A COMPETENT ALGORITHM TO FIND THE INITIAL · PDF fileAUTOMATICĂ şi CALCULATOARE ... FEASIBLE SOLUTION OF COST MINIMIZATION TRANSPORTATION PROBLEM ... of Cost Minimization Transportation

BULETINUL INSTITUTULUI POLITEHNIC DIN IAŞI Publicat de

Universitatea Tehnică „Gheorghe Asachi” din Iaşi Tomul LXI (LXV), Fasc. 2, 2015

SecŃia AUTOMATICĂ şi CALCULATOARE

A COMPETENT ALGORITHM TO FIND THE INITIAL BASIC

FEASIBLE SOLUTION OF COST MINIMIZATION

TRANSPORTATION PROBLEM

BY

AMINUR RAHMAN KHAN1,2*

, ADRIAN VILCU2,

MD. SHARIF UDDIN1 and FLORINA UNGUREANU

3

1Jahangirnagar University, Dhaka-1342, Bangladesh,

Department of Mathematics “Gheorghe Asachi” Technical University of Iasi, Romania,

2Department of Management Engineering 3Department of Automatic Control and Computer Engineering

Received: July 6, 2015 Accepted for publication: July 24, 2015

Abstract. In this paper, we propose a new algorithm along with MATLAB 7.7.0 code for determining the initial basic feasible solution of Cost Minimization Transportation Problem (CMTP). Comparative study is carried out between the proposed algorithm and the other existing algorithm by means of sample examples which shows that the proposed algorithm provides better result.

Key words: Cost Minimization Transportation Problem; MATLAB; distribution indicator; optimum solution.

2010 Mathematics Subject Classification: 90B50, 90C08.

*Corresponding author; e-mail: [email protected]

72 Aminur Rahman Khan et al.

1. Introduction

The branch of Linear Programming Problem (LPP) in which a single uniform commodity is shifted from several sources to different localities in such a way to minimize the total transportation cost while fulfilling all supply and demand limitations are CMTP. The basic CMTP was originally developed by Hitchcock (1941). Competent methods of solution derived from the simplex algorithm were developed, mainly by Dantzig (1963) and then by Charnes et al. (1953). The problem of CMTP has been studied since long and is well known by Abdur Rashid (2013), Aminur Rahman Khan (2011; 2012; 2015a; 2015b), Hamdy (2007), Juman & Hoque (2015), Kasana & Kumar (2005), Sharif Uddin et al. (2011; 2012), Md. Amirul Islam et al. (2012a, 2012b), Md. Ashraful Babu et al. (2013; 2014a; 2014b), Md. Main Uddin et al. (2013; 2015), Mollah Mesbahuddin Ahmed et al. (2014a; 2014b), Sayedul Anam et al. (2012) and Utpal Kanti Das et al. (2014a; 2014b). For determining the initial basic feasible solution of TP, Reinfeld and Vogel (1958) introduced Vogel’s Approximation Method (VAM) by defining penalty as the difference of lowest and next to lowest cost in each row and column of a transportation table and allocate to the minimum cost cell corresponding to the highest penalty; Kasana & Kumar (2005) presented Extremum Difference Method (EDM) by calculating the penalty as the difference between the highest and lowest unit transportation cost in each row and column and allocate as like as VAM; Aminur Rahman Khan (2011) proposed Highest Cost Difference Method (HCDM) by introducing pointer cost as the difference of highest and next to highest cost in each row and column of a transportation table and allocate to the minimum cost cell corresponding to the highest three pointer costs. Kirca and Satir (1990) first define the Total Opportunity Cost Matrix (TOCM) as the sum of Row Opportunity Cost Matrix (ROCM) and Column Opportunity Cost Matrix (COCM). Where, the ROCM is generated by subtracting the lowest cost of each row from the other cost elements in that row and, the COCM is generated by subtracting the lowest cost of each column from the other cost elements in that column. Kirca and Satir then essentially apply the Least Cost Method with some tie-breaking policies on the TOCM to determine the feasible solution of the transportation problem. Mathirajan & Meenakshi (2004) applied VAM on the TOCM, Md. Amirul Islam et al. applied EDM (2012a) and HCDM (2012b) on TOCM whereas Aminur Rahman Khan (2015a) calculate the pointer cost as the sum of all entries in the respective row or column of the TOCM to find the feasible solution of the transportation problem. Here, in this paper, we determine the distribution indicator for each cell of the TOCM by subtracting corresponding row and column highest element of every cell from the respective element. We then make maximum possible allocation to the cell having the smallest distribution indicator. The proposed

