 4 5 6 7 8 9 10

In iteration 1, we keep integer variables in 3 first periods the same, while relaxing integer

variables Y in the rest of periods. After solving iteration 1, we keep the results of 2 first periods to

fix in iteration 2. Generally, in each iteration, we solve to get results of 3 periods and fix the

results of 2 periods. Therefore, for 10 periods, we need to solve 10/2 = 5 iterations to get the

results of the whole problem.

In order to implement the heuristic, we need to make some changes in the problem

formulation. Firstly, the binary variables Y are now all relaxed to continuous variable that 0 ≤ Y

≤ 1. Binary variable is added into the problem with i,j = 1,2,..,N; e = 1,2,..,St; and s =

1,2,…,N. In the iteration, we add the following constraint into the problem:

With this constraint, periods from (k-1)*F +1 to (k-1)*F + St will remain integer while the

rest is relaxed.

Another trick is performed to fix the variables in each iteration. Let with i, j, s =

1,2,…,N; and t = 1,2,…,T be the result of variable Y in the k – 1 iteration. In iteration k, we will

add the following constraint to fix the variable.

25

In the numerical experiments chapter, various experiments will be carried out in order to

study the choices of these 2 parameters.

3.1. Heuristic 2

The second heuristic has the same spirit with the first one. Again, the main idea is to divide

the original problem into smaller sub-problem, so we will be able to solve it in reasonable time.

However, the approach of this heuristic is different from the fix and relax heuristic.

Looking at the objective function, we can see that the total cost is the sum of production costs,

inventory and backorder costs and changeover costs. We will deal with the problem by splitting it

into 2 sub-problems. The first one is a basic Capacitated Lot-Sizing Problem, while we deal with

Sequence-Dependent Setup Cost in the second sub-problem. In the first sub-problem, we assume

that the machine does not need to setup to produce. The problem therefore is presented below.

Indices

i Product types

t Time periods

Parameters


T Number of periods


Number of product i being returned for remanufacturing at the beginning of period t

Unit inventory holding cost of new product i in period t

Unit inventory holding cost of returned product i in period t

Cost to produce one unit of new product i in period t

26

Cost to reproduce one unit of product i in period t

Time required to produce or reproduce one unit of product i in period t



The amount of capacity we save for changeover times in the second sub-problem (the

development of this parameter will be discussed later after generating sub-problem 2)

Decision variables

Number of product i being produced in period t

Number of product i being reproduced in period t

Inventory level of finished-product i at the end of period t

Inventory level of returned-product i at the end of period t


{

Objective function:

∑∑

Subject to:

Inventory balance constraints

27

Capacity constraint

∑

Backorder level at the end

The relation between variable P and variable X

{

⁄ ∑

}

After solving the first sub-problem, in each period, we have information of which products to

produce and the quantity to produce. The only remain problem is to find a sequence which

minimizes the changeover costs. We will solve the second sub-problem for each period to get a

sequence with lowest changeover costs. The second sub-problem is formulated below.

Indices

i, j, k Product types

s Sub periods

Parameters

Results of variable in the first sub-problem

Results of variable in the first sub-problem

NoP Number of products being produced in the period

28

Time required to switch from product i to product j

Cost required to switch from product i to product j

The last product being produced in the previous period

Decision variables

{

Objective function

∑∑ ∑

Subject to

∑

∑

∑

∑ ∑

∑ ∑

29

The objective function minimizes the sum of changeover costs. Equation 1 and 2 makes sure

that the machine starts from the last product of the previous period. Equation forces the machine

to switch from one product to another. Equation 4 and 5 makes sure that the machine will be

setup for product i if and only if product i is produced in the results provided by the first sub-

problem. Looking at the sub-problem 2, it’s easy to notice that this problem is similar to the

Travelling Salesman Problem. The machine stands for the salesman, while the products can be

considered as the cities.

At this point, the only problem left is the amount of capacity we save for the changeover time.

We have saved capacity in the parameter . A safe way to make sure that we have enough

capacity for changeovers is to find an upper bound for the capacity we use in the second sub-

problem. Thus, is calculated by the following formula.

( )

It’s clear that the total changeover time will never exceed ( ).

After successfully formulate 2 sub-problems, the execution of this heuristic will follows 2

steps:

- Step 1: Solve the first sub-problem to get the results of which products to produce and the

quantity to produce.

- Step 2: Solve the second sub-problem T times to get the sequence of the changeover.

31

CHAPTER 4: NUMERICAL EXPERIMENTS

1.1. Development of instances

In order to test 2 heuristics developed in chapter 3, 48 instances are generated. These

instances are created randomly with the gradually increased number of products (parameter N).

