Optimal Solution of OFSTF, MDMA Methods with Existing Methods Comparison

Transcription

1 Volume 119 No , ISSN: (printed version); ISSN: (on-line version) url: ijpam.eu Optimal Solution of OFSTF, MDMA Methods with Existing Methods Comparison 1 A. Amaravathy, 2 K. Thiagarajan and 3 S. Vimala 1 Department of Mathematics, Mother Teresa Women s University, Kodaikanal, Tamil Nadu, India. amaravathy.sai@gmail.com 2 Department of Mathematics, PSNA College of Engineering and Technology, Dindigul, Tamil Nadu, India. vidhyamannan@yahoo.com 3 Department of Mathematics, Mother Teresa Women s University, Kodaikanal, Tamil Nadu, India. tvimss@gmail.com Abstract In this paper a different approach OFSTF (Origin, First, Second, Third, and Fourth quadrants) Method is applied for finding a feasible solution for transportation problems directly. The proposed method is a unique, it gives always feasible (may be optimal for some extant) solution without disturbance of degeneracy condition. This method takes least iterations to reach optimality. A numerical example is solved to check the validity of the proposed method and degeneracy problem is also discussed. Key Words:Assignment problem, transportation problem, degeneracy, Pay Off Matrix (POM), quadrants. 989

2 1. Introduction The transportation problem constitutes an important part of logistics management. In addition, logistics problems without shipment of commodities may be formulated as transportation problems [1]. For instance, the decision problem of minimizing dead kilometers (Raghavendra and Mathirajan, 1987) [11] can be formulated as a transportation problem (Vasudevan et al.,1993; Sridharan, 1991) [18],[20]. The problem is important in urban transport undertakings, as dead kilometres mean additional losses. It is also possible to approximate certain additional linear programming problems by using a transportation formulation (e.g., see Dhose and Morrison, 1996) [4]. Various methods are available to solve the transportation problem to obtain an optimal solution [16]. Typical/well-known transportation methods include the stepping stone method [2] (Charnes and Cooper, 1954), the modified distribution method (Dantzig, 1963), the modified stepping-stone method (Shih, 1987), the simplex-type algorithm (Arsham and Kahn, 1989) and the dual-matrix approach (Ji and Chu, 2002). Glover et al. (1974) presented a detailed computational comparison of basic solution algorithms for solving the transportation problems [6],[18]. Shafaat and Goyal (1988) proposed a systematic approach for handling the situation of degeneracy encountered in the stepping stone method [7][8][9],[14]. A detailed literature review on the basic solution methods is not presented. All the optimal solution algorithms for solving transportation problems need an initial basic feasible solution to obtain the optimal solution[3],[19]. There are various heuristic methods available to get an initial basic feasible solution, such as "North West Corner" rule, "Best Cell Method," "VAM Vogel's Approximation Method"[17] (Reinfeld and Vogel, 1958), Shimshaket a/.'s version of VAM (Shimshaket al.,1981)[13], Coyal's version of VAM (Goyal, 1984), Ramakrishnan's version of VAM (Ramakrishnan, 1988) etc [12]. Further, Kirca and Satir (1990) developed a heuristic, called TOM (Total Opportunity-cost Method), for obtaining an initial basic feasible solution for the transportation problem [10]. Gass (1990) detailed the practical issues for solving transportation problems [5] and offered comments on various aspects of transportation problem methodologies along with discussions on the computational results, by the respective researchers. Recently, Sharma and Sharma (2000) [15] proposed a new heuristic approach for getting good starting solutions for dual based approaches used for solving transportation problems. Even in the above method needs more iteration to arrive optimal solution. Hence the proposed method helps to get directly optimal solution with less iteration number of the proposed method [21],[22],[23] is given below. 990

