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INTERNATIONAL JOURNAL OF SCIENTIFIC & TECHNOLOGY RESEARCH VOLUME 9, ISSUE 03, MARCH 2020 ISSN 2277-8616
An Optimal Solution For Time Minimizing
Transportation Problems By Using Maximum
Range Method
N.Anandhi T.Geetha
Abstract: This article presents a new method named Maximum Range Method (MRM) for finding an optimal solution for Time Minimizing Transportation
Problems (TMTP).The main purpose of this method is to Minimize the maximum time of transportation for all availability to requirement, rather than
minimizing transportation cost. The procedure of MRM is developed by Transportation Problem with equality constraints. The proposed method is
determine an optimum solution for TMTP and compare with the obtained results of other conventional methods, and found that the proposed method
gives the better results from other traditional method. In this dissertation one more method of achieving a minimum time of transportation has been
developed which is very different from obtainable method. Finally, a numerical illustration has been presented for better understanding of the algorithm
Index Terms: Transportation problem, Transportation Time, Optimal Solution, Range and MRM
—————————— ——————————
1.INTRODUCTION The method revolves around the allocation of least time
The transportation problem for time minimization is one of cells which are determined by exploring the row/column
the special subclass of a transportation problem and it is consisting maximum range time of given transportation
defined as a Transportation Problem(TP) where instead problem. This text aims to present the method of
cost , time is need to be minimized. The objective is to determining the Optimum solution of TMTP. The proposed
diminish the time while transporting all available thinks to algorithm has very simple steps to reach solution and
the destinations. In this TP, the value of [t ] is given where hence implementation will be easier. It provides the best
solution for distribution problems which will be help full for
t is representing transporting time from i origins to managing persons. The numerical illustration for the
j destinations.For any feasible solution[x ] need to satisfy
proposed algorithms is shown to prove its efficiency to get
the capacity and demand condition , time of transportation the optimal solution for TMTP. The paper is arranged as
Max follows: Section 2 .explain the mathematical model of the
is { t ∶ x > 0}. The aim is to minimize the time take
(i. j) problem. In section 3, the algorithm of the proposed method
to bring the goods from source to destination. It is assumed is given .In section 4, a example problem is demonstrated
that the goods can be carried from source to target place in In section 5 Results has been Discussed. Section 6 is
single trip. The TMTP is extensively useful in many practical conclusion of this Paper.
situations such as military transportation for the time of
emergency, transport of all the fresh food items, fire service 1.1MATHEMATICAL FORMULATION OF THE TMTP
and hospital services …etc. In this paper a simple algorithm Let us consider the stranded balanced transportation
for solving a TMTP has been developed. The method problem, with originO (with availability a ) i = 1,2,…m, and
presented and discussed above gives us the optimal
destinations, D (with requirement b ),j = 1,2,…n. If x is
solution where minimum numbers of iterations are required.
the number of load (amount) units shifting from O toD ,the
The proposed algorithm is easy one to apply which is used
to derive the solution to a variety of distribution problems feasible solution *x + and set of feasible solution*x +
with equality constraints. Hammer[1], Garfinkel and Rao[2]
have first analyzed the time minimizing transportation X ={x ⁄∑x =a ,i=1,2,……m ; ∑x =b ,j
problem. The better solution and procedures are reached
by Szware[3] and Puri[4],. Some novel procedures are find
by Swarup[5], and Seshan[6] to minimize the time taken for =1,2,……n; x ≥0 ;∑a =∑b}
transportation. Recently, Khan et al. [7] defined and used
pointer cost to assign the cells for IBFS of the problem. In
this paper, the method of finding optimum solution of TMTP Let t be the time taken to transfer all x items using
is developed, with the same fundamental assumptions corresponding route (i,j) for all i = 1,2,……m and =
made by Gupta.P.K, and Hira D.S. 1,2,……n .
The problem is mathematically expressed as follow:
Max
( )
________________________________________ Minimize Total Time Z = { t ∶ x > 0} − −(1)
Subject to the constrains (i. j)
• N.Anandhi is currently working as Assistant Professor of
Mathematics, Anjalai Ammal- Mahalingam Engineering College,
( )
Kovilvenni – 614 403,Thiruvarur Dist., Tamil Nadu, India. ∑x =a ,i = 1,2,……m supply contrains −−−(2)
• Dr. T.Geetha is currently working as Assistant Professor of
Mathematics, kunthavai Naacchiyar Government Arts college
for–Dist.,Tamil Nadu, India
( )
∑x =b ,j=1,2,……n Demand contrains −−−(3)
918
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INTERNATIONAL JOURNAL OF SCIENTIFIC & TECHNOLOGY RESEARCH VOLUME 9, ISSUE 03, MARCH 2020 ISSN 2277-8616
and . . . .
x ≥0;a >0 ;b >0 for all i and j − . . . .
