^{1}

^{*}

^{2}

^{1}

^{1}

Transportation of products from sources to destinations with minimal total cost plays an important role in logistics and supply chain management. In this article, a new and effective algorithm is introduced for finding an initial basic feasible solution of a balanced transportation problem. Number of numerical illustration is introduced and optimality of the result is also checked. Comparison of findings obtained by the new heuristic and the existing heuristics show that the method presented herein gives a better result.

Transportation Problem (TP) is a special type of Linear Programming Problem (LPP). Transportation plan is mainly used to minimize transportation cost. Hitchcock [

Reinfeld and Vogel (1958) [

Kirca and Satir (1990) [

Mathirajan and Meenakshi (2004) [

Here in this article, the pointer cost has been calculated only one time by taking the difference between the highest and lowest cell cost for each row and column of the TOCM and make maximum possible allocation to the lowest cost cell corresponding to the highest pointer cost. It is to be mentioned that other allocation is obtained by Modified Extremum Difference Method without calculating the pointer cost time and again.

The Transportation Problem can be specified as an allocation problem in which there are m sources (suppliers) and n destinations (customers). Each of the m sources can allocate to any of the n destinations at a per unit carrying cost c i j (unit transportation cost from source i to destination j). Each sources has a supply of a i units, 1 ≤ i ≤ m and each destination has a demand of b j units, 1 ≤ j ≤ n . The objective is to determine which routes are to be opened and the size of the shipment on those routes, so that the total transportation cost of meeting demand, given the supply constraints, is minimized.

A transportation problem is a complete specification of how many units of the product should be transported from each source to each destination. So, the decision variables are:

x i j = The amount of the shipment from source i to destination j , where i = 1 , 2 , ⋯ , m and j = 1 , 2 , ⋯ , n .

Therefore we get, Minimize: Z = ∑ i = 1 m ∑ j = 1 n c i j x i j .

The proposed method named as “TOCM-MEDM Approach” comprises with the following steps:

A company manufacture television and it has four factories F 1 , F 2 , F 3 and F 4 whose daily production capacities are 30, 25, 20 and 15 pieces of television respectively. The company supplies televisions to its four showrooms located at D 1 , D 2 , D 3 and D 4 whose daily demands are 30, 30, 20 and 10 pieces of television respectively. The transportation costs per piece of televisions are given in the transportation

Iteration 1: In the first row 5 is the minimum element, so we subtract 5 from each element of the first row. Similarly, we subtract 3, 3 and 2 from each element of the 2nd, 3rd and 4th row respectively and place all the differences on the right-top of the corresponding elements in

Iteration 2: In the similar way, we subtract 2, 3, 7 and 3 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

Iteration 3: The right-top and left-bottom entry of each element of the transportation table obtained in Iteration 1 and Iteration 2 is added and forms the TOCM as in

Iteration 4: We determine the pointer cost for each row of the TOCM (

Showrooms | Production Capacity | |||||
---|---|---|---|---|---|---|

D_{1} | D_{2} | D_{3} | D_{4} | |||

Factories | F_{1} | 7 | 5 | 9 | 11 | 30 |

F_{2} | 4 | 3 | 8 | 6 | 25 | |

F_{3} | 3 | 8 | 10 | 5 | 20 | |

F_{4} | 2 | 6 | 7 | 3 | 15 | |

Demand | 30 | 30 | 20 | 10 |

Showrooms | Production Capacity | |||||
---|---|---|---|---|---|---|

D_{1} | D_{2} | D_{3} | D_{4} | |||

Factories | F_{1} | _{5}7^{2 } | _{2}5^{0 } | _{2}9^{4 } | _{8}11^{6 } | 30 |

F_{2} | _{2}4^{1 } | _{0}3^{0 } | _{1}8^{5 } | _{3}6^{3 } | 25 | |

F_{3} | _{1}3^{0 } | _{5}8^{5 } | _{3}10^{7 } | _{2}5^{2 } | 20 | |

F_{4} | _{0}2^{0 } | _{3}6^{4 } | _{0}7^{5 } | _{0}3^{1 } | 15 | |

Demand | 30 | 30 | 20 | 10 |

Showrooms | Production Capacity | |||||
---|---|---|---|---|---|---|

D_{1} | D_{2} | D_{3} | D_{4} | |||

Factories | F_{1} | 7 | 2 | 6 | 14 | 30 |

F_{2} | 3 | 0 | 6 | 6 | 25 | |

F_{3} | 1 | 10 | 10 | 4 | 20 | |

F_{4} | 0 | 7 | 5 | 1 | 15 | |

Demand | 30 | 30 | 20 | 10 |

Do the same for each column and place them in the bottom of the cost matrix below the corresponding columns [e.g. (7 − 0) = 7, (10 − 0) = 10, (10 − 5) = 5 and (14 − 1) = 13].

