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Method for Solving Unbalanced Transportation
Problems Using Standard Deviations 1T. Geetha and
2N. Anandhi
1Kunthavai Naacchiyar Government Arts College for Women,
Thanjavurt,
Thanjavur Dist.,
Tamilnadu, India. 2Anjalai Ammal- Mahalingam Engineering College,
Kovilvenni,
Thiruvarur Dist.,
Tamilnadu, India.
Abstract A transportation problems deals with two different problem (i) Balanced
TP (ii) Unbalanced TP. Here a new model for solving unbalanced TP is
proposed. This method is a systematic procedure both easy to understand
and to apply. In the present work a new method named Standard
Deviation Method (SDM) is proposed for finding an Initial Basic Feasible
Solution for unbalanced transportation problems provide comparatively a
best IBFS than the Vogel’s Approximation Method (VAM). This article
gives more options and very helpful to the decision makers who are
handling the unbalanced availability and demand. Finally the proposed
method presented in this paper claims its wide application in solving
transportation problems
Key Words:Transportation problem, unbalanced transportation problem
transportation cost, initial basic feasible solution, VAM and SDM.
Mathematics Subject Classification:90XX, 90CXX, 90BXX, 65KXX.
International Journal of Pure and Applied MathematicsVolume 119 No. 16 2018, 4971-4989ISSN: 1314-3395 (on-line version)url: http://www.acadpubl.eu/hub/Special Issue http://www.acadpubl.eu/hub/
4971
1. Introduction
The unbalanced transportation problem is a particular class of transportation
problem, which is associated with day-to-day activities in our real life and
mainly deals with logistics. It helps in solving problems on distribution and
transportation of resources from one place to another. The goods are transported
from „m‟ source to n designation and there capacities are 𝑎1, 𝑎2 ,…………𝑎𝑚
and 𝑏1, 𝑏2 ,………… 𝑏𝑚 respectively. In addition there is a penalty 𝑐𝑖𝑗
associated with transporting unit of product from source i to destination j. This
penalty may be cost or delivery time or safety of delivery etc. A variable
𝑥𝑖𝑗 represents the unknown quantity to be shipped from source i to destination j..
Insermann [2] introduced algorithm for solving this problem which provides
effective solutions. Lai and Hwang (3), Bellman and Zadeh [4] proposed the
concept of decision making in stranded deviation used by data analysis.. The
transportation solution problem can be found with a good success in the
improving the service quality of the public transport systems [1].Also it is found
in Zuhaimy Ismail at el. article [5]. As well as, the transportation solution
problem is used in the electronic commerce where the area of globalization the
degree of competition in the market article [6], and it can be used in a scientific
fields such as the simulated data for biochemical and chemical Oxygen
demands transport [7], and many other fields. A unbalanced condition (i.e.
Total demand is not equal to total supply) is assumed. Then finding an optimal
schedule of shipment of the commodity with the satisfaction of demands at each
destination is the main goal of the problem. The objective is to satisfy the
demand at destinations from the supply constraints at the minimum
transportation cost possible. To achieve this objective, we must know the
quantity of available supplies and the quantities demanded. In addition, we must
also know the location, to find the cost of transporting one unit of commodity
from the place of origin to the destination. The model is useful for making
strategic decisions involved in selecting optimum transportation routes so as to
allocate the production of various plants to several warehouses or distribution
centers. Basically, the solution procedure for the unbalanced transportation
problem consists of the following procedure
procedure 1: Mathematical formulation of the unbalanced transportation
problem.
procedure 2: Existence of feasible solution for unbalanced TP.
procedure 3: Transportation Table for unbalanced TP and modified balanced
TP.
procedure 4: Initial Basic Feasible Solution using Standard Deviation Method
(SDM).
procedure 5: Numerical Examples with Illustration
In this paper, procedure 4 has been focused in order to obtain IBFS for the
unbalanced transportation problems.
