A Solution Procedure To Solve Multi Objective Fractional Transportation Problem

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International Refereed Journal of Engineering and Science (IRJES)
ISSN (Online) 2319-183X, (Print) 2319-1821
Volume 7, Issue 1 (January 2018), PP. 67-72
www.irjes.com 67 | Page
A Solution Procedure to Solve Multi objective Fractional Transportation
Problem
Dr. Doke Dnyaneshwar Maruti, M.Sc .Ph.D.(Statistics)
Associate Professor and Head of Dept. , Mathematics and Statistics.
M.L. DahanukarCollege., VileParle (East), Mumbai
Corresponding Author: Dr. Doke Dnyaneshwar Maruti
Abstract: In decision making process if the objective function is ratio of two linear functions and objective function is to
be optimized. For example one may be interested to know the ratio of total cost to total time required for transportation.
This ratio is an objective function which is fractional objective function. When there are several such fractional objectives to
be optimized simultaneously then the problem becomes multi objective fractional programming problem (MOFLPP).
Initially Hungarian mathematician BelaMartos constructed such type of problem and named it as hyperbolic programming
problem. Same problem in general referred as Linear Fractional Programming Problem. Fractional programming problem
can be converted into linear programming problem (LPP) by using variable transformation given by Charnes and Cooper.
Then it can be solved by Simplex Method for Linear Programming Problem. .
In this paper we propose to solve multi objective fractional transportation problem. Initially will solve each of the
transportation problem as single objective and then using Taylor series approach expand each of the problem about its
optimal solution and ignoring second and higher order error terms each of the objective is converted into linear one. Then the
problem reduces to MOLTPP. Evaluate each of the objectives at every optimal solution and obtain evaluation matrix. Define
hyperbolic membership function using best and worst values of objective function with reference to evaluation matrix. These
membership functions are fuzzy functions Compromise solution is obtained using weighted a.m. of hyperbolic membership
functions and also weights quadratic mean of hyperbolic membership functions. Propose o solve problem at the end to
explain the procedure.
Keywords: Multi objective, compromise solution, hyperbolic membership function.
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Date of Submission: 19 -01-2018 Date of acceptance: 33-01-2018
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I. INTRODUCTION
It is well known fact that the transportation problem is special case of linear programming problem. In same way
multi objective fractional transportation programming problem is special case of multi objective fractional programming
problem. In this paper we will deal with multi objective fractional transportation programming problem. The problem is to
transfer goods from different origins to several destinations such that many fractional objective functions attain their optimal
values. Again optimizing several objectives simultaneously is not possible. Thus the golden mean is to find compromise
solution. For compromise solution there are several methods which are proved to be efficient. In next section linear
fractional transportation problem (LFTP) is introduced.
1. Formulation of Linear Fractional TP.(LFTP)
In general a problem is specified as under
i) There are m supply points from which goods are to be transported. Supply available at 𝑖𝑡𝑕point is at most 𝑎𝑖𝑓𝑜𝑟𝑖 =
1,2, … , 𝑚
ii) There are n demand centres where good are required. Minimum requirement of 𝑗𝑡𝑕 demand centre is 𝑏𝑗 𝑓𝑜𝑟𝑗 =
1,2, … , 𝑛.
iii) Profit matrix is 𝑃 = 𝑝𝑖𝑗 𝑖𝑠𝑚 × 𝑛 matrix where 𝑝𝑖𝑗 is profit of transporting one unit from 𝑖𝑡𝑕
origin to 𝑗𝑡𝑕
destination.
iv) Cost matrix is 𝐷 = 𝑑𝑖𝑗 𝑖𝑠𝑚 × 𝑛, matrix where 𝑑𝑖𝑗 is cost of transporting one unit from 𝑖𝑡𝑕
origin to 𝑗𝑡𝑕
destination
v) 𝑝0 and 𝑑0 be the fixed profit and cost respectively
vi) Suppose variable𝑥𝑖𝑗 denotes number units to be transported from 𝑖𝑡𝑕
origin to 𝑗𝑡𝑕
destination.
The objective function is,
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑜𝑟 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒𝑄 𝑥 =
𝑃(𝑥)
𝐷(𝑥)
=
𝑝𝑖𝑗
𝑛
𝑗=1
𝑚
𝑖=1 𝑥𝑖𝑗 + 𝑝0
𝑑𝑖𝑗
𝑛
𝑗 =1
𝑚
𝑖=1 𝑥𝑖𝑗 + 𝑑0
(1)
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠,
𝑥𝑖𝑗
𝑛
𝑗 =1
≤ 𝑎𝑖𝑓𝑜𝑟𝑖 = 1,2, … , 𝑚 (2)
𝑥𝑖𝑗
𝑚
𝑖=1
≤ 𝑏𝑗 𝑓𝑜𝑟𝑗 = 1,2, … , 𝑛 (3)
𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟𝑖 = 1,2, … , 𝑚𝑎𝑛𝑑𝑗 = 1,2, … . , 𝑛 (4)
A Solution Procedure to Solve Multi objective Fractional Transportation Problem
www.irjes.com 68 | Page
𝑄 𝑥 is called objective function which is fractional. Note that 𝑃(𝑥) and 𝐷(𝑥) are linear and (2) ,(3) and (4) are also linear.
Hence the name of the problem is linear fractional transportation Programming problem (LFTPP).
Here 𝐷 𝑥 > 0 , for every 𝑋 = 𝑥𝑖𝑗 𝑤𝑕𝑒𝑟𝑒𝑋 ∈ 𝑆, Where S is feasible set defined by constraint (2),(3) and (4).
Lastly we assume that𝑎𝑖 > 0 𝑎𝑛𝑑 𝑏𝑗 > 0 𝑓𝑜𝑟 𝑖 = 1,2, … , 𝑚 𝑎𝑛𝑑 𝑗 = 1,2, … . , 𝑛
Total supply is not less than total demand i.e. 𝑎𝑖 ≥
𝑚
𝑖=1 𝑏𝑗
𝑛
𝑗 =1 (5)
The problem is to find𝑋 = 𝑥𝑖𝑗 𝑤𝑕𝑒𝑟𝑒𝑋 ∈ 𝑆𝑎𝑛𝑑 satisfying (2-4) and having optimal value of (1).
This LFTPP has following properties,
1. The problem has a feasible solution, i.e. feasible set 𝑆 ≠ ∅.It means that solution set is non empty.
2. The set of feasible solutions is bounded.
3. The problem is always solvable.
2. Definitions.
i) If total demand equals to total supply, i.e.
𝑎𝑖 =
𝑚
𝑖=1
𝑏𝑗
𝑛
𝑗 =1
(6)
Then LFTP is said to be balanced transportation problem.
Non negativity condition (4) and demand and supply conditions (2-3) at the most gives following bound to decision
variables.
0 ≤ 𝑥𝑖𝑗 ≤ min𝑖,𝑗 𝑎𝑖, 𝑏𝑗 𝑓𝑜𝑟 𝑖 = 1,2, … , 𝑚 𝑎𝑛𝑑 𝑗 = 1,2, … , 𝑛.
ii) LFTP 𝑄 𝑥 =
𝑃(𝑥)
𝐷(𝑥)
=
𝑝𝑖𝑗
𝑛
𝑗=1
𝑚
𝑖=1 𝑥𝑖𝑗 +𝑝0
𝑑𝑖𝑗
𝑛
𝑗=1
𝑚
𝑖=1 𝑥𝑖𝑗 +𝑑0
(7)
is said to be in canonical form if
𝑥𝑖𝑗 = 𝑎𝑖𝑓𝑜𝑟𝑖 = 1,2, … , 𝑚
𝑛
𝑗 =1
(8)
𝑥𝑖𝑗 = 𝑏𝑗 𝑓𝑜𝑟𝑗 = 1,2, …, (9)
𝑚
𝑖=1
𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟𝑖 = 1,2, … , 𝑚𝑎𝑛𝑑𝑗 = 1,2, … . , 𝑛 (10)
iii) There is exactly one equality redundant constraint in (8) and (9). When any one of the constraint in (8) and (9) are
dropped, what remains is exactly 𝑚 + 𝑛 − 1 linearly independent equations.
iv) A solution of LFTPP is said to be basic solution if it satisfies conditions (8) and (9).In basic solution there are 𝑚 + 𝑛 −
1 non zero values of variables in basis.
v) A basic solution of the LFTPP is said to basic feasible (BFS) if all values in basis are non-negative i.e. it satisfies
condition (10).
vi) The basic solution is degenerate if at least one of its basic variable equal to zero. Otherwise it is said to be non-
degenerate.