Iteration 5: In

Iteration 6: Now complete the allocation along 4th row by making the allocation in the smallest cost (0) in the cell (4, 1). Here 4th row is exhausted for the allocation min (30, 5) = 5 units at the cell (4, 1).

Iteration 7: In this stage, the allocation along 1st column will be in the smallest cost (1) which is at the cell (3, 1). By making the allocation of min (25, 20) = 20 units in the cell (3, 1) the 3rd row is crossed out but 1st column is yet to exhaust. Then we will go for next smallest cost along the 1st column corresponding to this is 3 in the cell (2, 1). Therefore 1st column is exhausted for the allocation min (25, 5) = 5 units at the cell (2, 1).

Iteration 8: In this case, we allocate min (30, 20) = 20 units along 2nd row in the smallest cost corresponding to this is 0 in the cell (2, 2). Since the production capacity is satisfied by the cells (2, 1) and (2, 2) then we crossed out 2nd row.

Iteration 9: Now complete the allocation along 2nd column by making the allocation in the smallest cost (2) in the cell (1, 2). Here 2nd column is exhausted for the allocation min (10, 30) = 10 units at the cell (1, 2).

Iteration 10: Since only the 1st row is remaining with one unallocated cell (1, 3), so we allocate rest 20 units to the cell (1, 3). Thereby in

Iteration 11: Now according to algorithm of Step 8, all these allocations are transferred to the original Transportation

Showrooms | Production Capacity | Row Pointer | |||||
---|---|---|---|---|---|---|---|

D_{1} | D_{2} | D_{3} | D_{4} | ||||

Factories | F_{1} | 7 | 2 | 6 | 14 | 30 | (12) |

F_{2} | 3 | 0 | 6 | 6 | 25 | (6) | |

F_{3} | 1 | 10 | 10 | 4 | 20 | (9) | |

F_{4} | 0 | 7 | 5 | 1 | 15 | (7) | |

Demand | 30 | 30 | 20 | 10 | |||

Column Pointer | (7) | (10) | (5) | (13) |

Showrooms | Production Capacity | Row Pointer | |||||||||
---|---|---|---|---|---|---|---|---|---|---|---|

D_{1} | D_{2} | D_{3} | D_{4} | ||||||||

Factories | F_{1} | 10 | 20 | 30/20 | (12) | ||||||

7 | 2 | 6 | 14 | ||||||||

F_{2} | 5 | 20 | 25/20 | (6) | |||||||

3 | 0 | 6 | 6 | ||||||||

F_{3} | 20 | 20 | (9) | ||||||||

1 | 10 | 10 | 4 | ||||||||

F_{4} | 5 | 10 | 15/5 | (7) | |||||||

0 | 7 | 5 | 1 | ||||||||

Demand | 30/25/5 | 30/10 | 20 | 10 | |||||||

Column Pointer | (7) | (10) | (5) | (13) |

Showrooms | Production Capacity | |||||||||
---|---|---|---|---|---|---|---|---|---|---|

D_{1} | D_{2} | D_{3} | D_{4} | |||||||

Factories | F_{1} | 10 | 20 | 30 | ||||||

7 | 5 | 9 | 11 | |||||||

F_{2} | 5 | 20 | 25 | |||||||

4 | 3 | 8 | 6 | |||||||

F_{3} | 20 | 20 | ||||||||

3 | 8 | 10 | 5 | |||||||

F_{4} | 5 | 10 | 15 | |||||||

2 | 6 | 7 | 3 | |||||||

Demand | 30 | 30 | 20 | 10 | 90 |

Iteration 12: Finally according to Step 9, for a flow of 90 units, total transportation cost is

( 10 × 5 ) + ( 20 × 9 ) + ( 5 × 4 ) + ( 20 × 3 ) + ( 20 × 3 ) + ( 5 × 2 ) + ( 10 × 3 ) = 410

Consider the following transportation problem (Transportation Cost

After formulation and allocation on TOCM, we allocate the quantity on original cost table (i.e.

Hence for the flow of 135 units, the total transportation cost is

( 45 × 1 ) + ( 15 × 2 ) + ( 18 × 2 ) + ( 17 × 2 ) + ( 22 × 3 ) + ( 5 × 2 ) + ( 13 × 4 ) = 273

Consider that four products are produce in three machines and their per unit production costs are given in the following cost

After formulation and allocation on TOCM, we allocate the quantity on original cost table (i.e.