Bul. Inst. Polit. Iaşi, t. LXI (LXV), f. 2, 2015

73

method is also illustrated with several numerical examples. Comparative study shows that the proposed algorithm gives better result in comparison to the other existing heuristics available in the text. We also coded the presented algorithm by using MATLAB 7.7.0 and it is utilized via many randomly generated problems of different order in order to prove the exactness of the code. Based on the results we conclude that the coded algorithm for solving the transportation problem is accurate.

2. Formulation of Cost Minimization Transportation Problem

A general cost minimization transportation problem is represented by the network in the following Fig. 1.

Fig. 1 – Network representation of Cost Minimization Transportation Problem.

There are m sources and n destinations, each represented by a node. The arrows joining the sources and the localities represent the route through which the commodity is shifted. Suppose Si denotes the amount of supply at source i (i = 1, 2, … … , m), Dj represents the amount of demand at destination j (j=1, 2, … … , n), Cij denotes the unit transportation cost from sources i to destination j, Xij represents the amount transported from sources i to destination j. Then the LPP model of the balanced cost minimization transportation problem is

Min. ∑ ∑= ==

m

i

n

j ijij XCZ1 1

s/t, i

n

j ij SX =∑ =1; i=1,2, … … ,m

j

m

i ij DX =∑ =1; j=1,2, … … ,n

0≥ijX for all i, j. (1)

1111 : XC

mnmn XC :

1

2

m

1S

2S

mS

Source

⋮ Uni

ts o

f S

uppl

y

1

2

n

1D

2D

nD

Locality

Uni

ts o

f D

eman

d

⋮


3. Algorithm of Proposed Method

The proposed algorithm for determining the initial basic feasible solution consists of the following steps:

Step 1 : Subtract the smallest entry of every row from each of the

element of the subsequent row of the transportation table and place them on the right-top of the corresponding elements.

ikij CC

ijC−

, where ( )iniiik CCCC ,,,min 21 ⋯⋯= ,

mi ,......,2,1=

Step 2 : Apply the same operation on each of the column and place them on the left-bottom of the corresponding elements.

ijCC C

kjij −, where ( )mjjjkj CCCC ,,,min 21 ⋯⋯= ,

nj ,......,2,1=

Step 3 : Form the TOCM whose entries are the summation of right-top and left-bottom elements of Steps 1 and 2. ( ) ( )kjijikijij CCCCC −+−=

Step 4 : For each cell (i, j), calculate the distribution indicator, ∆ij = cij - ūi - ēj, where, ūi =largest unit time in the ith row and ēj = largest unit time in the jth column.

Step 5 : Make maximum possible allocation to the cell having the smallest value of ∆ij. If tie occurs in the distribution indicator, select any one of them arbitrarily.

Step 6 : No further deliberation is required for the row or column which is satisfied. If both the row and column are satisfied at a time, delete both of them assigning an extra zero supply (or demand) to any one cell of the satisfied row or column.

Step 7 : Calculate fresh distribution indicators for the remaining sub-matrix as in Step 4 and allocate following the procedure of Steps 5 and 6.

Step 8 : Continue the process until all rows and columns are satisfied.

Step 9 : Finally compute the total transportation cost as the sum of the product of original transportation cost and corresponding allocation obtained in step 5.


75

4. The Novelty of our Algorithm

Although we have used TOCM of Kirca and Satir in our proposed algorithm; we calculate the distribution indicator (in step 4) for each cell of the TOCM by subtracting corresponding row and column highest element of every cell from the respective element. Whereas Mathirajan and Meenakshi calculate the penalty as the difference of lowest and next to lowest entries of the TOCM; Md. Amirul Islam et al. calculate distribution indicator as the difference of highest and lowest entries of the TOCM and Aminur Rahman Khan et al. calculate pointer cost as the sum of all entries in the respective row or column of the TOCM.

