Internatinal Mathematical Frum, Vl. 6, 20, n. 40, 983-992 Flating Pint Methd fr Slving Transprtatin Prblems with Additinal Cnstraints P. Pandian and D. Anuradha Department f Mathematics, Schl f Advanced Sciences VIT University, Vellre-4, Tamilnadu, India pandian6@rediffmail.cm, anuradhadhanapal98@gmail.cm Abstract A new methd namely, flating pint methd is prpsed fr finding an ptimal slutin t a transprtatin prblem with additinal cnstraints. The prpsed methd is easy t understand and t apply fr finding an ptimal slutin t the abve said prblem and it differs frm the existing methds namely, simplex methd [7] and inverse matrix methd [6]. With the help f real life examples, the prpsed methd is illustrated. It can be very helpful fr the decisin makers t make ecnmical riented managerial decisins. Mathematics Subject Classificatins: 90C08, 90C90 Keywrds: Transprtatin prblem, zer pint methd, additinal cnstraints, flating pint methd. Intrductin The classical transprtatin prblem (TP) deals with shipping cmmdities frm surces t destinatins at minimum cst. The basic TP was develped by Hitchcck [5]. In these prblems it is assumed that the prduct is identical n matter which surce prduces it and that the custmers have n preference relating t the surce f supply. Hwever, in many practical instances, the prduct des vary in sme characteristics accrding t its surce and the final prduct mixture received by the destinatins, may
984 P. Pandian and D. Anuradha then be required t meet knwn specificatins. Fr example, it is assumed that the crude re cntains different amunts f an impurity, phsphrus, accrding t its surce and the actual time t prcess the re depends n bth its surce and destinatin. A TP f this type is described by Haley [4]. The simplex methd apprach fr slving cnstrained TPs was cnsidered by Klingman and Russell [7]. TP based n time criterin with additinal cnstraints was given by Nguyen Van Vy [8]. Kim [6] gave an inverse matrix algrithm fr slving a TP with additinal cnstraints. Reeta Gupta [] btained a slutin prcedure fr slving the generalized TP with cnstraints. Fractinal TP with impurities and time minimizing TP with impurities was presented by Chandra et al [,2]. Singh and Saxena [0] develped a methd fr slving multi bjective time TP with additinal impurity restrictins. Dutta and Murthy [3] develped a linear fractinal prgramming apprach fr slving fuzzy TP with additinal restrictins. In this paper, we prpse a new methd, namely flating pint methd t determine an ptimal slutin t TPs with additinal cnstraints. The flating pint methd takes advantage f the fact that the prblem has a special frm and it des nt require the use f the simplex methd [7] and inverse matrix methd [6]. The prpsed methd is illustrated with the help f tw real life examples. The flating pint methd prvides an apprpriate slutin which helps the decisin makers t analyze an ecnmic activities and t arrive at the crrect managerial decisins. 2. Transprtatin prblem with additinal cnstraints Cnsider the fllwing TP with additinal cnstraints: m n (P) Minimize z = subject t cij i= j= xij n xij = a, i,2,..., m i j= m x ij = b, j,2,..., n j i= m ( k ) (k) f x ij p, j =,2,..., n ; k,2,..., l ij j i= x ij 0, i,2,..., m = () = (2) = (3) = and j =,2,..., n (4) where a i is the tns available at i th surce; b j is the tns required at j th destinatin; ne (k) unit f the prduct cntains f units f l impurities k =,2,..., l when it is sent frm i t ij
Flating pint methd 985 (k ) j ; custmer j cannt receive mre than p j units f impurity k ; c ij is the unit cst f transprtatin frm the i th surce t j th destinatin and x ij is the tnnage transprted frm ith surce t jth destinatin. Any set f { x ij 0, i =,2,..., m; j =,2,.. n} which satisfies the equatins (), (2), (3) and (4) is called a feasible slutin t the prblem (P). A feasible slutin f prblem (P) which minimizes the ttal shipping cst is called an ptimal slutin t the prblem (P). 2. Flating pint methd Nw, we prpse a new methd namely, flating pint methd fr finding an ptimal slutin t a TP with additinal cnstraints. First, we prve the fllwing therem which is ging t be used in the flating pint methd. Therem : Let U = { u ij, i =,2,..., n and j =,2,..., m} where (Q) Minimize m n z = cij i= j= xij subject t (), (2) and (4) are satisfied, Then, X = {, i =,2,..., n and j =,2,..., m} which is btained frm the slutin x ij U = { u ij, i =,2,..., n and j =,2,..., m}, is an ptimal slutin t the prblem (P) Prf: If U = { u ij, i =,2,..., n and j =,2,..., m} satisfies all the additinal cnstraints f the prblem (P), then U = { u ij, i =,2,..., n ; j =,2,..., m} is an ptimal slutin f the prblem (P). If nt, U = { u ij, i =,2,..., n and j =,2,..., m} is nt satisfied atleast ne additinal cnstraint. Say, rth additinal cnstraint. That is, ( k) ( k) ( k) ( k) ( k) f ir uir = f r ur + f 2r u2r +... + f mr umr > p r i Nw, we flw the excess quantity in the rth clumn t ther quantity satisfied clumn such that rth clumn and cnsidered clumn are satisfied their additinal cnstraints and the alltment t rws and clumns are nt disturbed thrugh a clsed rectangular lp.