−−−(4) … …
. .
Where a is the quantity of load capacity at i origin ; b is
the quantity of load demand at j destination; x is the nt
e ∑
m
amount of commodity transporting from i origin to e … …
j destination. uir
q >∑
e
1.2 BALANCED AND UNBALANCED TRANSPORTATION TABLE(TT) R
Transportation problem is explicitly represented by the Figure 2.Transpotation table of unbalanced transportation
following transportation table problem
Modified Balanced transportation Table: ∑ a =∑ b
Destination
Or … … Ava Destination
g. . . . Or … … Ava
g. . . .
… …
. . …
…
.
.
…
.
. . . . . . …
.
. . . . . . =
… … . . . . . .
. . . . . . . .
. . . . . …
. . . . . .
…
.
… …
. . =
nt . . . . .
e ∑ . . . . .
m …
e … …
ri .
u …
q =∑ .
e
R =
Figure 1.Transpotation table of transportation problem t
n
e =
m ∑
The m×n squares are called cells. The transportation time e … … ∑ −
ri =
u
t from the i origin to the j the (i,j) cell. The solution ∑
eq
x is displayed in the upper left corner of the cell .The
R
variousa ‘s and b ‘s are called rim requirements. The ∑
feasibility of a solution can be verified by summing the Figure 3.Transpotation table of Modified balanced
values of x along the rows and down the columns (ie, x transportation problem
satisfying the rim conditions).
The TP feasible solution will have (m+n−1) positive Unbalanced transportation Table: (shortage in availability ie.
allocations.
∑a <∑b)
Unbalanced transportation Table: (Excess availability
Destination
∑ ∑
ie > )
Org … … Ava
Destination
Or … … . . . .
Ava.
g. . . … …
. .
… …
. .
…
… .
.
. . . . .
. . . . . . . . . .
. . . . .
… …
… …
. . . .
919
IJSTR©2020
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INTERNATIONAL JOURNAL OF SCIENTIFIC & TECHNOLOGY RESEARCH VOLUME 9, ISSUE 03, MARCH 2020 ISSN 2277-8616
. . . . 1 ;x > 0
. . . . { . Efficiency of Transportation E(x) =
0 ;x = 0
∑ ∑ t x .
… …
. .
Here the better IBFS is obtained by leaving basic cell which
tn have maximum time. If the basic cell transportation time is
e
greater than non basic cell transportation time then leave
me … … ∑ the basic cell and enter the non basic cell into the basis
ri <
u which rearrange the current allocation. Crossed out the
qe
cells which have larger time than the (r,s) cellThe better
R
∑ solution gives the allocated value {x } either partially or fully
Figure 4.Transpotation table of unbalanced transportation converted into other allocated cell or non allocated cell with
problem minimum time. The balanced TP form a closed loop , this
loop provide the suitable non basic cell enter into the basis.
Modified Balanced transportation Table: ∑ a =∑ b
The closed loop always having even number of cells. Let
Org. … … Ava. the amount φ is to be shifted and it is less than are equal to
basic variable x ,.choose the φ value only for the basic
… … allocation x such that x ± φ ≥ 0. Add the φ to the ‗+‘
sign and subtract the φ to the ‗-‗sign of the closed loop.
Iteratively all the non basic cell would be crossed out.
…
. . . . . 3. A INNOVATIVE APPROACH FOR SOLVING
. . . . . TRANSPORTATION PROBLEM
… … The process of proposed method is carried out in stepwise.
. . . . Step 1: To check the given time transportation table is
. . . . balanced. Suppose the given time transportation table is
unbalanced convert into balanced.
… …
Step 2: Calculate the range in each row and each column it
is display in the outside of the table. choose any one row or
column with highest range value. Assign a feasible value
with negligible unit time in the selected row or column.
… …
= ∑ ∑ Eliminate the satisfied row (or) column and modify the TT .
= =
= − Suppose row and column both are fulfilled concurrently,
eliminate any one and the remaining one has zero
nt availability( or) zero requirement . If tie occur then choose
e the cells with minimum time and utmost allocation can be
m made. Suppose the least time and highest allocation also
e … …
ri tie chose any one arbitrarily.
u ∑
qe ∑ =
R Step 3:
(i) Any one of the row (column) with zero (or) nonzero
Figure 5.Transpotation table of Modified balanced capacity (requirement) calculate the basic variable by using
transportation problem Matrix minima method then stop.
(ii) Otherwise go to step 2 and again the same process until
2.SOLUTION METHOD the requirement and capacity are drained.
