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Accepted Manuscript A heuristic solution technique to attain the minimal total cost bounds of transporting a homogeneous product with varying demands and supplies Z.A.M.S. Juman, M.A. Hoque PII: DOI: Reference: S0377-2217(14)00411-1 http://dx.doi.org/10.1016/j.ejor.2014.05.004 EOR 12304 To appear in: European Journal of Operational Research Please cite this article as: Juman, Z.A.M.S., Hoque, M.A., A heuristic solution technique to attain the minimal total cost bounds of transporting a homogeneous product with varying demands and supplies, European Journal of Operational Research (2014), doi: http://dx.doi.org/10.1016/j.ejor.2014.05.004 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain. A heuristic solution technique to attain the minimal total cost bounds of transporting a homogeneous product with varying demands and supplies Z. A. M. S. Juman and M. A. Hoque* Department of Mathematics, Faculty of Science, University of Brunei Darussalam, Brunei Darussalam. Abstract Transportation of a product from multi-source to multi-destination with minimal total transportation cost plays an important role in logistics and supply chain management. Researchers have given considerable attention in minimizing this cost with fixed supply and demand quantities. However, these quantities may vary within a certain range in a period due to the variation of the global economy. So, the concerned parties might be more interested in finding the lower and the upper bounds of the minimal total costs with varying supplies and demands within their respective ranges for proper decision making. This type of transportation problem has received attention of only one researcher, who formulated the problem and solved it by LINGO. We demonstrate that this method fails to obtain the correct upper bound solution always. Then we extend this model to include the inventory costs during transportation and at destinations, as they are interrelated factors. The number of choices of supplies and demands within their respective ranges increases enormously as the number of suppliers and buyers increases. In such a situation, although the lower bound solution can be obtained methodologically, determination of the upper bound solution becomes an NP hard problem. Here we carry out theoretical analyses on developing the lower and the upper bound heuristic solution techniques to the extended model. A comparative study on solutions of small size numerical problems shows promising performance of the current upper bound technique. Another comparative study on results of numerical problems demonstrates the effect of inclusion of the inventory costs. Keywords: Heuristics; transportation cost; inventory cost; total cost bound. * Corresponding Author- email: [email protected] , Tel: 673 2463001 1357, Fax: 673 2461502 2 1. Introduction In today’s competitive business environment, integrated suppliers-buyers supply chain management is a major concern. Two of the key issues in this supply chain management are the transportation and the inventory costs. To achieve significant savings, these two issues should be integrated instead of treating them separately. The transportation problem deals with transporting a homogeneous product from multi–source to multi–destination, with the minimal total cost of transportation subject to the satisfaction of the available supply and the demand quantities. However, each of the supply and demand quantities of a product may vary within a certain range in a period due to the variation of the global economy. Following this variation, the minimal total transportation cost also varies within a certain range. So, the concerned parties might be more interested in finding the lower and the upper bounds of the minimal total costs for better decision making specifically, for proper investment and return. But the number of choices of supply and demand quantities within their respective ranges increases enormously as the number of suppliers or buyers increases. However, this type of transportation problem can be reduced to a linear programming problem following Liu (2003)’s approach, and then it can be further reduced to the minimum cost flow problem (Ahuja et al., 1993). Thereafter, a polynomial time algorithm can be applied to this minimum cost flow problem for finding the lower bound of the minimal total costs. Thus, in such a situation, although the lower bound of the transportation problem can be found methodologically, determination of the upper bound of the minimal total costs becomes an NP hard problem. Inventory costs during transportation and in meeting demand from destinations are essential in this system, and hence these should be considered along with the transportation cost. The format of the extended transportation model including these inventory costs is equivalent to the original one of Liu (2003) (as shown later in sub-section 2.4). In this situation, development of a heuristic solution method to the extended transportation model in finding the upper bound of the minimal total costs is desirable. Although a solution method to the problem without taking into account the mentioned inventories is available, here we demonstrate that this does not lead to the correct upper bound solution always. Therefore, this study mainly considers an integration of the transportation and the inventory costs in transporting a homogeneous product from multiple suppliers to multiple buyers, and development of a better heuristic solution technique to this integrated supply chain problem in finding the upper bound of the minimal total costs. Hitchcock (1941) formulated the transportation problem initially. Then Charnes and Copper (1954) developed the stepping stone method for solution of the transportation problem. Srinivasan and Thompson (1977) described two new primal basic methods - the cell and area cost 3 operator algorithms for solving the transportation problem. Søren (1978) explained how to use the triple index and the threaded index for storing the basis-tree when applying the primal, dual or primal-dual simplex method to solve a classical transportation model. Donald and Haluk (1997) showed that general transportation algorithms automatically yield solution in integer values with integer demand and supply quantities. Sharma and Sharma (2000) proposed a new solution procedure to solve the incapacitated transportation problem. Sharma and Prasad (2003) presented a heuristic that obtained a very good initial basic feasible solution to the transportation problem in O ( n 3 ) time. Veena Adlakha & Kowalski (2009) proposed an alternative algorithm to obtain the minimal total cost solution. Saleem and Imad (2012) developed a hybrid two-stage algorithm (GA-RSM) to find the minimal total cost solution to the transportation problem. The first stage used genetic algorithm (GA) to find an improved initial basic feasible solution, and the second stage utilized this solution as a starting point in the revised