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In this paper, we discuss a new approach for solving an unbalanced assignment problem. A Lexi-search algorithm is used to assign all the jobs to machines optimally. The results of new approach are compared with existing approaches, and this approach outperforms other methods. Finally, numerical example (Table 1) has been given to show the efficiency of the proposed methodology.

Consider a problem which consists of a set of “n” machines

The objectives are to determine the optimal assignment cost, in such a way that all the jobs are to be allotted on the available machines in an optimum way. The mathematical formulation of the assignment problem [

300 | 290 | 280 | 290 | 210 | |

250 | 310 | 290 | 300 | 200 | |

180 | 190 | 300 | 190 | 180 | |

320 | 180 | 190 | 240 | 170 | |

270 | 210 | 190 | 250 | 160 | |

190 | 200 | 220 | 190 | 140 | |

220 | 300 | 230 | 180 | 160 | |

260 | 190 | 260 | 210 | 180 |

Minimize (Maximize):

Subject to

where

Problem definition:

Also, if the numbers of jobs are not equal to number of machines, then it is known as an unbalanced assignment problem. Now consider the assumptions of choosing an unbalanced assignment problem as:

• The completion of a program from computational point of view means that the all jobs are assigned to various machines and final optimal assignment cost has been obtained.

• The number of jobs are more than number of machines.

The variants of assignment problem are considered by various researchers like Kagade & Bajaj [

To determine the assignment cost as well as combination of job (s) Vs machine (s) of an unbalanced assignment problem for a set of “n” machines

180 | 190 | 300 | 190 | 180 | |

320 | 180 | 190 | 240 | 170 | |

270 | 210 | 190 | 250 | 160 | |

190 | 200 | 220 | 190 | 140 | |

220 | 300 | 230 | 180 | 160 |

290 | 290 | 210 | |

310 | 300 | 200 | |

190 | 210 | 180 |

defined assignment problem.Now we apply the Lexi-search approach to obtain the exact optimum solution of each sub problem (Tables 4-7). Finally, add the total assignment cost of each sub problem to obtain the optimal assignment cost along with assignment sets. And also we check the assignment cost for jobs clubbing problem (

Step-1: Consider “m” jobs on “n” machines costs given as a matrix (ACM), which is an unbalanced assign- ment problem where

Step-2:

Step-2.1: Obtain the sum of each row and column of the ACM and the store the results in the arrays namely

Step-2.2: Select the first m rows (jobs) on the basis of

Step-2.2.1: If there is no remaining jobs, i.e., (m-n = 0), then go to step-3.

Step-2.2.2: If the remaining (m-n) jobs are still more than n, then repeat step-2.2 for the remaining jobs to form next sub-problem (s), else, step-2.3.

Step-2.3: If remaining jobs are less than n then deleting (n-m) columns (machines) on the basis of

Step-3: If the total effectiveness of ACM is to be maximized, change the sign of each cost element in the effectiveness matrix and go to step-4, otherwise go directly to step-5 if ACM has the total value as minimum.

Step-4: Arrange all the jobs

Step-5: Include the job from the first machine in the partial solution value (psv) “w”. If the cost itself is greater than or equal to trial value (TRV) then stop. Otherwise go to next step.

Step-6: Calculate the bound.

Step-7: If the sum of bound and psv is greater than or equal to TRV then drop the job added in step 5, and go to step 5. Otherwise go to next step, i.e. go to Sub block (GS).

Step-8: Include the next available job (from the last job included in the partial solution “w”) into the partial solution.

A.B | Remark | |||||
---|---|---|---|---|---|---|

Bound = 870 | Bound = 870 | Bound = 870: | Bound = 870 | Bound = 870: | 870 | |

(180 + 180 + 190 + 180 + 140) | (180 + 180 + 190 + 180 + 140) | (180 + 180 + 190 +180 + 140) | (180 + 180 + 190 + 180 + 140) | (180 + 180 + 190 + 180 + 140) | ||

GNSB | ||||||

Bound: 900 > 870 | ||||||

(180 + 180 + 190 + 190 + 160) | ||||||

GNSB | ||||||

Bound: 920 > 870 | ||||||

(180 + 180 + 220 +180 + 160) | ||||||

GNSB | ||||||

Bound: 910 > 870 | ||||||

(180 + 200 + 190 + 180 + 160) | ||||||

GNSB | ||||||

Bound: 920 > 870 | ||||||

(190 + 180 + 190 + 180 + 180) | ||||||

Bound: 920 > 870 | ||||||

(220 + 180 + 190 + 190 + 140) | ||||||

Bound: 1030 > 870 | ||||||

(270 + 180 + 220 + 180 + 180) | ||||||

Bound: 1020 > 870 | ||||||

(320 + 190 + 190 + 180 + 140) |

A.B | Remark | |||
---|---|---|---|---|

Bound: 680 | 680 | |||

(190 + 290 + 200) | Bound: 680 | Bound: 680 | ||

(190 + 290 + 200) | (190 + 290 + 200) | |||

Bound: 700 > 680 | GNSB | |||

(190 + 300 + 210) | ||||

Bound: 700 > 680 | ||||

(290 + 210 + 200) | ||||

Bound: 730 > 680 | ||||

(310 + 210 + 210) |

520 | 590 | 510 | 470 | 370 | |

440 | 510 | 510 | 490 | 340 | |

180 | 190 | 300 | 190 | 180 | |

580 | 370 | 450 | 450 | 350 | |

270 | 210 | 190 | 250 | 160 |

A.B | Remark | |||||
---|---|---|---|---|---|---|

1650 | JB | |||||

Bound: 1650 | ||||||

(180 + 210 + 450 + 470 + 340) | Bound: 1650 | |||||

(180 + 210 + 450 + 470 + 340) | Bound: 1650 | Bound: 1650 | ||||

(180 + 210 + 450 + 470 + 340) | Bound: 1650 | (180 + 210 + 450 + 470 + 340) | ||||

