a disassembly line balancing problem with fixed number of workstations

ARTICLE IN PRESSJID: EOR [m5G;October 1, 2015;11:19]

European Journal of Operational Research 000 (2015) 1–13

Contents lists available at ScienceDirect

European Journal of Operational Research

journal homepage: www.elsevier.com/locate/ejor

Production, Manufacturing and Logistics

A disassembly line balancing problem with fixed number of workstations

Eda Göksoy Kalaycılar a, Meral Azizoglu b,∗, Sencer Yeralan c

a ASELSAN, Naval Systems Program Directorate, Ankara 06172, Turkeyb Department of Industrial Engineering, Middle East Technical University, Ankara 06800, Turkeyc Department of Industrial Engineering, Yasar University, Izmir, Turkey

a r t i c l e i n f o

Article history:

Received 30 June 2014

Accepted 3 September 2015

Available online xxx

Keywords:

Integer programming

Heuristics

Disassembly lines

Linear programming relaxation

a b s t r a c t

In this study, a Disassembly Line Balancing Problem with a fixed number of workstations is considered. The

product to be disassembled comprises various components, which are referred to as its parts. There is a speci-

fied finite supply of the product to be disassembled and specified minimum release quantities (possible zero)

for each part of the product. All units of the product are identical, however different parts can be released

from different units of the product. There is a finite number of identical workstations that perform the neces-

sary disassembly operations, referred to as tasks. We present several upper and lower bounding procedures

that assign the tasks to the workstations so as to maximize the total net revenue. The computational study

has revealed that the procedures produce satisfactory results.

© 2015 Published by Elsevier B.V.

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. Introduction

The environmental regulations, customer awareness and recent

dvances in technology all together have shifted the product recovery

rocess from the act of disposing to the act of remanufacturing and

ecycling. Recycling preserves the material content of the discarded

used) products via some manufacturing and disassembly operations.

emanufacturing, on the other hand, keeps the functional content of

he used products and improves their quality up to a desired usable

evel via disassembly operations and some manufacturing.

Disassembly is the first important step of product recovery activ-

ties (McGovern and Gupta, 2011) and it is methodical extraction of

aluable parts, operations involve the separation of the reusable parts

rom the discarded products. The parts are either subject to some re-

anufacturing process or sold to suppliers. Disassembly operations

re usually performed on a disassembly line that consists of a number

f serial workstations. The first workstation takes the product to be

isassembled. The cycle terminates, i.e., the product leaves the line,

henever all its required parts are disassembled.

The Disassembly Line Balancing Problem (DLBP) assigns the set

f tasks to each workstation for each product to be disassembled.

he problem is critical in minimizing the use of valuable resources

such as time and money) invested in disassembly, and maximizing

he level of automation of the disassembly process and the quality of

he parts or materials recovered.

∗ Corresponding Author. Tel.: +90 3122102281.

E-mail addresses: [email protected] (E.G. Kalaycılar), [email protected]

(M. Azizoglu), [email protected] (S. Yeralan).

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ttp://dx.doi.org/10.1016/j.ejor.2015.09.004

377-2217/© 2015 Published by Elsevier B.V.

Please cite this article as: E.G. Kalaycılar et al., A disassembly line balancin

Operational Research (2015), http://dx.doi.org/10.1016/j.ejor.2015.09.004

This study considers a DLBP with a fixed number of workstations.

t is assumed that there is a specified supply for the products to be

isassembled. For each part, a defined minimum release quantity

ust be met. The amount in excess of the minimum release quan-

ity can also be sold in the market. Hence the excess quantity is pro-

uced, provided that the part is profitable, and that there is enough

upply. The aim is to assign the tasks to the disassembly workstations

o as to maximize the total net revenue of the parts, while meeting

heir specified minimum release quantities, and without exceeding

he specified cycle time. To satisfy the minimum release quantities,

ifferent parts may be released from different units of the product.

he challenge is to use different line balances, hence use different

ask assignments to the already mounted workstations, while disas-

embling different units of the product. To the best of our knowledge,

he study is the first attempt to solve the DLBP with minimum release

uantities and fixed number of workstations.

Our study will have a direct impact on the industries that experi-

nce continuous advances in their technology. These advances nat-

rally affect the function and fashion oriented expectations of the

onsumers. One such application area that we take our motivation

rom is the electrical and electronic equipment industry. The con-

umers of the personal computers, televisions or cellular phones,

eplace their products within a few months to a few years, even

hen the products (thereby their parts) are functioning properly.

lectrical and electronic equipments consist of many different sub-

tances some of which may contain hazardous components and valu-

ble parts (Capraz, Polat, & Güngör, 2015). As the used products have

roper conditions, their valuable parts (like memories, CPUs, valuable

etals) can be used in manufacturing new models. The minimum re-

ease quantities for these parts (that we define in our models) are

g problem with fixed number of workstations, European Journal of

http://dx.doi.org/10.1016/j.ejor.2015.09.004

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associated to the demands from the third party like the producers of

the new models. Moreover, as we assume, the amount released ex-

cess of the minimum release quantities of the valuable parts can be

stored in the inventory for future orders. Another application area is

the automotive industry where the consumers of the luxurious cars

replace their not-so-much-used cars with new brands that make use

of the most recent technology.

The rest of the study is organized as follows. Section 2 reviews the

disassembly process and literature on the DLBP whereas Section 3

defines our problem. In Section 4 the mathematical models and

their use in finding optimal solutions are discussed. In Section 5,

we present upper bounds together with the mechanisms used to

strengthen them and present a heuristic procedure. The computa-

tional experiment is discussed in Section 6 and the study is concluded

in Section 7.

2. Disassembly process & the related literature

Güngör and Gupta (2001b) defined disassembly as a systematic

process of separating a product into its constituent parts, compo-

nents, subassemblies or other groupings. The issues in the area of

disassembly can be classified into two broad categories as design and

operational. Crowther (1999) considered a design for disassembly is-

sue and mentioned that a life cycle model that incorporates the stages

of a disassembly strategy can highlight the environmental advan-

tages of designing for disassembly, showing how it can extend service

life and thereby improve sustainability. As a practical application, he

discussed the construction industry and mentioned that experience

gained from disassemblable buildings can be used to create guide-

lines for other products. As discussed in Brennan, Gupta, and Taleb

(1994), the design aspects for disassembly have being recognized by

the industries that generate huge amount of ferrous and plastic waste

like motor cars and appliance sectors.

When the old products come to disassembly plant so that their

components can be recovered in the assembly plant or re-used, op-

erational problems arise. The operational problems that are likely to

arise are layout, resource allocation, process sequencing and disas-

sembly line balancing. As mentioned by Brennan et al. (1994), the op-

erational problems have environmental concerns that are even forced

by the governments like the recycling regulations, and limitations

on the energy consumptions. These regulations affect the operating

costs as extra costs are incurred for covering the expenses related to

the environmental matters.

In their review paper McGovern and Gupta (2011) discussed many

aspects of the operational problems with an emphasis on the disas-

sembly sequencing problem and the DLBP. The disassembly line se-

quencing problem decides on the disassembly process sequence of a

specified disassembly product and the DLBP decides on the assign-

ment of the tasks to each disassembly workstation.

Lambert (2003) provided a review of the disassembly sequencing

literature. Some noteworthy studies on disassembly sequencing are

due to Lambert (1997), Navin-Chandra (1994) and Lambert (2002).