simplex method (RSM) to get the minimal total cost solution to the problem. Aizemberg et al. (2014) studied tactical models of scheduling the shipments of a crude oil through routes linking platforms (offshore production sites) and terminals (onshore consumer sites) with the minimum transportation cost. Vancroonenburg et al. (2014) studied the Red-Blue Transportation Problem (Red-Blue TP), a generalization of the transportation problem where supply nodes are partitioned into two sets. Here, they provided two integer-programming formulations for Red-Blue TP and showed that one of them is strictly stronger than the other. They also presented a maximization variant of RedBlue TP (by modifying the objective function of the Red-Blue TP to maximization) and thus provided three approximation algorithms for Max-Red-Blue TP. All of these solution procedures to the transportation problem were developed with fixed supply and demand quantities. But a little attention has also been given in developing transportation models with variable parameter values. Das et al. (1999) proposed a solution for solving the multi-objective transportation problem, where the coefficients of terms in the objective functions and parameter values at the sources and the destinations were given in an interval. Safi and Razmjoo (2013) focused on the transportation problem where a fixed charge was added with the transportation cost per unit, and parameters values were given in intervals. They proposed two solution procedures to this problem. Note that both Das et al. (1999) and Safi and Razmjoo (2013) did not make any attempt to find the lower and the upper bounds of the minimal total cost following the parameter values in intervals. Liu (2003) investigated the transportation problem when the demand and supply quantities were varying within their respective ranges. Following these variations the minimal total cost were also varied within an interval. So, he constructed a pair of mathematical programs 4 where at least one of the supply or the demand was varying, to calculate the lower and the upper bounds of the total transportation cost. A considerable amount of research dealing with the management of the integrated supplier-buyer system involving joint inventory and transportation cost has emerged in the literature. Hill (1999), Hoque and Goyal (2000), Stanisław (2003), Ben-Daya and Hariga (2004), Hill and Omar (2006), Zhou and Wang (2007), Onur kaya et al. (2013) all have considered inventory and transportation cost in the single-vendor single-buyer integrated inventory system. Burns et al. (1985), Banerjee and Burton (1994), Lu (1995), Yang and Wee (2002), Shen et al. (2003), Chan and Kingsman (2007), Hoque (2008, 2011a, 2011b), Zavanella and Zanoni (2009), Darwish and Odah (2010), Kang and Kim (2010) have taken into account inventory and transportation cost in the integrated single-vendor multi-buyer system. Ben-khedher and Yano (1994) studied the problem of scheduling the delivery of multiple items from a single supplier to a manufacturer. Then, they proposed a heuristic solution approach to minimize the sum of the transportation and the inventory costs. Cetinkaya and Lee (2000) presented an analytical model for coordinating the inventory and the transportation decision in a vendor-managed inventory system. Chan et al. (2002) proposed a model to design simple inventory policies and transportation strategies to satisfy time-varying demands over a finite time horizon, while minimizing the system wide cost by taking advantage of quantity discounts in the transportation cost structure. Shu et al. (2005) studied stochastic transportation-inventory network design problem involving one supplier and multiple retailers. Berman and Wang (2006) considered the problem of selecting the appropriate distribution strategy for delivering a family of products from a set of suppliers to a set of plants so that the total transportation and inventory costs are minimized. Ertogral et al. (2007) incorporated the transportation cost explicitly into a model and developed optimal solution procedures for solving the integrated model. Kutanoglu and Lohiya (2008) presented an optimization-based model to gain insights into the integrated inventory and transportation problem for a singleechelon, multi-facility service parts logistics system with time-based service level constraints. Christoph (2011) focused on a single buyer sourcing a single product from a pool of heterogeneous suppliers. The author tackled the supplier selection and lot size decision with the objective of minimizing the total system cost of inventory, transportation, setup and ordering. Janeiro et al. (2013) proposed a new cost allocation rule for inventory transportation systems. Glock and Kim (2014) studied shipment consolidation in multiple vendors and a single buyer integrated inventory model. In developing the model, the buyer was assumed to consolidate deliveries by assigning vendors to groups to reduce transportation and handling costs. 5 Researchers have given considerable attention to the single-supplier single-buyer, single-suppliermulti-buyer and multi-supplier-single-buyer systems involving the joint inventory and transportation cost. Although the multi-source multi-destination system has received some attention, only Liu (2003) developed a method to find both the lower and the upper bounds of the minimal total costs of transporting a homogeneous product in this system with variable supply and demand quantities. However, we demonstrate here that Liu (2003)’s method is unable to provide the exact upper minimal total cost bound solution. Therefore, we intend to develop a better heuristic solution technique to find the upper minimal total cost bound to the problem along with a heuristic solution technique to obtain the lower minimal cost bound. In developing the heuristics, we have proved theoretically that it is possible to obtain the best upper minimal total cost bound by reduction in any pair of supplier-buyer’s supply-demand quantities by the same integral amount. Analogously, in case of finding the lower minimal total cost bound, it can be proved that it is possible to obtain the best lower minimal total cost bound by increment in any pair of supplier-buyer’s supply-demand quantities by the same integral amount. Based on these notions, two heuristic based algorithms for finding the lower and the upper minimal total cost bounds are developed separately, and coded in MATLAB 7.10. Algorithm 1 in finding the upper minimal total cost bound provides better results than Liu(2003)’s approach for small-size studied numerical problems, whereas Algorithm 2 in finding the lower minimal total cost bound provides the same result as found by Liu (2003). However, Algorithm 1 does not perform well on one of the large-size studied numerical problems. Although inventory costs during transportation and at destinations seem to be an important factor in deciding the minimal total cost bounds of the problem, Liu (2003) has not taken it into account in his model. We extend his model to include these inventory costs along with the transportation cost. The solution technique to the extended model is illustrated with numerical problems. Then we use the developed heuristic techniques to find the lower and the upper minimal total cost bounds to the extended model. Benefit of this extended model is shown by reasonable changes in the minimal total cost bounds for some numerical problems, obtained following Liu’ (2003) model. The remainder of this paper is organized as follows: Section 2 deals with the description of Liu’s (2003) model and its extension. Deficiency of his model is demonstrated in this section. In section 3 developments of two heuristic solution approaches are proposed to find the lower and the upper minimal total cost bounds. In section 4 the proposed heuristic technique in finding upper minimal total cost bound is compared to Liu’s (2003) approach. Conclusion by highlighting the limitations and future research scope on the topic is drawn in section 5. 