International Journal of Pure and Applied Mathematical Sciences.
ISSN 0972-9828 Volume 9, Number 2 (2016), pp. 219-229
© Research India Publications
http://www.ripublication.com/ijpams.htm
Assignment Programming Problem with Multi-Choice
Cost Coefficients
Jadunath Nayak
Baripada College, Baripada, Odisha.
E-mail: [email protected]
Sudarsan Nanda
KIIT University, Bhubaneswar, Odisha.
E-mail: [email protected]
Srikumar Acharya
KIIT University, Bhubaneswar, Odisha.
E-mail: [email protected]
Abstract
The general assignment problem is to assign various jobs to available persons so
as to minimize cost of assignment when cost of assigning i th person to j th job are
given. But in real life situation there exists several choices while assigning i th
job. In this paper we consider a multi choice assignment problem, which cannot
be solved directly. This paper presents an equivalent model of multi choice assignment problem with multi- choice cost coefficients using interpolating polynomial
approach, Binary variable approach and Least square approximation method. The
equivalent models are solved by existing methods and software. Numerical example is provided to illustrates the technique.
AMS subject classification:
Keywords: Multi-Choice decision making, Multi-choice Linear Programming, Assignment Problem.
1.
Introduction
In an Assignment Problem the decision maker has to assign various jobs to available
persons when cost of assigning i th to j th job are given and are unique. But in real life
2
Jadunath Nayak, et al.
situation even a single person for a particular job may very his efficiency in various
situations like with different machines. Thus there may be different choices of cost
value if i th person assign j th job. Several techniques are available to solve Assignment
problem like Hungarian method by Kuhn [3]. In uncertain condition various fuzzy [11]
approaches are also available.
We consider Assignment problem with multi-choice cost coefficients and named as
multi choice assignment problem. The multiple choice programming Problem (MCP)
was proposed by Healey [2]. Some remarkable progress are reflected in the article by
Synder [7] and Lin [4] in particular field Chang [1]. For multi-choice Linear Programming Problems Biswl and Acharya [5] [6], Acarya and Biswal [9] use binary variables in
order to transform a multi choice linear programming problem to an equivalent mathematical model. But in this paper we have used multi-choice cost coefficient and create an
equivalent model. For equivalent model we use three different approaches like Interpolating Polynomial Approach, Binary variable approach and Least Square approximation
method. The equivalent model was a non-linear programming problem and solved using
LINGO [10].
In section 2 basic preliminary is given, in section 3 we formulate the problem, in
section 3 transformation techniques are discussed, in section 4 equivalent model are
discussed and in section 5 numerical example illustrate the techniques.
2.
Basic Preliminary
A general Assignment Problem is
min Z =
n
n Cij Xij
(2.1)
i=1 j =1
subject to
n
Xij = 1 f or j = 1, 2, . . . n
i=1
n
(2.2)
Xij = 1 f or i = 1, 2, . . . n
(2.3)
Xij ∈ {0, 1}
(2.4)
j =1
Assignment Programming Problem with Multi-Choice Cost Coefficients
3.
3
Problem Formation
A mathematical programming model for assignment problem involving multi-choice
cost coefficients is presented as:
n
n (k )
(1) (2) (3)
{cij
, cij , cij , . . . , cij ij }xij , i = 1, 2, 3, . . . , n, j = 1, 2, 3, . . . , n
min : Z =
i=1 j =1
(3.5)
subject to
n
xij = 1 f or i = 1, 2, . . . , m
(3.6)
xij = 1 f or j = 1, 2, . . . , n
(3.7)
xij ∈ {0, 1}
(3.8)
j =1
n
i=1
(k )
(1) (2) (3)
Exactly one element from the set {cij
, cij , cij , . . . , cij ij }is to be selected in order to
minimize the objective function.
3.1.
Transformation technique
Since the coefficient is a finite set, the mathematical model cannot be solved directly,
thus we need an equivalent model to solve the problem.
min : Z =
n
n pkj −1 (z)xij
(3.9)
i=1 j =1
subject to
n
j =1
m
xij = 1 f or i = 1, 2, . . . , m
(3.10)
xij = 1 f or j = 1, 2, . . . , n
(3.11)
xij ∈ {0, 1}
(3.12)
i=1
where z is an integer
The main difficulty is to deal with multi choice parameter, to find the required equivalent model we use some different approaches.
4
Jadunath Nayak, et al.
4.
Equivalent Models
In this section, three different approaches are discussed.
4.1.
