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Minimization of transportation time is a great concern of the transportation problems like the cost minimizing transportation problems. In this writing, a transportation algorithm is developed and applied to obtain an Initial Basic Feasible Solution (IBFS) of transportation problems in minimizing transportation time. The developed method has also been illustrated numerically to test the efficiency of the method where it is observed that the proposed method yields a better result.

A special class of Linear Programming Problem is transportation problem, where the main objective is to minimize the cost of distributing a product from a number of sources to a number of destinations. This type of problem is known as cost minimizing transportation problem. The problem of minimizing transportation cost has been studied since long and is well known by [

Again, in the process of transporting of urgent material, such as weapons used in military operations, rescue equipments, equipments used for dealing with emergency, medical treatment things and the fresh foods with short storage period, where the speed of delivery is more important than the transportation cost. This type of transportation problem is known as time minimizing transportation problem where the objective is to minimize the transporting time rather than the cost of transportation.

An initial basic feasible solution for minimizing the time of transportation can be obtained by using any well known methods such as, North West Corner Method (NWCM), Least Cost Method (LCM), Vogel’s Approximation Method (VAM) and Extremum Difference Method (EDM). The time minimizing transportation problem had also been studied by a good number of researchers [

It is considered that the readers are well acquainted with transportation problem. Basing on this consideration, a new procedure of finding an initial basic feasible solution of the time minimizing transportation problems is been illustrated below:

Step-1: Identify the Largest Element (LE) of every row and every column in the transportation table and write down them in outer column of the supply and lower row of the demand.

Step-2: Determine the magnitude of the difference of row largest element and column largest element corresponding to each element.

Step-3: Place the value obtained in step-2 on the right bottom of the corresponding elements.

Step-4: Place the Row Distribution Indicators (RDI) and the Column Distribution Indicators (CDI) just after and below the supply limits and demand requirements respectively within first bracket, which are the difference of the largest and nearest-to-largest (Second largest) of right bottom entries. If there are two or more largest entries, the RDI or CDI is to be considered as zero.

Step-5: Identify the largest number among the RDI and CDI in each level. Choose the smallest time unit along the highest RDI or CDI. If there are two or more number of highest RDI or CDI; choose that highest indicator along which the smallest time element is present. If there are two or more number of smallest time element, choose any one of those arbitrarily.

Step-6: Allocate minimum of supply/demand values on the left top of the corresponding smallest time unit selected in Step-5.

Step-7: Check whether exactly one of the row/column corresponding to the selected cell has zero supply/ zero demand, respectively. If so, delete the row/column which has the zero supply/zero demand. Again if it is found that both the supply and demand has become zero corresponding to the selected cell, delete either the row or column but not both.

Step-8: Match the supply/demand of the left out row/column with the remaining demands/supplies of the undeleted columns/row.

Step-9: Repeat Step: 4 to 6 until all the rim requirements are satisfied.

Step-10: Determine the largest time T_{k} corresponding to basic cells.

Algorithm for finding the optimal time for time minimization transportation problems are described below:

Step-1: Determine an initial basic feasible solution using above method.

Step-2: Determine the largest time T_{k} corresponding to the basic cells and cross off all non-basic cells

Step-3: Construct a loop for the basic cells corresponding to largest time T_{k} including a non-basic cell in such a way that the allotment in the cell with T_{k} can be shifted to the non-basic cell in the loop by readjustment of row and column sum, the value at T_{k} cell is zero and no variables becomes zero or negative. If no such loop can be formed, the solution under test is optimum. Otherwise move to the next step.

Step-4: Repeat the procedure until an optimum basic feasible solution is obtained.