6 2. Liu’s (2003) model and its extension 2.1 The implicit assumptions and notations used by Liu (2003) Assumptions (i) A homogeneous product is transported from multi-supplier to multi-buyer. (ii) Demand of the product is deterministic integer and variable over time. (iii) The total supply is an integer and must be greater than or equal to the total demand (iv) Each of the buyers has enough storage capacity to accommodate the required inventory. (v) A transport vehicle is available to transport required shipment quantities. Notation m Total number of supply nodes (suppliers); n Total number of demand nodes (buyers); ∧ si Supply quantity (in units) from i th supplier; ∧ dj The demand (in units) per unit time to j th buyer; ci j Unit transportation cost from i th supplier to j th buyer; Si Lower bound of supplies from the i th supplier; − − Si Upper bound of supplies from the i th supplier; Dj Lower bound of demands of the j th buyer; − − Dj Upper bound of demands of the j th buyer; xi j Number of units transported from i th supplier to j th buyer. 2.2 Liu’s (2003) model and its solution − Liu (2003) dealt with finding respectively the lower bound, Z and the upper bound, Z of the − minimal total transportation costs of transporting a homogenous product from multi-source to multi-destination with varying supplies and demands within the ranges, ∧ − ∧ − S i ≤ s i ≤ S i ; ∀ i = 1, 2, ..., m and D j ≤ d j ≤ D j ∀ j = 1, 2, ..., n as follows: − − 7 m − x ∑x ij xi j i =1 j =1 ∧ n Subject to n ∑∑ c Z = Min ≤ si ; i =1, 2, ..., m ij j =1 ∧ m ∑x = d j ; j = 1, 2, ... , n ij (1) i =1 m ∧ ∑ n ∧ ∑d si ≥ i =1 xi j ≥ − Z = Max − v, w 0, ∀ i, j . n m ∧ ∑ j j =1 s i vi + i =1 ∧ ∑d wj j j =1 Subject to − vi + w j ≤ ci j , i = 1, 2, ... , m , m ∧ ∑s ≥ i i =1 j = 1, 2, ... , n , ∧ n ∑d (2) j j =1 vi ≥ 0 , w j is unrestricted in sign ∀ i, j . Then the author used LINGO solver to solve the mathematical models (1) and (2) for obtaining the lower and the upper minimal total transportation cost bounds. 2.3 Demonstration of the deficiency in Liu’s (2003) method Here we demonstrate that the Liu’s approach does not provide the exact upper minimal total cost bound all the time. For the 2-supplier 3-buyer numerical problem, where − − − D1 = 45 , D1 = 90 ; D2 = 30 , D2 = 60 ; D3 = 60 , D3 = 120 ; − − − − − S 1 = 60 , S 1 = 120 ; S 2 = 75 , S 2 = 150 ; − − C11 = 15 , C12 = 90 , C13 = 88 ; C 21 = 75 , C 22 = 80 , C 23 = 8 ; − the upper minimal total transportation cost bound ( z ) obtained by applying Liu’s approach is 7410 ∧ s1 and ∧ = 120, s2 ∧ occurs = 150, but with s 1 = 60, ∧ s2 ∧ d1 x 11 = 9 0 , x 12 = 30, x 13 = 0 , x 21 = 0 , x 22 = 30, x 23 = 120 at = 90 , = 150, ∧ d1 ∧ d2 with ∧ = 60, and d 3 = 120. However, for the same numerical problem = 90, ∧ d2 = 60, and ∧ d3 = 60, LINGO solver leads to the minimal total transportation cost, 8430 along with x11 = 6 0 , x12 = 0, x13 = 0 and x 21 = 30 , x 22 = 60, x 23 = 60. Thus, Liu’s approach misses certain supply and demand parameter values within their respective ranges for which the higher upper total transportation cost bound solution could be found. This is 8 demonstrated by placing the results of four 2-suppliers 3-buyers numerical problems in Table 1. For these numerical problems, the lower and the upper bounds of the demands and the supply quantities are the same as given previously. The concerned values of ci,j are given in Table 1. Table 1: Demonstration of the deficiency of Liu’s (2003) approach with 4 numerical problems Ex. No. Minimal T. cost solution with different s∧ & d∧ by Lingo Upper bound solutions by Liu’s (2003) approach Values for Ci j i j Upper minimal T.cost bound Minimal T.cost − ( z) C 11 = 25 , C 12 = 95 , 1 C 22 = 85 , C 23 = 15 . x21 = 0 , x22 = 30, x23 =120 , ∧ S 1 =120 , ∧ C 11 = 15 , C 12 = 10 , x21 = 0 , x22 = 0, x23 =105 ∧ ∧ S 1 =120, S 2 =105, ∧ d1 ∧ = 90 , d 2 = 60 , d 3 =120. C 22 = 10 , C 23 = 40 . x21 = 90 , x22 = 60, x23 = 0 , ∧ S 1 =120, ∧ C 11 = 11 , C 12 = 50 , C 13 = 10 , C 21 = 55 , C 22 = 79 , C 23 = 80 . ∧ C 11 = 11 , C 12 = 50 , C 22 = 25 , C 23 = 40 . , d1 10890 ∧ ∧ ∧ d1 4725 ∧ ∧ 11565 S 2 =150, ∧ ∧ = 45 , d 2 = 60 , d 3 =120. x11 = 60 , x12 = 0, x13 = 0 , x21 = 30 , x22 = 30, x23 = 90 x21 = 0 , x22 = 30, x23 =120 , S 2 =150, ∧ d1 x11 = 90 , x12 = 0, x13 = 0 , ∧ ∧ x11 = 0 , x12 = 0, x13 = 75 , x21 = 45 , x22 = 60, x23 = 45 ∧ ∧ ∧ S 2 =150, = 45, d 2 = 45, d 3 =120. S 1 = 75 , = 90, d 2 = 60 , d 3 =120. S 1 = 90 , = 60 , 9870 ∧ d 3 =120. x21 = 45 , x22 = 45, x23 = 60 ∧ S 1 = 60 , ∧ ∧ x11 = 0 , x12 = 0, x13 =120 , ∧ C 13 = 10 , C 21 = 55 , 4350 = 90, d 2 = 60 , d 3 =120. x21 = 90 , x22 = 60, x23 = 0 ∧ ∧ S 1 =120, S 2 =150, = 45, ∧ d2 x11 = 0 , x12 = 0, x13 = 60 , ∧ S 2 =150, ∧ d1 4 9450 x11 = 0 , x12 = 0, x13 =120 , C 13 = 20 , C 21 = 15 , d1 3 ∧ S 2 =150, ∧ d1 2 x11 = 45 , x12 = 60, x13 =15 , x11 = 9 0 , x12 = 30, x13 = 0 , C 13 = 98 , C 21 = 55 , 6540 ∧ = 90 , d 2 = 30 , d 3 =120. ∧ S 1 = 60 , ∧ d1 = 90 , ∧ S 2 =150, ∧ d2 = 30 , 6660 ∧ d 3 = 90. For each of the numerical problems in Table 1, the minimal total transportation cost found with different values of ∧ ∧ si & dj keeping them within their respective ranges by LINGO solver, is higher than the upper minimal total transportation cost bound obtained by Liu’s (2003) approach. These results clearly demonstrate that Liu’s approach is unable to provide the correct upper minimal total transportation cost bound. 2.4 The Extended model In extending Liu’s (2003) model we make an additional assumption as follows: Supplies from suppliers reach to a buyer at the same time Additional notations: ti hj j Transportation time from ith supplier to jth buyer Holding cost for the jth buyer per unit per unit time 9 Total cost = transportation cost + inventory cost during transportation + inventory cost at the n m buyers. Inventory cost during transportation is given by ∑∑ x ij ti j h j i =1 j =1 Inventory of the product created at the buyer j is given by the area of the triangle below: Quantity m ∑x ij i =1 Time m ∑x ij /dj =1 i =1 Fig 1. Inventory at the jth buyer during a cycle Thus the inventory cost at the buyer j is 1 m ∑ xi j h j , and hence the total inventory cost at all 2 i =1 n m buyers is given by 1 ∑∑ xi j h j . 