Interpolating Polynomial Approach
A polynomial that passes through a given set of points is called an interpolating polynomial. There are n points in a plane (i, yi ) , i = 0, 1, 2, 3, . . . , n − 1 with distinct i then
there exists a unique polynomial in x whose degree is n − 1.
Let the polynomial be
Pkj −1 (z) = a0 + a1 z + a2 (z)2 + a3 (z)3 + · · · + an−2 (z)n−2 + an−1 (z)n−1 (4.13)
after substituting the n points in the polynomial we have,
P (0) = a0 + a1 (0) + a2 (0)2 + a3 (0)3 + · · · + an−2 (0)n−2 + an−1 (0)n−1 = y0
P (1) = a0 + a1 (1) + a2 (1)2 + a3 (1)3 + · · · + an−2 (1)n−2 + an−1 (1)n−1 = y1
P (2) = a0 + a1 (2) + a2 (2)2 + a3 (2)3 + · · · + an−2 (2)n−2 + an−1 (2)n−1 = y2
..
.
P (n − 1) = a0 + a1 (n − 1) + a2 (n − 1)2 + a3 (n − 1)3 + . . . + an−1 (n − 1)n−1 = yn−1
(4.14)
(4.15)
(4.16)
(4.17)
(4.18)
The system of equations can be written as AX = B where

1
1


A = 1
 ..
.
0
1
2
..
.
0
12
22
..
.
···
···
···
...
0
0
1n−2
2n−2
..
.
1n−1
2n−1
..
.
1 (n − 1) (n − 1)2 · · · (n − 1)n−2 (n − 1)n−1




a0
y0
 a1 
 y1 








X =  a2  B =  y2 
 .. 
 .. 
 . 
 . 
an−1
yn−1







(4.19)
The value of a1 , a2 , · · · , an can be found out by solving AX = B. If A is non singular
then the system can be solved.
Now we formulate a multi-choice Assignment programming model by using Interpolating Polynomial as:
n
n Pkj −1 xij
(4.20)
min Z :
i=1 j =1
Assignment Programming Problem with Multi-Choice Cost Coefficients
5
subject to
n
j =1
n
xij = 1 f or i = 1, 2, . . . , m
(4.21)
xij = 1 f or j = 1, 2, . . . , m
(4.22)
xij ∈ {0, 1}
0≤z≤1
(4.23)
(4.24)
i=1
4.2.
Binary Variable Approach
As every natural number can be expressed as sum of 2k number of terms and each term
is a power of 2, where k ∈ N ∪{0}. By using this phenomenon we can formulate an
equivalent model for multi-choice Assignment Problem as
min : Z =
n
n (2)
P ((1)
ij , θij )Xij
(4.25)
i=1 j =1
subject to
n
j =1
m
xij = 1 f or i = 1, 2, . . . , m
(4.26)
xij = 1 f or j = 1, 2, . . . , n
(4.27)
i=1
(4.28)
xij
∈
{0, 1}
θij(1) , θij (2) ∈ {0, 1}
4.3.
(4.29)
(4.30)
Linear Least Square Approximation Approach
Let a function f (x) be given by the following table at a discrete set of points i; i =
0(1)n − 1
Then the function can be replace by a least square line such that the sum of the square
of the vertical distance of the points (i, fi ), i = 0(1)n − 1 from the line is the minimum.
In this case the degree of the least square polynomial is 1 and
li (zij ) = a0 + a1 zij
(4.31)
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Jadunath Nayak, et al.
Table 1: nodes
x
0
f (x) f0
1
f1
2
f2
… n−1
… fn−1
the least square error E is given by
E=
n−1
(fi − a0 − a1 i)2
(4.32)
i=0
The necessary conditions for E to be minimum are
n−1 n−1 n−1
i + a1
i2 =
ifi
a0
i=0
i=0
i=0
n−1 n−1
i =
fi
a0 n + a 1
i=0
(4.33)
(4.34)
i=0
which is a system of two linear equations in a0 , a1 and are the normal equations for
the least square linear polynomial. By solving the normal equation given by (4.33)–
(4.34). We get the two coefficients a0 and a1 and the linear least square polynomial
li (zij ) = a0 + a1 zij .
Now we formulate a multi-choice Assignment model for the multi-choice cost coefficient as linear programming problem by using Linear Least square approach as
min Z :
n
n Pkj −1 xij
(4.35)
i=1 j =1
subject to
m
xij = 1 f or i = 1, 2, . . . , n
(4.36)
xij = 1 f or j = 1, 2, . . . , m
(4.37)
xij ∈ {0, 1}
zij ∈ N
0 ≤ zij ≤ kj − 1
(4.38)
(4.39)
(4.40)
j =1
n
i=1
Assignment Programming Problem with Multi-Choice Cost Coefficients
5.