The number of time periods is always 20 for all instances.

The demand patterns and return patterns are generated following the research Dynamic lot

sizing with product returns and remanufacturing (Teunter et al., 2006). Demands and Returns are

generated according to the formula below:

(

)

In this formula, μ is the starting level of the pattern; τ is the trend level; a is the amplitude of

the cycle; d determines the peak of the cycle and (t = 1,2,…,T) are independently normally

distributed random variables with standard deviation δ. By monitoring parameters in the formula,

we are able to create different types of demands and return. Indeed, 3 types of demand and return

are created including Stationary, Positive Trend and Peak in Middle. These graphs below show

examples of 3 types of demand being tested.

μ 100

δ 10

τ 0

a 0

c 20

d 3

0

20

40

60

80

100

120

140

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Stationary

Dt

32

μ 100

δ 10

τ 10

a 0

c 20

d 3

μ 100

δ 10

τ 0

a 40

c 20

d 3

Other parameters are randomly generated following the table below:

Parameter Name From - to

Unit inventory holding cost of new product i in period t 2 - 3.5

Unit inventory holding cost of returned product i in period t 1 - 3

Cost to produce one unit of new product i in period t 10 - 25

0

50

100

150

200

250

300

350

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Positive Trend

Dt

0

50

100

150

200

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Peak in Middle

Dt

33

Cost to reproduce one unit of product i in period t 5 - 15

Time required to produce or reproduce one unit of product i in period t 15 - 25

Time required to switch from product i to product j in period t 55 - 100

Cost required to switch from product i to product j in period t 41 - 61

Unit backorder cost in one period for one unit of product i 5 -10

The starting configuration of the machine 1 - N

It should be noticed that demand and return are rounded up to zero decimal places, while the

other parameters are rounded up to one decimal place. As can be seen, the instances tend to

simulate the real world as much as possible.

So far with the instances, we have the capacity left to decide. The capacity is kept constant all

time periods. In order to secure the feasibility of the instance, the capacity is chosen by try and

error. We start with a capacity which makes the problem infeasible; increase the capacity 5%

each time the problem results infeasible. We stop as soon as we get a feasible problem, and mark

the instance as type A. Thus, we always get the capacity that is not larger than the minimum

required capacity by 5%. In order to test the influence of capacity, instances type B and type C

are also created. In instance type B, we keep all the parameters the same as instance type A but

increase the capacity by 10%. We do the same with instances type C, but the percentage this time

is 25%.

A system of notation is presented below in order to track the instances.

The first number X = {1,2,…,10} Number of products = 5*X

A,B,C The generating of capacity as described above

St Stationary

Po Positive trend

Pe Peak in middle

34

For example, 3.B.St means the instance has 15 products, capacity type B and the demand and

return generated following stationary model.

1.2. Computational results

All computational experiments are performed on a personal computer with Window 8.1,

processor Intel core i5-43000U and 8 Gb memory. Gurobi 6.5 was used as the mixed integer

programming solver.

In total, 48 instances were created to test the performance of the two heuristics. All raw

results can be found in the Appendix 1. In the results, shows the difference in percentage

between the results found by heuristics and the optimal objective value, while compares the

differences between the results found by heuristics and the lower bound. In the fix and relax

heuristic, the sign St-F is used. For example, 3-1 means St = 3 and F = 1 in this run. The sign ―-‖

means the computer ran more than 2 hours without finding the results.

Optimally solving the problem

Instance Number of

constraints Number of variables Number of binary variables

5x20 1441 3800 2500

10x20 4911 24600 20000

15x20 10431 77400 67500

20x20 18001 177200 160000

25x20 27621 339000 312500

30x20 39291 577800 540000

35x20 53011 908600 857500

40x20 68781 1346400 1280000

45x20 86601 1906200 1822500

50x20 106471 2603000 2500000

The table above shows the size of the problem with different sets of NxT. As can be seen, size

of the problem increases very quickly when we increase the number of products. Since the

35

hardness of the problem depends mostly on the number of binary variables, the problem becomes

NP-hard very quickly. Indeed, with the time limitation of 2 hours, we are able to solve optimally

instances with size up to 15x20. Beyond this size, it takes too much time to solve optimally the

problem.

The lower bound

As expected, the lower bound is found within very short CPU time. The quality of this lower

bound is quite good. In average, the difference between the lower bound and the optimal value is

only 7.35%.

Fix and relax heuristic

Similar to the optimal value, we are only able use this heuristic to solve the instances with

size up to 15x20. Beyond this size, even the set 1-1 - the set which theoretically should take least

time to solve – runs more than 2 hours without finding the solution. The performance of the first

heuristic can be seen in the table below.