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The SDM provides comparatively a best initial basic feasible solution than the
results obtained by the traditional algorithm VAM. Efficiency of SDM has also
been tested by solving several number of cost minimizing unbalanced
transportation problems and it is found that the SDM method yields
comparatively a best result. In this paper, the new method SDM was developed
to find the IBFS of unbalanced TP. It is found that SDM is better method to get
good results compare to VAM. SDM provides an optimal solution directly, (or)
a better optimal solution with minimum number of iterations for unbalanced
transportation problem. But other traditional methods VAM gives IBFS and
optimum solution with more number of iterations. This problem has been
studied since long and is well known by Abdur Rashid et al. [8], Aminur
Rahman Khan et al. [9]-[12], Hamdy A. T. [13], Kasana & Kumar [14], Kirca
and Satir [15], M. Sharif Uddin et al. [16], Md. Amirul Islam et al. [17] [18],
Md. Ashraful Babu et al. [19]-[21], Md. Main Uddin et al. [22], Mollah
Mesbahuddin Ahmed et al. [23]-[25], Pandian & Natarajan[26], Reinfeld &
Vogel [27], Sayedul Anam et al. [28], Shenoy et al. [29] and Utpal Kanti Das et
al. [30] [31].Again, some of the well reputed methods for finding an initial basic
feasible solution of transportation problems developed and discussed by them
are North West Corner Method (NWCM) [13], Least Cost Method (LCM) [13],
Vogel’s Approximation method (VAM) [13] [26]. A comparative study is also
carried out by solving transportation problems which shows that the proposed
method gives best result in comparison to VAM
2. Mathematical Formulation of the Unbalanced Transportation Problem
Let there be m sources of supply, 𝑆1, 𝑆2, … . . 𝑆𝑚 having 𝑎𝑖 ( 𝑖 = 1,2, …𝑚) units
of supply (or capacity) respectively, to be transported among n destinations,
𝐷1, 𝐷2 , … . . 𝐷𝑛 with (𝑗 = 1,2, …𝑛) units of demand ( or requirement)
respectively. Let 𝑐𝑖𝑗 be the cost of shipping one unit of the commodity from
source 𝑖 to destination 𝑗 for each route. If 𝑥𝑖𝑗 represents number of units shipped
per route from source 𝑖 to destination 𝑗, the problem is to determine the
transportation schedule so as to minimize the total transportation cost while
satisfying the supply and demand conditions. Mathematically, the problem, in
general, may be stated as follows:
Minimize Total cost Z = 𝐶𝑖𝑗𝑥𝑖𝑗 (1)
n
j=1
m
i=1
Subject to the constrains
𝑥𝑖𝑗 = 𝑎𝑖 , 𝑖 = 1,2, ……𝑚 𝑠𝑢𝑝𝑝𝑙𝑦 𝑐𝑜𝑛𝑡𝑟𝑎𝑖𝑛𝑠 (2)
𝑛
𝑗=1
𝑥𝑖𝑗 = 𝑏𝑗 , 𝑗 = 1,2, ……𝑛 𝐷𝑒𝑚𝑎𝑛𝑑 𝑐𝑜𝑛𝑡𝑟𝑎𝑖𝑛𝑠 (3)
𝑚
𝑗=1
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And
𝑥𝑖𝑗 ≥ 0 for all 𝑖 and 𝑗 (4)
In a transportation problem the sum of all available quantities is not equal to the
sum of requirements that is 𝑎𝑖𝑚𝑖=1 ≠ 𝑏𝑗
𝑛𝑗=1 , then such problem is called an
unbalanced transportation problem.
To modify unbalanced transportation problem
An unbalanced TP may occur in two different forms (i) Excess of availability
(ii) shortage in availability. Now discuss these two cases by considering our
usual m-origin, n-destination TP with the condition that 𝑎𝑖𝑚𝑖=1 ≠ 𝑏𝑗
𝑛𝑗=1 .