Origins | Destinations | Supply | ||
---|---|---|---|---|

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

O_{1} | 13 | 21 | 14 | 13 |

O_{2} | 8 | 12 | 21 | 20 |

O_{3} | 15 | 17 | 19 | 5 |

Demand | 12 | 15 | 11 |

From the above table it is seen that total supply and total demand are equal. Hence the given transportation problem is a balanced one. Obtaining IBFS for Example-1 using the Proposed Algorithm is shown in

Origins | Destinations | Supply | LE | RDI | ||||
---|---|---|---|---|---|---|---|---|

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

O_{1} | 13_{6} | ^{2}21_{0} | ^{11}14_{0} | 13 | 21 | (6) | (0) | (0) |

O_{2} | ^{12}8_{6} | ^{8}12_{0} | 21_{0} | 20 | 21 | (6) | (0) | -- |

O_{3} | 15_{4} | ^{5}17_{2} | 19_{2} | 5 | 19 | (2) | (0) | (0) |

Demand | 12 | 15 | 11 | |||||

LE | 15 | 21 | 21 | |||||

CDI | (0) | (2) | (2) | |||||

-- | (2) | (2) | ||||||

-- | (2) | (2) |

We see that the number of basic variables is 5 = (3 + 3 − 1) and the set of basic cells do not contain a loop. Thus the solution obtained is a basic feasible solution and the initial basic feasible solution is x_{12} = 2, x_{13} = 11, x_{21} = 12, x_{22} = 8, x_{32} = 5 and the shipping time of basic cells are t_{12} = 21, t_{13} = 14, t_{21} = 8, t_{22} = 12, t_{32} = 17.

Therefore, in order to complete the shifting it takes the time T_{1} = max{t_{12}, t_{13}, t_{21}, t_{22}, t_{32}} = max{21, 14, 8, 12, 17}= 21 units of time.

Iteration-1: From the IBFS, it is found that the largest time is T_{1} = 21, therefore we cross off the non-basic cell (2, 3) which contains equal time unit to T_{1}. Let us construct a loop for the basic cells corresponding to the largest time T_{1} = 21 including the non-basic cell (1, 1) which is called entering cell. The constructed loop in this iteration is shown in

Origins | Destinations | Supply | ||
---|---|---|---|---|

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

O_{1} | 13 + | ^{2}21 − | ^{11}14 | 13 |

O_{2} | ^{12}8 − | ^{8}12 + | 21 | 20 |

O_{3} | 15 | ^{5}17 | 19 | 5 |

Demand | 12 | 15 | 11 |

The allocation is readjusted in the corner of the loop and a new solution is obtained which is shown in

Origins | Destinations | Supply | ||
---|---|---|---|---|

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

O_{1} | ^{2}13 | 21 | ^{11}14 | 13 |

O_{2} | ^{10}8 | ^{10}12 | 21 | 20 |

O_{3} | 15 | ^{5}17 | 19 | 5 |

Demand | 12 | 15 | 11 |

Iteration-2: Now largest time is T_{2} = 17, therefore we cross off the cells (1, 2) and (3, 3) which contain larger time units than T_{2}. Let us construct a loop for the basic cells corresponding to the largest time T_{2} = 17 including the non-basic cell (3, 1). The constructed loop in this iteration is shown in

Origins | Destinations | Supply | ||
---|---|---|---|---|

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

O_{1} | ^{2}13 | 21 | ^{11}14 | 13 |

O_{2} | ^{10}8 − | ^{10}12 + | 21 | 20 |

O_{3} | 15 + | ^{5}17 − | 19 | 5 |

Demand | 12 | 15 | 11 |

Readjust the allocation in the corner of the loop, we obtain a new solution which is shown in

Origins | Destinations | Supply | ||
---|---|---|---|---|

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

O_{1} | ^{2}13 | 21 | ^{11}14 | 13 |

O_{2} | ^{5}8 | ^{15}12 | 21 | 20 |

O_{3} | ^{5}15 | 17 | 19 | 5 |

Demand | 12 | 15 | 11 |

Iteration-3: Now largest time is T_{3} = 15, therefore the cell (3, 2) which contains larger time unit than T_{3} is crossed off. Now all the crossed off cells including the crossed off cell in this stage is shown in

Origins | Destinations | Supply | ||
---|---|---|---|---|

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

O_{1} | ^{2}13 | 21 | ^{11}14 | 13 |

O_{2} | ^{5}8 | ^{15}12 | 21 | 20 |

O_{3} | ^{5}15 | 17 | 19 | 5 |

Demand | 12 | 15 | 11 |

Now no loop can be formed originating from the cell (3, 1). Thus the obtained solution x_{11} = 2, x_{13} = 11, x_{21} = 5, x_{22} = 15, x_{31} = 5 is optimum and the optimum shipment time is max {13, 14, 8, 12, 15} = 15 time units.