Consider a balanced TMTP with m origin and n destination
The hauler can able to transport the goods from the starting Step 4:
point to end place in a single time. The transportation time Display all the basic variable x to the corresponding cells
t is not depend upon the goods carried. There are two type in the given transportation table.
of solution in the TMTP first one is IBFS and second one Test for Optimality:
upgrading of the IBFS. (i) If all the basic cells transportation time is less than or
equal to the non basic cells transportation time in its
The solution of IBFS is denoted by Optimum (Minimum) row(column) then cross out all the non allocated cells and
( ) Max
time of Transportation Z (x) = { t ∶ x >0}=
(i, j) also the non - degenerate basic cells are not form a loop
t ( ), where n (n=1,2,3,4…..) is number of iteration. Note then the present BFS is optimum.
(ii) Suppose some of the non basic cells transportation time
that (r,s) cell may or may not be unique. is less than equal to the basic cells transportation time in its
And also calculate Total transportation time T(x) =
row(column). Selected uncrossed out non basic cells
∑ ∑ t u where the auxiliary function u =
without generate the loop then the current basic feasible
920
IJSTR©2020
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INTERNATIONAL JOURNAL OF SCIENTIFIC & TECHNOLOGY RESEARCH VOLUME 9, ISSUE 03, MARCH 2020 ISSN 2277-8616
solution is optimum. To find optimum
valueZ(x),T(x) and E(x).Otherwise go to the improved Iteration :I
optimum solution From the Transportation Table 6.1.1, Sum of availability and
Sum of requirement are equal. Therefore ,the above time
Step 5: Improved optimum solution TT is a balanced. To find optimum solution by applying
Selected non basic cell with generate a closed loop time MRM allocations are obtained as follows
transportation table, put a ‗+‘ sign and ‗-‗ sign to the closed
loop. Calculate following values
Destination
Entering value i D D D il il
O ri g n A v a a b it y
∅={x /x lies in ′− sign of cells in closed loop} 8 7
O 10 20 15
Add the ∅ to ‗+‘ sign of closed loop ie, x + ∅ > 0 and 0 11
12 13
subtract the ∅ to the ‗-‗ sign of the closed loop ie, x − ∅ ≥ 0 O 25
, the cell with x − ∅ = 0 , is leaving the basis and cross out 1 7 9 20
the cell. 2 3
O 12 8 5
Value of loop V = t −t +t +t ……. Where t is 16 18
12 t 8 15 10 45
entering non -basic cell transportation time and R e q u ir e m e n
t , t , t , t ….etc are basic cell transportation time in the
Transportation table 6.1.2
closed loop.
Weightage of the loop W = t −t Where t is entering
( )
( ) ( )
non -basic cell transportation time and t leaving basic cell Hence, Initial basic feasible solution Z x =18,T x =
( )
transportation time. 55 minitues and E x = 292
By examine the value of V and W and it is conclude that
Iteration :II
(i) If all V >0 W > 0 ,then the loop is rejected and
current solution is optimum and unique. Improved Optimality
(ii) If all V ≥0 and W ≥0 ,then the loop is rejected and
current solution is optimum and alternate optimum exists. n bil
(iii) If V >0 or V < 0 and at least one W < 0, then loop is i
g D D D D ila y
accepted .suppose more than one loop is accepted then Ori va it
select loop with most negative value of W .Therefore the A
5 10
current solution is not optimum .Rearrange the allocation O 10 20 15
and go to the step 6. 0 11
O 12 3 10 20 25
Step 6: Do again the steps 4 and 5 awaiting all non basic 1 7 9
cell crossed out and the optimum solution is reached O 12 8 5 18 5
calculate optimum value of Z(x),T(x) and E(x). 16
ire nt
4.NUMERICAL ILLUSTRATION qu 12 8 15 10 45
Re me
Example-6.1
A air force tools is to be transported from three origins to Transportation table 6.1.3
four destinations the availability of the origins, the ( )( )
requirement of the destination and time of shipment is Hence, Improved basic feasible solution Z x =
( ) ( )
shown in the table below and find the total time required for 16,T x = 44 minitues and E x = 313
shipment is minimum Z(x), Total time of transportation T(x)
and Transportation efficiency E(x) by using maximum range Iteration :III
method.
Destination
bi
Transportation table 6.1.1 in D D D D ila ty
Oigr va li
A
ni ila O 10 5 20 10 15
g D D D D y
ri va ilit 0 11
O A b O 7 3 15 20 25
O 10 0 20 11 15 1 7 9
5
O 1 7 9 20 25 O 8 16 18 5
12
O 12 14 16 18 5
q e nt
rui nt Re uir e 12 8 15 10 45
q e 12 8 15 10 45 m
eR em Transportation table 6.1.4
Solution of Example 6.1 Hence, Optimum solution
( )( ) ( ) ( )
Z x =12,T x =40 minitues and E x = 333
921
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