Case (i) (Excess availability 𝒂𝒊 > 𝒃𝒋)
The general TP may be stated as follows
𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑧 =
𝑚
𝑖=1
𝑐𝑖𝑗𝑥𝑖𝑗
𝑛
𝑗=1
(5)
Subject to the constrains
𝑥𝑖𝑗
𝑛
𝑗=1
< 𝑎𝑖 , 𝑓𝑜𝑟𝑖 = 1,2, … . . 𝑚 (6)
𝑥𝑖𝑗 = 𝑏𝑗 , 𝑓𝑜𝑟 𝑗 = 1,2, ……𝑛
𝑚
𝑖=1
(7)
And 𝑥𝑖𝑗 ≥ 0 (8)
Where 𝑐𝑖𝑛+1 = 0 𝑓𝑜𝑟 𝑖 = 1,2, … . . 𝑚 𝑎𝑛𝑑 𝑎𝑖𝑚𝑖=1 = 𝑏𝑗
𝑛+1𝑗=1
The problem with possess a feasible solution if 𝑎𝑖 > 𝑏𝑗 .In the first
constrain, the introduction of slack variable 𝑥𝑖𝑛+1 𝑖 = 1,2, … . 𝑚 gives
𝑥𝑖𝑗 + 𝑥𝑖𝑛+1 = 𝑎𝑖
𝑛
𝑗=1
, 𝑓𝑜𝑟 𝑖 = 1,2, … . . 𝑚
𝑥𝑖𝑗
𝑛
𝑗=1
+ 𝑥𝑖𝑛+1 = 𝑎𝑖
𝑚
𝑖=1
𝑚
𝑖=1
𝑥𝑖𝑗
𝑚
𝑖=1
+ 𝑥𝑖 ,𝑛+1 = 𝑎𝑖
𝑚
𝑖=1
𝑚
𝑖=1
𝑛
𝑗=1
𝑥𝑖𝑛+1 = 𝑎𝑖
𝑚
𝑖=1
− 𝑏𝑗
𝑛
𝑗=1
𝑚
𝑖=1
= Excess of availability 𝑏𝑛+1 , 𝑠𝑎𝑦 𝑆𝑖𝑛𝑐𝑒 𝑥𝑖𝑗 = 𝑏𝑗
𝑚
𝑖=1
If this excess availability is denoted by 𝑏𝑛+1 the modified general TP can be
reformulated as
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𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑧 =
𝑚
𝑖=1
𝑐𝑖𝑗𝑥𝑖𝑗
𝑛+1
𝑗 =1
(9)
Subject to constrains
𝑥𝑖𝑗 + 𝑥𝑖𝑛+1 = 𝑎𝑖 , 𝑓𝑜𝑟 𝑖 = 1,2, …𝑚
𝑛
𝑗=1
(10)
𝑥𝑖𝑗
𝑚
𝑖=1
= 𝑏𝑗 , 𝑓𝑜𝑟 𝑗 = 1,2, ……𝑛 + 1 (11)
and
𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟 𝑖 = 1,2, ………𝑚 ; 𝑗 = 1,2, ……𝑛 (12)
Hence the above TP is balanced 𝑎𝑖𝑚𝑖=1 = 𝑏𝑗
𝑛+1𝑗=1
Case(ii): (shortage in availability ie. 𝒂𝒊 < 𝒃𝒋)
In the case the general TP becomes
𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑧 = 𝑐𝑖𝑗𝑥𝑖𝑗𝑛𝑗=1
𝑚𝑖=1 (13)
Subject to the constrains
𝑥𝑖𝑗 = 𝑎𝑖 , 𝑓𝑜𝑟 𝑖 = 1,2, … . . 𝑚
𝑛
𝑗=1
(14)
𝑥𝑖𝑗 < 𝑏𝑗 , 𝑓𝑜𝑟 𝑗 = 1,2, ……𝑛
𝑚
𝑖=1
(15)
𝑥𝑖𝑗 ≥ 0 𝑓𝑜𝑟 𝑖 = 1,2, ………𝑚 ; 𝑗 = 1,2, ……𝑛 (16)
Now introducing the slack variable 𝑥𝑚+1𝑗 , 𝑓𝑜𝑟 𝑗 = 1,2, …… . 𝑛 in the second
constrained, We get
𝑥𝑖𝑗 + 𝑥𝑚+1𝑗 = 𝑏𝑗 , 𝑓𝑜𝑟 𝑗 = 1,2, ……… . 𝑛
𝑚
𝑖=1
𝑥𝑖𝑗
𝑚
𝑖=1
+ 𝑥𝑚+1𝑗 = 𝑏𝑗
𝑛
𝑗=1
𝑛
𝑗=1
𝑥𝑖𝑗
𝑛
𝑗=1
+ 𝑥𝑚+1𝑗
𝑛
𝑗=1
= 𝑏𝑗
𝑛
𝑗=1
𝑚
𝑖=1
𝑥𝑚+1𝑗
𝑛
𝑗=1
= 𝑏𝑗
𝑛
𝑗=1
− 𝑎𝑖
𝑚
𝑖=1
= 𝑆𝑜𝑟𝑡𝑎𝑔𝑒 𝑖𝑛 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑎𝑚+1 , 𝑠𝑎𝑦
Thus the modified TP can be reformulated as
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𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑧 = 𝑐𝑖𝑗 𝑥𝑖𝑗
𝑛
𝑗=1
𝑚+1
𝑖=1
(17)
Subject to the constrains
𝑥𝑖𝑗 = 𝑎𝑖 , 𝑓𝑜𝑟 𝑖 = 1,2, ……… . 𝑚 + 1
𝑛
𝑗=1