Navin-Chandra (1994) used a modified traveling salesman problem

so as to minimize total costs while meeting obligatory reclamation of

definite parts. Lambert (1997) proposed a graph based method so as

to maximize the economic performance of the disassembly process

within given technical and environmental constraints. Both meth-

ods were applicable to the disassembly of complex consumer prod-

ucts like automotive vehicles, consumer electronics and mechani-

cal assemblies and were applied to a headlamp and ballpoint pen

disassembly products. Lambert (2002) proposed a linear program-

ming based solution procedure for the minimum cost disassembly

sequence of an electronic equipment. His method was applicable all

products having hierarchical modular structures.

The DLBP was first introduced in Güngör and Gupta (2001b).

Güngör and Gupta (2001a) mentioned several complications like



arly leaving, self-skipping, skipping, disappearing and revisiting

ork pieces that could be faced in disassembly line practices. To re-

uce the effects of the complications on the disassembly process,

hey proposed a shortest path based solution algorithm. McGovern

nd Gupta (2007a) showed that the decision version of the DLBP

s NP-complete. McGovern and Gupta (2007b) considered a DLBP

o minimize the number of workstations while balancing the idle

imes between the workstations. They proposed an exhaustive search

ethod that returns optimal solutions for small sized problem in-

tances and a genetic algorithm that finds high quality solutions, for

he large sized problem instances.

Güngör and Gupta (2002) proposed a heuristic procedure for the

LBP under complete disassembly. They assumed an infinite supply

f a single product and considered the efficient utilization of the re-

ources while satisfying the minimum release quantity. They illus-

rated the proposed heuristic on an eight task personal computer dis-

ssembly example. Koc, Sabuncuoglu, and Erel (2009) considered a

omplete disassembly DLBP so as to minimize the number of work-

tations. They introduced AND/OR graphs and proposed Integer and

ynamic Programming formulations.

Altekin, Kandiller, and Özdemirel (2008) studied the DLBP un-

er partial disassembly and an infinite supply of a single product.

hey formulated their net revenue maximization model as a mixed-

nteger linear program and used its relaxations to find lower and up-

er bounds. They stated that their approach could be used in design-

ng and operating remanufacturing systems where large volumes of

imilar products should be disassembled. Altekin and Akkan (2012)

roposed a predictive–reactive approach based on a mixed-integer

odel to improve the profitability of the disassembly line. A predic-

ive balance was created and then given a failure, the line was re-

alanced. They stated that their algorithm could be used as a guide

y the disassembly workers about how to proceed in case of a task

ailure.

The recent disassembly lines research considered the uncer-

ainty of the task times and product quality. Bentaha, Battaïa, and

olgui (2014a, 2014b, 2014c, 2015) and Bentaha, Battaïa, and Dolgui

2014d) studied complete and partial disassembly models, respec-

ively. Bentaha et al. (2014a) modeled uncertainty using the notion of

esource cost and proposed a sample average approximation method.

entaha et al. (2014b) studied the joint problem of disassembly line

alancing and sequencing. Bentaha et al. 2014c proposed a lagrangian

elaxation approach to maximize the total profit. Bentaha et al. (2015)

onsidered the problem of minimizing the workstation operation

osts and additional costs for handling the hazardous parts. They de-

eloped several lower and upper bounding mechanisms. Bentaha et

l. (2014d) addressed workload balancing problem with fixed num-

er of workstations. They developed a stochastic binary program.

Ding, Feng, Tan, and Gao (2010) stated that a successful disas-

embly line often requires an integrated consideration of many ob-

ectives and proposed an ant colony algorithm to generate the effi-

ient set. They tested their algorithm with three objectives: number

f the workstations, workload balancing between the workstations

nd demand rating. They illustrated the heuristic on a 25 task cellular

hone disassembly example. Paksoy, Güngör, Özceylan, and Hancilar

2013) mentioned that in real world applications the objectives could

e fuzzy due to incomplete information and proposed fuzzy goal pro-

ramming and fuzzy multi-objective programming. They considered

hree objectives: number of the workstations, workload balancing

etween the workstations and cycle time. The approaches were ap-

lied on 10 tasks flash light and 30 tasks radio examples. Hezer and

ara (2014) introduced parallel line dissassembly balancing problem

nd proposed a network model based approach for its solution. For

ore information on the dissassembly line balancing and planning,

ne may refer to Ullerich (2014).

The most closely related published work to ours is Altekin et al.

2008)’s study. Our problem environment differs from theirs in the



E.G. Kalaycılar et al. / European Journal of Operational Research 000 (2015) 1–13 3


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Fig. 1. AND precedence relations.

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Fig. 2. POR precedence relations.

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ense that we take the number of workstations and cycle time as pa-

ameters whereas they decide on the number of workstations and

n the cycle time. We assume that there is a finite supply of disas-

embly product and that there is a lower bound on the quantity of

he part releases whereas they assume infinite supply and take min-

mum release quantities as upper bounds on the quantity of the part

eleases. Hence their solution considers a single line balance for all

nits, whereas, to satisfy the minimum release quantities, we use dif-

erent task assignments to disassemble different units.

. Problem definition

We consider a disassembly line balancing model with a finite sup-

ly of a single product. There are N tasks where task i is specified

y its cost ci and processing time ti. The tasks that release parts are

alled part releasing tasks. The minimum release quantity of each

art-releasing task i is di units. di is the associated demand for part

from the third party like the producers of the new product mod-

ls. The part-releasing task i has a revenue ri that includes its mar-

et value (for example when the part is sold to the third party) and

ecycled material value. The parts may have negative revenues that

epresent the disposal costs. Ri is the net revenue of task i, Ri = ri – ci.

t the end of the period, each unit of part i that is released excess of

i units can be held in the inventory for future orders and generates a

nit revenue ri when it is sold. Such excess releases would be favored

f they lead to increases in the total net revenue. Our problem is to

ssign the tasks to the workstations of the disassembly line for each

nit of the product so as to maximize the total net revenue. We make

he following additional assumptions.

• The workstations are already set up and there are K workstations.• All workstations are identically equipped and are capable of per-

forming any one of the tasks at the same pace.• There is a single disassembly product with a finite supply of S

units.• There is a single disassembly period of length L, that is sufficient

to disassemble all S units. That is, there is a single batch of size S

to be disassembled in L time units.• The cycle time, C, is the time between the completion of two units

consecutively leaving the disassembly line, hence it is the process-

ing workload that can be allocated to each workstation. One may

view the cycle time to be the ratio L/S .• The cycle time, C, is known with certainty and is not subject to

change as the period length L and supply S, are known with cer-

tainty and are not subject to change.• All units of the product are identical, hence contain all disassem-

bly parts.• A partial disassembly is done, i.e., a subset of -but not- all parts is

released.• The subset of parts released from one unit of the product may be

different from another unit, hence different task assignments are

possible while disassembling different units of the product.

The network comprises AND and OR type relations. Predecessor

ND (PAND) relations imply that a task cannot start before all of its

redecessor tasks are finished. The following figure illustrates the

ND precedence relations.

According to Fig. 1, tasks a, b, c and d are predecessors of task e.

ask d is the immediate predecessor of task e as there is no task in

etween.

We use AND(i) to denote the set of all AND predecessors of task i

nd PAND(i) to denote the set of all AND immediate predecessors of

ask i.

Predecessor OR (POR) relations imply that at least one task in a

pecified set should be complete before the successor task may begin.

ig. 2 illustrates POR type relations.



According to the above figure, at least one task in set { f, g, h} must

e completed before task i may begin. We let POR(i) denote the set of

R predecessors of task i.

. Mathematical model and properties

The mathematical formulation given below determines the opti-

um task assignments to the workstations while disassembling each

nit that maximizes the total net revenue. The decision variables of

he model are,

iks ={

1 if task i is assigned to workstationk for s th unit of the product

0 otherwise

Constraint set (1) deals with the minimum release quantities. The

mount of each part released should be higher than its minimum re-

ease quantity.