180 + 210 + 450 + 470 + 340) | ||||||

GNSB | ||||||

GNSB | ||||||

Bound: 1700 > 1650 | ||||||

(180 + 210 + 450 + 490 + 370) | ||||||

JB | ||||||

Bound: 1690 > 1650 | ||||||

(180 + 210 + 510 + 450 + 340) | ||||||

GNSB | ||||||

Bound: 1720 > 1650 | ||||||

(180 + 210 + 510 + 450 + 370) | ||||||

1550 | JB | |||||

Bound: 1550 < 1650 | Bound: 1550 < 1650 | |||||

(180 + 370 + 190 + 470 + 340) | (180 + 370 + 190 + 470 + 340) | |||||

Bound: 1550 < 1650 | ||||||

Bound: 1550 < 1650 | (180 + 370 + 190 + 470 + 340) | |||||

(180 + 370 + 190 + 470 + 340) | ||||||

GNSB | ||||||

GNSB | ||||||

Bound: 1600 > 1550 | ||||||

(180 + 370 + 190 + 490 + 370) |

Step-9: If partial solution value is greater than or equal to the TRV then drop the job added in step-8, and go to step-7. Otherwise go to step-10.

Step-10: If the sum of bound and psv is greater than or equal to TRV then drop the newly added job in step-8, and go to step -7.otherwise go to step 11.

Step-11: If the partial solution contains n − 1 jobs add the dummy job to the partial solution if it is greater than or equal to TRV then drop the dummy job and last two jobs from the partial solution. That is Jump out to the next higher order blocks (JO). If “w” contains only one job, go to step-5, otherwise go to step-8. Otherwise go to the next step.

Step-12: Now calculate the bound.

Step-13: If the sum of bound and psv is greater than or equal to TRV then drop the dummy job and also last job from “w”, and go to step-8. Otherwise go to step-14.

A.B | Remark | |||||
---|---|---|---|---|---|---|

JB | ||||||

Bound: 1650 > 1550 | ||||||

(180 + 370 + 510 + 250 + 340) | ||||||

GNSB | ||||||

Bound: 1680 > 1550 | ||||||

(180 + 370 + 510 + 250 + 370) | ||||||

Bound: 1700 > 1550 | ||||||

(180 + 510 + 190 + 450 + 370) | ||||||

Bound: 1750 > 1550 | ||||||

(180 + 590 + 190 + 450 + 340) | ||||||

Bound: 1720 > 1550 | ||||||

(270 + 190 + 450 + 470 + 340) | ||||||

Bound: 1640 > 1550 | ||||||

(440 + 190 + 190 + 450 + 370) | ||||||

Bound: 1690 > 1550 | ||||||

(520 + 190 + 190 + 450 + 340) | ||||||

Bound: 1770 > 1550 | ||||||

(580 + 190 + 190 + 470 + 340) |

Step-14: Include the latest possible job from the dummy job in“w”

Step-15: If psv is greater than or equal to TRV then drop the last dummy job and also the job from which the

Step-16: Now calculate the bound.

Step-17: If sum of bound and psv is greater than or equal to TRV then drop the recently added job in “w” and go to step-14. Otherwise go to next step.

Step-18: Include the latest available job from the last job in “w”

Step-19: Now calculate the bound.

Step-20: If the sum of bound and psv is greater than or equal to TRV then drop the latest job, and go to step- 18. Otherwise go to next step.

Step-21: If the number of elements in “w” is less than “n” go to step-18. Otherwise go to next step.

Step-22: Replace TRV by partial solution value and trial solution by w. Now go to step-18.

A company is faced with the problem of assigning five different machines to eight different jobs (

Solve the problem assuming that the objective is to minimize the total cost. Now obtain the sum of each row and column of

We partition the matrix

Sub-Problem-I:

Sub-Problem-II:

RP: Repitition, GNSB or JB: Go to next super block (JB), A.B or TRV = Absolute bound or Trail bound.

The final optimal assignments of

The final optimal assignments

The final optimal assignments assigned cost matrix (ACM) is :

The final optimal assignments

Total assignment cost = 1550.

The above illustration was taken by the defined algorithm and implemented on several sizes of the problems to test the effectiveness of the algorithm. This approach was implemented on different sizes of unbalanced assignment problems. From the above, we notice that the standard Hungarian method uses the dummy assignment which may not be possible in some applications, whereas this new approach never assigns the dummy machine in getting the optimum value. The time complexity with the Lexi-search method is verified and found that they are the same in getting optimum. Here, the optimum value of the original unbalanced assignment problems varies from that of balanced assignment problems either in Hungarian method or Lexi-search approach. The only advantage is that the Lexi-search method gives an exact optimum value with the same time complexity. Therefore the present paper suggests a new approach of clubbing the jobs for solving the unbalanced assignment problem with Lexi-search methodology.

VentepakaYadaiah,V. V.Haragopal, (2016) A New Approach of Solving Single Objective Unbalanced Assignment Problem. American Journal of Operations Research,06,81-89. doi: 10.4236/ajor.2016.61011