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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.

International Journal of Pure and Applied Mathematics Special Issue

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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)


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)


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

= 𝑆𝑕𝑜𝑟𝑡𝑎𝑔𝑒 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)


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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Hence, the total transportation cost(IBFS) is 1010.

To find IBFS by applying SDM allocations are obtained as follows.


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


involving four sources and three destinations.


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



balanced one.



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.


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




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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



balanced one.




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





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




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



balanced one.



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



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



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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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