Some equipment is to be transported from three origins to three destinations. The supply at the origins, the demand at the destinations and the time of shipment are shown in

Origins | Destinations | Supply | ||
---|---|---|---|---|

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

O_{1} | 15 | 7 | 25 | 12 |

O_{2} | 8 | 12 | 14 | 17 |

O_{3} | 17 | 19 | 21 | 7 |

Demand | 12 | 10 | 14 |

It is required to find the optimum plan for shipment to minimize the total time of shipment.

From

Origins | Destinations | Supply | LE | RDI | ||||
---|---|---|---|---|---|---|---|---|

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

O_{1} | ^{2}15_{8} | ^{10}7_{6} | 25_{0} | 12 | 25 | (2) | (2) | (2) |

O_{2} | ^{3}8_{3} | 12_{5} | ^{14}14_{11} | 17 | 14 | (6) | (2) | -- |

O_{3} | ^{7}17_{4} | 19_{2} | 21_{4} | 7 | 21 | (0) | (2) | (2) |

Demand | 12 | 10 | 14 | |||||

LE | 17 | 19 | 25 | |||||

CDI | (4) | (1) | (7) | |||||

(4) | (1) | -- | ||||||

(4) | (4) | -- |

It is seen that the number of basic variables is 5 (=3 + 3 − 1) and the set of basic cells do not contain a loop. Thus the solution obtained is a basic feasible solution and the initial basic feasible solution is x_{11} = 2, x_{12} = 10, x_{21} = 3, x_{23} = 14, x_{31} = 7 and the shipping time of basic cells are t_{11}= 15, t_{12} = 7, t_{21}= 8, t_{23} = 14, t_{31} = 17.

Therefore, in order to complete the shipment, it takes the time T_{1} = max{t_{11}, t_{12}, t_{21}, t_{23}, t_{31}} = max{15, 7, 8, 14, 17} = 17 units of time.

Iteration-1: Now largest time is T_{1} = 17, therefore the non basic cells, (1, 3), (3, 2) and (3, 3) containing larger time units than T_{1} are crossed off. These crossed off cells are shown in

Origins | Destinations | Supply | ||
---|---|---|---|---|

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

O_{1} | ^{2}15 | ^{10}7 | 25 | 12 |

O_{2} | ^{3}8 | 12 | ^{14}14 | 17 |

O_{3} | ^{7}17 | 19 | 21 | 7 |

Demand | 12 | 10 | 14 |

Now no loop can be formed originating from the cell (3, 1). Thus the obtained solution x_{11} = 2, x_{12} = 10, x_{21} = 3, x_{23} = 14, x_{31} = 7 is optimum and the optimum shipment time is max {15, 7, 8, 14, 17} = 17 units of time.

After obtaining an IBFS by the proposed method, the obtained result is compared with the solution obtained by other existing methods that how many number of iterations are to be required to obtain an optimum solution. This is shown in

Example | Optimum Time | NWCM | LCM | VAM | EDM | Proposed Method |
---|---|---|---|---|---|---|

1 | 15 | 4 | 3 | 3 | 3 | 3 |

2 | 17 | 3 | 3 | 2 | 1 | 1 |

In this writing, an algorithm has been discussed to solve time minimizing transportation problems. Efficiency of the developed algorithm is also been justified by solving numerical problems. During this process, a comparative study among the proposed method and the other existing methods is also carried out; where it is observed that the proposed method requires minimum number of iterations to obtain the optimal transportation time. Therefore, the proposed algorithm claims its wide application in the field of optimization in solving the time minimizing transportation problem.

Mollah MesbahuddinAhmed,Md. AmirulIslam,MomotazKatun,SabihaYesmin,Md. SharifUddin, (2015) New Procedure of Finding an Initial Basic Feasible Solution of the Time Minimizing Transportation Problems. Open Journal of Applied Sciences,05,634-640. doi: 10.4236/ojapps.2015.510062