K

k=1

S∑s=1

Xiks ≥ di i = 1, . . . , N (1)

Constraint set (2) states the cycle time limit should not be ex-

eeded at any workstation for any unit.

N

i=1

tiXiks ≤ C k = 1, . . . , K; s = 1, . . . , S (2)

Constraint sets (3) and (4) ensure that the precedence require-

ents are observed. The AND precedence relations should be satis-

ed. Thus task i cannot be assigned to a workstation that is before

he workstation of its AND predecessor task l.

K

k=1

kXlks ≤K∑

k=1

kXiks i = 1, . . . , N, l ∈ PAND(i), s = 1, . . . , S

(3)

Similarly, the OR precedence relations should be satisfied. Thus

ask i can be assigned to workstation k if and only if at least one of its

R predecessors is assigned to workstations 1 through k.

iks ≤k∑

h=1

∑j∈POR(i)

Xjhs i = 1, . . . , N; k = 1, . . . , K; s = 1, . . . , S

(4)





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Constraint set (5) ensures that a task can be assigned to at most

one workstation.

K∑k=1

Xiks ≤ 1 i = 1, . . . , N; s = 1, . . . , S (5)

The assignment variables should be nonnegative as by constraint

set (6).

Xiks ≥ 0 and integer i = 1, . . . , N; k = 1, . . . , K; s = 1, . . . , S

(6)

It follows from Constraint Sets (5) and (6) that Xiks is binary.

The model, hereafter referred to as Model I, is defined by objective

function (7) below, together with constraint sets (1) through (6).

Max

N∑i=1

K∑k=1

S∑s=1

RiXiks (7)

Model I has N ∗ K ∗ S binary decision variables.

McGovern and Gupta (2007b) showed that the decision version of

the DLBP is strongly NP-complete. It follows that our DLBP problem is

NP-hard in the strong sense. We thus resort to several approaches to

solve the DLBP approximately or heuristically. We also develop theo-

retical instruments, more specifically, valid constraints, that improve

the solution effort. These topics are discussed in the remainder of the

paper. In this section, we introduce the key insight into reducing the

problem size.

In typical implementations, the model solution yields different

disassembly tasks and different part-releasing assignments for the

batch of S units. Typically, first a subset of the batch is disassem-

bled to satisfy the minimum release quantity constraints. Recall that

minimum quantities for parts are novel and distinguishing aspect of

the current work. Once the minimum release quantities are satisfied,

though, the solution understandably favors task assignments that

yield the maximum revenue. Note that, once all of the part minimum

release quantities have been met, the specific task assignments to at-

tain the maximum revenue is independent of the remaining quantity

to be disassembled. In fact, the maximum revenue yielding task as-

signments can be found quite easily by solving Model I with S = 1. The

maximum revenue yielding task assignments could be replicated as

many times as needed for the remaining parts.

The above insight immediately suggests an approach to make

problems with large S more tractable. If one had access to the number

of units needed, say Y ≤ S, to be disassembled in order to satisfy the

minimum release quantities, then Model I could be solved using Y in-

stead of S. Since Y ≤ S, and sometimes even Y << S, the computations

are simplified. The remaining task assignments are then computed by

solving Model I with S = 1 and then replicating the solution for the

remaining S–Y units.

Clearly, the above approach would also work for an upper bound

of Y. Even an approximate value of Y may be useful. If the chosen

value of Y gives an infeasible solution, then we could investigate ways

to increase Y to obtain a feasible solution. In the following sections,

we discuss how the quantity Y may be computed, exactly or approxi-

mately, and how bounds for Y may be obtained.

4.1. Computing the number of units to be disassembled to satisfy the

minimum release quantities

The formulation given below, henceforth referred to as Model II,

finds the minimum number of units to be disassembled to satisfy all

minimum release quantities.

Model II can be defined by the following additional variable and

constraint.

s ={

1 if s th unit is disassembled to satisfyminimum release quantity

0 otherwise



N

i=1

K∑k=1

Xiks ≤ μ∗Ys s = 1, . . . , S (8)

Constraint Set (9) assigns Ys to 1, if there is at least once task as-

igned to any workstation while disassembling sth unit of the prod-

ct. Note that∑N

i=1

∑Kk=1 Xiks is bounded from above by μ. We use μ

N∗K. If∑N

i=1

∑Kk=1 Xiks ≥ 1 then Ys = 1 otherwise Ys is zero as we are

inimizing Y = ∑Ss=1 Ys. Y = ∑S

s=1 Ys gives the total number of units,

ith at least one task assignment. Hence it is the minimum number

f disassembly units to satisfy all minimum release quantities.

The objective function of Model II is expressed as below.

in

S∑s=1

Ys − εP

N∑i=1

K∑k=1

S∑s=1

RiXiks (9)

The above objective primarily minimizes the number of disassem-

ly units to satisfy all minimum release quantities. Among the solu-

ions that attain the minimum number of disassembly units Y, the

odel selects the one with the maximum net revenue. To maximize

he total net revenue while keeping the minimum number of disas-

embly units, some parts are released more than their minimum re-

ease requirements. The remaining S–Y units are handled by Model III

Section 4.2).

The magnitude of εp used in (9) is important in the sense that it

hould not increase Y = ∑Ss=1 Ys en route to increasing total net rev-

nue. εp should be set small enough to satisfy constraint (10).

S

s=1

Ys − εP∗Pmin ≤

S∑s=1

Ys + 1 − εP∗Pmax (10)

here Pmin (Pmax) = minimum (maximum) possible total net rev-

nue.

Hence any solution with a larger Y = ∑Ss=1 Ysvalue should not be

avored even if it leads to the maximum increase in the total net rev-

nue value. Constraint (10) implies that εP∗Pmax − εP

∗Pmin ≤ 1. This

ollows εP ≤ 1Pmax−Pmin

We use Pmax = S∗ ∑Ni=1 Ridi and Pmin = 0. Hence

P = 1

S∗ ∑Ni=1 Ridi

guarantees the maximum revenue solution among

nes with the minimum number of disassembled units.

Model II is completely defined by the objective function expressed

n (9) together with constraint sets (1)–(6) and (8).

The model satisfies all minimum release requirements by disas-

embling Y = ∑Ss=1 Ys units. For the remaining (S–Y) units, the same

ask assignments are used. To find these task assignments we mod-

fy Model I by setting S = 1, and then replicating the solution for the

emaining S–Y units. This solution, that is, Model I with S = 1 and

hen replicated as needed, is referred to as Model III. In other words,

odel III finds the maximum net revenue without considering mini-

um number of part releases.

.2. Maximum revenue solution in the absence of part minimum

elease quantities

In the following model, Model III, we drop subscript s from our

ecision variable Xiks as we are considering a single unit. The decision

ariable becomes Xik where

ik ={

1 if task i is assigned to workstation k0 otherwise

The objective function is as expressed in (11).

ax

N∑i=1

K∑k=1

RiXik (11)

We also drop the minimum release quantity constraint as they are

et by Model II. Model III is defined by the objective function ex-

ressed in (11) together with constraint sets (2) through (6) by setting

= 1.





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.3. Solution procedure using Model II and Model III

Model I given above yields the optimum solution but suffers from

omputational difficulties as the problem size increases. Alterna-

ively, Model II and Model III may now be used to find an optimal

olution as given by the following procedure.

Procedure 1. Optimal Solution via Model II and Model III

Step 1. Use Model II to find the task assignments for the first Y =∑Ss=1 Ys disassembly units.

Step 2. Use Model III to find the task assignments for the last (S–Y)

disassembly units.

