A GENERALIZED INTEGR PROGRAMMING MODEL WITH SPECIAL FORMS
FOR OPTIMAL PROVISION OF MULTIPLE MANUFACTURES
Said A. Hassan*, and Seraj Y. Abed**
*Prof., Industrial Engineering Dept., King Abdulaziz University
** Chairman, Industrial Engineering Dept., King Abdulaziz University
ABSTRACT
In a previous paper written by the same authors, the provision of
manufactured components of only one type was presented, the provision can be
done
using
three
different
methods,
production,
importing/storing,
and
subcontracting. The problem is to determine how many components should be
provided using different method in order to accomplish the required demand and to
minimize the total cost of provision. The method usually used to solve such a
problem is to consider that the provision of all the requirements by only one method,
evaluate each method, and then choose the best one. This may be hardly possible in
case of one type of manufactures, but for multiple manufacture mix, the try and error
method is very hard to be used and it will never approach even a near optimal
solution. A mathematical model for such a problem was formulated and used to
obtain the optimal solution.
The problem will be more complicated when more than one manufactured
components are to be considered. The decision maker can not evaluate all possible
alternatives, and so, he can not obtain the optimal solution. This paper deals with the
design of a general integer programming model for such a problem. The objective
function is to minimize the total fixed and variable costs of provision of all needed
types of components. A special (If-Then) form for the objective function is
generated due to the addition of a term corresponding to the fixed cost associated
with some of the provision methods, this term should be deleted in case of not using
that method in the final solution. Another special form for the objective function is
generated due to the stepped unit prices associated with the provision using the
subcontracting method.
The problem contains many different constraints: The provision of the total demand,
maximum available time for each production machine in both normal and over times, the
limited area for storing of imported components of all types, the overtime can not be
performed unless the total capacity in normal time is totally exhausted, and the constraints
corresponding to the subcontracting stepping prices conditions. Some of these constraints are
if-then conditions that can be transformed into an equivalent set of constraints without the
conditional relation with the addition of auxiliary 0-1 variables.
The steps of the used algorithm are summarized, and an illustrated small dimension example
is presented due to the limitations imposed by the used computer package. The corresponding
mathematical model has 18 construction integer decision variables and 9 additional 0-1
variables. The model is solved using LINGO 9.0, and the final solution is obtained in 1030
iterations with an optimal value of 0.8369308E+08. Of course any different feasible solution
for this problem will give a larger associated total cost.
1
1. INTRODUCTION
Many practical problems are concerned with the provision of manufactured
components of different types, the provision of these components can be done using
three different methods: production, importing/ storing, and subcontracting with a
third party. The total market demand from each type can be estimated, and also the
total fixed and variable cost elements for each method can be evaluated.
In practice, it can be easy to evaluate the total fixed and variable costs associated
with one of the candidate methods, but because each method has a maximum limited
capacity, then it may be required to establish additional lines to meet the required
demands. In such cases, when the required demand is too much, the number of
feasible solutions will be huge and cumbersome for evaluation especially when the
problem contains multiple types of products.
In a previous paper written by the same authors [1], it was shown that the number of
possible alternatives is huge and can not be exhaustively evaluated. An Integer
programming model was presented to represent the complete problem with the
objective of minimizing the total provision cost, and to cover all possible choices of
the provision mix for components of only one type. A real example of application
was presented for the Power Transmission & Distribution (PTD) equipment in one
of the famous companies in the kingdom of Saudi Arabia. The mentioned company
offers complete solutions ranging from products, to systems, to turn key solutions,
all related after sales support, training, and consulting services. In order to locally
support the ever-growing demand of the market, the company has also started its
manufacturing. Only one of the very important and highly costly components was
chosen to apply the procedure.
This paper describes an integer programming algorithm to formulate the problem of
provision of manufactured components of various types. The provision can be done
by three methods: production in normal time, production in overtime, import and
storing, and subcontracting with an external supplier. Due to the amounts required,
many production and storing lines may be required to satisfy the required demand
that will complicate and increase the possible solution methods. Each provision
method is associated with fixed and variable cost elements that should be considered
when evaluating different methods of provisions.
The second part of the paper is devoted to the problem definition concerning the
objective function, and the problem constraints. The third part explains the decision
variables which represent the number of components of different types to be
provided using various methods of provision. The fourth part explains the objective
function coefficients concerning the fixed and variable costs corresponding to
different methods of provision. It explains also some If-then constraints associated
with the fixed cost coefficients that should be added or deleted according to the
value of the corresponding decision variable. This was done by a linear
minimization model constructed without the conditional constraints and by the
addition of 0-1 variables.
2
The fifth part is devoted to the problem constraints: fixed cost coefficients constraints,
provision of the total demand, maximum capacity for production lines in both normal and
over times constraints, maximum storing capacity for importing lines, different ranges for
stepping prices, If-then for overtime, constraints, If-then constraints for subcontracting
stepping prices, integral constraints, and the 0-1 constraints. In the sixth part, the steps of the
used algorithm are summarized: the decision maker should make estimation or forecasting of
the demand quantity for the needed manufactured component, estimate the maximum needed
number of lines for production and importing methods of provision, estimate the needed times
for manufacturing of each component type in each machine, calculate the corresponding fixed
and variable cost elements, design the corresponding mathematical model, then solve the
model and obtain the optimal solution.
The seventh part gives an example of illustration with two types of manufactured
components, two production lines, two import/ storing lines, and three stepped
subcontracting prices for each type of manufactures. The complete mathematical
model is formulated, and the optimal solution is given using LINGO 9.0 computer
program. The last part of the paper has the main conclusions and the proposed points
for future researches.
2. PROBLEM DEFINITION
The problem is to determine how many components of each type should be provided
using individual method with the aim to minimize the total cost of provision of the
whole requirements, while to accomplish the required demand under the constraints
of maximum provision capacity for each plant in every method of provision. The
provision methods that are:
production in normal and / or overtime, importing/
storing, and subcontracting. The total cost comprises fixed and variable cost
elements associated with different methods of provision for all the needed
components. The provision of needed manufactures is not limited to using only one
method of provision, but any mix of the available methods can be used so that the
minimum total cost is achieved.
The problem contains many different constraints: The provision of the total demand, the
maximum capacity constraints for different machines in the production lines for normal time
and for overtime, importing / storing maximum area capacity for different plants, and
subcontracting constraints with stepped unit prices that decreases with the committing for a
certain level of components. Another type of constraints is the overtime condition which
constitutes that the overtime can not be performed unless the total capacity in normal time is
totally exhausted. The problem contains another additional type of constraints corresponding
to the construction of more provision lines, these addition lines can not be established unless
the total capacity of the previous lines is totally exhausted. At last, the problem contains also
the subcontracting stepping prices. The unit price for subcontracting are not constant, it has
the shape of stepping prices related to the amount of components subcontracted.
3. DECISION VARIABLES
r
Let: x t j represents the number of components of type t , to be provided using the
method j, in plant No. r , where r =1 to n j (the number of lines of for method j),
and j represents a provision method that can be: production line (p), overtime (o),
3
importing (i), and subcontracting (s). So, we can have the following decision
variables, for example:
1
x1 p =The number of components of type 1 to be produced in production line No.1,
1
x1 o =The number of components of type 1to be produced in overtime of production line No.1,
3
x2 i = The number of components of type 2 to be imported and stored in the third store,
1
x3 s = The number of components of type 3 to be subcontracted with price in the first interval,
4. OBJECTIVE FUNCTION
4.1 Objective Function Coefficients
Let f jr = the fixed cost coefficient corresponding to the method of provision j,
(production, and importing and subcontracting), and the line r,
r
vt j = represent the variable cost coefficients corresponding to the variable
xtr j .
4.2 If-Then Constraints for the Fixed Cost Part:
To minimize the total costs of provision of all needed parts:
nj
 