3. Theorems
i) The linear fractional transportation problem is solvable if and only if itis balanced LFTPP.
The problem is balanced means 𝑎𝑖 =
𝑚
𝑖=1 𝑏𝑗
𝑛
𝑗=1
ii) If all 𝑎𝑖 𝑎𝑛𝑑 𝑏𝑗 in LFTPP are positive integers, then every basic solution of LFTPP is integer vector.
Hence if all 𝑎𝑖𝑎𝑛𝑑𝑏𝑗 of LFTPP are positive integers and the problem is in the canonical form then LFTPP has an optimal
solution x* with all elements as positive integers.
4. Multi Objective Linear Fractional Transportation Programming Problem.
Suppose 𝑄1 𝑥 =
𝑃1 𝑥
𝐷1 𝑥
, 𝑄2 𝑥 =
𝑃2 𝑥
𝐷2 𝑥
, … . . , 𝑄𝑘 𝑥 =
𝑃𝑘 𝑥
𝐷𝑘 𝑥
arek linear fractional transportation problems such that all objective functions to be optimised simultaneously subject to
constraint (2) ,(3),(4) and (5) and 𝐷𝑘 𝑥 ≥ 0 for every k then the problem is called Multi Objective Linear Fractional
Transportation Programming Problem (MOLFTPP). Where
𝑄𝑙 𝑥 =
𝑝𝑖𝑗
𝑙
𝑥𝑖𝑗
𝑛
𝑗=1
𝑚
𝑖=1
𝑑𝑖𝑗
′
𝑥𝑖𝑗
𝑛
𝑗=1
𝑚
𝑖=1
𝑓𝑜𝑟 𝑙 = 1,2, … 𝑘
𝑝𝑖𝑗
𝑙
is profit of transporting one unit from 𝑖𝑡𝑕
origin to 𝑗𝑡𝑕
destination in case of 𝑙𝑡𝑕
objective function.
𝑑𝑖𝑗
𝑙
is cost of transporting one unit from 𝑖𝑡𝑕
origin to 𝑗𝑡𝑕
destinationin case of 𝑙𝑡𝑕
objective function.
5. Algorithm To Solve MOLFTPP
A Solution Procedure to Solve Multi objective Fractional Transportation Problem
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i) Consider MOLFTPP,𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 { 𝑄1 𝑥 , 𝑄2 𝑥 , … , 𝑄𝑘 𝑥 } subject to constraint (8-10).
ii) Find optimal solution of each of the linear fractional transportation problem as single objective function subject to
constraint. LINGO software can be used to find optimal solution of each of the fractional transportation problem.
iii) Suppose 𝑋𝑙
∗
is optimal solution of 𝑄𝑙 𝑥 𝑓𝑜𝑟𝑙 = 1,2, … , 𝑘
iv) Expand objective function 𝑄𝑙 𝑥 about𝑋𝑙
∗
= 𝑥𝑖𝑗 using Taylor’s theorem and ignoring second and higher order
terms convert 𝑄𝑙 𝑥 into linear function.
Consider𝑄𝑙(𝑥) =
𝑃𝑙(𝑥)
𝐷𝑙(𝑥)
and 𝑋𝑙
∗
be the optimal solution of 𝑄𝑙(𝑥) ,then using Taylor Series approach
differentiate𝑄𝑙 𝑥 with respect to each of the variable partially.
𝑄𝑙(𝑥) ≅ 𝑄𝑙 𝑋𝑙
∗
+ 𝑥11 − 𝑥11𝑙
𝜕𝑄𝑙 𝑋𝑙
𝜕𝑥11 𝑎𝑡𝑋𝑖
∗
+ 𝑥12 − 𝑥12𝑙
𝜕𝑄𝑙 𝑋𝑙
𝜕𝑥12 𝑎𝑡𝑋𝑖
∗
+ ⋯ + 𝑥𝑚𝑛 − 𝑥𝑚𝑛𝑙
𝜕𝑄𝑙 𝑋𝑙
𝜕𝑥𝑚𝑛 𝑎𝑡𝑋𝑖
∗
+ 𝑂(𝑕2)
𝑊𝑕𝑒𝑟𝑒𝑂 𝑕2
Represents second and higher order terms of the expansion which we will ignore. Using this expansion each of
the objective function becomes linear function. To avoid complexity of notations we write,𝑄𝑙(𝑥) = 𝑍𝑙 𝑥 𝑓𝑜𝑟𝑙 = 1,2, … , 𝑘
The problem is to 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 {𝑍1 𝑥 , 𝑍2 𝑥 , … . , 𝑍𝑘 𝑥 } subject to (7-10) i.e.Multi Objective Linear Transportation
Programming Problem. (MOLTPP).It can be solved by using fuzzy compromise approach as stated in next section.
v) Fuzzy Compromise approach for MOLTPP.
Consider MOLTPP
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 {𝑍1 𝑥 , 𝑍2 𝑥 , … . , 𝑍𝑘 𝑥 } ---------------- (11)
Subject to (7-10) and 𝑋 ∈ 𝑆, where𝑆is set of feasible solutions.
Solution to (11) is often conflicting as several objectives cannot be optimized simultaneously. To find compromise
solutions first solve each of the objective function as marginal or single objective function. Here we have converted Multi-
objective Linear Fractional Transportations Programming problem to Multi-objective Linear Programming problem using
Taylor series approach. Thus each of the MOLTPP can be solved using any one of the techniques of LPP. Once we solve
each of the problems individually then evaluate each of the objective function at optimal solutions of all the objectives.
Suppose 𝑋𝑙
∗
is optimal solution of 𝑙𝑡𝑕
objective function. Find values of each objective at optimal solution of 𝑙𝑡𝑕
objective
function for all 𝑙 = 1,2, … , 𝑘.Thus we have evaluation matrix of order 𝑘 × 𝑘 as given below.
𝑍1 𝑥1
∗
𝑍1 𝑥2
∗
… … … . . 𝑍1 𝑥𝑘
∗
.
𝑍2 𝑥1
∗
𝑍2 𝑥2
∗
… … … 𝑍2 𝑥𝑘
∗
:
:
𝑍𝑘 𝑥1
∗
𝑍𝑘 𝑥2
∗
… … . … 𝑍𝑘 ( 𝑥𝑘
∗
).
6. Fuzzy Programming Approach For Compromise Solution Of MOLTPP
Define Hyperbolic Membership function for each of the linearized objective function as under,.
Consider multi objective linear transportation problem.
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 𝑥 = 𝑍1 𝑥 , 𝑍2 𝑥 , … , 𝑍𝑘 𝑥 ′
𝑖. 𝑒. 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍𝑘 𝑥 = cij
k
xij
n
j=1
𝑚
𝑖=1
, 𝑘 = 1,2, … , 𝐾
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜
𝑥𝑖𝑗 =
𝑛
𝑗 =1
𝑎𝑖 , 𝑖 = 1,2, … . , 𝑚 , 𝑥𝑖𝑗
𝑚
𝑖=1
= 𝑏𝑗 , 𝑗 = 1,2, … . , 𝑛
𝑎𝑖 = 𝑏𝑗
n
j=1
𝑚
𝑖=1 And 𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖 𝑎𝑛𝑑 𝑗
Subject to x ∈ 𝑿
Where 𝑿 set of feasible solutions.