986 P. Pandian and D. Anuradha Cnstruct a rectangular lp ABCDA where AC are in the rth clumn and BD are in the sth clumn which satisfies its additinal cnstraint such that A and D are alltted cells. A ( ytr = utr θ ) B ( yts = uts + θ ) C ( y = pr u pr + θ ) D ( y = ps u ps θ ) We flw the minimum amunt θ thrugh the lp ABCDA such that rth clumn and sth clumn additinal cnstraints are satisfied and the ttal alltment t rws and clumns are nt disturbed. Then, we btain a new alltment fr A, B, C and D are y tr, y ts, y pr and ps y and cmpute c + + + + + + tr ytr cts yts c pr y pr c ps y ps ctrutr ctsuts c pru pr c psu ps called the excess flw f the lp ABCDA. Thus, we btain a feasible slutin Y = { y ij, i =,2,..., n and j =,2,..., m} frm U = { u ij, i =,2,..., n and j =,2,..., m } t the prblem (Q) with the rth additinal cnstraint where u ij : i t, p and j r, s u = = tr θ : i t and j r y = u ij ts + θ : i = t and j = s u ps θ : i = p and j = s u + i = p j = r pr θ : and Cnstruct all pssible lps frm an alltted cell in the rth clumn, flw the minimum amunt θ thrugh each f the lps such that rth clumn additinal cnstraint is satisfied and the cnsidered additinal cnstraint satisfied clumn, the ttal alltment t rws and clumns are nt disturbed and cmpute their excess flw. Such lps are called admissible lps. Select a lp in the set f admissible lps frm an alltted cell in the rth clumn having a minimum excess flw. After flwing the minimum amunt thrugh the selected lp, a feasible slutin t the prblem (Q) with rth additinal cnstraints can be btained. Since the flw in the lp is minimum, the btained feasible slutin t the prblem (Q) with rth additinal cnstraint is ptimal. If the current slutin is satisfied all the additinal cnstraints f (P), it is ptimal t the given prblem (P).
Flating pint methd 987 If nt, we cntinue the abve prcess until the current slutin is satisfied all the additinal cnstraints f (P). Nw, after a finite number f abve said steps, we btain an ptimal slutin X = { x ij, i =,2,..., n ; j =,2,..., m} frm U = { u ij, i =,2,..., n ; j =,2,..., m } t the given prblem (P). Hence the therem. Remark : If an admissible lp can nt be frmed fr any alltted cell in any clumn which is nt satisfied its additinal cnstraint, then the given prblem (P) has n feasible slutin. We nw intrduce the flating pint methd fr finding an ptimal slutin t the prblem (P). The prpsed methd prceeds as fllws. Step: Identify all impssible alltment cells using additinal cnstraints and remve thse cells frm the given TP. Step 2: Obtain an ptimal slutin t the reduced TP btained frm the Step. withut cnsidering the additinal cnstraints using the zer pint methd [9]. Step 3: Check that all additinal cnstraints are satisfied fr the ptimal slutin btained in the Step 2. If s, g t the Step 7. If nt, g t the Step 4. Step 4: Identify the clumn(s) which is (are) nt satisfied the additinal cnstraints. Step 5: Select a lp in a clumn which is nt satisfied the additinal cnstraint and flw the minimum quantity θ accrding t the Therem.. Then, find a new slutin t the transprtatin prblem with rth additinal cnstraints. Step 6: Check that the slutin btain frm the Step 5. satisfies all additinal cnstraints. If s, g t the Step 7. If nt, g t the Step 4. Step 7: The current slutin is an ptimal slutin t the TP with all additinal cnstraints by the Therem.. Nw, the prpsed methd can be illustrated with the help f fllwing examples.