2 j =1 i =1 So, the extended objective function becomes m n ∑∑ c i =1 j =1 ij m n x 2 ∑∑ xi j + 1 i =1 j =1 ij hj + m m n ∑∑ x ij ti j h j n ∑∑ [c ij + h j / 2 + ti j h j ] xi j i =1 j =1 i =1 j =1 m = n = ∑ ∑ A i j xi j ; where, Ai j = ci j + h j / 2 + ti j h j . i =1 j =1 So, our extended model is to minimize the modified objective function subject to the constraints in Liu’s model. Note that this extended model transforms to Liu’s model when ti j = 0, h j = 0. 3. Development of the heuristic solution techniques to the extended model to find the upper and the lower minimal total cost bounds It has been proved in Appendix A that if the total supply of the suppliers is higher than the total demand of the buyers, then it is possible that the minimal total cost obtained by reducing both the supply and the demand of any pair of a supplier and a buyer by the same integral amount is greater than or equal to the minimal total cost obtained by reducing them by different integral amounts. Based on this notion, the heuristic solution technique (Algorithm 1) for finding the upper minimal total cost bound is described below. 10 Algorithm 1: A Heuristic to the extended model in finding the upper minimal total cost bound. Step 1: ∧ − ∧ Initially, by setting s i = S i − dj = Dj & ; i = 1, 2, . . . , m j = 1, 2, . . . , n , & the minimal total cost "C 00 " of the extended transportation model is calculated. Then set UB (temp) : = C 00 and set − X (temp) : = (X ) m x n which is the upper bound solution associated with C 00 . Step 2: Set i = 1 & j = 1 . Step 3: Set k = 1 . Step 4: If both s i ≠ S i ∧ ∧ d j ≠ D j , then reduce both & − ∧ ∧ by 1 and solve the & dj si − resulted mathematical model and find the minimal total cost Ci j k (say) & its corresponding solutions ⎛ ∧ Else, If ⎜⎜ s i = S i ⎝ & − (X ij k )m x n (say); and go to step 5. ∧ ⎞ d j ≠ D j ⎟⎟ or − ⎠ ⎛∧ ⎜⎜ s i = S i − ⎝ ∧ ⎞ d j = D j ⎟⎟ , then − ⎠ & go to step 10 Else go to step 9. Step 5: If k < Step 6: Find ⎧⎛ − ⎞ ⎫ ⎛ − ⎞ min ⎨ ⎜ S i − S i ⎟ , ⎜ D j − D j ⎟ ⎬, then set − ⎠ − ⎠ ⎭ ⎝ ⎩⎝ ⎧ ⎧⎛ − ⎞ ⎞ ⎛ − ⎪ ⎪ C i j = Max ⎨C i j k : k = 1, 2, 3, ... , min ⎨⎜ S i − Si ⎟ , ⎜ D j − D j ⎟ ⎜ ⎟ ⎜ ⎟ ⎪⎩⎝ ⎪ − ⎠ ⎝ − ⎠ ⎩ (X ) i j mxn ai & b j Step 7: its corresponding solutions ; and the corresponding supply and the demand parameter values as respectively. ∧ − s i = a i (< S i ) & ( ) X (temp) = X i j mxn ∧ If both s i ≠ S i − Step 9: and go to step 4 If Ci j > UB (temp), then set UB (temp) := Ci j ; set Step 8: ⎫⎫ ⎪⎪ ⎬⎬ & ⎪⎭⎪ ⎭ k = k +1 ∧ − d j = b j ( < D j ) where a i & b j are associated with Ci j ; set which is the upper bound solutions associated with Ci j . ∧ & d j ≠ D j , then go to the next step. Else go to step 4. − If j < n then set j = j + 1 and go to step 3. Step 10: If i < m then set i = i + 1 and go to step 3. 11 Step 11: Stop. Find the current UB (temp) corresponding as ∧ X (temp) as the upper minimal total cost bound, the its solution, ∧ s i & d j ; i = 1, 2, . . . , m & j = 1, 2, . . . , n and the corresponding as its supply and demand quantities. Note that this algorithm becomes an algorithm to Liu’s model when ti j = 0, h j = 0. Similarly, in finding the lower minimal total cost bound it can be proved that if the total supply of the suppliers is higher than the total demand of the buyers, then it is possible that the minimal total cost obtained by increasing both the supply and the demand of any pair of a supplier and a buyer by the same integral amount is less than or equal to the minimal total cost obtained by increasing them by different integral amounts. This can easily be proved by following the proof of theorem 1 shown in the appendix A. Based on this notion, the heuristic solution technique (Algorithm 2) for finding the lower minimal total cost bound is described below. Algorithm 2: A heuristic to the extended model in finding the lower minimal total cost bound. Step 1: Initially, by setting ∧ si = Si ∧ dj = Dj & − ; i = 1, 2, . . . , m j = 1, 2, . . . , n , the & − minimal total cost " D 00 " of the extended transportation model is calculated. Set LB (temp) = D 00 and set associated with D 00 X (temp) : = ( X ) m x n − which is the lower bound solution . i = 1 & j = 1. Step 2: Set Step 3: Set k = 1. Step 4: If both s i ≠ S i ∧ − ∧ − d j ≠ D j , then increase both & ∧ si ∧ & dj by 1 and solve the resulted mathematical model and find the minimal total cost Ci j k (say) & its corresponding solutions Else, If (X ij k )m x n (say) ; and go to step 5. − ∧ − ⎞ − ⎛∧ ⎛∧ ⎜ s i = S i & d j ≠ D j ⎟ or ⎜ s i = S i ⎟ ⎜ ⎜ ⎠ ⎝ ⎝ ∧ − ⎞ & d j = D j ⎟⎟ ⎠ then go to step 10 Else go to step 9. Step 5: ⎧ − ⎞ ⎫ ⎛ − If k < min ⎨ ⎛⎜ S i − S i ⎞⎟ , ⎜ D j − D j ⎟ ⎬, then set k = k + 1 and go to step 4. − ⎠ − ⎠ ⎭ ⎝ ⎩⎝ 12 Step 6: Find ⎧ ⎧⎛ − ⎞ ⎞ ⎛ − ⎪ ⎪ C i j = Min ⎨C i j k : k = 1, 2, 3, ... , min ⎨⎜ S i − Si ⎟ , ⎜ D j − D j ⎟ ⎟ ⎜ ⎜ ⎟ ⎪ ⎪⎩⎝ − ⎠ ⎝ − ⎠ ⎩ ⎫⎫ ⎪⎪ ⎬⎬ ⎪⎭⎪ ⎭ & its corresponding solution (X i j ) m x n ; and the corresponding supply and demand parameter values as respectively. pi & q j Step 7: If Ci j < LB (temp) , then set LB (temp) := Ci j ; ∧ ∧ si = pi ( > Si ) & d j = q j ( > D j ) where p i & q j are − ( ) X (temp) = X i j mxn ∧ associated with Ci j ; − − which is the lower bound solution associated with Ci j . ∧ si ≠ Si − & dj ≠ D j then go to the next step. Else go to step 4. Step 8: If both Step 9: If j < n then set j = j + 1 and go to step 3. Step 10: If i < m then set i = i + 1 and go to step 3. Step 11: If the current LB (temp) ≠ D 00 then go to step 19. Else go to the next step. Step 12: Set i = 1 . Step 13: Set g = 1 . Step 14: By keeping ∧ dj = D j ∀ j as fixed, increase ∧ si = Si mathematical model and find the minimal total cost ⎞ ⎛ − g < ⎜ Si − Si⎟ ⎟ ⎜ − ⎠ ⎝ Step 15: If Step 16: Find then set g = g +1 with C i ; ⎫ ⎪ ⎬ ⎪⎭ ( ) (say). & its corresponding solution (X i ) m x n . then set LB (temp) := C i ; X (temp) = X i mxn Ci g and go to step 14. ⎧ ⎞ ⎛ − ⎪ C i = Min⎨C i g : g = 1, 2, 3, ... , ⎜ S i − S i ⎟ ⎟ ⎜ ⎪⎩ − ⎠ ⎝ Step 17: If C i < LB (temp), by 1 and solve the resulted − − ∧ s i = h i ( > S i ) where h i is associated − which is the lower bound solution associated with C i . Step 18: If i < m then set i = i + 1 and go to step 13. Step 19: Stop. Take the current corresponding ∧ X (temp) LB (temp) as as the lower minimal total cost bound, the its solution, and the ∧ corresponding s i & d j ; i = 1, 2, . . . , m & j = 1, 2, . . . , n as its supply and demand quantities. Note that this algorithm becomes an algorithm to Liu’s model when ti j = 0, h j = 0. 13 4. Numerical Studies Following the Algorithm 1 developed in the previous section a MATLAB computational program for finding the upper minimal total cost bound of the Transportation Problem with varying demand and supply (TPVDS) is developed, which is given in Appendix B. Utilizing the provided MATLAB computational program in Appendix B with the repeatedly called function “Solve_LP(a,b,A,B,f,lb) ” designed as in Appendix C, we find the upper minimal total cost bound solutions by this computer program to all the problems in Tables 2 and 3. 