St-F

1-1 0.128% 24.4

3-1 0.012% 57.2

5-1 0.004% 125.7

3-2 0.023% 26.3

5-2 0.005% 74.4

5-3 0.005% 51.8

shows the mean of the difference between the solution found by heuristic 1 and the

optimal value. Overall, heuristic 1 delivers very good solution. We get solutions which are so

close to the optimal value. In some cases, which can be found in the Appendix, we get the same

results with the optimal objective value.

We can see clearly the trade-off between quality and CPU time in the results. The set 1-1 took

least time to find a solution, but found the worst solution. On the opposite, the set 5-1 found the

best solution with the longest running time. This results make sense since the quality of the fix

and relax heuristic depends mostly on the parameter St. The higher St is, the harder the sub-

36

problem becomes. The harder the sub-problem is, the longer time we need to run the heuristic.

However, the harder the sub-problem is, the better solution the heuristic gets.

In some cases, heuristic 1 results infeasible. In one iteration, the sub-problem is infeasible,

which prevents the heuristic to go on. In this case, we have to choose another set of parameters to

solve the problem. The infeasible situation seems to appear unpredictably, which leads to the

assumption that it depends on the instance itself rather than the set of parameters.

Heuristic 2

Heuristic 2, in comparison to heuristic 1, took less time to solve the problem. In average, the

second heuristic takes 10.4 seconds of CPU time to get the solution. However, the difference

between solutions found by heuristic 2 and the optimal values is 4.15%, which is much higher

than heuristic 1.

Since it’s not able to find the optimal solution with instances larger than 15x20, we can use

the lower bound to evaluate the quality of the heuristic. If we include results of instances from 25

to 45, and compare them with the lower bound, the difference is 10.4%. This number in heuristic

1 is 7.9%. Generally, in comparison with heuristic 1, heuristic 2 takes less time to run but

produce worse solutions. One strong point of this heuristic is that it’s able to solve very large

instance. While heuristic 1 stops at 15x20, heuristic 2 is able to solve instances up to size 50x20.

However, in some instance type A, the heuristic results infeasible. This phenomenon is

because of the capacity we save for the second sub-problem in this heuristic. The

remaining capacity is not enough to produce all the demand, which leads to infeasible result in

the first sub-problem. That’s the reason why infeasible result often appears in instances type A in

which the capacity is set very tightly.

The effects of demand patterns

Demand

pattern

Heuristic 1 Heuristic 2

Stationary 0.029% 11.06%

Peak in middle 0.040% 10.64%

Positive trend 0.028% 5.24%

37

The table above shows the mean of the difference between the results found by heuristic 1

and the optimal value. The performance of heuristic 1 does not seem to be affected by the

demand patterns. Heuristic 2 seems to produce better results with the Positive trend pattern.

However, with the limited number of instances, it’s hard to conclude anything about the effect of

demand patterns on the results of the two heuristics.

The effect of capacity

Instance type


A 1395.4 0.059% 11.928%

B 479.7 0.020% 3.246%

C 325 0.016% 0.721%

This table shows the running time to find optimal solution in average of different types of

instances. The tightness of capacity obviously affects the time we use to solve the problem. It

seems that tightening the capacity makes the problem become harder to solve. It’s also clear that

the tightness of capacity has an effect on the result quality of both heuristics. The heuristics seem

to work better with the capacity which is not tightened too much.

Comparison between the two heuristics


0.032% 4.145%

7.90% 10.42%

151.7 10.2

38

Strengths and Weaknesses


Strength Ability to find very good solutions Ability to find quite good solutions even

for large instances

Weakness

Takes longer time to run

Could be infeasible in some cases

Have limit in solvable size (15x20)

Solution’s quality is not as good as

heuristic 1

Could be infeasible if the capacity is

tightened too hard

The tables above summarize and compare the performance of 2 heuristics. Although both of

them have weaknesses, it can be generally said that two heuristics are able to find reasonable

good solutions. While heuristic 1 may be preferable to solve small size problems, heuristic 2 may

be a good choice to deal with super large size problems.

39

CHAPTER 5: CONCLUDING REMARKS

Overall, the Capacitated Lot-sizing and Scheduling Problem with Sequence-Dependent Setup

Times and Return has been successfully formulated in the thesis. Both heuristics, although come

with different strengths and weaknesses, show promising results. However, there are some

aspects in the thesis that I wish to improve but could not due to the limitation in time as well as

knowledge. Firstly, the number of instances could be higher in order to deeply test the two

heuristics. The generating of instances could be done automatically in the code instead of

manually by excel. Due to the lack of experience in coding, I also feel that there is much room to

improve in the codes. Secondly, due to the limitation in the number of instances, I am aware that

the results seem to be unreliable sometimes. Although the results do not show any conflict with

theory, it also needs more experiments to confirm.