Min z =
t
j
r 1
r
r
vt j xt j
+

f jr'
,  j ; j' = p, i, and s ,
xtr j '
> 0
,
xtr j '
=0
j'
0
Where:
n j = number of lines for method of provision j.
This should be understood as:
r
r
1. For j' = p, and I: If any xt j ' > 0, t = 1, 2, …… nt , then a corresponding term f j ' of the
total fixed cost for the products provided with the provision mean j' in line No. r, should be
added, this term should be deleted in case when all
xtr j '
= 0.
Where: nt = the total number of types for the products.
r
2. For j' = s: If: any xt s > 0, r =1, 2, …., ns , then a term corresponding to the total fixed cost
r
for subcontracting f s , should be added, this term should be deleted in case when all xt s = 0,
r =1, 2, …., ns .
Where: ns = the total number of stepped prices for subcontracting.
For the first case: These are nonlinear in
It is known that:
0 ≤
xtr j '
xtr j ' ,
r
≤ N t j ' for all t, j, and n.
4
because the discontinuity at the origin.
r
Where: N t j ' = The maximum capacity for a product t provided by method j' in line r.
The nonlinearity can be transformed into linear minimization model as follows:
nj
 
Min. z =
t
r 1
j
r
r
vt j xt j
+
f jr' y rj ' ,
Subject to:
nt
nt
x
t 1
y rj '
r
t j'
≤
N
t 1
t
y rj ' ,
 r, j = production and import,
 r.
= 0, 1.
r
If any xt j ' > 0, then the corresponding
y rj '
is forced to be 1, to eliminate the
infeasibility, and:
r
If all xt j ' = 0, t = 1, 2, …., nt , then no restrictions are imposed on
y rj '
, it can be
equal to 0, or 1, but according to the minimization of the objective function, it will
be forced to have the value of 0.
The second case will be discussed in the following section.
4.3 Combined If-Then / Or Constraints for subcontracting
For the subcontracting variant:
nt s
IF: ( xt s OR xt s OR …. OR xt s OR any combinations of them) > 0 , t = 1, 2, , nt , then
a term corresponding to the total fixed cost for subcontracting the components f s , should be
1
2
added, this term should be deleted in case when all the x ts . Where nts = the number of
price steps for subcontracting components of type t.
This can be reduced to:
1
IF: xt s > 0 , t = 1, 2, …..., nt then a term corresponding to the total fixed cost for
subcontracting the panels f s , should be added, this term should be deleted in case when all
1
r
xt s = 0 . Since for any t, none of the other variables xt s , r >1, can have nonzero values
1
unless xt s > 0.
This could be done by adding the following constraint:
nt

t 1
x y
1
ts
1
s
nt

t 1
N t1s ,
5
y 1s = 0 , 1 .
4.4 Objective Function for the Whole Problem
nj
nt
 
Min. z =
t 1
r 1
j
r
tj
r
tj
v x
nq
+

f jr' y rj ' , j = p, o, i, s; j' = p, i, s ….…..… (1)
j ' r 1
5. CONSTRAINTS
5.1 Fixed cost coefficients Constraints
nt
nt
x
t 1
nt
r
tp
N
t 1
t
y rp , r = 1, 2, …., n p , …………………………………….…….…… (2)
nt
x
t 1
≤
r
ti
≤
N
t 1
t
y ir , r = 1, 2, …., n p , ……………………………….…….…….…… (3)
y rp = 0, 1, r = 1, 2, ……,
n p . ………………....................................……….………….. (i.1)
y ir = 0, 1 , r = 1, 2, ……, ni . ………………....................................……….………….. (i.2)
For the stepped price subcontracting, we have:
nt

x 
1
ts
t 1
nt

t 1
Nt1s . y1s
, ……………………………….……………………………… (4)
y 1s = 0 , 1 . ………………………..…………….………………………..…………….….. (i.3)
5.2 Provision of the Total Demand
nj
 x
j
r 1
r
tj
= N t , j = p, o, i, and s ; for t = 1, 2, …, nt ……...…..……………………… (5)
Where:
N t = The total No. of required components of type t .
5.3 Maximum capacity for Production Lines in normal and overtime
The total time for production of components in different machines in each of the
normal and overtime should not exceed the maximum available time for that
machines, so for normal time:
6
nt
x
t 1
r
tp
. tm  T pm r , r = 1, 2, …., n p ; m = 1, 2, …., n pr , ……......……...…………… (6)
Where:

m
t = needed time to produce one component of type t in machine no. m,
T
mr
p
= Total available time for production in machine m and line r in normal time.
n pr = number of kinds of machines in production line number r.
And for overtime:
nt
x
t 1
. tm  Tom r , r = 1, 2, …., n o ; m = 1, 2, …., nor , ……...…...……...…………… (7)
r
to
Where:
T
mr
o
= Total available time for production in machine m and line r in overtime.
nor = number of kinds of machines for overtime in line number r.
5.4 Maximum Capacity for Importing/ Storing Lines
The occupied area by different components of all types should not exceed the total
available area of the individual plants:
nt
x
t 1
r
ti
. at  A ir , r = 1, 2, …., ni , ………………….………………………………… (8)
Where:
at = area occupied by one unit of manufactures of type t,
A ir = Total available storing area number i .
ni = number of import/ storing lines.
5.5 Range for stepped subcontracting constraints
r
xt s ≤ N t rs  t , r , …………………………………………………………………………(9)
Where:
r
N t s = maximum possible quantity by subcontracting for product of type t in level price r .
5.6. If-Then for Overtime Constraints
The overtime in any line can not be performed unless the corresponding line is already
established. And since the unit variable cost in overtime is greater than that in normal one due
7
to the addition of the labor cost, the overtime will not be performed unless the total capacity
in normal time is exhausted. The If-Then constraints are as follows:
nt

If for a given r:
x
t 1
Where:
r
tp
nt
= 0, then

t 1
r
xt o = 0 , r = 1, 2, ….., n p ,
n p = number of production lines considered.
To transform these conditional constraints into an equivalent set of constraints
r
r
without the conditional relation, then knowing that: xt o ≤ N t o ,
The following formulation can be used, for each production line r:
nt
nt