Solve each of the objectives as single objective function and then find evaluation matrix. Find best possible value
of each of the objective and worst possible value for each of the objective. Define these values as under,
𝑈𝑘 = 𝑀𝑎𝑥 𝑍𝑘 𝑥 , 𝑘 = 1,2, … , 𝐾 ,𝐿𝑘 = 𝑀𝑖𝑛 𝑍𝑘 𝑥 , 𝑘 = 1,2, … , 𝐾
Hyperbolic membership function for 𝑘𝑡𝑕 objective function is defined as under, 𝜑𝑘 : 𝑋 → 0,1 . As given below,
𝜑𝑘 𝑥 =
1 𝑖𝑓 𝑍𝑘 (𝑥) ≤ 𝐿𝑘
1
2
tanh⁡
(
𝑈𝑘 + 𝐿𝑘
2
− 𝑍𝑘 𝑥 ) 𝛼𝑘 +
1
2
0 𝑖𝑓 𝑍𝑘 (𝑥) ≥ 𝑈𝑘
𝑖𝑓 𝐿𝑘 ≤ 𝑍𝑘 (𝑥) ≤ 𝑈𝑘
Where 𝛼𝑘 =
6
𝑈𝑘 −𝐿𝑘
7. Compromise Solution of MLOFTPP
A Solution Procedure to Solve Multi objective Fractional Transportation Problem
www.irjes.com 70 | Page
a) Aggregation operator as weighted arithmetic mean of hyperbolic membership functions. Thus the MOLFPP becomes
single objective nonlinear transportation problem.
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝜇 𝑥 = 𝜑𝑖
𝑘
𝑖=1 𝑤𝑖
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑡 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠
b) Aggregation operator as weighted quadratic mean of hyperbolic membership functions. Thus the MOLFPP becomes
single objective nonlinear transportation problem.
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝜇 ( 𝑥)= 𝜑𝑖
2
𝑘
𝑖=1 𝑤𝑖
1
2
= 𝜑𝑖
2
𝑘
𝑖=1 𝑤𝑖
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑡 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠
Thus to find compromise solution is nothing but single objective nonlinear programming problem. In this method objective
function is nonlinear and constraints are linear. Hence the problem is solved by LINGO techniques of nonlinear
optimization.
II. CONCLUSION
Above method dives efficient solutions of multi objective fractional transportation problem using fuzzy programming
approach.
Example: Consider following two objective fractional transportation problem,
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑄1 𝑥 =
10𝑥11 + 14𝑥12 + 8𝑥13 + 12𝑥14 + 8𝑥21 + 12𝑥22 + 14𝑥23 + 8𝑥24 + 9𝑥31 + 6𝑥32 + 15𝑥33 + 9𝑥34
15𝑥11 + 12𝑥12 + 16𝑥13 + 8𝑥14 + 10𝑥21 + 6𝑥22 + 13𝑥23 + 12𝑥24 + 13𝑥31 + 15𝑥32 + 12𝑥33 + 10𝑥34
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑄2 𝑥 =
14𝑥11 + 9𝑥12 + 11𝑥13 + 9𝑥14 + 12𝑥21 + 9𝑥22 + 6𝑥23 + 15𝑥24 + 6𝑥31 + 9𝑥32 + 12𝑥33 + 10𝑥34
12𝑥11 + 14𝑥12 + 7𝑥13 + 17𝑥14 + 6𝑥21 + 11𝑥22 + 13𝑥23 + 10𝑥24 + 9𝑥31 + 15𝑥32 + 12𝑥33 + 16𝑥34
𝑆𝑢𝑏𝑗𝑒𝑐𝑡𝑡𝑜
𝑥11 + 𝑥12 + 𝑥13 + 𝑥14 = 15, 𝑥21 + 𝑥22 + 𝑥23 + 𝑥24 = 25
𝑥31 + 𝑥32 + 𝑥33 + 𝑥34 = 20, 𝑥11 + 𝑥21 + 𝑥31 = 15
𝑥12 + 𝑥22 + 𝑥32 = 25 , 𝑥13 + 𝑥23 + 𝑥33 = 5 , 𝑥14 + 𝑥24 + 𝑥34 = 15
, 𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟𝑖 = 1,2,3 𝑎𝑛𝑑𝑗 = 1,2,3,4
a) Step1: A LINGO code to solve 𝑄1 𝑥 is as under,
MAX
= (10*X11+14*X12+8*X13+12*X14+8*X21+12*X22+14*X23+8*X24+
9*X31+6*X32+15*X33+9*X34)/(15*X11+12*X12+16*X13+8*X14+10*X21+6*X22+13*X23+12*X24+13*X31+15*X3
2+12*X33+10*X34);
X11+X12+X13+X14=15; X21+X22+X23+X24=25;
X31+X32+X33+X34=20; X11+X21+X31=15; X12+X22+X32=25;
X13+X23+X33=5;X14+X24+X34=15;
X11>=0; X12>=0;X13>=0;X14>=0;X21>=0; X22>=0;X23>=0;X24>=0;
X31>=0;X32>=0;X33>=0;X34>=0;
Optimal Solution of 𝑄1 𝑥 is,
𝑋1
∗
= 𝑥14 = 15, 𝑥22 = 25 , 𝑥31 = 15, , 𝑥33 = 5, 𝑎𝑛𝑑𝑄1 𝑋1
∗
= 1.314286
Similarly Optimal Solution of 𝑄2 𝑥 is,
𝑋2
∗
= 𝑥11 = 5, 𝑥12 = 5, 𝑥13 = 5, 𝑥21 = 10, 𝑥24 = 15, 𝑥32 = 20 𝑎𝑛𝑑𝑄2 𝑋2
∗
= 1.02996
b) Following MATLAB program finds coefficients of xij’s to convert it into linear function using Taylor Series.
% A program to find coefficients of linear objective function.
P1 = [10;14;8;12;8;12;14;8;9;6;15;9];D1 = [15;12;16;8;10;6;13;12;13;15;12;10];
X1 = [0;0;0;15;0;25;0;0;15;0;5;0];
C1P = P1'*(D1'*X1);C1D = D1'*(P1'*X1);Z1 = (C1P-C1D)/((D1'*X1)^2);
P2 = [14;9;11;9;12;9;6;15;6;9;12;10];D2 = [12;14;7;17;6;11;13;10;9;15;14;16];
X2 = [5;5;5;0;10;0;0;15;0;20;0;0];
C2P = P2'*(D2'*X2);C2D = D2'*(P2'*X2);Z2 = (C2P -C2D)/((D2'*X2)^2);
U1 = (P1'*X1) /(D1'*X1); L1 = (P1'*X2)/(D1'*X2);
U2 = (P2'*X2)/(D2'*X2); L2 = (P2'*X1)/(D2'*X1);
c) LinearObjective Functions.