For further research, the research could be extended from single machine to parallel machine

with or without different speed. The fix and relax heuristic could be tested with more set of

parameters, or even a different setup. The quality of heuristic 2 could also be improved with a

better upper bound for the saved capacity.

41

REFERENCES

Almada-Lobo, B., Klabjan, D., Antónia carravilla, M., & Oliveira, J. F. (2007). Single machine

multi-product capacitated lot sizing with sequence-dependent setups. International

Journal of Production Research, 45(20), 4873-4894.

Barany, I., Van Roy, T. J., & Wolsey, L. A. (1984). Strong formulations for multi-item

capacitated lot sizing. Management science, 30(10), 1255-1261.

Beltrán, J. L., & Krass, D. (2002). Dynamic lot sizing with returning items and disposals. IIe

transactions, 34(5), 437-448.

Belvaux, G., & Wolsey, L. A. (2000). bc—prod: A specialized branch-and-cut system for lot-

sizing problems. Management science, 46(5), 724-738.

Belvaux, G., & Wolsey, L. A. (2001). Modelling practical lot-sizing problems as mixed-integer

programs. Management science, 47(7), 993-1007.

Drexl, A., & Kimms, A. (1997). Lot sizing and scheduling—survey and extensions. European

Journal of Operational Research, 99(2), 221-235.

Eppen, G. D., & Martin, R. K. (1987). Solving multi-item capacitated lot-sizing problems using

variable redefinition. Operations Research, 35(6), 832-848.

Golany, B., Yang, J., & Yu, G. (2001). Economic lot-sizing with remanufacturing options. IIe

transactions, 33(11), 995-1003.

Gopalakrishnan, M. (2000). A modified framework for modelling set-up carryover in the

capacitated lotsizing problem. International Journal of Production Research, 38(14),

3421-3424.

Gopalakrishnan, M., Miller, D., & Schmidt, C. (1995). A framework for modelling setup

carryover in the capacitated lot sizing problem. International Journal of Production

Research, 33(7), 1973-1988.

Gupta, D., & Magnusson, T. (2005). The capacitated lot-sizing and scheduling problem with

sequence-dependent setup costs and setup times. Computers & Operations Research,

32(4), 727-747.

Kang, S., Malik, K., & Thomas, L. J. (1999). Lotsizing and scheduling on parallel machines with

sequence-dependent setup costs. Management science, 45(2), 273-289.

42

Mohammadi, M., Ghomi, S. F., Karimi, B., & Torabi, S. A. (2010). Rolling-horizon and fix-and-

relax heuristics for the multi-product multi-level capacitated lotsizing problem with

sequence-dependent setups. Journal of Intelligent Manufacturing, 21(4), 501-510.

Sox, C. R., & Gao, Y. (1999). The capacitated lot sizing problem with setup carry-over. IIe

transactions, 31(2), 173-181.

Teunter, R. H., Bayindir, Z. P., & Den Heuvel, W. V. (2006). Dynamic lot sizing with product

returns and remanufacturing. International Journal of Production Research, 44(20), 4377-

4400.

Wagner, H. M., & Whitin, T. M. (1958). Dynamic version of the economic lot size model.

Management science, 5(1), 89-96.

43

APPENDIX

Computational results

Instance Method CPU

Time Value Backorder ∆1 ∆2

1.A.St

(5x20)

Lower bound 3 € 162,203.32

Optimal 5 € 190,087.23 4878.5

Fix

and R

elax

1-1 3 € 190,398.71 4897.7 0.16% 14.81%

3-1 6 € 190,087.23 4878.5 0.00% 14.67%

5-1 6 € 190,115.87 4879.7 0.02% 14.68%

3-2 3 € 190,115.87 4879.7 0.02% 14.68%

5-2 4 € 190,087.23 4878.5 0.00% 14.67%

5-3 3 € 190,115.87 4879.7 0.02% 14.68%

Heuristic 2 3 € 229,394.86 8084.7 17.14% 29.29%

1.B.St

(5x20)

Lower bound 3 € 134,382.04

Optimal 3 € 140,513.91 934.9

Fix

and R

elax

1-1 4 € 140,630.68 945.6 0.08% 4.44%

3-1 4 € 140,537.34 932.7 0.02% 4.38%

5-1 5 € 140,535.26 932.4 0.02% 4.38%

3-2 3 € 140,547.93 934.2 0.02% 4.39%

5-2 3 € 140,540.07 933.7 0.02% 4.38%

5-3 3 € 140,542.14 933.9 0.02% 4.38%

Heuristic 2 3 € 145,304.97 1582.3 3.30% 7.52%

1.C.St

(5x20)