(18)
𝑥𝑖𝑗 + 𝑥𝑚+1𝑗
𝑚
𝑖=1
= 𝑏𝑗 , 𝑓𝑜𝑟𝑗 = 1,2, ……… . 𝑛 (19)
𝑥𝑖𝑗 ≥ 0 , 𝑓𝑜𝑟 𝑖 = 1,2, ………… . . 𝑚 + 1 , 𝑗 = 1,2, …………… . . 𝑛 (20)
Where 𝑐𝑚+1𝑗 = 0 for 𝑗 = 1,2, …… . 𝑛 and 𝑎𝑚+1 = 𝑏𝑗 − 𝑎𝑖
Hence, the above TP is balanced 𝑎𝑖 = 𝑏𝑗𝑛𝑗=1
𝑚+1𝑖−1
3. Existence of Feasible Solution for Unbalanced TP
A necessary and sufficient condition for the existence of a feasible solution to
the unbalanced transportation problem is 𝑎𝑖𝑚𝑖=1 = 𝑏𝑗
𝑛+1𝑗=1 (or)
or 𝑎𝑖 = 𝑏𝑗𝑛𝑗=1
𝑚+1𝑖−1 .That is ,the total supply must equal to demand.
4. Un Balanced Transportation Table (TT)
Unbalanced transportation Table: (Excess availability ie 𝑎𝑖 > 𝑏𝑗 )
Figure 1: Transpotation Table of Unbalanced Transportation Problem
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Modified Balanced transportation Table: 𝑎𝑖𝑚𝑖=1 = 𝑏𝑗
𝑛+1𝑗=1
Figure 2: Transpotation Table of Modified Balanced Transportation
Problem
The 𝑚 × 𝑛 squares are called cells. The transportation cost 𝑐𝑖𝑗 from the 𝑖𝑡
source to the 𝑗𝑡 destination is displayed in the lower right side of the 𝑖, 𝑗 𝑡
cell. Any feasible solution is shown in the table by entering the values of 𝑥𝑖𝑗 in
the upper left side of the 𝑖, 𝑗 𝑡 cell.
The various 𝒂𝒊’s and 𝒃𝒋’s are called rim requirements. The feasibility of a
solution can be verified by summing the values of 𝑥𝑖𝑗 along the rows and down
the columns(ie, 𝑥𝑖𝑗 satisfying the rim conditions).
Any feasible solution for transportation problem must have exactly (𝑚 + 𝑛 −1) non –negative basic variables (or) allocations).
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Unbalanced transportation Table: (shortage in availability ie. 𝑎𝑖 < 𝑏𝑗 )
Figure 3: Transpotation Table of Unbalanced Transportation Problem
Modified Balanced transportation Table: 𝑎𝑖𝑚+1𝑖=1 = 𝑏𝑗
𝑛𝑗=1
Figure 4: Transpotation Table of Modified Balanced Transportation Problem
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5. Proposed Standard Deviation Methods to Find an Initial Basic Feasible Solution
The stepwise procedure of SDM is carried out as follows.
Step 1: Check the given TP is unbalanced either excess availability or shortage
in availability.