3 2. Transport Problem through OFSTF (Origin, First, Second, Third, and Fourth quadrants) Method We now introduce a new method called the Transport Problem through OFSTF method for finding an feasible solution to a transportation problem. The OFSTF method proceeds as follows. Step1 Construct the TT for the given POM. Step 2 Choose the MAE & MIE in the constructed TT. Step 3 Find the difference between the MAE & MIE from Step 2. If the RE matched with anyone of the element in the POM, then find the difference between each element in the TT with the RE. That is, MAE MIE = RE If R.E = an element in TT Every element in TT RE. If R.E an element in TT select next MIE in TT and repeat the Step Repeat the process until the condition satisfied. Step 4 In the Reduced POM, there will be at least one zero in the TT, select a particular zero based on the MADE from the given zeros. Step 5 Case 1 Fix zero as origin, and find the MADE from the selected zero. Case 2 Fix zero as origin, and find the MADE in the first quadrant (+, +) from the selected zero. Case 3 Fix zero as origin, and find the MADE in the second quadrant (-, +) from the selected zero. 991

4 Case 4 Fix zero as origin, and find the MADE in the Third quadrant (-, -) from the selected zero. Case 5 Fix zero as origin, and find the MADE in the fourth quadrant (+, -) from the selected zero. Step 6 Compare and fulfill the demand of the MADE with the supply in the TT. Step 7 Calculate the total cost for each cases, the feasible solution is obtained in the origin area for all kind of transportation Problem. Hence we say that in transportation problem, calculating the cost from origin will lead to a feasible solution by our OFSTF. Example Consider the following cost minimizing transportation problems Demand Now Using the OFSTF method, TC in origin = 777, TC1Q = 789, TC2Q = 255, TC3Q = 327, TC4Q= 825. MDMA Method Consider the following CMITP. Step Demand Here the MAE is 40, Divide all elements by ME = 40 3/40 4/40 6/40 8/40 9/ /40 10/40 1/40 5/40 8/ /40 11/40 20/40 1 3/ /40 1/40 9/40 14/40 16/40 13 Demand

5 Select the MIE (1/40) allot the MID and allot [D(6),S(13)] =6 unit in the cell(4,2) Step2 Choose the next MAE is 20/40, Divide all elements by ME=20/40 3/20 6/20 8/20 9/ /20 1/20 5/20 8/ / /20 3/ /20 9/20 14/20 16/20 7 Demand Pick the MIE and assign the MID [D(8),S(30)] =8 unit in the cell(2,2) Step3 Choose the next MAE is 40/20, Divide all elements by ME= 40/20 3/40 8/40 9/ /40 5/40 8/ /40 1 3/ /40 14/40 16/40 7 Demand The MIE and Assign the MID [D(40),S(7)]=7 unit in the cell(4,1) Step4 Choose the next MAE is 9/40, Divide all elements by ME=9/40 3/9 8/ /9 5/9 8/9 22 7/9 40/9 3/9 15 Demand Pick the MIE and assign the MID [D(33),S(22)]= 22 unit in the cell(2,1) Step5 Choose the next MAE is 40/9, Divide all elements by ME=40/9 3/40 8/40 9/ /40 1 3/40 15 Demand Select the MIE and assign the MID [D(6),S(15)]= 6 unit in the cell(2,3) 993

6 Step6 Choose the next MAE is 8/40, Divide all elements by ME=8/40 3/ /8 40/8 9 Demand Select the MIE and assign the MID [D(11),S(20)]=11 unit in the cell(1,1) Step /8 9 Demand Finally Choose the MIS and allot 9 unit in cell(1,1) and 9 unit in cell(2,1). Which leads to the solution satisfying all the constraints. The resulting initial basic feasible solution is given below. Sept Demand units : Cost 11 x 3 = 33 9 units : Cost 9 x 8 = units : Cost 22 x 2 = 44 8 units : Cost 8 x 1 = 8 9 units : Cost 9 x 40 = units : Cost 6 x 3 = 18 7 units : Cost 7 x 2 = 14 6 units : Cost 6 x 1 = Total cost = North West corner rule : 918 LCM : 555 MDMA : 555 Origin : 777 First Quadrant :

7 Second Quadrant : 255 Third Quadrant : 327 Fourth Quadrant : 825 Second Third LCM MDMA Origin First Fourth North West corner rule Problem Demand North west corner rule : 4400 LCM : 2850 MDMA : 2850 Origin : 2850 First Quadrant : 3800 Second Quadrant : 2900 Third Quadrant : 4850 Fourth Quadrant : 4150 Problem Demand North west corner rule : 464 LCM : 450 MDMA : 450 Origin : 450 First Quadrant : 700 Second Quadrant : 450 Third Quadrant : 490 Fourth Quadrant : 480 Problem Demand North West corner rule : 1076 LCM : 781 MDMA :