Lower bound 1 € 131,063.92

Optimal 1 € 134,908.58 302.8

Fix

and R

elax

1-1 3 € 135,010.94 302.8 0.08% 2.92%

3-1 4 € 134,908.58 302.8 0.00% 2.85%

5-1 4 € 134,908.58 302.8 0.00% 2.85%

3-2 2 € 134,913.99 302.8 0.00% 2.85%

5-2 2 € 134,908.58 302.8 0.00% 2.85%

5-3 2 € 134,908.58 302.8 0.00% 2.85%

Heuristic 2 3 € 135,653.01 310.4 0.55% 3.38%

Table 1: 5 products and stationary demand

44

Instance Method CPU


1.A.Pe

(5x20)

Lower bound 1 € 184,740.74

Optimal 1 € 214,151.05 6686.5

Fix

and

Rel

ax 1-1 4 € 215,184.77 6743.2 0.48% 14.15%

3-1 5 € 214,219.85 6692.5 0.03% 13.76%

5-1 6 € 214,151.05 6686.5 0.00% 13.73%

3-2 3 € 214,720.85 6726.3 0.27% 13.96%

5-2 3 € 214,151.05 6686.5 0.00% 13.73%

5-3 3 € 214,151.05 6686.5 0.00% 13.73%

Heuristic 2 3 € 301,650.03 13582.9 29.01% 38.76%

1.B.Pe

(5x20)

Lower bound 1 € 152,972.58

Optimal 1 € 163,512.16 1987.3

Fix

and R

elax

1-1 4 € 163,606.03 1999.3 0.06% 6.50%

3-1 3 € 163,544.39 1996.6 0.02% 6.46%

5-1 5 € 163,512.16 1987.3 0.00% 6.45%

3-2 3 € 163,540.64 2001.2 0.02% 6.46%

5-2 3 € 163,512.16 1987.3 0.00% 6.45%

5-3 3 € 163,512.16 1987.3 0.00% 6.45%

Heuristic 2 3 € 169,726.52 2760.6 3.66% 9.87%

1.C.Pe

(5x20)

Lower bound 1 € 139,633.60

Optimal 1 € 144,777.72 477.9

Fix

and R

elax

1-1 3 € 144,816.17 477.9 0.03% 3.58%

3-1 5 € 144,784.20 478.9 0.00% 3.56%

5-1 5 € 144,779.81 479.4 0.00% 3.55%

3-2 3 € 144,784.20 478.9 0.00% 3.56%

5-2 3 € 144,787.34 478.4 0.01% 3.56%

5-3 3 € 144,794.64 479.0 0.01% 3.56%

Heuristic 2 3 € 145,565.24 443.1 0.54% 4.07%

Table 2: 5 product and peak in middle trend

45

Instance Method CPU


1.A.Po

(5x20)

Lower

bound 1 € 389,247.01

Optimal 1 € 437,503.51 5201.0

Fix

and R

elax

1-1 5 € 439,408.03 5467.9 0.43% 11.42%

3-1 5 € 437,764.43 5232.5 0.06% 11.08%

5-1 7 € 437,503.51 5201.0 0.00% 11.03%

3-2 2 € 437,775.27 5232.5 0.06% 11.09%

5-2 4 € 437,503.51 5201.0 0.00% 11.03%

5-3 3 € 437,503.51 5201.0 0.00% 11.03%

Heuristic 2 Infeasible

1.B.Po

(5x20)

Lower

bound 1 € 322,590.97

Optimal 1 € 332,466.25 69.0

Fix

and R

elax

1-1 4 € 332,660.75 69.0 0.06% 3.03%

3-1 4 € 332,466.25 69.0 0.00% 2.97%

5-1 5 € 332,476.93 69.0 0.00% 2.97%

3-2 3 € 332,489.33 69.0 0.01% 2.98%

5-2 3 € 332,476.93 69.0 0.00% 2.97%

5-3 3 € 332,468.55 69.0 0.00% 2.97%

Heuristic 2 2 € 336,691.12 69.0 1.25% 4.19%

1.C.Po

(5x20)