r
to
x ≤
t 1
all
r

Nt .
t 1
r
If
nt
xt p

t 1
r
xt p
, r = 1, 2, …..,
n p , ………………...………………… (10)
r
= 0, then accordingly all xt o will be equal to 0, and if any one of the
r
xt p
is greater than o, then no constraints are imposed for any of the variables xt o .
5.7 Subcontracting Stepping Prices Constraints
Consider the decision variables
r
xt s
representing the subcontracting amount with
r
ts
unit price = v , r = 1, ns = number of steps of unit prices. The unit price for the
subcontracting variant are not constant, it has the shape of stepping prices related to
the amount of components subcontracted x t s as shown in Figure No. 1 drawn for
three steps prices for a product of type t.
The subcontracting prices are:
1
vt s for subcontracting quantities
1
1
0 ≤ xt s ≤ N t s ,
1
2
vt s for subcontracting quantities N t 1s < xt 1s + xt 2s ≤ N t s + N t 2s ,
1
1
3
2
1
2
3
2
3
vt s for subcontracting quantities N t s + N t s < xt s + xt s xt s ≤ N t s + N t s + N t s .
The constraints are as follows:
If xt s < N t s , then xt s = 0 , t = 1, 2,…., nt , r = 1, 2, ……,
ns - 1, w = r + 1, r +2, …., nt s .
r
r
Knowing that: xt s ≤ N t s , t = 1, 2,…., nt , r = 1, 2, ……,
nt s ,
r
r
w
The conditional constraints can be transformed as follows:
r
r
xt s ≥ N t s
t rs ,
r = 1, 2, ……,
nt s -1, t = 1, 2,…., nt , …………..………….….….……. (11)
8
w
w
xt s ≤ N t s
t rs
t rs ,
r = 1, 2, ……,
nt s -1, t = 1, 2,…., nt , w = r + 1, r +2, …., nt s , …..…. (12)
= 0 , 1 , t = 1, 2,…., nt , r = 1, 2, ……,
nt s -1, ……..……….……………………… (i.4)
5.8 Integral Constraints
x t j  t, j , r Integers. ………………………………….……….……….………….……. (13)
r
5.7 0-1 Constraints
y rp , yir , y 1s , = 0, 1  r. ……….……………………………..…….…..……….……. (14)
t rs = 0 , 1 , r = 1, 2, ……, nt s -1, t = 1, 2,…., nt , ……….……...…….………….……. (15)
Unit
subcontracting
price
1
vt s
2
vt s
3
vt s
1
1
N t s + N t 2s
Nt s
1
N t s + N t 2s + N t 3s
Quantity subcontracted
Figure 1: Subcontracting stepping prices for type t product
6. STEPS OF THE ALGORITHM
The steps of the used algorithm can be summarized as shown in Figure 2. The
decision maker should make estimation or forecasting of the demanded quantity for
the needed manufactured components ( N t , t = 1, 2, …….., nt ), estimate the
maximum needed number of production lines, and import/storing facilities, calculate
the maximum capacity for each provision method, calculate all the associated fixed
and variable cost elements, design the corresponding mathematical model, solve the
model using a computer solution package, and finally obtain the optimal solution.
7. AN EXAMPLE OF ILLUSTRATION
7.1 Problem Data
9
Due to the limitations imposed by the solution package (LINGO 9.0), we will
choose only 2 types of components, two production lines with overtime, two import/
storing lines, and three stepped subcontracting prices for both types of manufactures.
The fixed and variable cost elements for the production system lines are shown in
Tables 1 and 2 respectively.
The variable cost elements for the production system lines in over-time is shown in
Table 3. The fixed and variable cost elements for the importing and storing lines are
shown in Tables 4 and 5 respectively. The fixed and variable cost elements for the
importing and storing lines are shown in Tables 6 and 7 respectively.
The unit, maximum area for importing/ storing, and the total available production
times are shown in Table 8. The required and the maximum capacity of production
time are shown in Table 9. The maximum quantity for subcontracting intervals, and