8 Origin : 891 First Quadrant : 902 Second Quadrant : 876 Third Quadrant : 906 Fourth Quadrant : 891 Problem Demand North West corner rule : 3051 LCM : 2953 MDMA : 2849 Origin : 2953 First Quadrant : 2953 Second Quadrant : 4588 Third Quadrant : 2877 Fourth Quadrant : 3428 Problem Demand North West corner rule : 102 LCM : 83 MDMA : 101 Origin : 85 First Quadrant : 104 Second Quadrant : 81 Third Quadrant : 120 Fourth Quadrant : 82 Problem Demand North West corner rule : 1095 LCM : 685 MDMA :

9 Origin : 850 First Quadrant : 1020 Second Quadrant : 955 Third Quadrant : 1150 Fourth Quadrant : 1105 Analysis PROBLEM No. 1 LCM MDMA Origin Second First Fourth Third North West corner Rule 2 LCM MDMA Origin Second Fourth North West Corner Rule Third First 3 LCM MDMA Second Origin Fourth Third North West Corner Rule First 4 MDMA LCM Origin First Third North West Corner Rule Fourth Second 5 Second Fourth LCM Origin MDMA North West Corner Rule First Third 6 MDMA LCM Origin Second First North West Corner Rule Fourth Third Variable portrayal in this paper. Nomenclature MAE - Maximum Element MIE - Minimum Element MADE - Maximum Deviated Element TC - Total Cost TC1Q - Total Cost in First Quadrant TC2Q - Total Cost in Second Quadrant TC3Q - Total Cost in Third Quadrant TC4Q - Total Cost in Fourth Quadrant CMITP -Cost Minimizing Transportation Problem MAD -Maximum Demand MID -Minimum Demand MIS -Minimum LCM -Least Cost Method MDMA Maximum Divide Minimum Allotment TT -Transportation Table POM -Pay off Matrix RE -ant Element 3. Conclusion and Future Work This paper provides the comparative analysis with best solution of the given problem in prescribed method here the compared and conclude the results of the different methods. These methods are important tool for the decision makers when they are handling various types of logistic problems to make the decision optimally. 997

10 Acknowledgement The authors would like to thank Dr. Ponnammal Natarajan, Former director of Research, Anna University, Chennai, India for her intuitive ideas and fruitful discussions with respect to the paper s contribution and support to complete this paper. References [1] Arsham H., Kahn A.B., A simplex-type algorithm for general transportation problems: An alternative to Stepping-Stone. Journal of Operational Research Society 40(6) (1989), [2] Charnes A., Cooper W.W., The Stepping-Stone method for explaining linear programming calculations in transportation problems, Management Science 1(1) (1954), [3] Dantzig G.B., Linear Programming an Extensions, Princeton, NJ: Princeton University Press (1963). [4] Dhose E.D., Morrison K.R., Using transportation solutions for a facility location problem, Computers and Industrial Engineering 31(1/2) (1996), [5] Gass S.I., On solving the transportation problem. Journal of Operational Research Society 41(4) (1990), [6] Glover F., Karney D., Klingman D., Napier A., A computation study on start procedures, basis change criteria, and solution algorithms for transportation problems, Management Science 20(5) (1974), [7] Goyal S.K., Improving VAM for unbalanced transportation problems, Journal of Operational Research Society 35(12) (1984), [8] Goyal S.K., A note on a heuristic for obtaining an initial solution for the transportation problem. Journal of Operational Research Society 42(9) (1991), [9] Ji P., Ghu K.F., A dual-matrix approach to the transportation problem, Asia- Pacific Journal of Operations Research 19(1) (2002), [10] Kirca O., Satir A., A heuristic for obtaining an initial solution for the transportation problem, Journal of Operational Research Society 41(9) (1990), [11] Raghavendra B.G., Mathirajan M., Optimal allocation of buses to depots a case study, Opsearch 24(4) (1987),

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