Lower

bound 1 € 285,030.66

Optimal 1 € 291,121.78 156.0

Fix

and R

elax

1-1 3 € 291,295.82 156.0 0.06% 2.15%

3-1 5 € 291,144.52 156.0 0.01% 2.10%

5-1 5 € 291,122.46 156.0 0.00% 2.09%

3-2 3 € 291,139.05 156.0 0.01% 2.10%

5-2 3 € 291,122.46 156.0 0.00% 2.09%

5-3 3 € 291,129.17 156.0 0.00% 2.09%

Heuristic 2 2 € 292,495.18 156.0 0.47% 2.55%

Table 3: 5 products and positive trend demand

46

Instance Method CPU


2.A.St

(10x20)

Lower bound 1 € 389,954.91

Optimal 212 € 483,982.03 15191.8

Fix

and R

elax

1-1 48 € 484,876.76 15242.2 0.18% 19.58%

3-1 147 € 484,019.74 15198.4 0.01% 19.43%

5-1 602 € 483,991.79 15193.2 0.00% 19.43%

3-2 85 € 484,030.45 15196.0 0.01% 19.44%

5-2 456 € 483,996.20 15196.5 0.00% 19.43%

5-3 337 € 484,010.60 15196.8 0.01% 19.43%


2.B.St

(10x20)

Lower bound 1 € 292,778.16

Optimal 95 € 318,636.31 2553.3

Fix

and R

elax

1-1 61 € 318,853.57 2574.3 0.07% 8.18%

3-1 208 € 318,658.55 2556.5 0.01% 8.12%

5-1 209 € 318,636.31 2553.3 0.00% 8.12%

3-2 62 € 318,665.47 2556.2 0.01% 8.12%

5-2 100 € 318,645.08 2553.8 0.00% 8.12%

5-3 98 € 318,645.08 2553.8 0.00% 8.12%

Heuristic 2 3 € 335,897.76 3837.6 5.14% 12.84%

2.C.St

(10x20)

Lower bound 1 € 276,516.75

Optimal 28 € 285,151.14 895.4

Fix

and R

elax

1-1 30 € 285,341.47 899.0 0.07% 3.09%

3-1 64 € 285,152.51 894.5 0.00% 3.03%

5-1 146 € 285,161.66 897.1 0.00% 3.03%

3-2 50 € 285,172.55 896.5 0.01% 3.04%

5-2 78 € 285,152.20 896.6 0.00% 3.03%

5-3 70 € 285,151.67 896.6 0.00% 3.03%

Heuristic 2 3 € 285,165.47 847.5 0.01% 3.03%


47

Instance Method CPU


2.A.Pe

(10x20)

Lower bound 1 € 431,460.23

Optimal 246 € 557,892.71 27514.6

Fix

and R

elax

1-1 65 € 559,370.36 27603.5 0.26% 22.87%

3-1 168 € 557,999.43 27522.9 0.02% 22.68%

5-1 385 € 557,892.71 27514.6 0.00% 22.66%

3-2 70 € 557,965.03 27516.0 0.01% 22.67%

5-2 162 € 557,951.52 27515.0 0.01% 22.67%

5-3 80 € 557,892.71 27514.6 0.00% 22.66%


2.B.Pe

(10x20)

Lower bound 1 € 304,316.52

Optimal 154 € 335,351.28 8355.3

Fix

and R

elax

1-1 58 € 335,660.84 8385.6 0.09% 9.34%

3-1 140 € 335,390.07 8354.7 0.01% 9.26%

5-1 405 € 335,357.87 8354.1 0.00% 9.26%

3-2 48 € 335,388.66 8358.1 0.01% 9.26%

5-2 225 € 335,366.69 8355.2 0.00% 9.26%

5-3 132 € 335,372.87 8352.4 0.01% 9.26%

Heuristic 2 3 € 358,668.18 10472.6 6.50% 15.15%

2.C.Pe

(10x20)

Lower bound 1 € 269,418.30

Optimal 68 € 279,629.31 631.8

Fix

and R

elax

1-1 30 € 279,774.20 618.0 0.05% 3.70%

3-1 62 € 279,691.41 619.0 0.02% 3.67%

5-1 105 € 279,672.55 624.2 0.02% 3.67%

3-2 33 € 279,683.34 624.6 0.02% 3.67%

5-2 82 € 279,687.21 627.5 0.02% 3.67%

5-3 48 € 279,660.78 627.3 0.01% 3.66%

Heuristic 2 3 € 281,158.55 894.5 0.54% 4.18%

Table 5: 10 products and peak in middle demand

48

Instance Method CPU


2.A.Po

(10x20)