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍1 𝑥 𝑎𝑛𝑑 𝑍2 𝑥 , 𝑊𝑕𝑒𝑟𝑒,
𝑍1 𝑥 = 1.3143 − 0.185𝑥11 − 0.0034𝑥12 − 0.0248𝑥13 + 0.0028𝑥14 − 0.0098𝑥21 + 0.0078𝑥22 − 0.0059𝑥23
− 0.0148𝑥24 − 0.0154𝑥31 − 0.0261 𝑥32 − 0.0015𝑥33 − 0.00079𝑥34
𝑍2 𝑥 = 1.0296 + 0.0024𝑥11 − 0.0080𝑥12 + 0.0056𝑥13 − 0.0126𝑥14 + 0.0086𝑥21 − 0.0034𝑥22 − 0.0109𝑥23
+ 0.0070𝑥24 − 0.0048𝑥31 − 0.0095 𝑥32 − 0.0036𝑥33 − 0.0096𝑥34
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑥11 + 𝑥12 + 𝑥13 + 𝑥14 = 15,𝑥21 + 𝑥22 + 𝑥23 + 𝑥24 = 25
𝑥31 + 𝑥32 + 𝑥33 + 𝑥34 = 20,𝑥11 + 𝑥21 + 𝑥31 = 15, 𝑥12 + 𝑥22 + 𝑥32 = 25 ,
𝑥13 + 𝑥23 + 𝑥33 = 5 ,𝑥14 + 𝑥24 + 𝑥34 = 15,
,𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟 𝑖 = 1,2,3 𝑎𝑛𝑑 𝑗 = 1,2,3,4
c) Evaluation Matrix:
A Solution Procedure to Solve Multi objective Fractional Transportation Problem
www.irjes.com 71 | Page
𝒁𝟏 𝑿𝟏
∗
𝒁𝟏 𝑿𝟐
∗
𝒁𝟐 𝑿𝟏
∗
𝒁𝟐 𝑿𝟐
∗ =
1.3143 0.6038
0.6939 1.0296
Thus 𝑈1 = 1.3143 , 𝐿1 = 0.6038 ,𝑈2 = 1.0296 , 𝐿2 = 0.6939
d) Hyperbolic Membership Function:
𝜑𝑘 𝑥 =
1
2
tanh⁡
(
𝑈𝑘 + 𝐿𝑘
2
− 𝑍𝑘 𝑥 ) 𝛼𝑘 +
1
2
Where𝛼𝑘 =
6
𝑈𝑘 −𝐿𝑘
, for k=1,2
𝑈1+𝐿1
2
= 0.9590, 𝛼1 =
6
𝑈1−𝐿1
= 8.44475
𝑈2+𝐿2
2
= 0.86175,𝛼2 =
6
𝑈2−𝐿2
= 17.8731
𝜑1 𝑥 =
1
2
tanh⁡
𝑈1+𝐿1
2
− 𝑍1 𝑥 ) 𝛼1 +
1
2
=
1
2
tanh⁡0.9590 − 8.4475 𝑍1 𝑥 +
1
2
𝜑2 𝑥 =
1
2
tanh⁡
𝑈2+𝐿2
2
− 𝑍2 𝑥 𝛼2 +
1
2
=
1
2
tanh⁡0.86175 − 17.8731 𝑍2 𝑥 +
1
2
e) Compromise Solution
i) Weighted A.M. mean of hyperbolic membership functions.
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝜇 𝑥 = 𝜑𝑖
𝑘
𝑖=1 𝑤𝑖𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑡 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠
MAX
=(0.5*@TANH(0.9590 -8.44475*(1.3143-0.185*X11-0.0034*X12
-0.0248*X13+0.0028*X14-0.0098*X21+0.0078*X22-0.0059*X23
-0.0148*X24-0.0154*X31-0.0261*X32-0.0015*X33-0.0079*X34))+0.5)*0.1+
(0.5*@TANH(0.86175-17.8731*(1.0296+0.0024*X11-0.0080*X12+0.0056*X13-
.0126*X14+0.0086*X21-0.0034*X22-0.0109*X23+0.007*X24-0.0048*X31
-0.0095*X32-0.0036*X33-0.0096*X34))+0.5)*0.9;
! ENTER DEMAND AND SUPPLY CONSTRAINTS;
X11+X12+X13+X14=15;X21+X22+X23+X24=25;X31+X32+X33+X34=20; X11+X21+X31=15;
X12+X22+X32=25;X13+X23+X33=5; X14+X24+X34=15;
! ENTER NON NEGATIVITY CONSTRAINTS;
X11>=0; X12>=0; X13>=0; X14>=0;X21>=0; X22>=0; X23>=0; X24>=0;
X31>=0; X32>=0; X33>=0; X34>=0;
@GIN(X11);@GIN(X12);@GIN(X13);@GIN(X14);
@GIN(X21);@GIN(X22);@GIN(X23);@GIN(X24);
@GIN(X31);@GIN(X32);@GIN(X33);@GIN(X34);
For different values of weights we have following solutions.
X1* (0.2,0.6) (0.3,0.7) 0.6,0.4) (0.8,0.2) (0.9,0.1) X2*
X11 0 0 15 15 11 14 5
X12 0 0 0 0 0 0 5
X13 0 1 0 0 4 0 5
X14 15 14 0 0 0 1 0
X21 0 0 0 0 0 1 10
X23 25 23 5 6 10 5 0
X33 0 2 5 4 0 5 0
X24 0 0 15 15 15 14 15
X31 15 15 0 0 4 0 0
X32 0 2 20 19 15 20 20
X33 5 2 0 1 1 0 0
X34 0 1 0 0 0 0 0
Q1 1.3143 1.20871 0.65 0.66709 0.68997 0.65992 0.60377
Q2 0.70345 0.6831 0.92 0.93423 0.98701 0.90812 1.0296
ii) Weighted Quadratic Mean of hyperbolic membership functions.
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝜇 ( 𝑥)= 𝜑𝑖
2
𝑘
𝑖=1 𝑤𝑖
1
2
= 𝜑𝑖
2
𝑘
𝑖=1 𝑤𝑖
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑡 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠
MAX
=@SQRT(((0.5*@TANH(0.9590 -8.44475*(1.3143-0.185*X11-0.0034*X12
-0.0248*X13+0.0028*X14-0.0098*X21+0.0078*X22-0.0059*X23
-0.0148*X24-0.0154*X31-0.0261*X32-0.0015*X33-0.0079*X34))+0.5)^2)*0.1+
((0.5*@TANH(0.86175-17.8731*(1.0296+0.0024*X11-0.0080*X12+0.0056*X13-
.0126*X14+0.0086*X21-0.0034*X22-0.0109*X23+0.007*X24-0.0048*X31
-0.0095*X32-0.0036*X33-0.0096*X34))+0.5)^2)*0.9);
! ENTER DEMAND AND SUPPLY CONSTRAINTS;
X11+X12+X13+X14=15;X21+X22+X23+X24=25;X31+X32+X33+X34=20;
X11+X21+X31=15; X12+X22+X32=25;X13+X23+X33=5; X14+X24+X34=15;
! ENTER NON NEGATIVITY CONSTRAINTS;
X11>=0; X12>=0; X13>=0; X14>=0;X21>=0; X22>=0; X23>=0; X24>=0;
X31>=0; X32>=0; X33>=0; X34>=0;
@GIN(X11);@GIN(X12);@GIN(X13);@GIN(X14);
@GIN(X21);@GIN(X22);@GIN(X23);@GIN(X24);
A Solution Procedure to Solve Multi objective Fractional Transportation Problem
www.irjes.com 72 | Page
@GIN(X31);@GIN(X32);@GIN(X33);@GIN(X34);
For different values of weights we have following solutions.
X1* (0.1,0.9) (0.2,0.8) (0.4,0.6) (0.6,0.4) (0.9,0.1) X2*
X11 0 0 15 12 15 14 5
X12 0 0 0 0 0 0 5
X13 0 2 0 3 0 0 5
X14 15 13 0 0 0 1 0
X21 0 0 0 0 0 1 10
X23 25 21 6 19 5 5 0
X33 0 2 4 2 5 5 0
X24 0 2 15 4 15 14 15
X31 15 15 0 3 0 0 0
X32 0 4 19 6 20 20 20
X33 5 1 1 0 0 0 0
X34 0 0 0 11 0 0 0
Q1 1.3143 1.1088 0.6671 0.9069 0.6500 0.6599 0.60377
Q2 0.70345 0.7065 0.9342 0.8540 0.9200 0.9081 1.0296
REFERENCES
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Alexandra I. Tkacenko“The Multiple criteria transportation model” . (Special case ) Recent advances in applied
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Charnes,A., Cooper W.W. “Programming with linear fractional”. Naval Res. Logistics Quart. Vol. 9, No.3 and No.4, 1962,
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Diaz JA Finding a complete description of all efficient solutions to a multi objective transdportation.