the total needed quantities from the two manufactures are shown in Table 10.
Input: N t ,
t= 1, 2, ….., nt
Design the candidate production lines, and
r
r
calculate: v t p , f pr , vt o , and f or
r = 1, 2, … n p , t= 1, 2, ….., nt
Design the candidate import/ store plants, and
r
calculate: vt i and f i r ,
r = 1, 2, … n p , t= 1, 2, ….., nt
For subcontracting,
Input:
nt s , and N t rs , vt rs ,
t= 1, 2, ….., nt , r = 1, 2,……, nt s
Input:  tm , for all machine kinds
at , t = 1, 2, …., nt
Design the corresponding
mathematical model
Solve the model, and obtain the optimal solution
by using an appropriate computer program
10
Figure 2: Flowchart for the steps of the algorithm
Table 1: Fixed cost elements for the production system lines:
Cost Elements
Cost
Headquarter office
Land cost
Building cost
Cutting machine (m = 1)
Punching machine (m = 2)
Bending machine (m = 3)
Labeling machine (m = 4)
Engraving machine (m = 5)
Notching machine (m = 6)
Degreasing machine (m = 7)
Computerized Production system
Advertising
Labor cost (120 labor)
Security cost
Electricity cost
Insurance cost
2,000,000 SR
1,500,000 SR
5,500,000 SR
500,000 SR
500,000 SR
600,000 SR
700,000 SR
700,000 SR
500,000 SR
200,000 SR
1,000,000 SR
100,000 SR/ year
5,760,000 SR/year
180,000 SR/year
100,000 SR/year
130,000 SR/year
Discounted
period
(years)
20
20
20
10
10
10
10
10
10
10
5
1
1
1
1
1
Total f p1 = f p2
Discounted
rate
(SR/year)
200,000
75,000
325,000
50,000
50,000
60,000
70,000
70,000
50,000
20,000
200,000
100,000
5,760,000
180,000
100,000
130,000
7,440,000
Table 2: Variable cost elements for the production system lines (normal time):
Cost Elements
First type
(SR)
Second type
(SR)
Powder coating
Material
Components
88,000
352,000
8,800,000
40,000
120,000
3,500,000
Total
Unit cost (SR/unit)
type
2nd. type
100
73
400
218
18,000
6,364
1st.
18,500
1
2
v1 p = v1 p
Table 3: Variable costs for the production system in over-time:
Cost Elements
Powder coating
Unit cost (SR/unit)
1st. type
2nd. type
100
73
11
6,655
1
2
v2 p = v2 p
Material
Components
Labor cost
400
218
18,000
6,364
4,000
3,000
Total
22,500
9,655
1
2
1
2
v2 o = v2 o
v1o = v1o
Table 4: Fixed cost elements for the importing and storing lines:
Cost Elements
Cost
Headquarter office
Advertising
Land cost
Building cost
Security cost
Electricity cost
Labor cost (10 workers)
4,000,000 SR
100,000 SR
500,000 SR
2,000,000 SR
80,000 SR/year
50,000 SR/year
480,000 SR/ year
Discounted
time (years)
20
1
20
20
1
1
1
Discounted rate
(SR/year)
200,000
100,000
25,000
100,000
80,000
50,000
480,000
Total, f i1 = f i 2
1,035,000
Table 5: Variable cost elements for the importing and storing lines:
Cost Elements
Total cost
(SR)
Unit import price
Transportation cost
Insurance cost
No. of units/ year
1st type 2nd type
836
180,000
30,000
Unit cost (SR/unit)
1st type
2nd type
40,000
25,000
215
210
35
25
40,250
25,235
1
2
2
1
v1i = v1i
v2 i = v 2 i
420
Total
Table 6: Fixed cost elements for subcontracting:
Normal case
Headquarter office
Advertising
Discounted
time (years)
20
1
Total f s
2,000,000
100,000
Discounted rate
(SR/year)
200,000
100,000
300,000
Table 7: Variable cost elements for subcontracting:
1
Value (SR/unit)
2
3
2
1
3
v1 s
v1 s
v1 s
v2 s
v2 s
v2 s
45,000
40,000
37,000
35,000
30,000
26,000
Table 8: Unit, maximum area for importing/ storing, and total available production times:
a1
a2
A 1i
A i2
3 m2
2 m2
4000 m2
3000 m2
T
mr
p
T
2640 hrs/years
mr
o
3960 hrs/years
Table 9: Required production times at different machines:


1
1
 12
 13
 14

hrs
3.1
3.5
4.2
2.7
3.8
5
1

6
1
4.0
 17
 12

2.5
2.6
2.8
12
2
2

3
2
3.2

4
2
4.1

5
2
1.9

6
2
2.2

7
2
1.8
Table 10: Maximum quantity for subcontracting intervals, and total needed quantities:
1
2
N1 s
500
Number (units)
3
N1 s
1000
1
N1 s
2000
N2s
400
2
N2s
800
3
N2s
1500
N1
N2
2000
1800
7.2 Problem Mathematical Model
nj
nt
 
Min. z =
t 1
r
tj
v x
r 1
j
r
tj
nq
+

f jr' y rj ' , j = p, o, i, s; j' = p, i, s
j ' r 1
Subject to:
nt
nt
x
r
tp
t 1
nt
≤
x
nt

x
t 1
nt
x
t 1
x
t 1
r
tj
r 1
nt
nt

t 1
 x
nt
t 1
x 
nj
j
N
≤
1
ts
t 1
t 1
y rp , r = 1, 2, …., n p ,
t
nt
r
ti
t 1
N
t
y ir , r = 1, 2, …., n p ,
Nt1s . y1s
,
= N t , j = p, o, i, and s ; for t = 1, 2, …, nt
r
tp
. tm  T pm r , r = 1, 2, …., n p ; m = 1, 2, …., n pr ,
r
to
. tm  Tom r , r = 1, 2, …., n o ; m = 1, 2, …., nor ,
r
ti
. at  A ir ,
r = 1, 2, …., ni ,
r
xt s ≤ N t rs  t , r ,
nt

t 1
r
nt
r
to
x ≤
t 1
t rs
r
xt s ≥ N t s
w

w
xt s ≤ N t s
nt
Nt .