5. Material and Methods

Table 1, 10 and 11 shows three sample cost minimizing transportation problem, selected at random to solve by using proposed algorithm and the existing algorithms. Example 1:

Table 1

Cost Matrix for the Numerical Example

Destination

1 2 3 4 Supply

1 3 6 8 4 20

2 6 1 2 5 28

Fac

tory

3 7 8 3 9 17

Demand 15 19 13 18

Step 1: 3 is the minimum element of the first row, so we subtract 3 from each element of the first row. Similarly, we subtract 1 and 3 from each element of the 2nd and 3rd row respectively and place all the differences on the right-top of the corresponding elements in Table 2.

Step 2: In a similar fashion, we subtract 3, 1, 2 and 4 from each element of the 1st, 2nd, 3rd and 4th column respectively and place the result on the left-bottom of the corresponding elements in Table 2.


Table 2

Formation of Total Opportunity Cost Matrix

Destination

1 2 3 4 Supply

1 030

563

685

041 20

2 365

010

021

154 28

Fac

tory

3 474

785

130

596 17

Demand 15 19 13 18

Step 3: We add the right-top and left-bottom entry of each element of

the transportation table obtained in Iteration 1 and Iteration 2 and formed the TOCM as in Table 3.

Table 3

Total Opportunity Cost Matrix (TOCM) Destination

1 2 3 4 Supply

1 0 8 11 1 20

2 8 0 1 5 28

Fac

tory

3 8 12 1 11 17

Demand 15 19 13 18

Step 4: We determine the distribution indicator for each cell of the TOCM by subtracting corresponding row and column highest element of every cell from the respective element. Here, c11 =0, highest entry in the first row is 11 and in the first column is 8, so distribution indicator, ∆11 =0-8-11=-19.

Do the same for each entry and place them in the right-top of every cell of the cost matrix.

Table 4

Determination of Distribution Indicator after Step 4 1 2 3 4 Supply

-19 -15 -11 -21 1

0 8 11 1 20

-8 -20 -18 -14 2

8 0 1 5 28

-12 -12 -22 -12 3

8 12 1 11 17

Demand 15 19 13 18


77

Step 5: Here, smallest value of ∆ij = -22 corresponding to the cell (3, 3). So we allocate 13 units (minimum of 17 and 13) to the cell (3, 3). We adjust the supply and demand requirements corresponding to the cell (3, 3) and since the demand for the cell (3, 3) is satisfied, we delete the third column and calculate the distribution indicator again for the resulting reduced transportation table.

Table 5


-16 -12 -18 1

0 8 11 1 20

-8 -20 -14 2

8 0 1 5 28

-12 -12 13 -12 3

8 12 1 11 4

Demand 15 19 0 18

Step 6: Here, smallest value of ∆ij = -20 corresponding to the cell (2, 2). So we allocate 19 units (minimum of 28 and 19) to the cell (2, 2). We adjust the supply and demand requirements corresponding to the cell (2, 2) and since the demand for the cell (2, 2) is satisfied, we delete the third column and calculate the distribution indicator again for the resulting reduced transportation table.

Table 6


-9 -12 -11 1

0 8 11 1 20

-8 19 -20 -14 2

8 0 1 5 9

-11 -12 13 -11 3

8 12 1 11 4

Demand 15 0 0 18

Step 7: Here, smallest value of ∆ij = -14 corresponding to the cell (2, 4). So we allocate 9 units (minimum of 9 and 18) to the cell (2, 4). We adjust the supply and demand requirements corresponding to the cell (2, 4) and since the supply for the cell (2, 4) is satisfied, we delete the second row and calculate the distribution indicator again for the resulting reduced transportation table.


Table 7


-9 -12 -11 1 0 8 11 1

20

-8 19 -20 9 -14 2 8 0 1 5

0

-11 -12 13 -11 3 8 12 1 11

4

Demand 15 0 0 9

Step 8: Here, smallest value of ∆ij = -11 corresponding to the cell (1, 4),

(3, 1) and (3, 4). So we allocate 9 units (minimum of 20 and 9) to the cell (1, 4).

We adjust the supply and demand requirements corresponding to the cell (1, 4)

and since the demand for the cell (1, 4) is satisfied, we delete the fourth column

and calculate the distribution indicator again for the resulting reduced

transportation table.