7
Numerical Example
A departmental head has three subordinate and there are three tasks to be performed.
Subordinate are differ in efficiency for different work and also with different machine.
The table give below show the multi choice time value of each subordinate for each work.
Find a suitable assignment.
T ask
A
B
C
5.1.
III
I
II
{8, 10, 15} {26, 24} {17, 19}
{13, 12} {28, 30} {4, 8}
{29, 19} {18, 4}
{36, 24}
Method-1 Binary Variable Approach
As there are multi-choice for each cost value i.e. Cij we use binary variable for each cost
(2)
value using binary variable(1)
11 , 11 12 , 13 , 21 , 22 , 23 , 31 , 32 , 33 we have
C11
C12
C13
C21
C22
C23
C31
C32
C33
=
=
=
=
=
=
=
=
=
(1) (2)
(1)
(2)
(1) (2)
8θ11
θ11 + 10θ11
(1 − θ11
) + 15(1 − θ11
)θ11
26θ12 + 24(1 − θ12 )
17θ13 + 19(1 − θ13 )
13θ21 + 12(1 − θ21 )
28θ22 + 30(1 − θ22 )
4θ23 + 8(1 − θ23 )
36θ31 + 24(1 − θ31 )
29θ32 + 19(1 − θ32 )
18θ33 + 4(1 − θ33 )
(5.41)
(5.42)
(5.43)
(5.44)
(5.45)
(5.46)
(5.47)
(5.48)
(5.49)
then the problem reduce to
(1) (2)
(1)
(2)
(1)
(2)
min : Z = [8θ11 θ11 + 10θ11 (1 − θ11 ) + 15(1 − θ11 )θ11 ]X11 + [26θ12 + 24(1 − θ12 )]X12
+[17θ13 + 19(1 − θ13 )]X13 + [13θ21 + 12(1 − θ21 )]X21
(5.50)
+[28θ22 + 30(1 − θ22 )]X22 + [4θ23 + 8(1 − θ23 )]X23 + [36θ31 + 24(1 − θ31 )]X31
+[29θ32 + 19(1 − θ32 )]X32 + [18θ33 + 4(1 − θ33 )]X33
8
Jadunath Nayak, et al.
subject to
X11 + X12 + X13
X21 + X22 + X23
X31 + X32 + X33
X11 + X21 + X31
X12 + X22 + X32
X13 + X23 + X33
=1
=1
=1
=1
=1
=1
(5.51)
(5.52)
(5.53)
(5.54)
(5.55)
(5.56)
(1)
(2)
θ11
+ θ11
≥1
(5.57)
(1)
θ11
≤2
θ12 ≤ 1
θ13 ≤ 1
θ21 ≤ 1
θ22 ≤ 1
θ23 ≤ 1
θ31 ≤ 1
θ32 ≤ 1
θ33 ≤ 1
Xij ∈ {0, 1}
θij ∈ {0, 1}
(5.58)
(5.59)
(5.60)
(5.61)
(5.62)
(5.63)
(5.64)
(5.65)
(5.66)
(5.67)
(5.68)
(1) (2)
, θ11 ∈ {0, 1}
θ11
(5.69)
(5.70)
(2)
+ θ11
This problem is solve using LINGO 12.0 [8] and the result is X11 = 1, X12 = 0, X13 =
0, X21 = 0, X22 = 0, X23 = 1, X31 = 0, X32 = 1, X33 = 0 and the maximum value of
Z is 31.
5.2.
Method-2 Interpolating Polynomial Approach
We can use Interpolating Polynomial to convert multi-choice cost value to a single polynomial. if Cij are replace by Pkj −1 (ij ) then the value of Pkj −1 (ij ) can be find out using
Polynomial (Newton Divided Difference in this case) to replace Cij .