t 1
r
xt p
, r = 1, 2, …..,
, t = 1, 2,…., nt , r = 1, 2, ……,
t rs , t = 1, 2,…., nt ,
r = 1, 2, ……,
x t j  t, j , r Integers.
r
13
np ,
nt s -1,
nt s -1, w = r + 1, r +2, …., nt s ,
1
y rp , yir , y s ,
= 0, 1  r.
t rs
= 0 , 1 , r = 1, 2, ……, nt s -1, t = 1, 2,…., nt .
7.4 The Complete LINGO 9.0 Program:
For easier interpretation, the subscripted variables are abbreviated by simple variables as
shown in Tables 10, and 11:
Table 10: Abbreviation of the problem structured variables:
x1 p
1
x1 p
2
x1 o
1
x1 o
x1i
x1i
2
x1 s
1
x1 s
2
x1 s
X1
X2
X3
X4
X5
X6
X7
X8
X9
x2 p
1
x2 p
2
x2 o
1
x2 o
2
x2 i
1
x2 i
2
x2 s
1
x2 s
2
x2 s
X10
X11
X12
X13
X14
X15
X16
X17
X18
2
1
3
3
Table 11: Abbreviation of the problem additional variables:
…….
………
y1p
x19
y 2p
x20
y i1
y i2
y 1s
x21
x22
x23
11o
21o
1 o2
2 o2
11p
 2 1p
1 2p
 2 2p
x24
x25
x26
x27
x28 x29
x30
x31 x32 x33 x34 x35 x36 x37
11i
 21i
11s
21s
1 2s
2 2s
The complete LINGO is as follows:
MIN = 18500*X1 + 18500*X2 + 22500*X3 + 22500*X4 + 40250*X5 + 40250*X6 +
45000*X7 + 40000*X8 + 37000*X9 + 6655*X10 + 6655*X11 + 9655*X12 + 9655*X13 +
25235*X14 + 25235*X15 + 35000*X16 + 30000*X17 + + 26000*X18 + 7440000*X19 +
7440000*X20 + 1035000*X21 + 1035000*X22 + 300000*X23;
X1 + X10 - 4000 * X19 <= 0;
X2 + X11 - 4000 * X20 <= 0;
X5 + X14 - 4000 * X21 <= 0;
X6 + X15 - 4000 * X22 <= 0;
X7 + X16 - 3500 * X23 <= 0;
X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 = 2100;
X10 + X11 + X12 + X13 + X14 + X15 + X16 + X17 + X18 = 1900;
3.1 * X1 + 2.6 * X10 <= 2640;
3.5 * X1 + 2.8 * X10 <= 2640;
4.2 * X1 + 3.2 * X10 <= 2640;
14
2.7 * X1 + 4.1 * X10 <= 2640;
3.8 * X1 + 1.9 * X10 <= 2640;
4.0 * X1 + 2.2 * X10 <= 2640;
2.5 * X1 + 1.8 * X10 <= 2640;
3.1 * X2 + 2.6 * X11 <= 2640;
3.5 * X2 + 2.8 * X11 <= 2640;
4.2 * X2 + 3.2 * X11 <= 2640;
2.7 * X2 + 4.1 * X11 <= 2640;
3.8 * X2 + 1.9 * X11 <= 2640;
4.0 * X2 + 2.2 * X11 <= 2640;
2.5 * X2 + 1.8 * X11 <= 2640;
3.1 * X3 + 2.6 * X12 <= 3960;
3.5 * X3 + 2.8 * X12 <= 3960;
4.2 * X3 + 3.2 * X12 <= 3960;
2.7 * X3 + 4.1 * X12 <= 3960;
3.8 * X3 + 1.9 * X12 <= 3960;
4.0 * X3 + 2.2 * X12 <= 3960;
2.5 * X3 + 1.8 * X12 <= 3960;
3.1 * X4 + 2.6 * X13 <= 3960;
3.5 * X4 + 2.8 * X13 <= 3960;
4.2 * X4 + 3.2 * X13 <= 3960;
2.7 * X4 + 4.1 * X13 <= 3960;
3.8 * X4 + 1.9 * X13 <= 3960;
4.0 * X4 + 2.2 * X13 <= 3960;
2.5 * X4 + 1.8 * X13 <= 3960;
3 * X5 + 2 * X14 <= 4000;
3 * X6 + 2 * X15 <= 3000;
X7 <= 500;
X8 <= 1000;
X9 <= 2000;
X16 <= 400;
X17 <= 800;
X18 <= 1500;
X3 + X12 - 4000 * (X1 + X10) <= 0;
X4 + X13 - 4000 * (X2 + X11) <= 0;
X7 - 500 * X24 >= 0;
X16 - 400 * X25 >= 0;
X8 - 1000 * X26 >= 0;
X17 - 800 * X27 >= 0;
X8 - 1000 * X24 <= 0;
X9 - 2000 * X24 <= 0;
X17 - 800 * X25 <= 0;
X18 - 1500 * X25 <= 0;
X9 - 2000 * X26 <= 0;
X18 - 1500 * X27 <= 0;
@GIN(X1); @GIN(X2); @GIN(X3); @GIN(X4); @GIN(X5); @GIN(X6); @GIN(X7);
@GIN(X8); @GIN(X9); @GIN(X10); @GIN(X11); @GIN(X12); @GIN(X13);
@GIN(X14); @GIN(X15); @GIN(X16); @GIN(X17); @GIN(X18);
15
@BIN(X19); @BIN(X20); @BIN(X21); @BIN(X22);@BIN(X23); @BIN(X24);
@BIN(X25); @BIN(X26); @BIN(X27);
7.5 Model Solution
The model is solved using LINGO 9.0, and the final solution is obtained in 6588 iterations.