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12
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problem”, International Journal of Statistika and MathematikaVol.18,issue 1 ,pp21-26,2016
13
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coefficients”, International Journal of Mathematics and Computer Research., Volume 4 ,Issue 7 ,pp 1554-1562,2016
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(2003)
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A Solution Procedure To Solve Multi Objective Fractional Transportation Problem

  • 1. International Refereed Journal of Engineering and Science (IRJES) ISSN (Online) 2319-183X, (Print) 2319-1821 Volume 7, Issue 1 (January 2018), PP. 67-72 www.irjes.com 67 | Page A Solution Procedure to Solve Multi objective Fractional Transportation Problem Dr. Doke Dnyaneshwar Maruti, M.Sc .Ph.D.(Statistics) Associate Professor and Head of Dept. , Mathematics and Statistics. M.L. DahanukarCollege., VileParle (East), Mumbai Corresponding Author: Dr. Doke Dnyaneshwar Maruti Abstract: In decision making process if the objective function is ratio of two linear functions and objective function is to be optimized. For example one may be interested to know the ratio of total cost to total time required for transportation. This ratio is an objective function which is fractional objective function. When there are several such fractional objectives to be optimized simultaneously then the problem becomes multi objective fractional programming problem (MOFLPP). Initially Hungarian mathematician BelaMartos constructed such type of problem and named it as hyperbolic programming problem. Same problem in general referred as Linear Fractional Programming Problem. Fractional programming problem can be converted into linear programming problem (LPP) by using variable transformation given by Charnes and Cooper. Then it can be solved by Simplex Method for Linear Programming Problem. . In this paper we propose to solve multi objective fractional transportation problem. Initially will solve each of the transportation problem as single objective and then using Taylor series approach expand each of the problem about its optimal solution and ignoring second and higher order error terms each of the objective is converted into linear one. Then the problem reduces to MOLTPP. Evaluate each of the objectives at every optimal solution and obtain evaluation matrix. Define hyperbolic membership function using best and worst values of objective function with reference to evaluation matrix. These membership functions are fuzzy functions Compromise solution is obtained using weighted a.m. of hyperbolic membership functions and also weights quadratic mean of hyperbolic membership functions. Propose o solve problem at the end to explain the procedure. Keywords: Multi objective, compromise solution, hyperbolic membership function. --------------------------------------------------------------------------------------------------------------------------------------- Date of Submission: 19 -01-2018 Date of acceptance: 33-01-2018 --------------------------------------------------------------------------------------------------------------------------- I. INTRODUCTION It is well known fact that the transportation problem is special case of linear programming problem. In same way multi objective fractional transportation programming problem is special case of multi objective fractional programming problem. In this paper we will deal with multi objective fractional transportation programming problem. The problem is to transfer goods from different origins to several destinations such that many fractional objective functions attain their optimal values. Again optimizing several objectives simultaneously is not possible. Thus the golden mean is to find compromise solution. For compromise solution there are several methods which are proved to be efficient. In next section linear fractional transportation problem (LFTP) is introduced. 1. Formulation of Linear Fractional TP.(LFTP) In general a problem is specified as under i) There are m supply points from which goods are to be transported. Supply available at 𝑖𝑡𝑕point is at most 𝑎𝑖𝑓𝑜𝑟𝑖 = 1,2, … , 𝑚 ii) There are n demand centres where good are required. Minimum requirement of 𝑗𝑡𝑕 demand centre is 𝑏𝑗 𝑓𝑜𝑟𝑗 = 1,2, … , 𝑛. iii) Profit matrix is 𝑃 = 𝑝𝑖𝑗 𝑖𝑠𝑚 × 𝑛 matrix where 𝑝𝑖𝑗 is profit of transporting one unit from 𝑖𝑡𝑕 origin to 𝑗𝑡𝑕 destination. iv) Cost matrix is 𝐷 = 𝑑𝑖𝑗 𝑖𝑠𝑚 × 𝑛, matrix where 𝑑𝑖𝑗 is cost of transporting one unit from 𝑖𝑡𝑕 origin to 𝑗𝑡𝑕 destination v) 𝑝0 and 𝑑0 be the fixed profit and cost respectively vi) Suppose variable𝑥𝑖𝑗 denotes number units to be transported from 𝑖𝑡𝑕 origin to 𝑗𝑡𝑕 destination. The objective function is, 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑜𝑟 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒𝑄 𝑥 = 𝑃(𝑥) 𝐷(𝑥) = 𝑝𝑖𝑗 𝑛 𝑗=1 𝑚 𝑖=1 𝑥𝑖𝑗 + 𝑝0 𝑑𝑖𝑗 𝑛 𝑗 =1 𝑚 𝑖=1 𝑥𝑖𝑗 + 𝑑0 (1) 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠, 𝑥𝑖𝑗 𝑛 𝑗 =1 ≤ 𝑎𝑖𝑓𝑜𝑟𝑖 = 1,2, … , 𝑚 (2) 𝑥𝑖𝑗 𝑚 𝑖=1 ≤ 𝑏𝑗 𝑓𝑜𝑟𝑗 = 1,2, … , 𝑛 (3) 𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟𝑖 = 1,2, … , 𝑚𝑎𝑛𝑑𝑗 = 1,2, … . , 𝑛 (4)