Step 3. The optimal total net revenue over all S units is:

Total Net revenue by Model II + (S – Y) ∗ Total Net revenue by

odel III

Model II is more complex than Model I due to the added Ys vari-

bles, hence in finding the optimal solutions, Procedure 1 should not

e favored to Model I. However, Procedure 1 is of importance, since

e use the procedure to derive upper and lower bounds on the opti-

al total net revenue value.

.4. Solution procedure using Model I and Model III

Model I and Model III together may be used to find optimal solu-

ions as well. Recall that Y = ∑Ss=1 Ys returned by Model II is the min-

mum number units to be disassembled to satisfy all minimum part

equirements. If an upper bound on Y, UB(Y), is available, then Model

can be solved with UB(Y) in place of S units. Thereafter, Model III

ptimally solves the remaining S–UB(Y) units. Procedure 2, below, is

sed to find an optimal solution using Model I and Model III. The up-

er bound UB(Y) may be guessed, or sought using a simple algorithm

uch as bisection search. We will revisit the task of finding UB(Y) in

he following sections.

Procedure 2. Optimal Solution via Model I and Model III

Step 1. Solve Model I with S = UB(Y).

Step 2. Solve Model III to find the task assignments for the remain-

ing S –UB(Y) units.

Step 3. Find the optimal total net revenue over all S units as:

Total net revenue over UB(Y) units by Model I+(S–UB(Y))

nits∗Total net revenue over a single unit by Model III

Note that there are three ways of finding an optimal solution: the

riginal Model I, Procedure 1, and Procedure 2. Procedure 2 should be

avored to original Model I, in particular when compared to S, a small

B(Y) is available.

In our experiments, we use Model I to find the optimal solutions.

e develop two upper bounds to the optimum objective function

alue that use the Linear Programming Relaxations (LPRs) of the orig-

nal model. We benefit from the ideas used in Model II and III to find

third upper bound and to develop a heuristic solution procedure.

hese are presented next.

. Solution approaches

As mentioned, our DLBP is NP-hard in the strong sense. Hence

here is little chance of finding polynomial even a pseudo-polynomial

lgorithm that solves it optimally. In the section we present our solu-

ion approaches that provide upper bounds and a heuristic solution

o our DLBP.

.1. Upper Bounds

Upper Bound 1 (UB1): An optimal solution to any relaxation pro-

ides an upper bound on the optimal objective function value of our

aximization problem.



Model I is solved to optimality by relaxing the integrality con-

traints on the Xiks values. The resulting optimal objective function

alue, ZLP, provides an upper bound on the optimal total net revenue.

he only difference between the original model and its LPR is the in-

egrality requirement of the Xiksvalues. Consequently, ZLP is an upper

ound on the optimal objective function value Z∗.

Upper Bound 2 (UB2): UB2 is found by adding valid cuts to the LPR

f Model I. The valid cuts are found by investigating the properties of

he optimal solutions that may not be satisfied by the optimal LPR.

Cut1–Fixing the minimum release quantities

heorem 1. There exists an optimal solution in which di units of part i

re released from the first di disassembly units.

roof. Assume an optimal solution in which the first di units are re-

eased from di units of the product. If the tasks in any two units, say

and b, are changed, i.e., Xika is set to Xikb and Xikb is set to Xika for

ll i , then the optimal net revenue is not affected. Assume an optimal

olution in which part i is released in units a+1, a+ di . Setting Xjkr =jk(a+r) for all j, r ≤ dj and Xjk(ta+r) = Xjkr does not change the revenue

alue. Then, there is an optimal solution in which dj units of part i is

eleased in the first di units. �

Using Theorem 1, dr units for any part r can be obtained from the

rst dr disassembly units. Constraint set (12) below ensures that dr

nits of part r are released from the first dr disassembly units. To in-

rease the power of (12), we select task r such that dr=Maxi{di }.

K

k=1

dr∑s=1

Xrks = dr s = 1, . . . , dr (12)

In computational studies, we observe that the constraint set (12)

liminates many alternate optimal solutions, thereby decreasing the

omputational effort. The reduction in computational effort was re-

ected as significantly reduced CPU times.

Cut2 –Lower Bound on the workstation positions

Cut2 provides a lower bound on the workstation index that any

articular task.

heorem 2. In all optimal solutions, task i cannot be assigned to work-

tations 1, 2, . . . , Ei − 1 where Ei = � ti+Min j∈POR(i){t j}+∑j∈AND(i) t j

C �.

roof. Task i cannot start before its predecessors and at least one

f its immediate POR predecessors are completed. Hence it should

ait at least Min j∈POR(i){t j} units for its POR predecessors and

j∈AND(i) t j units for its AND predecessors to start and at least

i + Min j∈POR(i){t j} + ∑j∈AND(i) t j units to complete. When task split-

ing is allowed and other tasks are ignored, ti + Min j∈POR(i){t j} +j∈AND(i) t j units require � ti+Min j∈POR(i){t j}+∑

j∈AND(i) t j

C � workstations.

hen task splitting is not allowed and other tasks are consideredti+Min j∈POR(i){t j}+∑

j∈AND(i) t j

C � is a lower bound on the earliest station

or task i. �

In the absence of POR relations, Theorem 2 reduces to the one de-

ived in Patterson and Albracht (1975) for the classical assembly line

alancing problem.

Constraint set (13) below supports Theorem 2.

i−1∑k=1

S∑s=1

Xiks = 0 i = 1, . . . , n (13)

Constraint set (13) ensures that task i is not assigned to work-

tations 1, 2, …, Ei–1. The network in Fig. 3 is used to illustrate Ei

omputations.

Note that { t1 + t2 } is the minimum total processing load to start

ask A and Min {t3, t4}+ t5 is the minimum total processing load to

tart task B. Then Min{(t1 + t2+ tA), (Min {t3, t4} + t5 + tB)} is the





Min {SA+tA , SB+tB } = Sj

t1 + t2 = SA

Min {t3 , t4 } + t5 = SB

A

j

B

1

2

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5

Fig. 3. An example network to illustrate Ei computations.

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minimum total processing load to start task j.

Note that Ei =⌈

ti + Minimum total processing load tostart task i

C

⌉.

Hence Ei =⌈

ti + Min{(t1 + t2 + tA), (Min{t3, t4} + t5 + tB)}C

⌉.

The network in Fig. 4 is adapted from Bourjault (1984) to illustrate

Theorem 2.

We take K = 4, C = 63, S = 80 and the task times are as given

below:

Tasks 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

ti 13 7 1 6 7 5 15 8 17 3 14 12 17 4 11 18 15 19 4 18 16 19

E13 =⌈

t13 + Min{(t1 + t3 + t4), (t1 + t2)} + t22 + t7

C

⌉

E13 =⌈

17 + Min{(13 + 1 + 6), (13 + 7)} + 19 + 15

63

⌉=

⌈71

63

⌉=2

The theorem states that task i cannot be assigned to workstations

1, 2, …, Ei – 1. Accordingly, task 13 cannot be assigned to workstation

1 (E13– 1 = 1).

Cut3–Projecting minimum release quantities

Altekin et al. (2008) stated that in the optimal solution, the total

fraction of any task assigned is never greater than the total fraction

of each of its assigned AND predecessors and the total fractional as-

signment of all of its OR predecessors, and moreover the constraints

may not be satisfied for the optimal LPR. Their results hold for any

problem that contains PAND and POR relations and an infinite supply.

We generalize their results to our problem where the supply is finite

and impose the constraint sets (14) and (15).

K∑k=1

S∑s=1

Xiks ≤K∑

k=1

S∑s=1

Xjks i = 1, . . . , n; j ∈ PAND(i). (14)

K∑k=1

S∑s=!