Hence for the flow of 1200 units, the total production cost is

( 300 × 1 ) + ( 250 × 2 ) + ( 150 × 5 ) + ( 50 × 3 ) + ( 250 × 3 ) + ( 200 × 2 ) = 2850

Showrooms | Supply | ||||||
---|---|---|---|---|---|---|---|

D_{1} | D_{2} | D_{3} | D_{4} | D_{5} | |||

Factories | F_{1} | 4 | 1 | 2 | 4 | 4 | 60 |

F_{2} | 2 | 3 | 2 | 2 | 2 | 35 | |

F_{3} | 3 | 5 | 2 | 4 | 4 | 40 | |

Demand | 22 | 45 | 20 | 18 | 30 |

Showrooms | Supply | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|

D_{1} | D_{2} | D_{3} | D_{4} | D_{5} | ||||||||

Factories | F_{1} | 45 | 15 | 60 | ||||||||

4 | 1 | 2 | 4 | 4 | ||||||||

F_{2} | 18 | 17 | 35 | |||||||||

2 | 3 | 2 | 2 | 2 | ||||||||

F_{3} | 22 | 5 | 13 | 40 | ||||||||

3 | 5 | 2 | 4 | 4 | ||||||||

Demand | 22 | 45 | 20 | 18 | 30 |

Products | ||||||
---|---|---|---|---|---|---|

P_{1} | P_{2} | P_{3} | P_{4} | Capacity | ||

Machines | M_{1} | 3 | 1 | 7 | 4 | 300 |

M_{2} | 2 | 6 | 5 | 9 | 400 | |

M_{3} | 8 | 3 | 3 | 2 | 500 | |

250 | 350 | 400 | 200 |

Products | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|

P_{1} | P_{2} | P_{3} | P_{4} | Capacity | ||||||

Machines | M_{1} | 300 | 300 | |||||||

3 | 1 | 7 | 4 | |||||||

M_{2} | 250 | 150 | 400 | |||||||

2 | 6 | 5 | 9 | |||||||

M_{3} | 50 | 250 | 200 | 500 | ||||||

8 | 3 | 3 | 2 | |||||||

Demand | 250 | 350 | 400 | 200 |

The following

The objective of the transportation problem is to determine the shipping schedule or supply route that minimizes the total shipping cost while satisfying the demand and supply limit. The proposed method can be one of the solution procedures to select this route.

Here in this article, we have used mainly two methods, the Extremum Difference Method (EDM) and Modified Extremum Difference Method (MEDM). Basically the proposed algorithm is set up by applying MEDM on TOCM. It is observed that the proposed method performs either same or better than EDM and MEDM. The uniqueness of this method is the computational procedure which is easier (less iteration) than the existing VAM, EDM, HCDM as the

Method | Total Transportation/Production Cost | ||
---|---|---|---|

Ex 1 | Ex 2 | Ex 3 | |

North West Corner Method | 540 | 363 | 4400 |

Row Minimum Method | 470 | 278 | 2850 |

Column Minimum Method | 435 | 295 | 3600 |

Least Cost Method | 435 | 278 | 2900 |

Vogel’s Approximation Method | 415 | 273 | 2850 |

Extremum Difference Method | 415 | 273 | 2850 |

Modified Extremum Difference Method | 410 | 273 | 2850 |

Highest Cost Difference Method | 415 | 273 | 2900 |

Average Cost Method | 455 | 273 | 2900 |

TOCM-MMM Approach | 435 | 278 | 2900 |

TOCM-VAM Approach | 430 | 273 | 2850 |

TOCM-EDM Approach | 415 | 278 | 2850 |

TOCM-HCDM Approach | 415 | 273 | 2900 |

TOCM-SUM Approach | 455 | 275 | 2850 |

TOCM-MEDM Approach (Proposed) | 410 | 273 | 2850 |

Optimum Solution | 410 | 273 | 2850 |

penalty is required to find only for the first allocation. But in case of VAM, EDM and HCDM the penalty is needed to bring out for each and every allocation. By using the proposed method we conclude that we obtain an efficient and modified algorithm for finding an initial basic feasible solution which is optimal or close to optimal as compared to existing methods. It will help to calculate cost related to transportation which plays a vital role to minimize cost or maximize profit.

The authors declare no conflicts of interest regarding the publication of this paper.

Hossain, Md.M., Ahmed, M.M., Islam, Md.A. and Ukil, S.I. (2020) An Effective Approach to Determine an Initial Basic Feasible Solution: A TOCM-MEDM Approach. Open Journal of Optimization, 9, 27-37. https://doi.org/10.4236/ojop.2020.92003