4.1 A comparative study of the Liu’s approach and the technique developed here The 2-supplier 3-buyer numerical problem originally solved by Liu (2003) is again solved by following the algorithms developed in this paper. The lower and the upper bounds are found to be the same as obtained by Liu. However, we find the upper bounds of the minimal total costs with varying supply and demand quantities for additional eight 2-suppliers 3-buyers numerical problems, by applying Algorithm 1 of this paper and the Liu’s (2003) approach. For these numerical problems, the lower and the upper bounds of the demands and the supply quantities are the same as given previously in section 2.3. In all cases, our method leads to the significantly higher upper bound solutions. Comparative upper bound solutions along with the concerned values of ci,j are given in Table 2. In the table the difference between the upper bounds obtained by our method and the corresponding one by Liu (2003), is expressed as a percentage of the Liu’s one. This is shown at the right hand column of Table 2. 14 Table 2: Comparative results of eight numerical problems obtained by Algorithm 1 and Liu’s approach Upper bound solutions by Liu’s approach Ex. No . Values for Ci j C 11 = 15 , C 12 = 90 , 1 C 13 = 88 , C 21 = 75 , C 22 = 80 , C 23 = 8 . x11 = 9 0 , x12 = 30, x13 = 0 , x21 = 0 , x22 = 30, x23 =120 ∧ 8.60 11565 6.20 8730 25.4 10725 19.20 7725 77.60 12630 16.00 8010 9.90 ∧ S 1 = 60 , ∧ ∧ ∧ S 2 =150 , ∧ d1 ∧ = 90 , d 2 = 60 , d 3 = 60. x11 = 0 , x12 = 0, x13 =120 , x11 = 0 , x12 = 0, x13 = 60 , x21 = 90 , x22 = 60, x23 = 0 , x21 = 45 , x22 = 45, x23 = 60 C 22 = 10 , C 23 = 40 . ∧ ∧ S 1 =120 , ∧ x21 = 90 , x22 = 60, x23 = 0 , C 13 = 50 , C 21 = 85 , C 22 = 80 , C 23 = 8 . ∧ ∧ S 1 =120, ∧ ∧ x11 = 9 0 , x12 = 30, x13 = 0 , x21 = 0 , x22 = 30, x23 =120 , ∧ ∧ S 1 =120 , ∧ x11 = 0 , x12 = 0, x13 =120 , x21 = 90 , x22 = 60, x23 = 0 , ∧ ∧ S 1 =120, ∧ ∧ 4350 ∧ ∧ ∧ x11 = 0 , x12 = 0, x13 = 60 , x21 = 45 , x22 = 45, x23 = 60 ∧ ∧ ∧ S 2 = 75, = 45, d 2 = 30 , d 3 =120. S 1 = 60 , = 90 , d 2 = 60 , d 3 =120. ∧ = 90 , d 2 = 60 , d 3 = 60. ∧ d1 S 2 =150, ∧ d1 ∧ S 2 =150, x11 = 45 , x12 = 30, x13 = 45 , x21 = 0 , x22 = 0, x23 = 75 ∧ ∧ C 13 = 20 , C 21 = 15 , ∧ S 1 =120, = 90 , d 2 = 60 , d 3 =120. C 11 = 30 , C 12 = 40 , ∧ x11 = 6 0 , x12 = 0, x13 = 0 , x21 = 30 , x22 = 60, x23 = 60 ∧ S 2 =150, ∧ d1 ∧ d1 9000 ∧ S 2 =150, = 45, d 2 = 60 , d 3 =120. S 1 = 60 , ∧ C 13 = 125 , C 21 = 20 , C 22 = 10 , C 23 = 90 . 6960 = 90 , d 2 = 60 , d 3 =120. C 11 = 25 , C 12 = 95 , ∧ d1 S 2 =150, ∧ x11 = 0 , x12 = 0, x13 = 75 , x21 = 45 , x22 = 60, x23 = 45 ∧ ∧ ∧ S 1 =120, ∧ S 1 = 75 , = 90, d 2 = 60, d 3 =120. x11 = 9 0 , x12 = 30, x13 = 0 , x21 = 0 , x22 = 30, x23 =120 ∧ 10890 ∧ S 2 =150, = 45, d 2 = 45, d 3 =120. d1 S 2 =150, ∧ d1 C 22 = 70 , C 23 = 15 . ∧ ∧ x11 = 0 , x12 = 0, x13 =120 , ∧ S 1 = 60 , = 90 , d 2 = 60 , d 3 =120. C 13 = 10 , C 21 = 55 , C 22 = 79 , C 23 = 80 . 4350 S 2 =150, C 11 = 11 , C 12 = 50 , C 11 = 15 , C 12 = 75 , ∧ S 2 =150, ∧ d1 ∧ = 45, d 2 = 45, d 3 =120. C 11 = 38 , C 12 = 80 , x11 = 0 , x12 = 0, x13 =120 , x11 = 0 , x12 = 0, x13 = 60 , C 13 = 10 , C 21 = 55 , x21 = 90 , x22 = 60, x23 = 0 , x21 = 45, x22 = 45, x23 = 60 C 22 = 79 , C 23 = 100 . ∧ ∧ S 1 =120 , ∧ d1 8 4725 x11 = 6 0 , x12 = 0, x13 = 0 , x21 = 30 , x22 = 60, x23 = 60 C 13 = 20 , C 21 = 15 , ∧ 7 13.80 Upper minimal T.cost bound C 11 = 15 , C 12 = 10 , d1 6 8430 7410 S 2 =150, ∧ ∧ 5 − ( z) = 90 , d 2 = 60 , d 3 =120. d1 4 ∧ S 1 =120, ∧ 3 − ( z) % Increa se in the Upper Bound Upper minimal T.cost bound d1 2 Upper bound solutions by Algorithm 1 10890 ∧ ∧ S 1 = 60 , S 2 =150, ∧ ∧ = 90 , d 2 = 60 , d 3 =120. d1 ∧ S 2 =150, ∧ ∧ = 45, d 2 = 45, d 3 =120. C 11 = 11 , C 12 = 25 , x11 = 90 , x12 = 30, x13 = 0 , x11 = 60, x12 = 0, x13 = 0 , C 13 = 45 , C 21 = 115 , x21 = 0 , x22 = 30, x23 =120 , x21 = 30 , x22 = 60, x23 = 60 C 22 = 25 , C 23 = 40 . S 1 =120 , ∧ ∧ d1 ∧ S 2 =150, ∧ ∧ = 90 , d 2 = 60 , d 3 =120. 7290 ∧ S 1 = 60 , ∧ d1 ∧ S 2 =150, ∧ ∧ = 90 , d 2 = 60 , d 3 = 60. The right hand column of Table 2 clearly demonstrates that the Algorithm 1 in this paper leads to the significant percentage increase in the upper bound of the minimal total cost in each case. However, the lower bounds of the minimal total costs for each of these eight numerical problems found by applying Algorithm 2 and the Liu’s approach are found to be the same. Comparative 15 upper bounds of the minimal total transportation costs obtained by Algorithm 1 and Liu’s approach are depicted by bar charts in Fig 2. Fig 2. Bar chart plot of upper minimal total transportation cost bound vs. numerical examples in case of solving by Liu’s approach and Algorithm 1. The bar charts in Fig 2 clearly show that the Algorithm 1 leads to higher upper bound of the minimal total transportation costs. Thus the Algorithm 1 performs much better than Liu’s (2003) approach in getting upper bound of the minimal total transportation costs with varying supply and demand quantities for the studied numerical problems. In addition, both the Algorithm1 and the Liu’s approach were applied to many problem instances. In all cases, Algorithm1 either leads to the same upper bound solution as the corresponding one obtained by Liu’s approach or better than that. But, the Algorithm 2 performs the same as Liu’s approach in all those cases. To evaluate the performance of the developed heuristics in this paper in obtaining the upper bound minimal total cost solution to large size problems, here we also carry out comparative study of our heuristic method with Liu’s one on the upper total cost bounds found for four large size numerical problems. Data for these numerical problems are given in Appendix D and the comparative results are presented in Table 3. 