Table 8

Determination of Distribution Indicator after Step 8

1 2 3 4 Supply

-9 -12 9 -11 1

0 8 11 1 11

-8 19 -20 9 -14 2

8 0 1 5 0

-11 -12 13 -11 3

8 12 1 11 4

Demand 15 0 0 0

Step 9: Since only the first column is remaining with two unallocated

cell in this case, we allocate 11 units (minimum of 11 and 15) to the cell (1, 1)

and 4 units (minimum of 4 and 4) to the cell (3, 1).

We adjust the supply and demand requirements again and we see that

all supply and demand values are exhausted.


79

Table 9


11 -9 -12 9 -11 1 0 8 11 1

11

-8 19 -20 9 -14 2 8 0 1 5

0

4 -11 -12 13 -11 3 8 12 1 11

4

Demand 15 0 0 0

Step 10: Since all the rim conditions are satisfied and total number of allocation is 6. Therefore, the solution for the given problem is

1111 =x , 914=x , 1922 =x , 924 =x , 431 =x and 1333 =x .

for a flow of 65 units with the total transportation cost

200

313745911949311

=

×+×+×+×+×+×=z

Example 2:

Table 10


Destination

1 2 3 Supply

1 2 7 4 5

2 3 3 1 8

3 5 4 7 7

Fac

tory

4 1 6 2 14

Demand 7 9 18

Example 3: Table 11


Destination

1 2 3 4 Supply

1 6 1 9 3 70

2 11 5 2 8 55

Fac

tory

3 10 12 4 7 90

Demand 85 35 50 45


6. Result

Table 12 shows a comparison among the solutions obtained by our

Proposed Approach and the other existing methods and also with the optimal solution by means of the above three sample examples and it is seen that our proposed method gives better results.

Table 12

A Comparative Study of Different Solutions

Total Transportation Cost Solution obtained by Ex. 1 Ex. 2 Ex. 3

North West Corner Method 273 102 1265 Row Minimum Method 231 80 1165

Column Minimum Method 231 111 1220

Least Cost Method 231 83 1165

Vogel’s Approximation Method 204 80 1220

Extremum Difference Method 218 83 1165

Highest Cost Difference Method 231 83 1165

TOCM-MMM Approach 231 83 1165

TOCM-VAM Approach 204 76 1165

TOCM-EDM Approach 204 83 1165

TOCM-HCDM Approach 255 81 1165

TOCM-SUM Approach (Proposed) 200 76 1280

Average Cost Method 218 80 1280

Proposed Approach 200 76 1165

Optimum Solution 200 76 1160

We also solve randomly selected transportation problem of order 3×3, 3×4, 3×5, 4×3, 4×4, 4×5, 4×6, 6×6 and see that the MATLAB code presented by us gives identical result as the manual solution which proves the correctness of our code.

7. Conclusion

A new algorithm along with MATLAB 7.7.0 code for finding an initial basic feasible solution of cost minimization transportation problem is introduced. We also illustrate this algorithm numerically to test the efficiency of the proposed method. Comparative study among the solution obtained by proposed method and the other existing methods by means of sample examples show that our proposed method gives better result.


81

Acknowledgements. The first author acknowledges the financial support provided by the EU Erasmus Mundus Project-cLINK, Grant Agreement No: 212-2645/001-001-EM, Action 2.

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ALGORITM PENTRU DETERMINAREA SOLUłIEI

INIłIALE ÎN CAZUL MINIMIZĂRII COSTULUI PROBLEMEI DE TRANSPORT

(Rezumat)

Lucrarea propune un nou algoritm pentru determinarea unei soluŃii iniŃiale

pentru o problemă clasică de transport bazată pe cost. Se prezintă problema, este descrisă metoda, se aplică algoritmul pe un exemplu numeric şi se compară, pe trei instante, rezultatele algoritmului propus cu cele furnizate de alŃi algoritmi citaŃi în literatura de specialitate.

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A COMPETENT ALGORITHM TO FIND THE INITIAL · PDF fileAUTOMATICĂ şi CALCULATOARE ... FEASIBLE SOLUTION OF COST MINIMIZATION TRANSPORTATION PROBLEM ... of Cost Minimization Transportation