988 P. Pandian and D. Anuradha Example : Cal Shipping Prblem: Barat Cking Cal, a cal manufacturing unit f the State has different types f cal pulverizatin units in each f the three wrk centers (j) situated in varius parts f the State. The wrk centers (j) are receiving a fixed quantity f cal (i), which has three different grades. The basic gal is t determine a feasible transprtatin cst, while satisfying the extra requirement that the amunt f sulphur impurity present in cal is less than a certain critical level. The fllwing table displays the transprtatin cst, availabilities, the impurities, requirements and the maximum sulphur cntents.. Tns Sulphur Wrk centers j available a i cntents 4 3 2 4 2 Cal i, 2 4 6 7 5 3 7 4 6 6 0 Tns required b j 5 5 5 Max. Sulphur p j 4 9 f i Nw, using the Step we have the fllwing reduced TP table Wrk centers j Tns available a i Sulphur cntents 4 X 2 4 2 Cal i, 2 4 6 7 5 3 7 4 6 6 0 Tns required Max. Sulphur b j 5 5 5 p j 4 9 Nw, we btain the fllwing ptimal slutin t the TP by the zer pint methd. Wrk centers j Tns available Nw, the current slutin des nt satisfy the first additinal cnstraint, but it satisfies secnd and third additinal cnstraints. Then, by using the Step 5, we have nly ne minimum flw admissible lp L (2,)-(2,3)-(3,3)-(3,)-(2,). After flwing unit thrugh L, we btain the fllwing slutin t the TP with first additinal cnstraint: a i f i Sulphur cntents f i 4 X 2 (4) 4 2 Cal i, 2 4 (5) 6 7 5 3 7 4 (5) 6 () 6 0 Tns required b j 5 5 5 Max. Sulphur p j 4 9
Flating pint methd 989 Wrk centers j Tns available a i Sulphur cntents 4 X 2 (4) 4 2 Cal i, 2 4 (4) 6 7 () 5 3 7 () 4 (5) 6 6 0 Tns required b j 5 5 5 5 Max. Sulphur p j 4 9 f i Nw, the abve slutin satisfies all the additinal cnstraints. Therefre, by the Therem., it is an ptimal t the given TP with the additinal cnstraints. Thus, the ptimal slutin t the given prblem is x 3 = 4, x 2 = 4, x 23 =, x 3 =, x 32 = 5 and the ttal minimum transprtatin cst is 58. Example 2: A cmpany making irn has a different type f furnace in each f three wrks. The wrks must receive a fixed weight f re which is available in three different grades. Fr technical reasns the prcessing time f re depends n its grade and the wrks t which it is sent. The furnaces are nly available fr a certain time in each week. The prblem is t find the allcatin which minimizes the prductin cst while satisfying the extra requirement that the amunt f phsphrus is less than a certain critical level. The fllwing table displays the cst c ij (lwer right crner) f transprt, purchase and prcessing re per tn and prcess time t ij (upper left crner) in hurs. 2 3 Ore, i 2 3 9 Tns required Max. phs Ttal time available Wrks, j 4 7 4 2 5 6 8 6 7 8 b j 5 0 0 p j 3.5 7 7 T j 30 60 40 9 5 3 Tns available Phs. a i Cntent f i 7 0.4 2 0.8 6 0.7 Nw, we btain the fllwing ptimal slutin t the TP by the zer pint methd.