16 Table 3: Comparative upper bound solutions of 4 large size numerical problems obtained by Algorithm 1 and Liu’s approach Upper bound solutions by Liu’s approach Ex. No. Problem Size (m x n) Upper bound solutions by Algorithm 1 Upper minimal T.cost bound Upper minimal T.cost bound − ( z) 1 5 x 10 x12 = 20 , x17 = 40, x21 = 40 , x23 = 60 , x210 = 20 , x46 = 20 , x49 =120, x54 = 80 , x55 = 40 , x58 = 60 , x12 = 20 , x17 = 40, x21 = 40 , x23 = 60 , x210 = 20 , x46 = 20 , x49 =120, x54 = 80 , x55 = 40 , x58 = 60 , [ S i ] 5 x 1 = [400 ; 500 ; 600 ; 300 ; 800 ] 300 ; 800 ] 0.00 5920 0.00 5320 0.00 3840 - 9.4 ^ [d j ] 1 x 10 = [40 20 60 80 [d j ] 1 x 10 = [40 20 60 80 40 20 40 60 120 20 ] 40 20 40 60 120 20 ] x12 = 20 , x17 = 40, x21 = 40 , x23 = 32 , x210 = 20 , x46 = 20 , x49 =120 , x54 = 80 , x78 = 60 , x85 = 40 , x93 = 28, x12 = 20, x17 = 40, x21 = 40, x23 = 60, x210 = 20, x46 = 20, x49 =120, x54 = 80, x78 = 60, x85 = 40, 7520 ^ [ S i ] 5 x 1 = [400 ; 500 ; 600 ; ^ 10 x 10 ( z) 7520 ^ 2 − % Increa se in the Upper Bound 5920 ^ ^ [ S i ] 10 x 1 = [400 ; 500 ; 600 ; [S i ] 10 x 1 = [800;1000;1200; 300 ; 800 ; 400 ; 500 ; 600 ; 300 ; 600;1600; 800; 1000;1200; 600; 1600] 800 ] ^ [d j ] 1 x 10 = [40 20 60 80 ^ [d j ] 1 x 10 = [40 20 60 80 40 20 40 60 120 20 ] 40 20 40 60 120 20 ] x12 = 20 , x17 = 40, x21 = 40 , x23 = 32 , x210 = 20 , x46 = 20 , x49 =120 , x54 = 80 , x78 = 60 , x85 = 40 , x93 = 28, x12 = 20 , x17 = 40, x 21 = 40 , x 23 = 60 , x 210 = 20 , x 46 = 20 , 3 10 x 10 x 49 = 120 , x54 = 80 , x 78 = 60 , 5320 x85 = 40 , ^ [S i ] 10 x 1 = [800;1000;1200; ^ [S i ] 10 x 1 = [400 ; 500 ; 600 ; 600;1600; 800; 1000;1200; 600; 1600] 300 ; 800 ; 400 ; 500 ; 600 ; 300 ; 800 ] ^ [d j ] 1 x 10 = [40 20 60 80 ^ [d j ] 1 x 10 = [40 20 60 80 40 20 40 60 120 20 ] 40 20 40 60 120 20 ] 4 10 x 10 x12 = 20 , x21 = 40 , x23 = 20 , x29 = 45 , x29 = 45 , x210 = 20 , x36 = 20 , x49 = 75 , x54 = 80 , x78 = 60 , x85 = 40 , x87 = 40 , x93 = 40 , ^ [ S i ] 10 x 1 = [100 ; 125 ; 150 ; 75 ; 200 ; 100 ; 125 ; 150 ; 75 ; 200 ] 4240 x12 = 20 , x21 = 40, x23 = 40 , x210 = 20 , x46 = 20 , x49 =120 , x54 = 80 , x78 = 60 , x85 = 40 , x87 = 40 , x93 = 20 , ^ [S i ] 10 x 1 = [200; 250; 300; 150; 400; 200; 250; 300;150; 400 ] ^ [d j ] 1 x 10 = [40 20 60 80 ^ [d j ] 1 x 10 = [40 20 60 80 40 20 40 60 120 20 ] 40 20 40 60 120 20 ] 17 It can easily be seen from Table 3 that the upper bound solutions to each of the problems 1, 2 and 3 found by both the methods are the same, whereas the solution to problem 4 found by our method is inferior to the corresponding solution obtained by Liu’s approach. However, we have solved many large- size numerical problems by both the methods and the obtained solutions are found to be the same in all cases except problem 4 in Table 3. It should be pointed out here that the same lower bound of the minimal total costs for each of the numerical problems is found by Algorithm 2 of this paper and Liu’s approach. 4.2 Comparative Study of the extended model and the extended model without including inventories 4.2.1 Significant changes in the minimal total cost bounds To show the benefit of the extended model, here we solve both the extended model and the extended model without including inventories on many numerical problems following Algorithms 1 & 2. For each of the numerical problems, the minimal total cost bounds for the extended model are found to vary from the original one up to a certain range. These variations are shown for the three numerical problems in Table 4. Data for these numerical problems are as follows: Numerical problems 1: Data for 2-suppliers 3-buyers numerical problem. ci,j and the lower and the upper bounds of the demands and the supply quantities are the same as given in the Liu’s (2003) paper. Numerical problems 2: Data for 3-supplier 4-buyer numerical problem − − − − D1 = 45 , D1 = 90 ; D2 = 40 , D2 = 80 ; D3 = 60 , D3 = 120 ; D4 = 90 , D4 = 160 ; − − − − − − − S 1 = 60 , S 1 = 120 ; S 2 = 75 , S 2 = 150 ; S 3 = 100 , S 3 = 180 ; − − − C11 = 4 , C21 = 6 , C31 = 5 ; C12 = 5 , C22 = 6 , C32 = 6 ; C13 = 8 , C23 = 10 , C33 = 5 ; C14 = 8 , C24 = 7 , C34 = 8 ; Numerical problems 3: Data for 2-suppliers 3-buyers numerical problem. C11 =15 , C12 =10 , C13 = 20 ; C21 =15 , C22 =10 , C23 = 40 ; the lower and the upper bounds of the demands and the supply quantities are the same as given in the Liu’s (2003) paper. 18 Table 4: Comparative results of numerical problems found for the extended model and the model without including inventories The Extended Model Ex. No. Values for Lower bound solutions by Algorithm 2 h j & ti j Upper bound solutions by Algorithm 1 Lower minimal T.cost bound (Z ) Upper minima l T.cost bonnd − ( z) − 1 h j = 0 ; j = 1, 2 , 3 t ij = 0 ; i = 1, 2 & j = 1, 2, 3 x11 = 0 , x12 = 0, x13 = 60 , ∧ S 1 = 60 , ∧ d1 ∧ 1575 x11 = 0 , x12 = 0, x13 = 60 , x21 = 45 , x22 = 30, x23 = 0 , t 12 = 0.2 , t 13 = 0.9, ∧ S 1 = 60 , d1 1988.55 d1 h j = 0 ; j = 1, 2 , 3 , 4 x11 = 45 , x12 = 40, x13 = 0, x14 = 0, x21 = 0 , x22 = 0, x23 = 0, x24 = 90, x31 = 0 , x32 = 0, x33 = 60, x34 = 0. ∧ ∧ S 1 =120, ∧ d1 ∧ 1310 ∧ = 45, d 2 = 40, d 3 = 60 , t 13 = 0.03, t 14 = 0.3 , t 21 = 0.4 , t 22 = 0.3 , t 23 = 0.6 , t 24 = 0.9 , t 31 = 0.3 , t 32 = 0.07 , x21 = 0 , x22 = 0, x23 = 0, x24 = 0, x31 = 0 , x32 = 0, x33 = 60, x34 = 90 . ∧ ∧ S 1 =120, ∧ d1 = 45, 1725.87 ∧ = 40, ∧ ∧ d 3 = 60 , d 4 ∧ ∧ ∧ ∧ = 90 , d 2 = 80 , d 3 =120, d1 2540 ∧ d 4 =160 x11 = 90 , x12 = 0, x13 = 0, x14 = 30, x21 = 0 , x22 = 80, x23 = 0, x24 = 70, x31 = 0 , x32 = 0, x33 =120, x34 = 60 . ∧ = 90 ∧ S 2 =150, S 3 =180 ∧ S 1 =120, S 2 = 75, S 3 =180 ∧ d2 ∧ ∧ d1 4947.4 = 45 , d 2 = 45 , d 3 =120. x11 = 68 , x12 = 52, x13 = 0, x14 = 0, x21 = 0 , x22 = 0, x23 = 0, x24 =150, x31 = 22 , x32 = 28, x33 =120, x34 = 10 . ∧ ∧ d 4 = 90 h 1 =1.1 , h 2 = 1.15 , h 3 = 1.25 , x11 = 45 , x12 = 40, x13 = 0, x14 = 0, h 4 = 4.5 ; t 11 = 0.07 , t 12 = 0.6 , ∧ ∧ S 2 =150, S 1 =120, S 2 =150, S 3 =100 ∧ ∧ x21 = 45 , x22 = 45, x23 = 60 , ∧ ∧ = 45, d 2 = 30 , d 3 = 60. d1 & j = 1, 2, 3 , 4 4350 = 45 , d 2 = 45 , d 3 =120. ∧ S 1 = 60 , t 23 = 0.6 ; t ij = 0 ; i = 1, 2 , 3 ∧ S 2 =150, x11 = 0 , x12 = 0, x13 = 60 , ∧ S 2 = 75, ∧ ∧ S 1 = 60 , ∧ ∧ = 45, d 2 = 30 , d 3 = 60. h 3 = 1.95 ; t 11 = 0.1, ∧ x21 = 45 , x22 = 45, x23 = 60 , ∧ S 2 = 75, h 1 =1.55 , h 2 = 1.75 , t 21 = 1.5 , t 22 = 1.6 , 2 x11 = 0 , x12 = 0, x13 = 60 , x21 = 45 , x22 = 30, x23 = 0 , 3547.73 ∧ ∧ S 2 =150, S 3 =180 ∧ ∧ = 90, d 2 = 80, d 3 =120, ∧ d 4 =160 t 33 = 0.04 , t 34 = 0.01; 3 h j = 0 ; j = 1, 2 , 3 x11 = 0 , x12 = 0, x13 = 60 , t ij = 0 ; i = 1, 2 x 21 = 45 , x 22 = 30, x 23 = 0 , & j = 1, 2, 3 ∧ S 1 = 60 , ∧ d1 h 1 =1.1 , h 2 = 1.15 , h 3 = 1.25 ; t 11 = 0.2 , t 12 = 0.5 , t 13 = 0.8, t 21 = 3.7 , t 22 = 4. 0, t 23 = 4.3 ; 2175 ∧ ∧ d1 = 45, d 2 = 30 , d 3 = 60. x11 = 30 , x12 = 30, x13 = 60 , x 21 = 15 , x 22 = 0, x 23 = 0 , ∧ S 1 =120, ∧ d1 ∧ S 2 = 75, ∧ ∧ S 1 = 60 , ∧ S 2 = 75, ∧ x11 = 0 , x12 = 0, x13 = 60 x21 = 45 , x22 = 45, x23 = 60 ∧ = 45, d 2 = 30 , d 3 = 60. 2399.40 ∧ ∧ ∧ = 45 , d 2 = 45 , d 3 =120. x11 = 0 , x12 = 0, x13 = 60 x21 = 45 , x22 = 45, x23 = 60 ∧ S 1 = 60 , ∧ d1 4725 S 2 =150, ∧ S 2 =150, ∧ ∧ = 45, d 2 = 45, d 3 =120. 5622.40 19 As can be seen from Table 4, the lower bound of the minimal total costs obtained for the extended model increases by 413.55, 415.87 and 224.4 respectively from the corresponding one in the extended model without including inventories. The upper bound of the minimal total costs for these example problems found following the extended model also increases by 597.4, 1007.73 and 897.4 respectively from the corresponding one in the extended model without including inventories. These increases in the lower and the upper bounds of the minimal total costs are due to the inclusion of the inventory costs during transportation and at destinations with the transportation cost in the extended model. Thus these results demonstrate that the inventory cost during transportation and at destinations plays a significant role in deciding the lower and the upper bounds of the minimal total costs. 