  • 2. A Solution Procedure to Solve Multi objective Fractional Transportation Problem www.irjes.com 68 | Page 𝑄 𝑥 is called objective function which is fractional. Note that 𝑃(𝑥) and 𝐷(𝑥) are linear and (2) ,(3) and (4) are also linear. Hence the name of the problem is linear fractional transportation Programming problem (LFTPP). Here 𝐷 𝑥 > 0 , for every 𝑋 = 𝑥𝑖𝑗 𝑤𝑕𝑒𝑟𝑒𝑋 ∈ 𝑆, Where S is feasible set defined by constraint (2),(3) and (4). Lastly we assume that𝑎𝑖 > 0 𝑎𝑛𝑑 𝑏𝑗 > 0 𝑓𝑜𝑟 𝑖 = 1,2, … , 𝑚 𝑎𝑛𝑑 𝑗 = 1,2, … . , 𝑛 Total supply is not less than total demand i.e. 𝑎𝑖 ≥ 𝑚 𝑖=1 𝑏𝑗 𝑛 𝑗 =1 (5) The problem is to find𝑋 = 𝑥𝑖𝑗 𝑤𝑕𝑒𝑟𝑒𝑋 ∈ 𝑆𝑎𝑛𝑑 satisfying (2-4) and having optimal value of (1). This LFTPP has following properties, 1. The problem has a feasible solution, i.e. feasible set 𝑆 ≠ ∅.It means that solution set is non empty. 2. The set of feasible solutions is bounded. 3. The problem is always solvable. 2. Definitions. i) If total demand equals to total supply, i.e. 𝑎𝑖 = 𝑚 𝑖=1 𝑏𝑗 𝑛 𝑗 =1 (6) Then LFTP is said to be balanced transportation problem. Non negativity condition (4) and demand and supply conditions (2-3) at the most gives following bound to decision variables. 0 ≤ 𝑥𝑖𝑗 ≤ min𝑖,𝑗 𝑎𝑖, 𝑏𝑗 𝑓𝑜𝑟 𝑖 = 1,2, … , 𝑚 𝑎𝑛𝑑 𝑗 = 1,2, … , 𝑛. ii) LFTP 𝑄 𝑥 = 𝑃(𝑥) 𝐷(𝑥) = 𝑝𝑖𝑗 𝑛 𝑗=1 𝑚 𝑖=1 𝑥𝑖𝑗 +𝑝0 𝑑𝑖𝑗 𝑛 𝑗=1 𝑚 𝑖=1 𝑥𝑖𝑗 +𝑑0 (7) is said to be in canonical form if 𝑥𝑖𝑗 = 𝑎𝑖𝑓𝑜𝑟𝑖 = 1,2, … , 𝑚 𝑛 𝑗 =1 (8) 𝑥𝑖𝑗 = 𝑏𝑗 𝑓𝑜𝑟𝑗 = 1,2, …, (9) 𝑚 𝑖=1 𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟𝑖 = 1,2, … , 𝑚𝑎𝑛𝑑𝑗 = 1,2, … . , 𝑛 (10) iii) There is exactly one equality redundant constraint in (8) and (9). When any one of the constraint in (8) and (9) are dropped, what remains is exactly 𝑚 + 𝑛 − 1 linearly independent equations. iv) A solution of LFTPP is said to be basic solution if it satisfies conditions (8) and (9).In basic solution there are 𝑚 + 𝑛 − 1 non zero values of variables in basis. v) A basic solution of the LFTPP is said to basic feasible (BFS) if all values in basis are non-negative i.e. it satisfies condition (10). vi) The basic solution is degenerate if at least one of its basic variable equal to zero. Otherwise it is said to be non- degenerate. 3. Theorems i) The linear fractional transportation problem is solvable if and only if itis balanced LFTPP. The problem is balanced means 𝑎𝑖 = 𝑚 𝑖=1 𝑏𝑗 𝑛 𝑗=1 ii) If all 𝑎𝑖 𝑎𝑛𝑑 𝑏𝑗 in LFTPP are positive integers, then every basic solution of LFTPP is integer vector. Hence if all 𝑎𝑖𝑎𝑛𝑑𝑏𝑗 of LFTPP are positive integers and the problem is in the canonical form then LFTPP has an optimal solution x* with all elements as positive integers. 4. Multi Objective Linear Fractional Transportation Programming Problem. Suppose 𝑄1 𝑥 = 𝑃1 𝑥 𝐷1 𝑥 , 𝑄2 𝑥 = 𝑃2 𝑥 𝐷2 𝑥 , … . . , 𝑄𝑘 𝑥 = 𝑃𝑘 𝑥 𝐷𝑘 𝑥 arek linear fractional transportation problems such that all objective functions to be optimised simultaneously subject to constraint (2) ,(3),(4) and (5) and 𝐷𝑘 𝑥 ≥ 0 for every k then the problem is called Multi Objective Linear Fractional Transportation Programming Problem (MOLFTPP). Where 𝑄𝑙 𝑥 = 𝑝𝑖𝑗 𝑙 𝑥𝑖𝑗 𝑛 𝑗=1 𝑚 𝑖=1 𝑑𝑖𝑗 ′ 𝑥𝑖𝑗 𝑛 𝑗=1 𝑚 𝑖=1 𝑓𝑜𝑟 𝑙 = 1,2, … 𝑘 𝑝𝑖𝑗 𝑙 is profit of transporting one unit from 𝑖𝑡𝑕 origin to 𝑗𝑡𝑕 destination in case of 𝑙𝑡𝑕 objective function. 𝑑𝑖𝑗 𝑙 is cost of transporting one unit from 𝑖𝑡𝑕 origin to 𝑗𝑡𝑕 destinationin case of 𝑙𝑡𝑕 objective function. 5. Algorithm To Solve MOLFTPP
  • 3. A Solution Procedure to Solve Multi objective Fractional Transportation Problem www.irjes.com 69 | Page i) Consider MOLFTPP,𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 { 𝑄1 𝑥 , 𝑄2 𝑥 , … , 𝑄𝑘 𝑥 } subject to constraint (8-10). ii) Find optimal solution of each of the linear fractional transportation problem as single objective function subject to constraint. LINGO software can be used to find optimal solution of each of the fractional transportation problem. iii) Suppose 𝑋𝑙 ∗ is optimal solution of 𝑄𝑙 𝑥 𝑓𝑜𝑟𝑙 = 1,2, … , 𝑘 iv) Expand objective function 𝑄𝑙 𝑥 about𝑋𝑙 ∗ = 𝑥𝑖𝑗 using Taylor’s theorem and ignoring second and higher order terms convert 𝑄𝑙 𝑥 into linear function. Consider𝑄𝑙(𝑥) = 𝑃𝑙(𝑥) 𝐷𝑙(𝑥) and 𝑋𝑙 ∗ be the optimal solution of 𝑄𝑙(𝑥) ,then using Taylor Series approach differentiate𝑄𝑙 𝑥 with respect to each of the variable partially. 𝑄𝑙(𝑥) ≅ 𝑄𝑙 𝑋𝑙 ∗ + 𝑥11 − 𝑥11𝑙 𝜕𝑄𝑙 𝑋𝑙 𝜕𝑥11 𝑎𝑡𝑋𝑖 ∗ + 𝑥12 − 𝑥12𝑙 𝜕𝑄𝑙 𝑋𝑙 𝜕𝑥12 𝑎𝑡𝑋𝑖 ∗ + ⋯ + 𝑥𝑚𝑛 − 𝑥𝑚𝑛𝑙 𝜕𝑄𝑙 𝑋𝑙 𝜕𝑥𝑚𝑛 𝑎𝑡𝑋𝑖 ∗ + 𝑂(𝑕2) 𝑊𝑕𝑒𝑟𝑒𝑂 𝑕2 Represents second and higher order terms of the expansion which we will ignore. Using this expansion each of the objective function becomes linear function. To avoid complexity of notations we write,𝑄𝑙(𝑥) = 𝑍𝑙 𝑥 𝑓𝑜𝑟𝑙 = 1,2, … , 𝑘 The problem is to 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 {𝑍1 𝑥 , 𝑍2 𝑥 , … . , 𝑍𝑘 𝑥 } subject to (7-10) i.e.Multi Objective Linear Transportation Programming Problem. (MOLTPP).It can be solved by using fuzzy compromise approach as stated in next section. v) Fuzzy Compromise approach for MOLTPP. Consider MOLTPP 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 {𝑍1 𝑥 , 𝑍2 𝑥 , … . , 𝑍𝑘 𝑥 } ---------------- (11) Subject to (7-10) and 𝑋 ∈ 𝑆, where𝑆is set of