The obtained optimal solution comprises two combined different provision methods, and the
Objective value = 83,693,080 SR. The nonzero variables and their corresponding structured
variables are shown in Table 12:
Table 12: Optimal solution for the example:
Nonzero
variables
X1
Structured
variable
x11 p
Value
Structured
variable
x12 p
Value
560
Nonzero
variables
X10
X2
x12 p
299
X11
x22 p
447
X3
1
x1o
424
X12
x12 o
681
X4
x1o2
424
X13
x22o
681
X6
x1i2
404
X15
x22 i
1
X19
y1p
1
X22
y i2
1
X20
y 2p
1
90
8. CONCLUSIONS AND POINTS FOR FUTURE RESEARCHES
8.1 CONCLUSIONS
This research has the following conclusions:
1- The problem of provision of multiple types of components using different
provision methods like: production mix in normal and overtime, importing, and
subcontracting with multiple lines for each method, has a huge number of alternative
solutions that are very difficult to search and to evaluate using intuition, or try and
error.
2- An integer / 0-1 programming mathematical model is formulated to represent
such problems. The formulation has some special forms such as: If-Then, combined
If-Then/Or, addition of fixed costs to the objective function, and stepped variable
unit costs.
3- The optimal solution may comprise different combined methods of provisions
that are very difficult to obtain by try and error or by exhaustive search methods.
8.2 POINTS FOR FUTURE RESEARCHES
1- To formulate a mathematical model for optimal design of the production and
storing facilities; this means to determine the optimum number of different machines
16
in the production system, and the area used for importing/ storing, beside the
determination of the number of components to be provided by each method.
2- To develop a decision support system for complete modeling and solving large
scale problems using a commercial modeling and/or general purpose languages.
REFERENCES
1- Said A. Hassan, and Seraj Y. Abed, " An Integer programming model with special forms
for the provision of needed manufactures with an application example", (to be published).
2- LINDO 9.0, Systems Inc., Chicago, http://www.lindo.com, 2004.
3- Hillier, and Lieberman, Introduction to Operations Research, eight Edition, McGraw-Hill
International Edition, 2005.
4- M Junger, Integer Programming and Combinatorial Optimization: 11th International Ipco
Conference, Berlin, Germany, 2005.
5- David Ray Anderson, Dennis J Sweeney, Thomas Arthur Williams, Introduction to
Management Science: Quantitative Approaches to Decision Making, 11th Edition, SouthWestern, 2004.
6- W J Cook, A S Schulz, Integer Programming and Combinatorial Optimization, Springer,
2002.
7- Gerard Sierksma, Sierksma Sierksma, Linear and Integer Programming: Theory and
Practice, Second Edition, Marcel Dekker Inc., New York, 2002.
8- K Aardal, Integer Programming and Combinatorial Optimization, Springer, 2001.
9- G. L. Nemhauser, and L. A. Wolsey, Integer and Combinatorial Optimization, Wiley Interscience, 1999.
10- Hamdy A. Taha, Operations Research, an Introduction, Sixth Edition, Printice Hall, New
Jersey, 1997.
11- John Edward Beasley, Advances in Linear and Integer Programming, Mathematics, 1996.
17