Xiks ≤∑

j∈POR(i)

K∑k=1

S∑s=1

Xjks i = 1, . . . , n (15)

Cut4–Existence of a feasible solution

Our last cut uses the results of the following two theorems.

Theorem 3. If there exists a feasible schedule that processes all tasks

other than task i and its successors in r workstations then there exists an

optimal schedule in which task i is processed in workstations 1 through r.

Proof. Assume task i is assigned to workstation k such that k ≤ r and

workstations r +1 through K process the successors of task i. Taking

task i from workstation k and placing to workstation l such that l ≥r + 1 cannot increase the total net revenue as such a replacement



annot allow more task assignments. This is due to the fact that all

asks in workstations r + 1 through K are successors of task i, hence

annot be processed in workstations 1 through r, since task i is to be

rocessed later. Then, there exists an optimal solution in which task i

s processed in the first r workstations. �

heorem 4. If there exists a feasible schedule that processes all tasks

ther than task i and its predecessors in the last r workstations, then

here is an optimal schedule in which task i is processed in workstations

– r + 1 through K.

roof. Assume task i is assigned to workstation k such that k ≥ K –

+ 1 and workstations 1 through K – r process the predecessors of

task j. Taking task i from workstation k and placing it to workstation

l such that l ≤ K – r cannot increase the total net revenue, as such a

eplacement cannot allow more Assignments. This is due to the fact

hat all tasks in workstations 1 through K–r are predecessors of task i,

ence cannot be processed in workstations K – r + 1 through K, since

ask i is processed earlier. Then, there exists an optimal solution in

hich task i is processed in the last r workstations. �

Note that Theorems 3 and 4 require a feasible solution. To find

uch a solution we develop the following simple rule.

Let A be a set of tasks that should be assigned to K workstations.

rder the tasks in set A according to non-decreasing task times. Take

he tasks from the order starting from the first feasible task. A task is

easible if its inclusion to the current workstation does not violate the

recedence relations and cycle time constraints. If no such task exists,

lose the current workstation and open a new one. Stop when all jobs

n set A is assigned. Let m(A) be the resulting number of workstations.

Following Theorem 3, we set A = A1 where A1 is the set of all tasks

xcept the successors of task i. Following Theorem 4, we set A = A2

here A2 is the set of all tasks except the predecessors of task j. After

(A1) and m(A2) are obtained, implementing the heuristic for sets A1

nd A2, we include one of the two cuts (16) and (17).

S

s=1

K∑k=m(A1)+1

Xiks = 0 i = 1, . . . , N (16)

S

s=1

m(A2)−1∑k=1

Xiks = 0 i = 1, . . . , N (17)

We select one of the cuts for the LPR using the following rule.

Rule. Use Cut (16) if m(A1) ≥ m(A2) – 1, else use Cut (17)

The rule selects the cut that prevents more assignments, hence

he stronger one.

We hereafter refer to the LPR using the cuts (12) through (17)

s strengthened LPR. UB2 is the total net revenue returned by the

trengthened LPR.

Upper Bound 3 (UB3): To obtain UB3, we use the idea used in

odel II. Recall that Model II aims to satisfy all minimum release

uantities (∑

i di) and minimizes Y = ∑Ss=1 Ys. Here in place of Y, we

se LB(Y), to get rid of the Ys binary variables.

First, we find LB(Y), and then increase it one by one, until a feasible

olution is reached. Theorem 5 states LB(Y).

We let dnewi = Max{di, Max j|i∈AND j{d j}}. Note thatdnewi is the

inimum release quantity of task i which is modified by its mini-

um release quantity and the minimum release quantities of all its

ND successors.

heorem 5. LB(Y) = �∑n

i=1 ti∗dnewi

K∗C � is a valid lower bound on the num-

er of units to satisfy all minimum release quantities.

roof. The total processing required to produce all minimum re-

ease quantity is∑n

i=1 ti∗dnewi Each unit can be disassembled in no

ore than K∗C time units as each workstation is available for C time

nits. �∑n

i=1 ti∗dnewi∗ � is the number of units disassembled to satisfy
K C




1

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Fig. 4. Precedence diagram adapted from Bourjault’s ball-point pen example.

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ll minimum release quantity if task splitting between the disas-

embly units and workstations were allowed. As no task splitting

s allowed �∑n

i=1 ti∗dnewi

K∗C � is a lower bound on the number of units

isassembled. �

We illustrate Theorem 5 and UB3 on the network in Fig. 4 with

= 4, S = 80 and C = 63. The minimum release quantities, task times

nd dnewi values are given in Table 1.

Using Theorem 5 we find LB(Y) = � 13∗4+7∗4+···+19∗44∗63 � = 3. We first

et the number of units to LB(Y) = 3. However a feasible solution

or the LPR is not found. There is no feasible solution with 4 units

s well. The LPR finds a feasible solution with 5 units. Then the mini-

um number of units to satisfy all minimum release quantities, Y = 5.

odel III is used to find the line balance for the remaining 75 units.

ormally we let,

ZA = maximum net revenue for the strengthened LPR problem

with Y units

ZB = maximum net revenue for a single unit problem with

Model III.

UB3 = ZA + (S–Y)∗ZB

In our example, the strengthened LPR finds ZA as 1626. Model III

nds ZB , as 326. Then UB3= 1626 + (80 – 5) ∗ 326 = 26.076.

Note that here, UB2 = UB3. ZA is found much easier than the LPR

ith S = 80. Moreover ZB is found easily as it considers a single unit.

ence one should expect to obtain UB3 much quicker than UB2.

For this instance we find that UB1=26.796 and the optimal net

evenue = 26.076. Note that UB3, improves UB1 by 720 money units

nd catches the optimal solution in considerably small solution

ime.



.2. Lower bound, heuristic solution

The lower bound uses UB3 as a stepping stone to find a feasible so-

ution. The motivation to use UB3 is its satisfactory behavior in terms

f its solution quality and time. The bound first solves UB3, then fixes

ts variables that are assigned to 1 and obtains a reduced problem.

he reduced problem has only the partially assigned or unassigned

asks of the UB3 solution. As the partial assignments are quite low

ompared to the original variables, the reduced problem is very small

n size. Since our MILP solver can handle small-sized problems very

uickly, we prefer to use it to find an optimal solution for partially

ssigned tasks.

After fixing the variables to 1, the resulting solution obtained by

ILP may violate the cycle time constraints or the precedence con-

traints. If the cycle time constraint is violated, then we increase the

umber of disassembly units one by one, until we obtain a feasible

olution. If the precedence relations are violated, the tasks whose fix-

ngs cause violation, are set to zero, and the MILP is solved again.

Below is the stepwise description of our lower bounding

rocedure.

Step 0. Let LB(Y) = �∑n

i=1 ti∗dnewi

K∗C � and let T = LB(Y)

Step 1. Solve the LPR with cuts for T units of the disassembly

product.

Fix the variables that have value ‘1’ in the LPR solution.

Step 2. Find the optimal solution by Model I for the non-fixed tasks.

If the solution is feasible then let ZC be the objective function

value and go to Step 4.

Step 3. If infeasibility is due to the cycle time constraints then let T

= T+1, go to Step 1.

If infeasibility is due to the precedence relations then set the

variable of a task to zero if one of its predecessor tasks is not

assigned to ‘1’, go to Step 2.





Table 1

The data for the Bourjault’s ball-point pen problem.