feasible solutions. Solution to (11) is often conflicting as several objectives cannot be optimized simultaneously. To find compromise solutions first solve each of the objective function as marginal or single objective function. Here we have converted Multi- objective Linear Fractional Transportations Programming problem to Multi-objective Linear Programming problem using Taylor series approach. Thus each of the MOLTPP can be solved using any one of the techniques of LPP. Once we solve each of the problems individually then evaluate each of the objective function at optimal solutions of all the objectives. Suppose 𝑋𝑙 ∗ is optimal solution of 𝑙𝑡𝑕 objective function. Find values of each objective at optimal solution of 𝑙𝑡𝑕 objective function for all 𝑙 = 1,2, … , 𝑘.Thus we have evaluation matrix of order 𝑘 × 𝑘 as given below. 𝑍1 𝑥1 ∗ 𝑍1 𝑥2 ∗ … … … . . 𝑍1 𝑥𝑘 ∗ . 𝑍2 𝑥1 ∗ 𝑍2 𝑥2 ∗ … … … 𝑍2 𝑥𝑘 ∗ : : 𝑍𝑘 𝑥1 ∗ 𝑍𝑘 𝑥2 ∗ … … . … 𝑍𝑘 ( 𝑥𝑘 ∗ ). 6. Fuzzy Programming Approach For Compromise Solution Of MOLTPP Define Hyperbolic Membership function for each of the linearized objective function as under,. Consider multi objective linear transportation problem. 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 𝑥 = 𝑍1 𝑥 , 𝑍2 𝑥 , … , 𝑍𝑘 𝑥 ′ 𝑖. 𝑒. 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍𝑘 𝑥 = cij k xij n j=1 𝑚 𝑖=1 , 𝑘 = 1,2, … , 𝐾 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑥𝑖𝑗 = 𝑛 𝑗 =1 𝑎𝑖 , 𝑖 = 1,2, … . , 𝑚 , 𝑥𝑖𝑗 𝑚 𝑖=1 = 𝑏𝑗 , 𝑗 = 1,2, … . , 𝑛 𝑎𝑖 = 𝑏𝑗 n j=1 𝑚 𝑖=1 And 𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖 𝑎𝑛𝑑 𝑗 Subject to x ∈ 𝑿 Where 𝑿 set of feasible solutions. Solve each of the objectives as single objective function and then find evaluation matrix. Find best possible value of each of the objective and worst possible value for each of the objective. Define these values as under, 𝑈𝑘 = 𝑀𝑎𝑥 𝑍𝑘 𝑥 , 𝑘 = 1,2, … , 𝐾 ,𝐿𝑘 = 𝑀𝑖𝑛 𝑍𝑘 𝑥 , 𝑘 = 1,2, … , 𝐾 Hyperbolic membership function for 𝑘𝑡𝑕 objective function is defined as under, 𝜑𝑘 : 𝑋 → 0,1 . As given below, 𝜑𝑘 𝑥 = 1 𝑖𝑓 𝑍𝑘 (𝑥) ≤ 𝐿𝑘 1 2 tanh⁡ ( 𝑈𝑘 + 𝐿𝑘 2 − 𝑍𝑘 𝑥 ) 𝛼𝑘 + 1 2 0 𝑖𝑓 𝑍𝑘 (𝑥) ≥ 𝑈𝑘 𝑖𝑓 𝐿𝑘 ≤ 𝑍𝑘 (𝑥) ≤ 𝑈𝑘 Where 𝛼𝑘 = 6 𝑈𝑘 −𝐿𝑘 7. Compromise Solution of MLOFTPP
  • 4. A Solution Procedure to Solve Multi objective Fractional Transportation Problem www.irjes.com 70 | Page a) Aggregation operator as weighted arithmetic mean of hyperbolic membership functions. Thus the MOLFPP becomes single objective nonlinear transportation problem. 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝜇 𝑥 = 𝜑𝑖 𝑘 𝑖=1 𝑤𝑖 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑡 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠 b) Aggregation operator as weighted quadratic mean of hyperbolic membership functions. Thus the MOLFPP becomes single objective nonlinear transportation problem. 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝜇 ( 𝑥)= 𝜑𝑖 2 𝑘 𝑖=1 𝑤𝑖 1 2 = 𝜑𝑖 2 𝑘 𝑖=1 𝑤𝑖 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑡 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠 Thus to find compromise solution is nothing but single objective nonlinear programming problem. In this method objective function is nonlinear and constraints are linear. Hence the problem is solved by LINGO techniques of nonlinear optimization. II. CONCLUSION Above method dives efficient solutions of multi objective fractional transportation problem using fuzzy programming approach. Example: Consider following two objective fractional transportation problem, 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑄1 𝑥 = 10𝑥11 + 14𝑥12 + 8𝑥13 + 12𝑥14 + 8𝑥21 + 12𝑥22 + 14𝑥23 + 8𝑥24 + 9𝑥31 + 6𝑥32 + 15𝑥33 + 9𝑥34 15𝑥11 + 12𝑥12 + 16𝑥13 + 8𝑥14 + 10𝑥21 + 6𝑥22 + 13𝑥23 + 12𝑥24 + 13𝑥31 + 15𝑥32 + 12𝑥33 + 10𝑥34 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑄2 𝑥 = 14𝑥11 + 9𝑥12 + 11𝑥13 + 9𝑥14 + 12𝑥21 + 9𝑥22 + 6𝑥23 + 15𝑥24 + 6𝑥31 + 9𝑥32 + 12𝑥33 + 10𝑥34 12𝑥11 + 14𝑥12 + 7𝑥13 + 17𝑥14 + 6𝑥21 + 11𝑥22 + 13𝑥23 + 10𝑥24 + 9𝑥31 + 15𝑥32 + 12𝑥33 + 16𝑥34 𝑆𝑢𝑏𝑗𝑒𝑐𝑡𝑡𝑜 𝑥11 + 𝑥12 + 𝑥13 + 𝑥14 = 15, 𝑥21 + 𝑥22 + 𝑥23 + 𝑥24 = 25 𝑥31 + 𝑥32 + 𝑥33 + 𝑥34 = 20, 𝑥11 + 𝑥21 + 𝑥31 = 15 𝑥12 + 𝑥22 + 𝑥32 = 25 , 𝑥13 + 𝑥23 + 𝑥33 = 5 , 𝑥14 + 𝑥24 + 𝑥34 = 15 , 𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟𝑖 = 1,2,3 𝑎𝑛𝑑𝑗 = 1,2,3,4 a) Step1: A LINGO code to solve 𝑄1 𝑥 is as under, MAX = (10*X11+14*X12+8*X13+12*X14+8*X21+12*X22+14*X23+8*X24+ 9*X31+6*X32+15*X33+9*X34)/(15*X11+12*X12+16*X13+8*X14+10*X21+6*X22+13*X23+12*X24+13*X31+15*X3 2+12*X33+10*X34); X11+X12+X13+X14=15; X21+X22+X23+X24=25; X31+X32+X33+X34=20; X11+X21+X31=15; X12+X22+X32=25; X13+X23+X33=5;X14+X24+X34=15; X11>=0; X12>=0;X13>=0;X14>=0;X21>=0; X22>=0;X23>=0;X24>=0; X31>=0;X32>=0;X33>=0;X34>=0; Optimal Solution of 𝑄1 𝑥 is, 𝑋1 ∗ = 𝑥14 = 15, 𝑥22 = 25 , 𝑥31 = 15, , 𝑥33 = 5, 𝑎𝑛𝑑𝑄1 𝑋1 ∗ = 1.314286 Similarly Optimal Solution of 𝑄2 𝑥 is, 𝑋2 ∗ = 𝑥11 = 5, 𝑥12 = 5, 𝑥13 = 5, 𝑥21 = 10, 𝑥24 = 15, 𝑥32 = 20 𝑎𝑛𝑑𝑄2 𝑋2 ∗ = 1.02996 b) Following MATLAB program finds coefficients of xij’s to convert it into linear function using Taylor Series. % A program to find coefficients of linear objective function. P1 = [10;14;8;12;8;12;14;8;9;6;15;9];D1 = [15;12;16;8;10;6;13;12;13;15;12;10]; X1 = [0;0;0;15;0;25;0;0;15;0;5;0]; C1P = P1'*(D1'*X1);C1D = D1'*(P1'*X1);Z1 = (C1P-C1D)/((D1'*X1)^2); P2 = [14;9;11;9;12;9;6;15;6;9;12;10];D2 = [12;14;7;17;6;11;13;10;9;15;14;16]; X2 = [5;5;5;0;10;0;0;15;0;20;0;0]; C2P = P2'*(D2'*X2);C2D = D2'*(P2'*X2);Z2 = (C2P -C2D)/((D2'*X2)^2); U1 = (P1'*X1) /(D1'*X1); L1 = (P1'*X2)/(D1'*X2); U2 = (P2'*X2)/(D2'*X2); L2 = (P2'*X1)/(D2'*X1); c) LinearObjective Functions. 