Lower bound 3 € 734,799.17

Optimal 100 € 772,878.77 2207.6

Fix

and R

elax

1-1 38 € 773,331.44 2214.7 0.06% 4.98%

3-1 83 € 772,886.00 2205.7 0.00% 4.93%

5-1 126 € 772,886.00 2205.7 0.00% 4.93%

3-2 37 € 772,903.34 2206.7 0.00% 4.93%

5-2 67 € 772,901.66 2206.9 0.00% 4.93%

5-3 43 € 772,896.10 2209.5 0.00% 4.93%

Heuristic 2 3 € 802,223.11 3195.6 3.66% 8.40%

2.B.Po

(10x20)

Lower bound 1 € 647,187.44

Optimal 52 € 666,937.87 579.0

Fix

and R

elax

1-1 27 € 667,524.45 579.0 0.09% 3.05%

3-1 72 € 666,994.29 579.0 0.01% 2.97%

5-1 168 € 666,956.13 579.0 0.00% 2.96%

3-2 42 € 667,066.58 579.0 0.02% 2.98%

5-2 112 € 667,011.67 579.0 0.01% 2.97%

5-3 70 € 666,957.09 579.0 0.00% 2.96%

Heuristic 2 3 € 673,093.90 570.4 0.91% 3.85%

2.C.Po

(10x20)

Lower bound 1 € 573,577.12

Optimal 38 € 587,656.86 462.0

Fix

and R

elax

1-1 28 € 588,098.04 462.0 0.08% 2.47%

3-1 44 € 587,690.37 462.0 0.01% 2.40%

5-1 68 € 587,665.96 462.0 0.00% 2.40%

3-2 22 € 587,672.54 462.0 0.00% 2.40%

5-2 30 € 587,662.69 462.0 0.00% 2.40%

5-3 28 € 587,676.20 462.0 0.00% 2.40%

Heuristic 2 3 € 590,161.58 464.7 0.42% 2.81%


49

Instance Method CPU


3.A.St

(15x20)

Lower bound 3 € 675,179.07

Optimal 6627 € 875,022.12 34523.6

Fix

and R

elax

1-1 606 € 877,446.27 34677.0 0.28% 23.05%

3-1 3451 € 875,060.40 34527.7 0.00% 22.84%

5-1 -

3-2 1595 € 875,060.40 34527.7 0.00% 22.84%

5-2 -

5-3 -

Heuristic 2 3 Infeasible

3.B.St

(15x20)

Lower bound € 478,518.89

Optimal 1856 € 536,040.41 10200.4

Fix

and R

elax

1-1 226 € 536,299.35 10229.2 0.05% 10.77%

3-1 2174 € 536,059.84 10219.2 0.00% 10.73%

5-1 -

3-2 727 € 536,062.38 10219.8 0.00% 10.73%

5-2 -

5-3 -

Heuristic 2 3 € 573,336.60 13505.8 6.51% 16.54%

3.C.St

(15x20)

Lower bound 5 € 408,215.41

Optimal 1562 € 424,357.05 1148.7

Fix

and R

elax

1-1 134 € 424,673.50 1153.2 0.07% 3.88%

3-1 1074 € 424,431.76 1152.2 0.02% 3.82%

5-1 -

3-2 Infeasible

5-2 -

5-3 -

Heuristic 2 3 € 428,831.65 1677.7 1.04% 4.81%


“-“ means it takes more than 2 hours to solve the problem

50

Instance Method CPU


3.A.Pe

(15x20)

Lower bound 4 € 413,254.64

Optimal 2501 € 430,367.64 1809.8

Fix

and R

elax

1-1 Infeasible

3-1 715 € 430,428.96 1810.4 0.01% 3.99%

5-1 -

3-2 423 € 430,454.47 1809.5 0.02% 4.00%

5-2 -

5-3 -

Heuristic 2 5 € 432,404.23 1997.3 4.43% 4.43%

3.B.Pe

(15x20)

Lower bound 5 € 409,490.26

Optimal 1940 € 423,288.81 1340.4

Fix

and R

elax

1-1 116 € 423,550.70 1341.0 0.06% 3.32%

3-1 Infeasible

5-1 -

3-2 514 € 423,327.16 1345.0 0.01% 3.27%

5-2 -

5-3 -

Heuristic 2 5 € 423,517.44 1442.0 0.05% 3.31%

3.C.Pe

(15x20)

Lower bound 5 € 407,009.27

Optimal 1111 € 419,227.40 1044.4

Fix

and R

elax

1-1 Infeasible

3-1 Infeasible

5-1 -

3-2 € 419,317.01 1044.6 0.02% 2.94%

5-2 -

5-3 -

Heuristic 2 5 € 429,959.76 1044.3 2.50% 5.34%

Table 8: 15 products and peak in middle demand


51

Instance Method CPU


3.A.Po

(15x20)