5. Conclusion Liu (2003) developed a model to transport a homogeneous product from multi-source to multidestination, to meet the demand of each destination. The total cost bounds of transportation for varying demand and supply quantities within given ranges was obtained by using LINGO solver. Here we have demonstrated a deficiency of Liu’s method in getting an upper bound of the minimal total costs of transportation. Then we extend this model to include the inventory costs of the product during transportation and at destinations. In addition, we developed two new efficient heuristic solution techniques - Algorithms 1 & 2 to find the upper and the lower minimal total cost bounds respectively. By comparative studies of the solution techniques on the solutions of small size numerical problems, it is observed that our proposed heuristic technique (Algorithm1) performs the same or significantly better in finding the upper bound of the minimal total cost as compared with Liu’s (2003) approach. However, it does not perform well in one out of many large– size numerical problems studied. Algorithm 2 provides the same lower bound of the minimal total costs to each of the numerical problems studied as the one found by Liu’s (2003) approach. Moreover, numerical studies demonstrate that the inclusion of inventory costs during transportation and at destinations with the transportation costs changes the lower and the upper minimal total cost bounds reasonably. Our technique of obtaining the upper minimal total cost bound has failed to attain the optimum upper bound in case of a large-size numerical problem out of many problems studied. Besides, our methods (of obtaining the lower and the upper bounds of the minimal total transportation costs) as heuristics provide optimal or near optimal bounds rather than the optimal bounds always. So, further research might be carried out in developing a technique to obtain the optimum upper bound of the minimal total costs. In this research, lead time of delivering an order quantity 20 is considered to be a deterministic constant value. Practically, it might not be the case. Lead time may vary because of the presence of some realistic factors like variations in loading, transportation and unloading times. Development of a model with variable lead time may produce more realistic results. So, the current model might be extended to include the variation in lead time. In our extended model we assume that supplies from suppliers reach to a buyer at the same time. In practice, it may create more inventory cost. This inventory cost can be reduced by delivering a single supply to each of the buyers at a time when the previous one finishes there. In developing the extended model, the demand is assumed to be variable. However, variation in the demand may follow a particular trend. Hence future research might be carried out in extending the extended model, to include variations in the lead time in delivering a single supply to each of the buyers when the previous one ends there, and assuming an appropriate trend of demand distribution. We intend to devote ourselves in this direction of future research. Appendix A Theorem 1: If the total supply of the suppliers is higher than the total demand of the buyers, then it is possible that the minimal total cost obtained by reducing both the supply and the demand of any pair of a supplier and a buyer by the same integral amount is greater than or equal to the minimal total cost obtained by reducing them by different integral amounts. Proof: To prove the theorem do the following changes to the constraints in the extended model. Set (i) Replace all the " ≤ " by " < " ∧ − Set (ii) s i = S i ; i = 1, 2, 3, ..., k − 1, k + 1, ... , m ⎡ ⎛ − ∧ − ⎞⎤ s k = S k − α ; α is an integer & α ∈ ⎢ 1, ⎜ S k − S k ⎟⎥ ⎟ ⎢⎣ ⎜⎝ − ⎠ ⎥⎦ ∧ Set (iii) d − j = Dj ; j = 1, 2, 3, ... , l − 1, l + 1, ... , n ⎡ ⎛ − ∧ − ⎞⎤ d l = D l − β ; β is an integer & β ∈ ⎢ 1, ⎜ D l − D l ⎟⎥ ⎟ ⎢⎣ ⎜⎝ − ⎠ ⎥⎦ m − ⎞ ⎛ − Si Set (iv) ⎜ S k − α ⎟ + ⎟ ⎜ ⎠ ⎝ i =1 ∑ i ≠ k n − ⎞ ⎛ − Dj ; > ⎜Dl − β ⎟ + ⎟ ⎜ ⎠ ⎝ j =1 Then the extended model becomes ∑ j ≠l i.e. the total supply > the total demand 21 m n ∑∑ A Min ij xi j (A.1) ; i =1, 2, ... , k −1, k + 1, ... , m (A.2) i =1 j =1 Subject to n − i j < Si ∑x j =1 m − ∑x ij = D j ; j =1, 2, ..., l − 1, l +1, ... , n (A.3) i =1 n ∑x − k j < Sk − ⎡ ⎛ − ⎞⎤ ; α is an integer & α ∈ ⎢1, ⎜ S k − S k ⎟⎥ ⎟⎥ ⎜ ⎢ − ⎠⎦ ⎣ ⎝ (A.4) ⎡ ⎛ − ⎞⎤ ; β is an integer & β ∈ ⎢1, ⎜ D l − D l ⎟⎥ ⎟ ⎢⎣ ⎜⎝ − ⎠ ⎥⎦ (A.5) α j =1 m ∑x − il = Dl − β i =1 ⎞ ⎛ − ⎜S k − α ⎟ + ⎟ ⎜ ⎠ ⎝ m − ∑S i =1 i ≠ k i ⎞ ⎛ − > ⎜⎜ D l − β ⎟⎟ + ⎠ ⎝ n − ∑D (A.6) j j =1 j ≠l xi j ≥ 0 ∀ i, j (A.7) Case (1) α > β (Reduction in the supply quantity of the k th supplier is greater than the reduction in the demand quantity of the l th buyer). Let the minimal total cost, V 1 of the objective function (A.1) subject to the constraints (A.2) to (A.7) is obtained with given α , β . Again, let the minimal total cost, W 1 of the objective function − (A.1) subject to these constraints is obtained by increasing S k − α to the other parameter values the same as before. Since − Sk − α − Sk − β , but by retaining − in (A.4) is increased to S k − β , the values of some of the variables x k 1 , x k 2 , x k 3 , ... , x k l , ... , x k n in (A.4) may or may not be increased in the latter minimal total cost solution. If the values of the variables are not increased, then the former and the latter minimal total cost solutions are the same. Now, let the values of some of the variables in (A.4) are increased. Note that the variables x k 1 , x k 2 , x k 3 , ... , x k l , ... , x k n are not present in (A.2), so the rest of the variables in (A.2) are not changed. Increment in some of the variables x k 1 , x k 2 , x k 3 , ... , x k l , ... , x k n will force to increase the left hand sides of (A.3) and (A.5) because of the presence of the respective variable/variables in these equations. But the left hand sides of (A.3) and (A.5) cannot be increased since − − − − − D 1 , D 2 , D 3 , ... , D l − β , ... , D n are fixed. Therefore, no changes can be made in 22 the variables x k 1 , x k 2 , x k 3 , ... , x k l , ... , x k n of the former solution due to the increment − on S k m −α in (A.4). Thus, the minimal total costs n ∑∑ A ij xi j in the extended model will be the i =1 j =1 same in both the solutions. Hence, W1 = V 1 . Case (2) α < β (Reduction in the supply quantity of the k th supplier is less than the reduction in the demand quantity of the l th buyer). Let the minimal total cost, V 2 of the objective function (A.1) subject to the constraints (A.2) to (A.7) is obtained with given α , β . Again, let the minimal total cost, W 2 of the objective function − − (A.1) subject to these constraints is obtained by increasing D l − β to retaining the other parameter values the same as before. Since to − Dl − α , − Dl − β D l − α , but by in (A.5) is increased at least the value of the one of the variables x 1 l , x 2 l , x 3 l , ..., x k l , ... , x m l in (A.5) must be increased in the latter minimal total cost solution. Thus, the latter minimal total m cost n ∑∑ A ij xi j in the extended model will be greater than its former minimal total cost. i =1 j =1 Hence, W2 > V2 . Appendix B MATLAB computational program for finding the upper minimal total cost bound of Transportation Problem with varying demand and supply (TPVDS) clc clear all % Data Input % ........................................................................ m=2; % number of suppliers n=3; % number of buyers S_low =[60;75]; % Supplier lower bound B_low =[45;30;60]; % Buyer lower bound S_high =[120;150]; % Supplier upper bound B_high =[90;60;120]; % Buyer upper bound a= [120; 150]; b= [90;60;120]; f=[15 90 88 75 80 8]; % Cost coefficient matrix A= [1 1 1 0 0 0; 0 0 0 1 1 1]; % Coefficient matrix for supply constraints. B= [1 0 0 1 0 0; 0 1 0 0 1 0; 0 0 1 0 0 1]; % coefficient matrix for demand constraints lb= [0;0;0;0;0;0]; % lower bound of the shipment quantities % ........................................................................ 23 Max_Cost = Solve_LP(a,b,A,B,f,lb); Max_a=a; Max_b=b; Max_alpha=0; Max_beta=0; Max_S_Count=1; Max_B_Count=1; tstart=tic; for i=1:m % supplier initialize to zero alpha(i)=0; end for j=1:n % buyer initialize to zero beta(j)=0; end for i=1:m for j=1:n Z(i,j)=0; end end for j=1:n buyer_counter(j)=1; end for i=1:m supplier_counter(i)=1; end for i=1:m step_S=0; step_B=0; prev_lim=1; for j=1:n lim=min((1-alpha(i))*(S_high(i)-S_low(i)),(1-beta(j))*(B_high(j)B_low(j))); step_s=0; step_b=0; ZZ=0; c=0; buyer_k=0; supplier_k=0; for k=1:lim step_s(k) = (supplier_counter(i)/(S_high(i)-S_low(i))); step_b(k) = (buyer_counter(j)/(B_high(j)-B_low(j))); R_s=S_high(i)-((S_high(i)-S_low(i))*step_s(k)); R_b=B_high(j)-((B_high(j)-B_low(j))*step_b(k)); alpha(i)=step_s(k); beta(j)=step_b(k); S_const=0; B_const=0; a(i)=R_s; b(j)=R_b; for mi=1:m 24 S_const = S_const + ((S_high(mi)-((S_high(mi)S_low(mi))*alpha(mi)))); end for nj=1:n B_const = B_const + ((B_high(nj)-((B_high(nj)B_low(nj))*beta(nj)))); end if (S_const >= B_const) [zz] = Solve_LP(a,b,A,B,f,lb); ZZ(i,k)=zz; else ZZ(i,k)=0; end buyer_counter(j) = buyer_counter(j) + 1; supplier_counter(i) = supplier_counter(i) + 1; supplier_k(k)=supplier_counter(i); buyer_k(k)=buyer_counter(j); end Z(i,j)=max(max(ZZ)); if ((Z(i,j))~=0 && (Z(i,j)>Max_Cost)) Max_Cost=Z(i,j); [r,c] = find(ZZ==max(max(ZZ(:)))); alhpa(i) = step_s(c); beta(j) = step_b(c); buyer_counter(j)= buyer_k(c); supplier_counter(i)= supplier_k(c); step_S=alpha(i); step_B=beta(j); Max_alpha=alpha(i); Max_beta=beta(j); prev_lim=lim; a(i)=S_high(i) - (step_s(c)*((S_high(i)-S_low(i)))); b(j)=B_high(j)- (step_b(c)*((B_high(j)-B_low(j)))); Max_a(i)=a(i); Max_b(j)=b(j); Max_S_Count=supplier_counter(i); Max_B_Count=buyer_counter(j); else alpha(i)=Max_alpha; beta(j)=Max_beta; supplier_counter(i)=Max_S_Count; buyer_counter(j)=Max_B_Count; a(i)=S_high(i) - (Max_alpha*((S_high(i)-S_low(i)))); b(j)=B_high(j)- (Max_beta*((B_high(j)-B_low(j)))); end end end obj=max(max(Z)); Max_Cost Max_a Max_b 25 [x,zz,exitflag,output] = ... linprog(f,A,Max_a,B,Max_b,lb,[],[],optimset('Display','iter')); x tend=toc(tstart) Appendix C function Max_Cost = Solve_LP(a,b,A,B,f,lb) [x,Max_Cost,exitflag,output] = ... linprog(f,A,a,B,b,lb,[],[],optimset('Display','iter')); end Where “Solve_LP(a,b,A,B,f,lb) ” is an user defined function. Here, “ a, b, A, B, f, lb ” are input vectors for the MATLAB computational program and they are defined as follows: a = Supply quantities of the suppliers (supply vector) b = Demand of the buyers (demand vector) A = Coefficient matrix for the supply constraints B = Coefficient matrix for the demand constraints f = Cost coefficient of the objective function lb = Lower bound of the variables (shipments quantities) and Linprog is a built in function of MATLAB that is used to solve Linear Programming Problems (LPP). The minimal total cost of the transportation problem (a special class of LPP) varies within a certain range due to the variation in the supply and demand quantities (different values of a and b). So, Linprog finds the Max_Cost (the upper bound of the minimal total cost) as an output. Appendix D Numerical problem 1: Data for 5-suppliers 10-buyers numerical problem. [C ] = i j 5 x 10 [25 14 34 46 45 48 11 26 45 16; 10 47 14 20 41 37 42 39 13 15; 22 42 38 21 46 12 38 28 31 20; 36 20 41 38 44 10 37 47 12 31; 34 33 30 14 34 32 41 19 39 33] − − − − − D1 = 20 , D1 = 40 ; D2 = 10 , D2 = 20 ; D3 = 30 , D3 = 60 ; D4 = 40 , D4 = 80 ; D5 = 20 , D5 = 40 ; − − − − − − − − − − D6 = 10 , D6 = 20 ; D7 = 20 , D7 = 40 ; D8 = 30 , D8 = 60 ; D9 = 60 , D9 = 120 ; D10 = 10 , D10 = 20 − − − − − − − − − − S 1 = 400 , S 1 = 800 ; S 2 = 500 , S 2 = 1000 ; S 3 = 600 , S 3 = 1200 ; S 4 = 300 , S 4 = 600 ; S 5 = 800 , S 5 = 1600 . − − − − − Numerical problems 2: Data for 10-suppliers 10-buyers numerical problem [C ] i j 10 x 10 = [ 25 14 34 46 45 48 11 26 45 16; 10 47 14 20 41 37 42 39 13 15; 22 42 38 21 46 12 38 26 28 31 20; 36 20 41 38 44 10 37 47 12 31; 34 33 30 4 34 32 41 19 39 33; 37 43 29 29 33 24 43 22 50 41; 21 42 18 28 26 47 14 17 27 16; 44 32 19 39 17 41 17 39 48 34 ; 26 40 14 38 43 18 36 38 43 26; 15 46 50 43 28 18 29 26 24 42] − − − − − D1 = 20 , D1 = 40 ; D2 = 10 , D2 = 20 ; D3 = 30 , D3 = 60 ; D4 = 40 , D4 = 80 ; D5 = 20 , D5 = 40 ; − − − − − − − − − − D6 = 10 , D6 = 20 ; D7 = 20 , D7 = 40 ; D8 = 30 , D8 = 60 ; D9 = 60 , D9 = 120 ; D10 = 10 , D10 = 20 . − − − − − − − − − − S 1 = 400 , S 1 = 800 ; S 2 = 500 , S 2 = 1000 ; S 3 = 600 , S 3 = 1200 ; S 4 = 300 , S 4 = 600 ; S 5 = 800 , S 5 = 1600 − − − − − − − − − − S 6 = 400 , S 6 = 800 ; S 7 = 500 , S 7 = 1000 ; S 8 = 600 , S 8 = 1200 ; S 9 = 300 , S 9 = 600 ; S 10 = 800 , S 10 = 1600 . − − − − − Numerical problems 3: Data for 10-suppliers 10-buyers numerical problem All ci,j (except c78 ) and the lower and the upper bounds of the demand and the supply quantities are the same as given in the above numerical problem 2. c78 = 7 . Numerical problems 4: Data for 10-suppliers 10-buyers numerical problem c21 = 3 , c49 = 5 , c85 = 9 , c87 = 10 ; The rest of the ci,j and the lower and the upper bounds of the demand quantities are the same as given in the above numerical problem 3. − − − − − S 1 = 100 , S 1 = 200 ; S 2 = 125 , S 2 = 250 ; S 3 = 150 , S 3 = 300 ; S 4 = 75 , S 4 = 150 ; S 5 = 200 , S 5 = 400 − − − − − − − − − − S 6 = 100 , S 6 = 200 ; S 7 = 125 , S 7 = 250 ; S 8 = 150 , S 8 = 300 ; S 9 = 75 , S 9 = 150 ; S 10 = 200 , S 10 = 400 . − − − − − Acknowledgment The authors are grateful to the referees for their valuable comments and suggestions. The authors also acknowledge that this research is supported by a graduate research scholarship of University Brunei Darussalam. References Adlakha, V., and Kowalski, K. (2009). Alternate solutions analysis for transportation problems. Journal of Business & Economics Research, 7(11), 41-49. Ahuja , R. K., Magnanti, T. K. , and Orlin, J. B. ,1993. 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