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍1 𝑥 𝑎𝑛𝑑 𝑍2 𝑥 , 𝑊𝑕𝑒𝑟𝑒, 𝑍1 𝑥 = 1.3143 − 0.185𝑥11 − 0.0034𝑥12 − 0.0248𝑥13 + 0.0028𝑥14 − 0.0098𝑥21 + 0.0078𝑥22 − 0.0059𝑥23 − 0.0148𝑥24 − 0.0154𝑥31 − 0.0261 𝑥32 − 0.0015𝑥33 − 0.00079𝑥34 𝑍2 𝑥 = 1.0296 + 0.0024𝑥11 − 0.0080𝑥12 + 0.0056𝑥13 − 0.0126𝑥14 + 0.0086𝑥21 − 0.0034𝑥22 − 0.0109𝑥23 + 0.0070𝑥24 − 0.0048𝑥31 − 0.0095 𝑥32 − 0.0036𝑥33 − 0.0096𝑥34 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑥11 + 𝑥12 + 𝑥13 + 𝑥14 = 15,𝑥21 + 𝑥22 + 𝑥23 + 𝑥24 = 25 𝑥31 + 𝑥32 + 𝑥33 + 𝑥34 = 20,𝑥11 + 𝑥21 + 𝑥31 = 15, 𝑥12 + 𝑥22 + 𝑥32 = 25 , 𝑥13 + 𝑥23 + 𝑥33 = 5 ,𝑥14 + 𝑥24 + 𝑥34 = 15, ,𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟 𝑖 = 1,2,3 𝑎𝑛𝑑 𝑗 = 1,2,3,4 c) Evaluation Matrix:
  • 5. A Solution Procedure to Solve Multi objective Fractional Transportation Problem www.irjes.com 71 | Page 𝒁𝟏 𝑿𝟏 ∗ 𝒁𝟏 𝑿𝟐 ∗ 𝒁𝟐 𝑿𝟏 ∗ 𝒁𝟐 𝑿𝟐 ∗ = 1.3143 0.6038 0.6939 1.0296 Thus 𝑈1 = 1.3143 , 𝐿1 = 0.6038 ,𝑈2 = 1.0296 , 𝐿2 = 0.6939 d) Hyperbolic Membership Function: 𝜑𝑘 𝑥 = 1 2 tanh⁡ ( 𝑈𝑘 + 𝐿𝑘 2 − 𝑍𝑘 𝑥 ) 𝛼𝑘 + 1 2 Where𝛼𝑘 = 6 𝑈𝑘 −𝐿𝑘 , for k=1,2 𝑈1+𝐿1 2 = 0.9590, 𝛼1 = 6 𝑈1−𝐿1 = 8.44475 𝑈2+𝐿2 2 = 0.86175,𝛼2 = 6 𝑈2−𝐿2 = 17.8731 𝜑1 𝑥 = 1 2 tanh⁡ 𝑈1+𝐿1 2 − 𝑍1 𝑥 ) 𝛼1 + 1 2 = 1 2 tanh⁡0.9590 − 8.4475 𝑍1 𝑥 + 1 2 𝜑2 𝑥 = 1 2 tanh⁡ 𝑈2+𝐿2 2 − 𝑍2 𝑥 𝛼2 + 1 2 = 1 2 tanh⁡0.86175 − 17.8731 𝑍2 𝑥 + 1 2 e) Compromise Solution i) Weighted A.M. mean of hyperbolic membership functions. 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝜇 𝑥 = 𝜑𝑖 𝑘 𝑖=1 𝑤𝑖𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑡 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠 MAX =(0.5*@TANH(0.9590 -8.44475*(1.3143-0.185*X11-0.0034*X12 -0.0248*X13+0.0028*X14-0.0098*X21+0.0078*X22-0.0059*X23 -0.0148*X24-0.0154*X31-0.0261*X32-0.0015*X33-0.0079*X34))+0.5)*0.1+ (0.5*@TANH(0.86175-17.8731*(1.0296+0.0024*X11-0.0080*X12+0.0056*X13- .0126*X14+0.0086*X21-0.0034*X22-0.0109*X23+0.007*X24-0.0048*X31 -0.0095*X32-0.0036*X33-0.0096*X34))+0.5)*0.9; ! ENTER DEMAND AND SUPPLY CONSTRAINTS; X11+X12+X13+X14=15;X21+X22+X23+X24=25;X31+X32+X33+X34=20; X11+X21+X31=15; X12+X22+X32=25;X13+X23+X33=5; X14+X24+X34=15; ! ENTER NON NEGATIVITY CONSTRAINTS; X11>=0; X12>=0; X13>=0; X14>=0;X21>=0; X22>=0; X23>=0; X24>=0; X31>=0; X32>=0; X33>=0; X34>=0; @GIN(X11);@GIN(X12);@GIN(X13);@GIN(X14); @GIN(X21);@GIN(X22);@GIN(X23);@GIN(X24); @GIN(X31);@GIN(X32);@GIN(X33);@GIN(X34); For different values of weights we have following solutions. X1* (0.2,0.6) (0.3,0.7) 0.6,0.4) (0.8,0.2) (0.9,0.1) X2* X11 0 0 15 15 11 14 5 X12 0 0 0 0 0 0 5 X13 0 1 0 0 4 0 5 X14 15 14 0 0 0 1 0 X21 0 0 0 0 0 1 10 X23 25 23 5 6 10 5 0 X33 0 2 5 4 0 5 0 X24 0 0 15 15 15 14 15 X31 15 15 0 0 4 0 0 X32 0 2 20 19 15 20 20 X33 5 2 0 1 1 0 0 X34 0 1 0 0 0 0 0 Q1 1.3143 1.20871 0.65 0.66709 0.68997 0.65992 0.60377 Q2 0.70345 0.6831 0.92 0.93423 0.98701 0.90812 1.0296 ii) Weighted Quadratic Mean of hyperbolic membership functions. 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝜇 ( 𝑥)= 𝜑𝑖 2 𝑘 𝑖=1 𝑤𝑖 1 2 = 𝜑𝑖 2 𝑘 𝑖=1 𝑤𝑖 𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑡 𝑜𝑓 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠 MAX =@SQRT(((0.5*@TANH(0.9590 -8.44475*(1.3143-0.185*X11-0.0034*X12 -0.0248*X13+0.0028*X14-0.0098*X21+0.0078*X22-0.0059*X23 -0.0148*X24-0.0154*X31-0.0261*X32-0.0015*X33-0.0079*X34))+0.5)^2)*0.1+ ((0.5*@TANH(0.86175-17.8731*(1.0296+0.0024*X11-0.0080*X12+0.0056*X13- .0126*X14+0.0086*X21-0.0034*X22-0.0109*X23+0.007*X24-0.0048*X31 -0.0095*X32-0.0036*X33-0.0096*X34))+0.5)^2)*0.9); ! ENTER DEMAND AND SUPPLY CONSTRAINTS; X11+X12+X13+X14=15;X21+X22+X23+X24=25;X31+X32+X33+X34=20; X11+X21+X31=15; X12+X22+X32=25;X13+X23+X33=5; X14+X24+X34=15; ! ENTER NON NEGATIVITY CONSTRAINTS; X11>=0; X12>=0; X13>=0; X14>=0;X21>=0; X22>=0; X23>=0; X24>=0; X31>=0; X32>=0; X33>=0; X34>=0; @GIN(X11);@GIN(X12);@GIN(X13);@GIN(X14); @GIN(X21);@GIN(X22);@GIN(X23);@GIN(X24);
  • 6. A Solution Procedure to Solve Multi objective Fractional Transportation Problem www.irjes.com 72 | Page @GIN(X31);@GIN(X32);@GIN(X33);@GIN(X34); For different values of weights we have following solutions. X1* (0.1,0.9) (0.2,0.8) (0.4,0.6) (0.6,0.4) (0.9,0.1) X2* X11 0 0 15 12 15 14 5 X12 0 0 0 0 0 0 5 X13 0 2 0 3 0 0 5 X14 15 13 0 0 0 1 0 X21 0 0 0 0 0 1 10 X23 25 21 6 19 5 5 0 X33 0 2 4 2 5 5 0 X24 0 2 15 4 15 14 15 X31 15 15 0 3 0 0 0 X32 0 4 19 6 20 20 20 X33 5 1 1 0 0 0 0 X34 0 0 0 11 0 0 0 Q1 1.3143 1.1088 0.6671 0.9069 0.6500 0.6599 0.60377 Q2 0.70345 0.7065 0.9342 0.8540 0.9200 0.9081 1.0296 REFERENCES 1 Alexandra I. Tkacenko“The generalized algorithm for solving the fractional multi objective transportation problem.” ROMAI J.,1(2006),197-202. 2 Alexandra I. Tkacenko“The Multiple criteria transportation model” . 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