Tasks 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

di(units) 0 0 0 4 0 4 0 3 4 4 4 4 4 3 2 1 0 4 2 1 0 0

ti(time units) 13 7 1 6 7 5 15 8 17 3 14 12 17 4 11 18 15 19 4 18 16 19

dnewi(units) 4 4 4 4 4 4 4 3 4 4 4 4 4 3 2 1 0 4 2 1 0 4

ci(money units) 17 18 14 7 14 8 8 15 16 7 15 12 6 17 17 6 9 13 12 16 7 15

ri(money units) 49 10 35 37 16 12 20 37 43 32 24 41 19 35 19 18 48 21 10 40 16 11

Ri (money units) 32 -8 21 30 2 4 12 22 27 25 9 29 13 18 2 12 39 8 -2 24 9 -4

1

25

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12

7

32

31

17

348

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16

33

23

24

28

29

21

20

22

19 25

26

27

18

Fig. 5. Precedence diagram adapted from Lambert’s radio example.

a

Step 4. Find an optimal solution to the single unit problem using

Model III.

Let ZD be the objective function value

The heuristic gives a total net revenue of ZC + (S – T) ∗ ZD.

6. Computational experiment

In this section the data generation scheme is discussed and the

computational results are evaluated. Two networks from the previous

works are used for network generation.

1st Network (22 tasks). Bourjault (1984) illustrated a disassembly

network for a 10-part ballpoint pen. We adapt this network by

replacing the successor OR relations with successor AND rela-

tions and give the resulting network in Fig. 4.



2nd Network (34 tasks). Lambert (1997) illustrated a disassembly

network for a 20-part radio. We adapt this network by replac-

ing the successor OR relations with successor AND relations

and give the resulting network in Fig. 5.

Three networks are found by extending the small networks using

dditional arcs.

3rd Network (47 tasks). 2nd network and a part of the 1st net-

work are combined with some arcs added.

4th Network (60 tasks). 3rd network and a part of the 1st network

are combined with some arcs added.

5th Network (73 tasks). 4th network and a part of the 1st network

are combined with some arcs added.

The following two sets for the number of workstations K are used.





Table 2

UB1 performances.

Variables Set C1 Set C2

Set D1 Set D2 Set D1 Set D2

Avg Max Avg Max Avg Max Avg Max

Set K1 N = 22 3,520 Percent DEV 1 2 1 2 1 2 1 2

# of frac. 722.4 1636 1120.7 1778 487.1 1356 470.6 1684


# of frac. 2437.5 3643 2497.4 3592 1628.1 3017 1900.4 3306


# of frac. 715.9 1950 1150.8 2251 786 2367 741 2210


# of frac. 3803.1 4864 4238.7 5136 3004.1 4307 3102.6 4386


# of frac. 1184 4011 2509.4 3924 584.2 2403 1152.4 3391


# of frac. 5492.4 6865 5910.8 6966 4041.1 5888 4654.9 5946

Set K1 N = 60 19,200 Percent DEV - ∗ - - - - - - -

# of frac. 8749.8 9579 8999.5 9791 6887.3 8344 7751.1 8935

Set K2 N = 60 38,400 Percent DEV - - - - - - - -

# of frac. 13515.3 14742 13853.1 15153 11399.3 13268 12307.1 14280


# of frac. 10876.3 11804 11249.8 11974 9137.3 10462 9716.3 10633


# of frac. 16730.7 18440 17291.3 18654 14045.5 15689 15019.8 16147

∗ ‘-‘ illustrates that optimal solution for large networks could not be obtained.

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Set K1. K = 2 for small networks with 20, 34, 47 tasks

K = 4 for large networks with 60, 73 tasks

Set K2. K = 4 for small networks with 20, 34, 47 tasks

K = 8 for large networks with 60, 73 tasks

The following two sets for the cycle times, C are used.

et C1.C =⌈∑

ti

K

⌉

et C2.C = 1.5∗⌈∑

ti

K

⌉The processing times are generated from a discrete uniform dis-

ribution [1, 20]. A task receives a minimum release quantity, hence

alled a part-releasing task, according to a random process. We gen-

rate a random number between 0 and 1. If the generated number is

elow 0.3, we set its minimum release quantity to zero.

Two distributions are used to generate their minimum release

uantities.

Set D1. Minimum release quantity is uniform [1, 5]

Set D2. Minimum release quantity is uniform [5, 10]

Note that Set D1 contains instances with a low minimum release

uantity, and Set D2, with a high minimum release quantity.

The number of units, S, is set to 80. The costs are generated from a

iscrete uniform distribution between 5 and 20 and the revenues are

enerated from a discrete uniform distribution between 10 and 50.

There are 40 combinations due to 5, 2, 2 and 2 alternatives for

, K, C and D respectively (3∗2∗2∗2 = 24 for small networks and∗2∗2∗2 = 16 for large networks). For each combination, 10 problem

nstances are generated. That is, a total of 400 problem instances are

sed in our experiments.

The mathematical models are solved by CPLEX 10 and the algo-

ithms are coded in Microsoft Visual C++ 2008. The experiments are

un in Intel Core2 Duo 2.00 gigahertz, 2 gigabyte RAM.

The execution of the MILP (Model I) is terminated if the optimal

olution is not returned in 3600 seconds. The optimal solutions are

ound within 3600 seconds when N = 22, 34 and 47. However, optimal

olutions were not found when N = 60 and 73 .

First the performance of our upper bounds is studied. The devia-

ion of an instance is found as %Dev = (UBi−OPT

) × 100 where OPT =
OPT


he optimal net revenue and UBi = the total net revenue by upper

ound i. The average and maximum deviations and number of frac-

ional variables are reported in Table 2 and Table 3, for UB1 and UB2,

espectively.

As can be observed from the tables, the deviations are consistently

ow over all problem sets. For UB1, the average deviations are below

percent and almost all maximum deviations are below 3 percent

hen N = 22, 34 and 47. Note that the deviations do not deteriorate

ith an increase in N. However they deteriorate with an increase in K

ue to the inflation of the number of binary variables. When N and K

alues are fixed, the minimum release quantities and cycle times do

ot affect the deviations significantly.

UB2 produces slightly smaller deviations due to the power of

ur cuts. For example, the maximum deviation for UB1 is 3 percent,

hereas it is nearly zero for UB2 in sets K1, C1 and D1 for N = 22.

e also give the number of fractional variables in Tables 2, 3 and 4.

he number of the fractional variables produced by the LPR is too

igh. For example when N = 22, for sets K2, C2 and D2 (i.e., 4 work-

tations, high minimum release quantity, and high cycle time cases),

306 out of 7040 variables are found to be fractional. This value

educes to 412 when cuts are added. On the average, the cuts re-

uce the number of fractional variables from 1900.4 to 180.3. The

umber of fractional variables is generally affected by the N val-

es. For example, from Tables 2–4, it can be seen that for sets K2,

2 and D2, increasing N increases the number of fractional vari-

bles of UB2, on the average. Note that the average number of frac-

ional variables are 1900.4, 3102.6 and 4654.9 for N = 22, 34 and 47

espectively.

It can also be observed from the tables that, for fixed N, an increase

n the number of workstations increases the number of the fractional

ariables. For example for N = 22 and sets C1 and D1, the average

umber of fractional variables is 722.4 when K = 2, and 2437.5 when

= 4. This is due to the fact that more workstations lead to higher

ask splits, leading to more fractional variables. For fixed N and K,

e observe that increasing C reduces the number of fractional vari-

bles (with few exceptions). For example, when N = 47, K = 2, and

1 combination, the average number of fractional variables are 1184

nd 584.2 for low and high C, respectively. This is due to the fact that

ncreasing C gives more room for complete task assignments, hence

educing the number of fractional variables. In our experiments, we





Table 3

UB2 performances.