Step 2: For each row and column reaming under consideration calculate its
Standard Deviations(SD). In that rows SD is denoted by 𝜎𝑠𝑖 =
1
𝑛 𝑐𝑖𝑗 − 𝑐𝑖𝑗
2
for 𝑖 = 1,2, … . . 𝑚 , where n is number of destinations and columns SD is
denoted by 𝜎𝐷𝑗 =
1
𝑚 𝑐𝑖𝑗 − 𝑐𝑖𝑗
2 for 𝑗 = 1,2, … . 𝑛, where m is number of
origins. Indentify the row (or) column with the largest SD and allocate
maximum possible to the variable with least unit TC in the selected row (or)
column. So as to exhaust either the supply at particular origin or satisfy demand
at a destination.
Step 3: Rules for ties 1. If a ties occurs in the value of SD, select that row(column) which has minimum
transportation cost.
2. If there is a ties in the minimum cost also, select that row(column) which will have
maximum possible quantity to the cell in that selected row (or) column so as to
exhaust either the supply at particular origin (or) satisfy demand at a destination.
Step 4: Adjust supply and demand and cross out the satisfied row (or)
column. If a row and a column are satisfied simultaneously only one of the two
is crossed out, and remaining row (column) is assigned zero supply (demand).
Any row (or) column with zero supply or demand should not be used in
computing future SD.
Step 5: Re-compute the row and column SD for the reduced TT, omitting
rows (column) crossed out in the preceding step. Repeat steps 2 to 4 for the
reduced table until exact one row (or) column with zero supply (demand)
remains uncrossed out step.
Step 6:
(i) If one row (column) with non-zero supply (demand) remains uncrossed out,
determine the basic variable in the row(column) by LCM step.
(ii) If all the variable in uncrossed out rows and columns have zero supply
(demand), determine the zero basic variables by LCM step.
Step 7: Given TP is unbalanced TP, there are two cases (i) 𝑎𝑖 > 𝑏𝑗
(ii) 𝑎𝑖 < 𝑏𝑗
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Case(i) Whenever 𝑎𝑖 > 𝑏𝑗 , introduce a dummy destination 𝑛 + 1 𝑡
column in the transportation table. The unit transportation cost to this dummy destination are the value of 𝑐𝑖𝑛+1 = 0 for 𝑖 = 1,2, … . 𝑚 the demand of this dummy destination is 𝑏𝑛+1 = 𝑎𝑖 − 𝑏𝑗 . In this case
subject to the constrains is 𝑥𝑖𝑗 ≤ 𝑎𝑖𝑛+1𝑗=1 , 𝑖 = 1,2, ……𝑚. The unit of
amount shifting from 𝑖𝑡 origin to 𝑛 + 1 𝑡 destination is 𝑥𝑖𝑛+1 = 𝑎𝑖 − 𝑥𝑖𝑗 , for 𝑖 = 1,2, …… . . 𝑚 𝑛
𝑗=1 .
Case (ii) Whenever 𝑎𝑖 < 𝑏𝑗 , Introduce a dummy origin 𝑚 + 1 𝑡 row
in the transportation table. The unit transportation cost to this dummy origin are 𝑐𝑚+1𝑗 and the value of 𝑐𝑚+1𝑗 = 0 for 𝑗 = 1,2, … . . 𝑛. The supply
of this dummy origin is 𝑎𝑚+1 = 𝑏𝑗 − 𝑎𝑖 . The unit of amount shafting
from 𝑚 + 1 𝑡 origin to 𝑗𝑡 destination is 𝑥𝑚+1𝑗 = 𝑏𝑗 − 𝑥𝑖𝑗𝑚𝑖=1 , for 𝑗 =
1,2, … . 𝑛.
Step 8: The balanced transportation table having at least one basic cell in each row(column) including dummy row(column) and also the basic cell do not form loop. Hence the initial basic feasible solution is obtained and the minimum transportation cost equal to 𝑐𝑖,𝑗𝑥𝑖 ,𝑗
𝑛+1𝑗=1
𝑚𝑖=1 𝑜𝑟 𝑐𝑖 ,𝑗𝑥𝑖 ,𝑗 𝑜𝑟 𝑛
𝑗=1𝑚+1𝑖=1 = 𝑐𝑖 ,𝑗𝑥𝑖 ,𝑗 .𝑛
𝑗=1𝑚𝑖=1
6. Numerical Examples with Illustration
Example-6.1
Consider the following transportation problem (Transportation table 6.1.1)
involving three sources and four destinations.