990 P. Pandian and D. Anuradha Wrks, j Tns Phs. available a i Cntent f i 5 2 7 0.4 Ore, i 2 8 4 2 0.8 3 6 6 0.7 Tns required b j 5 0 0 Max. phs p j 3.5 7 7 Ttal time available T j 30 60 40 Nw, the current slutin des nt satisfy the secnd and third impurity cnstraint, but it satisfy the first impurity cnstraint and all the time cnstraints. Then, using Step 5, we have nly ne minimum flw admissible lp L (2,2)-(,2)-(,)-(2,)-(2,2).After flwing 0.5 unit thrugh L, we btain the fllwing new slutin t the prblem. Wrks, j Tns Phs. Cntent available a i f i 4.5 2.5 7 0.4 Ore, i 2 0.5 7.5 4 2 0.8 Tns required Max. phs Ttal time available 3 6 6 0.7 b j 5 0 0 p j 3.5 7 7 T j 30 60 40 Nw, the current slutin des nt satisfy the third impurity cnstraint, but it satisfy the first, secnd impurity cnstraints and all the time cnstraints. Repeating Step 3 t the Step 7 by fur times, we btain the fllwing ptimum slutin t the prblem. Wrks, j Tns available Phs. a i Cntent f i 3.0535 2.5.4465 7 0.4 Ore, i 2 0.605 7.5 4.3395 2 0.8 Tns required Max. phs Ttal time available 3.786 4.24 6 0.7 b j 5 0 0 p j 3.5 7 7 T j 30 60 40
Flating pint methd 99 Thus, the ptimal slutin t the given prblem is = 3. 0535 x 2 = 0.605, x 22 = 7. 5, x 23 = 4. 3395, x 3 =. 786, 33 = 4. 24 transprtatin cst is 85.537. x, x 2 = 2. 5, x 3 =. 4465, x and the ttal minimum Cnclusin In this paper, the transprtatin prblems with additinal cnstraints are slved by the flating pint methd which differs frm the existing methds namely, simplex methd and inverse matrix methd. The ptimal slutin btained by flating pint methd is nt necessarily t have ( m + n ) allted entries which is the main advantage f this methd. The prpsed methd enables the decisin makers t ptimize the ecnmical activities and make the crrect managerial decisins. In near future, we extend the prpsed methd t interval/fuzzy transprtatin prblems with additinal cnstraints. References [] S. Chandra and P.K. Saxena, Fractinal transprtatin prblem with impurities, Advances in Management Studies, 2, 983, 335-349. [2] S. Chandra, K.Seth and P.K. Saxena, Time minimizing transprtatin prblem with impurities, Asia-Pacific Jurnal f Operatinal Research, 4,987, 9-27. [3] D. Dutta and A. Satyanarayana Murthy, Fuzzy transprtatin prblem with additinal restrictins, ARPN Jurnal f Engineering and Applied Sciences, Vl. 5, N.2, 200. [4] K.B. Haley and A. J. Smith, Transprtatin prblems with additinal restrictins, JSTOR, 5, 966, 6-27. [5] F.L.Hitchcck, The distributin f a prduct frm several surces t numerus lcalities, J. Math. Phys. 20, 94, 224-230. [6] K.V. Kim, An inverse matrix algrithm fr slving a transprtatin prblem with additinal cnstraints, Eknmika I matematicheskie meldy, 967, 588 592. [7] D. Klingman and R. Russell, Slving cnstrained transprtatin prblems, Operatins Research, Vl.23, N., 975, 9-06.
992 P. Pandian and D. Anuradha [8] Nguyen Van Vy, Transprtatin prblem based n time criterin with additinal cnstraints, Cybernetics and Systems Analysis, Vl. 4, N. 4, 569 575 [9] P. Pandian and G.Natarajan, A new methd fr finding an ptimal slutin f fully interval integer transprtatin prblems, Applied Mathematical Sciences, 4, 200, 89-830. [0] Preetvanti Singh and P.K. Saxena, The multibjective time transprtatin Prblem with additinal restrictins, Eurpean Jurnal f Operatinal Research, 46, 2003, 460-476. [] Reeta Gupta, Slving the generalized transprtatin prblem with cnstraints, Jurnal f Applied Mathematics and Mechanics, Vl. 58, 45 458. Received: February, 20