# of frac. 65.3 160 67 160 0 0 0 0

Set K2 N = 22 7,040 Percent DEV 0 0 0 0.01 0 0 0 0

# of frac. 368.4 570 376 596 197.8 432 180.3 412


# of frac. 19 160 22.9 158 0 0 0 0


# of frac. 525.2 870 517.7 855 271.1 438 259.1 446


# of frac. 49 160 50 160 0 0 0 0


# of frac. 570.3 970 592 982 370.4 800 362.2 815

Set K1 N = 60 19,200 Percent DEV -∗ - - - - - - -

# of frac. 746 1524 765.7 1546 418.3 1068 424.8 1068


# of frac. 1778.7 2203 1725.2 2116 1192.1 1523 1179.2 1434


# of frac. 657.5 922 649.9 859 460.8 1027 462.4 1045


# of frac. 1789.1 2326 1769 2244 1197.1 1931 1201.8 2008

∗ ’-‘ illustrates that optimal solution for large networks could not be obtained.

Table 4

The number of units with different line balances (out of 80).

N C1 C2

D1 D2 D1 D2


K1 22 5.30 6.00 11.00 12.00 5.50 6.00 11.20 12.00

34 5.10 6.00 10.00 10.00 5.90 6.00 10.00 10.00

47 5.50 6.00 11.30 12.00 5.80 6.00 11.20 12.00

K2 22 5.10 4.00 11.80 13.00 5.30 6.00 10.70 11.00

34 5.00 5.00 10.90 11.00 5.10 6.00 10.00 10.00

47 5.30 6.00 11.10 12.00 5.40 6.00 11.00 12.00

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do not observe a significant effect of the minimum release quantity

figures on the number of fractional variables.

Table 4 reports the number of different line balances incurred in a

period, while S units are disassembled, for small networks.

Note that the number of different line balances returned by Model

II is quite small relative to the total number of disassembly units. For

D1, there are about 5 and 6 different line balances. Increasing D from

D1 to D2, almost doubles the number of different balances, as more

units have to be used to satisfy the minimum release quantities. UB3

is based on the idea of the different line balances, hence one should

expect that it runs much faster than UB2.

The solution times are measured in central processing unit (CPU)

seconds. The average and maximum CPU times for our three upper

bounds are given in Table 5 and Table 6 for small and large networks,

respectively. Table 5 includes the CPU times of the MILP, as the small

networks could be solved to optimality.

As can be observed from the tables, the upper bounds are pro-

duced very quickly. Compared to UB1, UB2 is computed more quickly,

due to the power of the cuts, i.e., their effectiveness in reducing the

solution space.

Moreover, compared to UB2, UB3 is computed more quickly as rel-

atively fewer units, hence fewer variables, are used by the LPRs. For

example, for N = 22 and the instances in sets K2, C2 and D2, the CPU

time decreases from 0.59 to 0.25 seconds by adding the cuts, and sim-

ilarly, the CPU time decreases from 0.25 to 0.08 seconds by using up

to 10 supply units instead of 80. The CPU times of finding the up-

per bounds increase with an increase in N and K. This is due to the



ncrease in the dimensions of the linear programs. For example, for

ets K1, C2 and D2, as N increases from 60 to 73, the average CPU

imes of UB1, UB2 and UB3 increase from 3.18, 0.72, 0.11 to 4.52, 0.83,

.13 respectively.

For N = 34 and sets C2 and D2, as K increases, the average CPU

imes increase from 0.43, 0.08, 0.06 to 1.18, 0.31, 0.09 seconds for UB1,

B2 and UB3 respectively.

For fixed N, K and D values, an increase in the C value decreases the

PU time. This is due to the fact that for large C, more tasks find an as-

ignment for a workstation, and hence assignment decisions are pro-

uced easier by linear programs. For example, for N = 34 and Sets K2

nd D2, increasing C from C1 to C2, decreases the average CPU times

rom 1.67, 0.46, 0.11 seconds to 1.18, 0.31, 0.09 seconds for UB1, UB2

nd UB3, respectively. When the other parameters are fixed, the in-

rease in the D value increases the CPU times slightly. This is because

hen D increases, the number of units with different assignments

ncreases. For example, for N = 34 and sets C1 and K2, increasing D

rom D1 to D2, increases the average CPU times from 1.27, 0.44, 0.08

econds to 1.67, 0.46, 0.11 seconds for UB1, UB2 and UB3 respectively.

he effect of K is more dominant for the MILP, due to the inflation of

he number of decision variables.

For fixed N, an increase in the K value increases the CPU times re-

arkably. For example for N = 22 and sets C1 and D1, an increase in

from K1 to K2 increases the CPU time from 3.83 seconds to 1078.32

econds. However the effect of N is not as significant as that of K.

or example, for K = 2, when N increases from 22 to 34 for sets C1

nd D1, the CPU time increases from 3.83 to 4.37 seconds. Note from

able 5 that for fixed N and K, an increase in the C value decreases

he CPU times remarkably. This is due to the fact that a workstation

as room to accommodate many tasks, thereby leading to easier de-

isions. However D values do not have a consistent effect on the solu-

ion time of the MILP.

Fig. 6 illustrates the CPU times of UB1 for sets K, C and D when N =0 and 73.

As can be observed from Fig. 6 the CPU times of UB1 increase with

ncreases in N and K values and decrease with an increase in C value.

To summarize, our experiments have revealed that high quality

pper bounds could be obtained even for large networks in less than

alf a minute. The upper bound deviations are not affected from

he increases in the problem sizes and CPU times to obtain them





Table 5

The CPU times (in second) of the Upper Bounds and MILP–small networks.

Set C1 Set C2



Set K1 N = 22 UB1 0.33 0.5 0.42 0.54 0.22 0.28 0.23 0.34

UB2 0.09 0.1 0.11 0.14 0.07 0.09 0.07 0.08

UB3 0.05 0.09 0.06 0.08 0.05 0.06 0.06 0.08

MILP 3.83 10.67 10.65 73.21 1.76 2 1.77 2.15

Set K2 N = 22 UB1 0.7 1.25 0.84 1.09 0.47 0.61 0.59 0.73

UB2 0.41 0.67 0.4 0.65 0.23 0.35 0.25 0.36

UB3 0.07 0.1 0.09 0.11 0.06 0.07 0.08 0.1

MILP 1078.32 3599.85 796.07 3600.62 363.9 3599.87 365.29 3600.3

Set K1 N = 34 UB1 0.57 0.86 0.78 0.92 0.36 0.44 0.43 0.53

UB2 0.11 0.12 0.11 0.13 0.09 0.11 0.08 0.11

UB3 0.06 0.08 0.07 0.1 0.06 0.08 0.06 0.07

MILP 4.37 6.04 5.93 8.15 3.22 4.32 3.41 4.44

Set K2 N = 34 UB1 1.27 1.95 1.67 2.21 0.91 1.14 1.18 1.34

UB2 0.44 0.53 0.46 0.58 0.32 0.39 0.31 0.38

UB3 0.08 0.1 0.11 0.12 0.08 0.09 0.09 0.11

MILP 1454.94 3600.34 939.75 3417.72 373.65 3599.81 19.76 27.86

Set K1 N = 47 UB1 1.05 1.36 1.42 1.63 0.48 0.69 0.65 0.87

UB2 0.16 0.18 0.14 0.16 0.11 0.12 0.12 0.13

UB3 0.07 0.11 0.08 0.12 0.06 0.07 0.07 0.09

MILP 22.77 118.85 14.38 64.45 5.65 6.88 5.62 7.02

Set K2 N = 47 UB1 2.46 3.59 2.94 3.51 1.52 1.9 2.17 2.86

UB2 0.82 1.17 0.79 1.05 0.55 0.83 0.56 0.84

UB3 0.09 0.11 0.16 0.2 0.09 0.1 0.12 0.17

MILP 784.49 3600.0 759.57 3600.0 376.38 3600.0 385.74 3600.0

Table 6

The CPU times (in seconds) of the Upper Bounds - large networks.