Transportation Table 6.1.1
Solution of Example 1
From the Transportation Table 6.1.1, it is seen that total supply and total
demand are not equal. Hence the given transportation problem convert into
balanced one.
To find IBFS by applying VAM allocations are obtained as follows.
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Transportation Table 6.1.2
Hence, the total transportation cost(IBFS) is 1010.
To find IBFS by applying SDM allocations are obtained as follows.
Transportation Table 6.1.3
Destination Source 𝐷1 𝐷2 𝐷3 𝐷4 Supply
𝑆1 35 35 9 3 70
6 1 𝑆2 5
5 50
8 55 11 2
𝑆3 25 12 4
45 70
10 7 𝑆3 20
0 0 0 20 0
Demand 85 35 50 45 215
Hence, the total transportation cost(IBFS) is 965.
Example-6.2
Consider the following transportation problem (Transportation table 6.2.1)
involving four sources and three destinations.
Transportation Table 6.2.1
Destination
Source 𝑫𝟏 𝑫𝟐 𝑫𝟑 Supply
𝑺𝟏 5 6 9 100
𝑺𝟐 3 5 10 75
𝑺𝟑 6 7 6 50
𝑺𝟒 6 4 10 75
Demand 70 80 120
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Solution of Example 2
From the Transportation Table 6.2.1, it is seen that total supply and total
demand are not equal. Hence the given transportation problem convert into
balanced one.
To find IBFS by applying VAM allocations are obtained as follows.
Transportation Table 6.2.2
Destination
Source 𝐷1 𝐷2 𝐷3 𝐷4 Supply
𝑆1 100 0 100 5 6 9
𝑆2 70 5 10 0 75 3 5
𝑆3 7
20 30 50 6 6 0
𝑆4 75 10 0 75 6 4
Demand 70 80 120 30 300
Hence, the total transportation cost(IBFS) is 1,555
To find optimum solution by applying SDM allocations are obtained as follows.
Transportation Table 6.2.3
Destination
Source 𝐷1 𝐷2 𝐷3 𝐷4 Supply
𝑆1 5 6
70 30 100 9 0
𝑆2 70 5 10 0 75 3 5
𝑆3 6 7
50 0 50 6
𝑆4 6
75 10 0 75 4
Demand 70 80 120 30
Hence, the total transportation cost(IBFS) is 1465
Example-6.3
Consider the following transportation problem (Transportation table 6.3.1)
involving three sources and four destinations.
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Transportation Table 6.3.1
Destination Source 𝑫𝟏 𝑫𝟐 𝑫𝟑 𝑫𝟒 Supply
𝑺𝟏 10 15 12 12 200 𝑺𝟐 8 10 11 9 150 𝑺𝟑 11 12 13 10 120
Demand 140 120 80 220 Solution of Example 3
From the Transportation Table 6.3.1, it is seen that total supply and total
demand are not equal. Hence the given transportation problem convert into
balanced one.
To find IBFS by applying VAM allocations are obtained as follows.
Transportation Table 6.3.2
Destination Source 𝐷1 𝐷2 𝐷3 𝐷4 Supply
𝑆1 10
12
200 200 15 12
𝑆2 140 10 11 9 150 8 10
𝑆3 11
100 20 120 12 13 10
𝑆4 0
10 80 0 90 0 0
Demand 140 120 80 220 560
Hence, the total transportation cost(IBFS) is 5020
To find optimum solution by applying SDM allocations are obtained as follows.
Transportation Table 6.3.3
Destination Source 𝐷1 𝐷2 𝐷3 𝐷4 Supply
𝑆1 110 15 12
90 200
10 12 𝑆2 30 120
11 9 150
8 10 𝑆3
11 12 13 120
120 10
𝑆4 0
80 10 90
0 0 0 Demand 140 120 80 220 560
Hence, the total transportation cost(IBFS) is4820
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Example-6.4
Consider the following transportation problem (Transportation table 6.4.1)
involving three sources and four destinations.