Lower bound 5 € 1,089,287.63

Optimal 2866 € 1,164,137.02 6390.0

Fix

and R

elax

1-1 Infeasible

3-1 Infeasible

5-1 -

3-2 675 € 1,164,316.67 6390.2 0.02% 6.44%

5-2 -

5-3 -

Heuristic 2 5 € 1,230,742.01 9743.3 5.41% 11.49%

3.B.Po

(15x20)

Lower bound 5 € 985,785.95

Optimal 215 € 1,026,581.57 1474.7

Fix

and R

elax

1-1 Infeasible

3-1 Infeasible

5-1 -

3-2 Infeasible

5-2 -

5-3 -

Heuristic 2 5 € 1,046,303.00 1773.7 1.88% 5.78%

3.C.Po

(15x20)

Lower bound 5 € 833,394.67

Optimal 115 € 854,358.51 531.2

Fix

and R

elax

1-1 Infeasible

3-1 Infeasible

5-1 -

3-2 Infeasible

5-2 -

5-3 -

Heuristic 2 5 € 857,905.35 427.2 0.41% 2.86%



52

Instance Lower bound Heuristic 2 CPU

time

4.A.St € 561,470.87 € 634,234.83 5 11.5%

4.A.Pe € 568,283.60 € 623,994.70 5 8.9%

4.A.Po € 1,222,340.62 € 1,269,345.62 5 3.7%

5.A.St € 783,201.76 € 930,880.82 10 15.9%

5.A.Pe € 832,697.29 € 949,238.47 15 12.3%

5.A.Po € 1,699,155.50 € 1,818,536.34 8 6.6%

6.A.St € 839,125.17 € 927,534.38 34 9.5%

6.A.Pe € 1,045,562.31 € 1,207,207.03 14 13.4%

6.A.Po € 1,767,444.91 € 1,825,751.50 40 3.2%

7.A.St € 1,025,926.78 € 1,165,204.70 34 12.0%

7.A.Pe € 1,171,496.10 € 1,317,030.19 25 11.1%

7.A.Po € 2,171,459.92 € 2,260,755.46 35 3.9%

8.A.St € 1,120,711.56 € 1,271,654.17 85 11.9%

8.A.Pe € 1,364,280.67 € 1,658,903.99 25 17.8%

8.A.Po € 2,508,056.20 € 2,609,939.37 73 3.9%

9.A.St € 1,553,421.59 € 1,971,491.18 25 21.2%

9.A.Pe € 1,564,840.10 € 1,861,137.39 55 15.9%

9.A.Po € 2,914,253.73 € 3,061,657.47 102 4.8%

10.A.St € 1,813,114.82 € 2,567,634.08 15 29.4%

10.A.Pe € 2,064,648.69 € 2,915,976.70 15 29.2%

10.A.Po € 3,127,343.90 € 3,262,872.64 267 4.2%

Table 10: Results found by heuristic 2 for instances from 20x20 to 50x20

53

Instruction to run the code

This thesis is summited along with 4 julia files which contain the code. The files are named

following their purpose. The Optimal.jpynb is to solve optimally the problem, while the

Lowerbound.jpynb is to find the lower bound. The codes for the two heuristis are named

Heuristic1.jpynb and Heuristic2.jpynb.

The instances are organized in text file. The parameters are organized in the order below:

- Demands

- Returns

- Inventory holding cost for new products

- Inventory holding cost for returns

- Unit production cost for new product

- Unit production cost for remanufacturing product

- Time needed to produce one unit of product

- Time required to switch from one product to another

- Cost required to switch from one product to another

- Backorder cost

- The starting configuration

The code is designed to reuse, which makes it easier to switch from an instance to another.

Therefore, in the beginning of the code, there are some parameters we need to fulfill. The

parameters include the number of products N, the number of time periods T, and the number of

instance. Number of instance is showed in the table below, along with detail information about

the instances. In the Heuristic1.jpynb, parameters St and F also can be changed. Detail

explanation of the code and instruction will be given in every box of the code.

54

Instance Number NxT Demand pattern Capacity

type

First test 0 3x3

1.A.St 1 5x20 Stationary A

1.B.St 2 5x20 Stationary B

1.C.St 3 5x20 Stationary C

1.A.Pe 4 5x20 Peak in middle A

1.B.Pe 5 5x20 Peak in middle B

1.C.Pe 6 5x20 Peak in middle C

1.A.Po 7 5x20 Positive trend A

1.B.Po 8 5x20 Positive trend B

1.C.Po 9 5x20 Positive trend C



















55






















heuristic solution approach for the capacitated lot …...researches focus on improving the mip...

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