Set C1 Set C2



Set K1 N = 60 UB1 4.27 5.03 4.89 5.82 2.47 2.94 3.18 3.66

UB2 1.21 1.69 1.26 1.98 0.72 1.12 0.72 1.21

UB3 0.10 0.12 0.16 0.23 0.08 0.10 0.11 0.15

Set K2 N = 60 UB1 13.87 18.6 19.59 25.51 6.82 8.67 9.44 11.55

UB2 5.36 6.79 4.97 6.76 2.52 3.13 2.67 3.44

UB3 0.17 0.21 0.38 0.54 0.15 0.18 0.24 0.33

Set K1 N = 73 UB1 6.27 8.8 7.59 9.42 3.85 4.37 4.52 5.21

UB2 1.57 2.23 1.49 2.51 0.79 1.13 0.83 1.14

UB3 0.11 0.14 0.17 0.22 0.11 0.15 0.13 0.16

Set K2 N = 73 UB1 19.99 23.34 32.14 42.58 9.35 10.88 13.59 17.4

UB2 5.83 8.80 6.92 10.83 3.63 5.06 3.31 4.16

UB3 0.21 0.31 0.56 0.99 0.15 0.17 0.29 0.45

05

101520253035

C1 D1K1

C2 D1K1

C1 D2K1

C2 D2K1

C1 D1K2

C2 D1K2

C1 D2K2

C2 D2K2

N = 60 73

Fig. 6. The average CPU times (in seconds) of UB1.

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t

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a

t

n

ncrease at a linear rate. On the contrary, the solution times of the

ILP model increase at an exponential rate and the model cannot

nd any solution in one hour, for large networks. Hence our upper

ounds could be used in place of the optimal solutions to evaluate

he performances of the lower bounds that are used to find feasible

olutions. t



We finally investigate the performances of the lower bound. For

mall networks with 22, 34 and 47 tasks, the optimal solutions

re found by the MILP and the deviations are measured relative to

he optimal solutions, i.e., Percent Dev = ( OPT−LBOPT ) × 100. For large

etworks, the optimal solutions are not available and the devia-

ions are measured relative to UB2 in place of optimal solutions, i.e.,





Table 7

The Lower Bound performances.





CPU 0.06 0.13 0.08 0.12 0.07 0.09 0.07 0.08


CPU 0.1 0.15 0.12 0.16 0.08 0.1 0.1 0.14


CPU 0.05 0.08 0.08 0.08 0.09 0.11 0.08 0.11


CPU 0.03 0.04 0.13 0.15 0.1 0.12 0.11 0.14


CPU 0.09 0.15 0.1 0.14 0.11 0.12 0.12 0.13


CPU 0.12 0.14 0.19 0.23 0.12 0.14 0.15 0.21

Set K1 N = 60 19,200 Percent GAP 0 1 0.01 1 0 1 0 1

CPU 0.12 0.14 0.18 0.25 0.1 0.13 0.13 0.18

Set K2 N = 60 38,400 Percent GAP 0 1 1 2 1 1 1 1

CPU 0.2 0.24 0.41 0.59 0.18 0.23 0.28 0.38


CPU 0.14 0.17 0.18 0.25 0.14 0.2 0.16 0.21


CPU 0.26 0.34 0.61 1.04 0.19 0.22 0.32 0.49

0

20

40

60

80

100

22 34 47

Average

Maximum

Fig. 7. The optimal total net revenue values.

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7

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Percent Gap = ( UB2−LBUB2

) × 100. The average and maximum deviations

of the lower bounds are reported in Table 7.

Note from the table that the lower bound solutions are quite satis-

factory over all problem sets. The solutions have small deviations and

are obtained in negligible times. The performance does not deterio-

rate with an increase in N. However the deviations slightly increase

with an increase in K. For example for N = 60 and sets C1 and D2, the

average deviations of K1 and K2 are 1 percent, whereas the maximum

deviations are 1 percent and 2 percent for K1 and K2, respectively.

The CPU times to find the lower bounds increase with an increase in

N and K with only a few exceptions. This is due to the fact that the

linear and mixed-integer linear programs have more decision vari-

ables for higher N and K. For example, for sets K1, C2 and D2, as N

increases from 60 to 73, the average CPU times increase from 0.13 to

0.16 seconds. For N = 47 and sets C2 and D2, increasing K from 2 to 4,

increases the average CPU times from 0.12 to 0.15 seconds.

We could not observe any notable effect of C on the lower bound

performances and CPU times. For fixed N, K and C, increasing D, does

not change the performance, but increases the CPU times slightly.

This increase can be attributed to the increase in LB(Y). For example

for N = 60 and sets C1 and K2, the average CPU times are 0.20 seconds

and 0.41 seconds for set D1 and set D2, respectively.

To summarize, our computational experiment has revealed the

excellent performances of the bounds that do not deteriorate as the

number of tasks increases. On the other hand, the optimal total rev-

enue values increase significantly with the increases in the number of

the tasks. Fig. 7 shows the optimal total net revenue values for each

n, over all other parameter combinations, for small networks. For the



arge networks, the optimal solutions, hence their total net revenue

alues are not available.

Note from the above figure that the average optimal total net rev-

nue values are around 30, 45 and 63 money units for 22, 34 and 47

asks, respectively. The respective maximum total net revenue values

re 37, 54 and 78 money units and the respective maximum lower

ound deviations are around 1 percent, 2 percent and 1 percent.

hese altogether imply that the practitioners that use the results of

ur study would lose no more than 1000 money units over the opti-

al total net revenue that are of magnitude of around 50.000 money

nits.

. Conclusions

This study considers disassembly systems that have gained signif-

cant importance in recent years. This heightened importance stems

rom the recognition of environmental issues and the advances in

anufacturing technologies. A disassembly line balancing problem,

here the disassembly line is configured with defined workstations,

s studied. The units of the product to be disassembled are identi-

al and the parts have defined minimum release quantity, cost and

evenue. It is assumed that the amount released excess of minimum

elease quantity can be sold at the end of the period. The aim is to

ssign the tasks to the workstations so that the total net revenue is

aximized.

A MILP model that could solve the problems with up to 50 tasks,

s developed. For larger sized instances, we propose upper and lower

ounds that are motivated by the satisfactory behavior of the LPR. The

PR is strengthened by same valid cuts.

The experimental results have revealed that the bounding mecha-

isms produce high quality solutions very quickly. For the maximum

rial size of 73 tasks, there is a gap of less than 5 percent between our

ower and upper bounds.

To the best of our knowledge, the study is the first attempt to

olve the DLBP with minimum release quantities and fixed number of

orkstations. The proposed model assumes that all units of a speci-

ed part bring the same revenue. The presented methods can be di-

ectly applied when the unit revenues of the first level (minimum

elease quantity) and second level (excess of minimum release quan-

ity) are different. Note that as the minimum release quantities are

xed, the contribution of the first level’s total revenue to the objec-

ive function is fixed, hence irrelevant for the model solutions.





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We hope that the results of our study will trigger some advances

n the disassembly literature. Some extensions of the study may con-

ider the Successor OR type precedence relations, non-identical prod-

cts (different units of the disassembly product with different parts)

nd a stochastic nature of the disassembly outcome (some parts may

e found to be defective or damaged during disassembly).

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