Transportation Table 6.4.1
Destination Source 𝑫𝟏 𝑫𝟐 𝑫𝟑 𝑫𝟒 Supply
𝑺𝟏 7 8 11 10 30 𝑺𝟐 10 12 5 4 45 𝑺𝟑 6 11 10 9 35
Demand 20 28 19 33 Solution of Example 4
From the Transportation Table 6.4.1, it is seen that total supply and total
demand are not equal. Hence the given transportation problem convert into
balanced one.
To find IBFS by applying VAM allocations are obtained as follows.
Transportation Table 6.4.2
Destination Source 𝐷1 𝐷2 𝐷3 𝐷4 𝐷5 Supply
𝑆1 7
20 11
10
10 30
8 0 𝑆2
10 12 12 33
0 45 5 4
𝑆3 20 8 7 9
0 35 6 11 10
Demand 20 28 19 33 10 110
Hence, the total transportation cost(IBFS) is 620
To find optimum solution by applying SDM allocations are obtained as follows.
Transportation table 6.4.3
Destination Source 𝐷1 𝐷2 𝐷3 𝐷4 𝐷5 Supply
𝑆1 7
28 11 10
2 30
8 0 𝑆2
10 12 12 33
0 45
5 4 𝑆3 20
11 7
9 8
35 6 10 0 Demand 20 28 19 33 10 110
Hence, the total transportation cost(IBFS) is 606
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7. Results and Discussion
After obtaining an optimum solution by the proposed “standard Deviation
method (SDM)”, the obtained result is compared with the results obtained by
vogel’s Approximation methods is shown in Table 7.1 Example Name of the methods Allocation Optimal Solution
6.1 VAM 𝑥11 = 65;𝑥12 = 5; 𝑥22 = 30 𝑥23 = 25; 𝑥33 = 25; 𝑥34 = 45 𝑥41 = 20
1010
SDM 𝑥11 = 35;𝑥12 = 3 5; 𝑥21 = 5 𝑥23 = 50; 𝑥31 = 25; 𝑥34 = 45 𝑥41 = 20
965
6.2 VAM 𝑥13 = 100;𝑥21 = 70; 𝑥22 = 5 𝑥33 = 20; 𝑥34 = 30; 𝑥42 = 75
1555
SDM 𝑥13 = 70;𝑥14 = 30; 𝑥21 = 70 𝑥22 = 5; 𝑥33 = 50; 𝑥42 = 75
1465
6.3 VAM 𝑥14 = 200;𝑥21 = 140; 𝑥22 = 10 𝑥32 = 100; 𝑥34 = 20; 𝑥42 = 10 𝑥43 = 80
5020
SDM 𝑥11 = 110;𝑥14 = 90; 𝑥21 = 30 𝑥22 = 120; 𝑥34 = 120; 𝑥44 = 10 𝑥43 = 80
4820
6.4 VAM 𝑥12 = 20;𝑥15 = 10; 𝑥23 = 12 𝑥24 = 33; 𝑥31 = 20; 𝑥32 = 8 𝑥33 = 7
620
SDM 𝑥12 = 28;𝑥15 = 2; 𝑥23 = 12 𝑥24 = 33; 𝑥31 = 20; 𝑥33 = 7 ; 𝑥35
= 8
606
As observed from Table 7.1, the SDM provides comparatively a best initial
basic feasible solution than the results obtained by the traditional algorithm
VAM. Efficiency of SDM has also been tested by solving several number of
cost minimizing unbalanced transportation problems and it is found that the
SDM method yields comparatively a best result.
8. Conclusion
In this paper , presented and discussed above gives an IBFS of the unbalanced
TP. From the example, illustrated here by this method it is found that SDM
takes less calculation and gives best optimum solution compare to VAM. SDM
provides an optimal solution directly, (or) a better optimal solution with
minimum number of iterations for unbalanced transportation problem. The
method developed here ensure a solution which is very good to find the IBFS of
unbalanced TP. Finally it can be claimed that the SDM may provides a
remarkable Initial Basic Feasible Solution by ensuring minimum transportation
cost. This will help to achieve the goal to those who want to maximize their
profit by minimizing the transportation cost.
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