Pkj −1 (11)
Pkj −1 (12)
Pkj −1 (13)
Pkj −1 (21)
=
=
=
=
2
7Z11
− 5Z11 + 8
20 + 4Z12
17 + 2Z13
13 − Z21
(5.71)
(5.72)
(5.73)
(5.74)
(5.75)
Assignment Programming Problem with Multi-Choice Cost Coefficients
=
=
=
=
=
Pkj −1 (22)
Pkj −1 (23)
Pkj −1 (31)
Pkj −1 (32)
Pkj −1 (33)
28 + 2Z22
4 + 4Z23
36 − 12Z13
29 − 10Z32
18 − 14Z33
9
(5.76)
(5.77)
(5.78)
(5.79)
(5.80)
Replacing value of Pkj −1 (ij ) in place of Cij we fine the equivalent model as
2
min : Z = [7Z11
− 5Z11 + 8]X11 + [20 + 4Z12 ]X12 + [17 + 2Z13 ]X13
+[13 − Z21 ]X21 + [28 + 2Z22 ]X22 + [4 + 4Z23 ]X23 + [36 − 12Z31 ]X31
(5.81)
+[29 − 10Z32 ]X32 + [18 − 14Z33 ]X33
Subject to
X11 + X12 + X13 = 1
X21 + X22 + X23 = 1
X31 + X32 + X33 = 1
X11 + X21 + X31 = 1
X12 + X22 + X32 = 1
X13 + X23 + X33 = 1
Xij ∈ {0, 1}
Z11 ∈ {0, 1, 2}
Zij ∈ {0, 1}(i, j ) = (1, 1)
(5.82)
(5.83)
(5.84)
(5.85)
(5.86)
(5.87)
(5.88)
(5.89)
(5.90)
The above problem can be solved using LINGO 12.0[8] and the result is X11 = 0, C11 =
8, X12 = 1, C12 = 24, X13 = 0, C13 = 19, X21 = 1, C21 = 12, X22 = 0, C22 =
30, X23 = 0, C23 = 8, X31 = 0, C31 = 24, X32 = 0, C32 = 19, X33 = 1, C33 = 4 and
the maximum value of Z is 31.
5.3.
Method-3 Least square Approximation Approach
We can use Least Square Approximation to fit multi-choice cost values to a single line.
if Cij are replace by l0 ij then the value of l0 ij can be find out using the relation
a0 (
n−i
i=0
i) + a1 (
n−1
i 2) =
i=0
n−1
a0 (n) + a1 (
i=0
i) =
n−1
(ifi )
i=0
n−1
(fi )
i=0
(5.91)
(5.92)
10
Jadunath Nayak, et al.
to replace Cij(i) .
l0 (Z(11) ) =
l0 (Z(12) )
l0 (Z(13) )
l0 (Z(21) )
l0 (Z(22) )
l0 (Z(23) )
l0 (Z(31) )
l0 (Z(32) )
l0 (Z(33) )
=
=
=
=
=
=
=
=
45 7
+ (Z11 )
2
2
20 + 4Z12
17 + 2Z13
13 − Z21
28 + 2Z22
4 + 4Z23
36 − 12Z13
29 − 10Z32
18 − 14Z33
Now the equivalent model to the given problem is
45 7
min : Z =
+ (Z11 ) X11 + [20 + 4Z12 ]X12 + [17 + 2Z13 ]X13
2
2
+[13 − Z21 ]X21 + [28 + 2Z22 ]X22 + [4 + 4Z23 ]X23 + [36 − 12Z13 ]X13
(5.93)
(5.94)
(5.95)
(5.96)
(5.97)
(5.98)
(5.99)
(5.100)
(5.101)
(5.102)
+[29 − 10Z32 ]X32 + [18 − 14Z33 ]X33
Subject to
X11 + X12 + X13
X21 + X22 + X23
X31 + X32 + X33
X11 + X21 + X31
X12 + X22 + X32
X13 + X23 + X33
=1
=1
=1
=1
=1
=1
Xij ∈ {0, 1}
Z11 ∈ {0, 1, 2}
Zij ∈ {0, 1}(i, j ) = (1, 1)
(5.103)
(5.104)
(5.105)
(5.106)
(5.107)
(5.108)
(5.109)
(5.110)
(5.111)
The above problem can be solve using LINGO 12.0[8] and the result is X11 = 0, C11 =
8, X12 = 1, C12 = 24, X13 = 0, C13 = 19, X21 = 1, C21 = 12, X22 = 0, C22 =
30, X23 = 0, C23 = 8, X31 = 0, C31 = 24, X32 = 0, C32 = 19, X33 = 1, C33 = 4 and
the maximum value of Z is 36.
6.
Conclusion
After solving the problem in three different approach we found that the result of Interpolating polynomial approach and Binary Variable approach are same but the Least
Assignment Programming Problem with Multi-Choice Cost Coefficients
11
square approximation method provide approximate solution. This may be because of
error involved in fitting the curve. This technique can be extended to multi choice fuzzy
transportation problem, multi choice fuzzy stochastic transportation problem etc.
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